Wednesday, December 31, 2008

Friday, December 26, 2008

jurnal MATEMatika

arXiv:0812.4550v1 [math.MG] 24 Dec 2008
Inequalities for mixed p-affine surface area ∗
Elisabeth Werner† Deping Ye
Abstract
We prove new Alexandrov-Fenchel type inequalities and new affine isoperimetric
inequalities for mixed p-affine surface areas. We introduce a new class
of bodies, the illumination surface bodies, and establish some of their properties.
We show, for instance, that they are not necessarily convex. We give
geometric interpretations of Lp affine surface areas, mixed p-affine surface areas
and other functionals via these bodies. The surprising new element is that
not necessarily convex bodies provide the tool for these interpretations.
1 Introduction
This article deals with affine isoperimetric inequalities and Alexandrov-Fenchel
type inequalities for mixed p-affine surface area. Mixed p-affine surface area was
introduced by Lutwak for p ≥ 1 in [28]. It has the dual mixed volume [25] and
the Lp affine surface area [28] as special cases. Lp affine surface area is at the core
of the rapidly developing Lp Brunn-Minkowski theory. Contributions here include
the study of solutions of nontrivial ordinary and, respectively, partial differential
equations (see e.g. Chen [10], Chou and Wang [11], Stancu [39, 40]), the study
of the Lp Christoffel-Minkowski problem by Hu, Ma and Shen [18], extensions of
Lp affine surface area to all p (see e.g., [34, 37, 38, 45]), a new proof by Fleury,
Gu´edon and Paouris [12] of a result by Klartag [19] on concentration of volume,
results on approximation of convex bodies by polytopes (e.g., [16, 24, 38]), results
on valuations (e.g., Alesker [2, 3], and Ludwig and Reitzner [22, 23]) and the affine
Plateau problem solved in R3 by Trudinger and Wang [41], and Wang [43].
∗Keywords: mixed p-affine surface area, affine isoperimetric inequality, Lp affine surface area,
Lp Brunn- Minkowski theory. 2000 Mathematics Subject Classification: 52A20, 53A15
†Partially supported by an NSF grant, a FRG-NSF grant and a BSF grant
1
The classical affine isoperimetric inequality, which gives an upper bound for
the affine surface area in terms of volume, is fundamental in many problems (e.g.
[14, 15, 29, 36]). In particular, it was used to show the uniqueness of self-similar
solutions of the affine curvature flow and to study its asymptotic behavior by
Andrews [4, 5], Sapiro and Tannenbaum [35]. More general Lp affine isoperimetric
inequalities were proved in [28] for p > 1 and in [45] for all p. These Lp affine
isoperimetric inequalities generalize the celebrated Blaschke-Santal´o inequality and
inverse Santal´o inequality due to Bourgain and Milman [6] (see also Kuperberg
[20]). We also refer to related works by Lutwak, Yang and Zhang [30] and Campi
and Gronchi [9].
For mixed p-affine surface area, Alexandrov-Fenchel type inequalities (for p =
1,±∞) and affine isoperimetric inequalities (for 1 ≤ p ≤ n) were first established by
Lutwak in [25, 26, 28]. Here we derive new Alexandrov-Fenchel type inequalities for
mixed p-affine surface area for all p ∈ [−∞,∞] and new mixed p-affine isoperimetric
inequalities for all p ∈ [0,∞]. Classification of the equality cases for all p in the
Alexandrov-Fenchel type inequalities for mixed p-affine surface area is related to the
uniqueness of solutions of the Lp Minkowski problem (e.g., [10, 11, 27, 29, 31, 32, 39,
40]), which is unsolved for many cases. This is similar to the classical Alexandrov-
Fenchel inequalities for mixed volume, where the complete classification of the
equality cases is also an unsolved problem.
We also give new geometric interpretations for functionals on convex bodies. In
particular, for Lp affine surface area, mixed p-affine surface area, and i-th mixed
p-affine surface area (see below for the definitions). To do so, we construct a new
class of bodies, the illumination surface bodies, and study the asymptotic behavior
of their volumes. We show that the illumination surface bodies are not necessarily
convex, thus introducing a novel idea in the theory of geometric characterizations
of functionals on convex bodies, where to date only convex bodies where used (e.g.
[34, 37, 38, 45]).
From now on, we will always assume that the centroid of a convex body K in
Rn is at the origin. We write K ∈ C2+
, if K has C2 boundary with everywhere
strictly positive Gaussian curvature. For real p ≥ 1, the mixed p-affine surface
area, asp(K1, · · · ,Kn), of n convex bodies Ki ∈ C2+
was introduced in [28] by
asp(K1, · · · ,Kn) =
Z
Sn−1

hK1(u)1−pfK1(u) · · · h1−p
Kn
fKn(u)
1
n+p
dσ(u). (1.1)
2
Here Sn−1 is the boundary of the Euclidean unit ball Bn
2 in Rn, σ is the usual
surface area measure on Sn−1, hK(u) is the support function of the convex body
K at u ∈ Sn−1, and fK(u) is the curvature function of K at u, i.e., the reciprocal
of the Gauss curvature κK(x) at this point x ∈ ∂K, the boundary of K, that has
u as its outer normal.
We propose here to extend the definition (1.1) for mixed p-affine surface area
to all p 6= −n. We also propose a definition for the (−n)- mixed affine surface area
(see Section 2).
We show that mixed p-affine surface areas are affine invariants for all p. Note
that for p = ±∞,
as±∞(K1, · · · ,Kn) =
Z
Sn−1
1
hK1(u) · · ·
1
hKn(u)
dσ(u)
= n ˜ V (K◦1 , · · · ,K◦n ) (1.2)
where K◦ = {y ∈ Rn, hx, yi ≤ 1, ∀x ∈ K} is the polar body of K, and ˜ V (K◦1 , · · · ,K◦n)
is the dual mixed volume of K◦1 , · · · ,K◦n , introduced by Lutwak in [25].
When all Ki coincide with K, then for all p 6= −n
asp(K, · · · ,K) =
Z
Sn−1
fK(u)
n
n+p
hK(u)
n(p−1)
n+p
dσ(u) = asp(K). (1.3)
asp(K) is the Lp affine surface area of K, which is defined for a general convex
body K as in [28] (p > 1) and in [38] (p < 1) by
asp(K) =
Z
@K
κK(x)
p
n+p
hx,NK(x)i
n(p−1)
n+p
dμK(x). (1.4)
NK(x) is the outer unit normal vector at x to ∂K, μK denotes the usual surface
area measure on ∂K, and h·, ·i is the standard inner product on Rn which induces
the Euclidian norm k · k. If K ∈ C2+
, (1.4) can be rewritten as (1.3). We show
in Section 2 that the corresponding formula (1.3) for p = −n also holds, where
as−n(K) is the L−n affine surface area of K introduced in [34].
Note further that the surface area of K can be written as (−1)-th mixed 1-affine
surface area of K and the Euclidean ball Bn
2 (see Section 2).
Thus, mixed p-affine surface area is an extension of dual mixed volume, surface
area, and Lp affine surface area.
3
Further notations. For sets A and B, [A,B] = conv(A,B) := {λx+(1−λ)y :
λ ∈ [0, 1], x, y ∈ A∪B} is the convex hull of A∪B. A subset K of Rn is star convex
if there exists x0 ∈ K such that the line segment [x0, x], from x0 to any point x in
K, is contained in K. A convex body K is said to be strictly convex if ∂K does
not contain any line segment.
For a convex body K in Rn, |K| stands for the n-dimensional volume of K.
More generally, for a set M, |M| denotes the Hausdorff content of its appropriate
dimension.
For u ∈ Sn−1, H(x, u) is the hyperplane through x with outer normal vector u,
H(x, u) = {y ∈ Rn, hy, ui = hx, ui}. The two half-spaces generated by H(x, u) are
H−(x, u) = {y ∈ Rn, hy, ui ≥ hx, ui} and H+(x, u) = {y ∈ Rn, hy, ui ≤ hx, ui}. For
f : ∂K → R+ ∪ {0}, μf is the measure on ∂K defined by μf (A) =
R
A fdμK.
The paper is organized as follows. In Section 2, we prove new Alexandrov-
Fenchel type inequalities and new isoperimetric inequalities for mixed p-affine surface
areas. We show monotonicity behaviour of the quotients

asp(K1, · · · ,Kn)
as∞(K1, · · · ,Kn)
n+p
and

asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p
.
We prove Blaschke-Santal´o type inequalities for mixed p-affine surface areas. Similar
results for the i-th mixed p-affine surface areas are also proved in Section 2.
In Section 3, we introduce the illumination surface body and describe some of its
properties. In Section 4, we derive the asymptotic behavior of the volume of the
illumination surface body, and geometric interpretations of Lp affine surface areas,
mixed p-affine surface areas, and other functionals on convex bodies.
2 Mixed p-affine surface area and related inequalities
2.1 Inequalities for mixed p-affine surface area
We begin by proving that mixed p-affine surface area is affine invariant for all p.
For p ≥ 1, this was proved by Lutwak [28]. We will first treat the case p 6= −n.
All the results concerning the case p = −n are at the end of this subsection.
It will be convenient to use the notation
fp(K, u) = h1−p
K (u)fK(u) (2.5)
4
for a convex body K in Rn and u ∈ Sn−1. We will also write asmp
(K1, · · · ,Kn)
for

asp(K1, · · · ,Kn)
m, and |det(T)| for the absolute value of the determinant of
linear transform T.
Lemma 2.1 Let T : Rn → Rn be an invertible linear transform. Then for all
p 6= −n,
asp(TK1, · · · , TKn) = |det(T)|
n−p
n+p asp(K1, · · · ,Kn).
In particular, if |det(T)| = 1, then asp(K1, · · · ,Kn) is affine invariant:
asp(TK1, · · · , TKn) = asp(K1, · · · ,Kn).
Proof.
Since K ∈ C2+
, for any u ∈ Sn−1, there exists a unique x ∈ ∂K such that u = NK(x)
and fK(u) = 1
K(x) . By Lemma 12 of [38]
fK(u) =
1
κK(x)
=
fTK (v)
det2(T) kT−1t(u)kn+1 , (2.6)
where v = T−1t(u)
kT−1t(u)k ∈ Sn−1 and where for an operator A, At denotes its usual
adjoint. On the other hand,
hK(u) = hx, ui = hTx, T−1t(u)i = kT−1t(u)k hTx, vi = kT−1t(u)k hTK(v).
Thus, with notation (2.5), for all p,
fp(K, u) =
fTK (v) h1−p
TK (v) kT−1t(u)k1−p
det2(T) kT−1t(u)kn+1 =
fp(TK, v)
det2(T) kT−1t(u)kn+p . (2.7)
Lemma 10 and its proof in [38] show that -up to a small errorfTK(
v) dσ(v) = |det(T)| kT−1t(u)kfK(u) dσ(u).
Together with (2.6), one gets that (again up to a small error) kT−1t(u)k−n dσ(u) =
|det(T)| dσ(v). Therefore, up to a small error,
[fp(K1, u) · · · fp(Kn, u)]
1
n+p dσ(u) = |det(T)|
p−n
n+p [fp(TK1, v) · · · fp(TKn, v)]
1
n+p dσ(v).
The lemma then follows by integrating over Sn−1.
5
A general version of the classical Alexandrov-Fenchel inequalities for mixed
volumes (see [1, 8, 36]) can be written as
mY−1
i=0
V (K1, · · · ,Kn−m,K| n−i, ·{·z· ,Kn−}i
m
) ≤ V m(K1, · · · ,Kn).
Here we prove the analogous inequalities for mixed p-affine surface area. For p =
±∞ and p = 1, the inequalities were proved by Lutwak [25, 26]. For p ≥ 1,
inequality (2.8) was proved by Lutwak in [28], with equality if and only if the
associated Ki are dilates of each other.
Proposition 2.1 Let all Ki be convex bodies in C2+
with centroid at the origin. If
p 6= −n, then for 1 ≤ m ≤ n
asmp
(K1, · · · ,Kn) ≤
mY−1
i=0
asp(K1, · · · ,Kn−m,K| n−i, ·{·z· ,Kn−}i
m
).
Equality holds if the Kk, for k = n − m + 1, · · · , n are dilates of each other. If
m = 1, equality holds trivially.
In particular, if m = n,
asnp
(K1, · · · ,Kn) ≤ asp(K1) · · · asp(Kn). (2.8)
Proof. Put g0(u) = [fp(K1, u) · · · fp(Kn−m, u)]
1
n+p and for i = 0, · · · ,m − 1, put
gi+1(u) = [fp(Kn−i, u)]
1
n+p . By H¨older’s inequality (see [17])
asp(K1, · · · ,Kn) =
Z
Sn−1
g0(u)g1(u) · · · gm(u) dσ(u)

mY−1
i=0
Z
Sn−1
g0(u)gm
i+1(u) dσ(u)
1
m
=
mY−1
i=0
as
1
m
p (K1, · · · ,Kn−m, |Kn−i ·{·z·Kn−}i
m
).
As Ki ∈ C2+
, fp(Ki, u) > 0 for all i and all u ∈ Sn−1. Therefore, equality in
H¨older’s inequality holds if and only if g0(u)gm
i+1(u) = λmg0(u)gm
j+1(u) for some
λ > 0 and all 0 ≤ i 6= j ≤ m − 1. This is equivalent to hKn−i(u)1−pfKn−i(u) =
λhKn−j (u)1−pfKn−j (u) for all 0 ≤ i 6= j ≤ m − 1. This condition holds true if the
Kk, for k = n −m + 1, · · · , n are dilates of each other.
6
Remark. It is an unsolved problem for many p whether fp(K, u) = λfp(L, u) guarantees
that K and L are dilates of each other. This is equivalent to the uniqueness
of the solution of the Lp Minkowski problem: for fixed α ∈ R, under which conditions
on a continuous function γ : Sn−1 → (0,∞), there exists a (unique) convex
body K such that hK(u)fK(u) = γ(u) for all u ∈ Sn−1. In many cases, the uniqueness
of the solution is an open problem. We refer to e.g., [11, 27, 29, 32, 39, 40]
for detailed information and more references on the subject. For p ≥ 1, p 6= n, the
solution to the Lp Minkowski problem is known to be unique and for p = n, the
solution is unique modulo dilates [27]. Therefore, we have the characterization of
equality in Proposition 2.1 for p ≥ 1.
Next, we prove affine isoperimetric inequalities for mixed p-affine surface areas.
Proposition 2.2 Let all Ki be convex bodies in C2+
with centroid at the origin.
(i) For p ≥ 0,
asnp
(K1, · · · ,Kn)
asnp
(Bn
2 , · · · ,Bn
2 ) ≤

|K1|
|Bn
2 | · · · |Kn|
|Bn
2 |
n−p
n+p
,
with equality if the Ki are ellipsoids that are dilates of one another.
(ii) For 0 ≤ p ≤ n,
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤

V (K1, · · · ,Kn)
V (Bn
2 , · · · ,Bn
2 )
n−p
n+p
with equality if the Ki are ellipsoids that are dilates of one another.
In particular, for p = n
asn(K1, · · · ,Kn) ≤ asn(Bn
2 , · · · ,Bn
2 ),
with equality if and only if the Ki are ellipsoids that are dilates of one another.
(iii) For p ≥ n,
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤

˜V (K1, · · · ,Kn)
˜ V (Bn
2 , · · · ,Bn
2 )
!n−p
n+p
,
with equality if and only if the Ki are ellipsoids that are dilates of one another.
7
In particular, for p = ±∞
˜ V (K1, · · · ,Kn) ˜ V (K◦1 , · · · ,K◦n) ≤ |Bn
2 |2,
with equality if and only if Ki are ellipsoids that are dilates of one another.
Remark. For 1 ≤ p ≤ n, inequality (ii) (with equality if and only if the Ki are
ellipsoids that are dilates of one another) was proved by Lutwak in [28]. If Ki = K
for all i, one recovers the Lp affine isoperimetric inequality proved in [45].
Remark. We cannot expect to get strictly positive lower bounds in Proposition
2.2. As in [45], we consider the convex body K(R, ε) ⊂ R2, obtained as the intersection
of four Euclidean balls with radius R centered at (±(R−1), 0), (0,±(R−1)), R
arbitrarily large. We then “round” the corners by putting there arcs of Euclidean
balls of radius ε, ε arbitrarily small. To obtain a body in C2+
, we “bridge” between
the R-arcs and ε-arcs by C2+
-arcs on a set of arbitrarily small measure. Then
asp(K(R, ε)) ≤ 16
R
p
2+p
+ 4π ε
2
2+p , which goes to 0 as R → ∞ and ε → 0. Choose
now Ri and εi, 1 ≤ i ≤ n, such that Ri → ∞ and εi → 0, and let Ki = K(Ri, εi)
for i = 1, 2 · · · , n. By inequality (2.8), asnp
(K1, · · · ,Kn) ≤
Qn
i=1 asp(K(Ri, εi)) and
thus asp(K1, · · · ,Kn) → 0 for p > 0. A similar construction can be done in higher
dimensions.
Proof of Proposition 2.2.
(i) Clearly asp(Bn
2 , · · · ,Bn
2 ) = asp(Bn
2 ) = n|Bn
2 | for all p 6= −n. By inequality
(2.8), one gets for all p ≥ 0
asnp
(K1, · · · ,Kn)
asnp
(Bn
2 , · · · ,Bn
2 ) ≤
asp(K1)
asp(Bn
2 ) · · ·
asp(Kn)
asp(Bn
2 ) ≤

|K1|
|Bn
2 | · · · |Kn|
|Bn
2 |
n−p
n+p
. (2.9)
The second inequality follows, for p ≥ 0, from the Lp affine isoperimetric inequality
in [45]. Equality holds true in the Lp isoperimetric inequality [45] if and only if
the Ki are all ellipsoids, and equality holds true in inequality (2.8) if the Ki are
dilates of one another. Thus, equality holds in (2.9) if the Ki are ellipsoids that
are dilates of one another.
(ii) A direct consequence of the classical Alexandrov-Fenchel inequality for mixed
volume (see e.g. [7, 21]) is that
|K1| · · · |Kn| ≤ V n(K1, · · · ,Kn).
8
If 0 ≤ p ≤ n, then n−p
n+p ≥ 0. Thus
􀀀
|K1| · · · |Kn|
n−p
n+p ≤

V n(K1, · · · ,Kn)
n−p
n+p .
As V (Bn
2 , · · · ,Bn
2 ) = |Bn
2 |, one gets together with (2.9)
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤

V (K1, · · · ,Kn)
V (Bn
2 , · · · ,Bn
2 )
n−p
n+p
,
with equality if the Ki are ellipsoids that are dilates of one another.
(iii) The analogous inequality for dual mixed volume [25] is
|K1| · · · |Kn| ≥ ˜ V n(K1, · · · ,Kn),
with equality if and only the Ki are dilates of one another. p > n implies n−p
n+p < 0.
Thus 􀀀
|K1| · · · |Kn|
n−p
n+p ≤

˜ V n(K1, · · · ,Kn)
n−p
n+p .
Together with (2.9) and ˜ V (Bn
2 , · · · ,Bn
2 ) = |Bn
2 |, one gets
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤

˜V (K1, · · · ,Kn)
˜ V (Bn
2 , · · · ,Bn
2 )
!n−p
n+p
.
As for p ≥ 1 equality in (2.8) holds if and only if the Ki are dilates of one another,
equality holds true here if and only if the Ki are ellipsoids that are dilates of one
another.
Proposition 2.3 Let E be a centered ellipsoid. If either all Ki ∈ C2+
are subsets
of E, for 0 ≤ p < n, or E is subset of all Ki for p > n, then
asp(K1, · · · ,Kn) ≤ asp(E).
For p = n, the inequality holds for all Ki in C2+
by Proposition 2.2 (ii).
Remark. This proposition was proved by Lutwak [28] if Ki = K for all i.
Proof. It is enough to prove the proposition for E = Bn
2 . For 0 ≤ p < n, one has
n−p
n+p > 0 and hence

|Ki|
|Bn
2 |
n−p
n+p ≤ 1 as Ki ⊂ Bn
2 . Similarly, p > n implies n−p
n+p < 0
and therefore

|Ki|
|Bn
2 |
n−p
n+p ≤ 1 as Bn
2 ⊂ Ki for all i. In both cases, the proposition
follows by inequality (2.9).
9
The next proposition gives a Blaschke-Santal´o type inequality for p-mixed affine
surface area. When Ki = K for all i, the proposition was proved in [45].
Proposition 2.4 Let all Ki be convex bodies in C2+
with centroid at the origin.
For all p ≥ 0,
asnp
(K1, · · · ,Kn)asnp
(K◦1 , · · · ,K◦n ) ≤ n2n|K1||K◦1 | · · · |Kn||K◦n |. (2.10)
Furthermore, asp(K1, · · · ,Kn)asp(K◦1 , · · · ,K◦n) ≤ as2
p(Bn
2 ) with equality if the Ki
are ellipsoids that are dilates of one another.
Proof. It follows from (2.8) that for all p 6= −n,
asnp
(K1, · · · ,Kn)asnp
(K◦1 , · · · ,K◦n ) ≤ asp(K1)asp(K◦1 ) · · · asp(Kn)asp(K◦n ),
with equality if the Ki are dilates of one another. By Corollary 4.1 in [45], for
p ≥ 0,
asnp
(K1, · · · ,Kn)asnp
(K◦1 , · · · ,K◦n ) ≤ n2n|K1||K◦1 | · · · |Kn||K◦n |.
Blaschke-Santal´o inequality states that |K||K◦| ≤ |Bn
2 |2 with equality if and only
if K is a 0-centered ellipsoid. We apply it to inequality (2.10), and obtain that for
p ≥ 0,
asp(K1, · · · ,Kn)asp(K◦1 , · · · ,K◦n ) ≤ as2
p(Bn
2 , · · · ,Bn
2 ).
Equality holds if the Ki are ellipsoids that are dilates of one another.
Theorem 2.1 Let s 6= −n, r 6= −n, p 6= −n be real numbers. Let all Ki be convex
bodies in C2+
with centroid at the origin.
(i) If (n+p)(r−s)
(n+r)(p−s) > 1, then
asp(K1, · · · ,Kn) ≤
􀀀
asr(K1, · · · ,Kn)
(p−s)(n+r)
(r−s)(n+p)
􀀀
ass(K1, · · · ,Kn)
(r−p)(n+s)
(r−s)(n+p) .
(ii) If n+p
n+r > 1, then
asp(K1, · · · ,Kn) ≤ (asr(K1, · · · ,Kn))
n+r
n+p

n ˜ V (K◦1 , · · · ,K◦n )
p−r
n+p . (2.11)
10
Remark. When all Ki coincide with K, (i) of Theorem 2.1 was proved in [45].
Proof.
(i) By H¨older’s inequality -which enforces the condition (n+p)(r−s)
(n+r)(p−s) > 1,
asp(K1, · · · ,Kn) =
Z
Sn−1

fp(K1, u) · · · fp(Kn, u)
1
n+p dσ(u)
=
Z
Sn−1

fr(K1, u) · · · fr(Kn, u)
1
n+r
(n+r)(p−s)
(n+p)(r−s)

fs(K1, u) · · · fs(Kn, u)
1
n+s
(n+s)(r−p)
(n+p)(r−s)
dσ(u)

􀀀
asr(K1, · · · ,Kn)
(p−s)(n+r)
(r−s)(n+p)
􀀀
ass(K1, · · · ,Kn)
(r−p)(n+s)
(r−s)(n+p) .
(ii) Similarly, again using H¨older’s inequality -which now enforces the condition
n+p
n+r > 1
asp(K1, · · · ,Kn) =
Z
Sn−1

fp(K1, u) · · · fp(Kn, u)
1
n+p dσ(u)
=
Z
Sn−1

fr(K1, u) · · · fr(Kn, u)
1
n+r
n+r
n+p

1
hK1(u) · · · hKn(u)
p−r
n+p
dσ(u)
≤ (asr(K1, · · · ,Kn))
n+r
n+p (as∞(K1, · · · ,Kn))
p−r
n+p .
Together with (1.2), this completes the proof.
Remark. The condition (n+p)(r−s)
(n+r)(p−s) > 1 implies 8 cases: −n < s < p < r, s <
−n < r < p, p < r < −n < s, r < p < s < −n, s < p < r < −n, p < s < −n < r,
r < −n < s < p and −n < r < p < s.
In [45], we proved monotonicity properties of

asr(K)
n|K|
n+r
r . Here we prove
similar results for mixed p-affine surface area.
Proposition 2.5 Let all Ki ∈ C2+
be convex bodies with centroid at the origin.
(i) If −n < r < p or r < p < −n, one has

asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n )
!n+p


asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
.
11
(ii) If 0 < p < r, or p < r < −n, or r < −n < 0 < p, or −n < p < r < 0, one has

asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p


asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+r
r
.
Proof.
(i) We divide both sides of inequality (2.11) by n ˜ V (K◦1 , · · · ,K◦n), and get for n+p
n+r >
1,
asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n ) ≤

asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
n+p
. (2.12)
Condition n+p
n+r > 1 implies that −n < r < p or p < r < −n. If −n < r < p,
n + p > 0 and therefore, inequality (2.12) implies inequality (i). If p < r < −n,
n + p < 0 and therefore,

asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n )
!n+p


asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
.
Switching r and p, one obtains the inequality in (i): for r < p < −n,

asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n )
!n+p


asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
.
(ii) Let s = 0 in inequality (i) of Theorem 2.1. Then for r(n+p)
p(n+r) > 1,
asp(K1, · · · ,Kn) ≤
􀀀
asr(K1, · · · ,Kn)
p(n+r)
r(n+p)
􀀀
as0(K1, · · · ,Kn)
(r−p)n
r(n+p) .
We divide both sides of the inequality by as0(K1, · · · ,Kn) and get
asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn) ≤

asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
p(n+r)
r(n+p)
.
The condition r(n+p)
p(n+r) > 1 implies that 0 < p < r, or p < r < −n, or −n < r < p <
0, or r < −n < 0 < p. In the cases 0 < p < r, or p < r < −n, or r < −n < 0 < p,
one has n+p
p > 0 and therefore inequality (ii) holds true. On the other hand, if
−n < r < p < 0, then n+p
p < 0 and hence,

asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+r
r


asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p
.
12
Switching r and p, one gets inequality (ii): if −n < p < r < 0, then

asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p


asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+r
r
.
Now we treat the case p = −n. The mixed (−n)-affine surface area of K1, · · · ,Kn
is defined as
as−n(K1, · · · ,Kn) = max
u∈Sn−1

fK1(u)
1
2n hK1(u)
n+1
2n · · · fKn(u)
1
2n hKn(u)
n+1
2n

. (2.13)
It is easy to verify that as−n(K, · · · ,K) equals to as−n(K), the L−n affine surface
area of K [34]. We have the following proposition.
Proposition 2.6 Let all Ki be convex bodies in C2+
with centroid at the origin.
Let p 6= −n and s 6= −n be real numbers.
(i) Let T : Rn → Rn be an invertible linear transform. Then
as−n(TK1, · · · , TKn) = |det(T)| as−n(K1, · · · ,Kn).
In particular, if |det(T)| = 1, then as−n(K1, · · · ,Kn) is affine invariant:
as−n(TK1, · · · , TKn) = as−n(K1, · · · ,Kn).
(ii) Alexandrov-Fenchel type inequalities
asm−
n(K1, · · · ,Kn) ≤
mY−1
i=0
as−n(K1, · · · ,Kn−m,K| n−i, ·{·z· ,Kn−}i
m
),
with equality if the Kj , for j = n − m + 1, · · · , n are dilates.
In particular, if m = n,
asn
−n(K1, · · · ,Kn) ≤ as−n(K1) · · · as−n(Kn).
(iii) If n(s−p)
(n+p)(n+s) ≥ 0, then
asp(K1, · · · ,Kn) ≤
􀀀
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn).
(iv) If n(s−p)
(n+p)(n+s) ≤ 0, then
asp(K1, · · · ,Kn) ≥
􀀀
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn).
13
Proof.
(i) By formula (2.7), f−n(TK, v) = det(T)2 f−n(K, u). Therefore
as−n(TK1, · · · , TKn) = maxv∈Sn−1 [f−n(TK1, v) · · · f−n(TKn, v)]
1
2n
= |det(T)| maxu∈Sn−1 [f−n(K1, u) · · · f−n(Kn, u)]
1
2n
= |det(T)| as−n(K1, · · · ,Kn).
(ii) Let ˜g0(u) = fK1(u)
1
2n hK1(u)
n+1
2n · · · fKn−m(u)
1
2n hKn−m(u)
n+1
2n and ˜gi+1(u) =
fKn−i(u)
1
2n hKn−i(u)
n+1
2n , for i = 0, · · · ,m − 1. Then
as−n(K1, · · · ,Kn) = max
u∈Sn−1
˜g0(u)˜g1(u) · · · ˜gm(u)

mY−1
i=0

max
u∈Sn−1
˜g0(u)˜gm
i+1(u)
1
m
=
mY−1
i=0
as
1
m
−n(K1, · · · ,Kn−m, |Kn−i · ·{·z ,Kn−}i
m
).
Equality holds if and only if for all i, 0 ≤ i ≤ m − 1, ˜g0(u)˜gm
i+1(u) attain their
maximum at the same direction u0. This condition holds true if the Kj , for j =
n − m + 1, · · · , n, are dilates.
(iii) and (iv)
asp(K1, · · · ,Kn) =
Z
Sn−1
[fp(K1, u) · · · fp(Kn, u)]
1
n+p dσ(u)
=
Z
Sn−1
[fs(K1, u) · · · fs(Kn, u)]
1
n+s

h
n+1
2n
K1
(u)f
1
2n
K1
(u) · · · h
n+1
2n
Kn
(u)f
1
2n
Kn
(u)
2n(s−p)
(n+p)(n+s)
dσ(u)
which is

􀀀
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn), if
n(s − p)
(n + p)(n + s) ≥ 0,
and

􀀀
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn), if
n(s − p)
(n + p)(n + s) ≤ 0.
14
2.2 i-th mixed p-affine surface area and related inequalities
For all p ≥ 1 and all real i, the i-th mixed p-affine surface area of K,L ∈ C2+
is
defined as [26, 42]
asp,i(K,L) =
Z
Sn−1
fp(K, u)
n−i
n+p fp(L, u)
i
n+p dσ(u).
Recall that fp(K, u) = fK(u)h1−p
K (u). Here we further generalize this definition to
all p 6= −n and all i. An analogous definition for the i-th mixed (−n)-affine surface
area of K and L is
as−n,i(K,L) = max
u∈Sn−1

fK(u)
n−i
2n hK(u)
(n+1)(n−i)
2n fL(u)
i
2n hL(u)
(n+1)i
2n

.
When i ∈ N, 0 ≤ i ≤ n, then, for all p, the i-th mixed p-affine surface area of K
and L is
asp,i(K,L) = asp(K| , ·{·z· ,K}
n−i
, |L, ·{·z· ,L}
i
).
Clearly, for all p, asp,0(K,L) = asp(K), and asp,n(K,L) = asp(L). When L = Bn
2 ,
we write asp,i(K) for asp,i(K,Bn
2 ). Thus
asp,i(K) =
Z
Sn−1
fp(K, u)
n−i
n+p dσ(u), for p 6= −n,
asp,i(K) = max
u∈Sn−1

fK(u)
n−i
2n hK(u)
(n+1)(n−i)
2n

, for p = −n.
In particular, as1,−1(K) =
R
Sn−1 fK(u) dσ(u) is the surface area of K.
The next proposition and its proof is similar to Proposition 2.1 and its proof.
Therefore we omit it.
Proposition 2.7 Let K and L be convex bodies in C2+
with centroid at the origin.
Let i ∈ R and s 6= −n, r 6= −n, and p 6= −n be real numbers.
(i) If (n+p)(r−s)
(n+r)(p−s) > 1, then
asp,i(K,L) ≤
􀀀
asr,i(K,L)
(p−s)(n+r)
(r−s)(n+p)
􀀀
ass,i(K,L)
(r−p)(n+s)
(r−s)(n+p) .
(ii) If n+p
n+r > 1, then
asp,i(K,L) ≤ (asr,i(K,L))
n+r
n+p

n ˜ Vi(K◦,L◦)
p−r
n+p
15
where ˜Vi(K◦,L◦) = 1
n
R
Sn−1
1
hK(u)n−i hL(u)i dσ(u) for all i.
(iii) If n(s−p)
(n+p)(n+s) ≥ 0, then
asp,i(K,L) ≤
􀀀
as−n,i(K,L)
2n(s−p)
(n+p)(n+s) ass,i(K,L).
(iv) If n(s−p)
(n+p)(n+s) ≤ 0, then
asp,i(K,L) ≥
􀀀
as−n,i(K,L)
2n(s−p)
(n+p)(n+s) ass,i(K,L).
The following proposition was proved in [26, 42] for p ≥ 1.
Proposition 2.8 Let K and L be convex bodies in C2+
with centroid at the origin.
If j < i < k or k < i < j (equivalently, k−j
k−i > 1 ), then for all p,
asp,i(K,L) ≤ asp,j(K,L)
k−i
k−j asp,k(K,L)
i−j
k−j ,
with equality if K and L are dilates of each other.
In particular,
asp,i(K) ≤ asp,j(K)
k−i
k−j asp,k(K)
i−j
k−j ,
with equality if K is a ball.
For p 6= −n, the proof is the same as the proof in [26, 42]. For p = −n, it is similar
to the proof of Proposition 2.1. Note that for i ∈ N, 0 < i < m, m = j and k = 0,
the proposition is a direct consequence of Proposition 2.1.
In Proposition 2.8, if j = 0 and k = n, then for all p and 0 ≤ i ≤ n
asnp,i(K,L) ≤ asn−i
p (K)asi
p(L). (2.14)
If we let i = 0 and j = n, then for all k ≤ 0 and for all p
asnp
,k(K,L) ≥ asn−k
p (K)askp
(L). (2.15)
Let i = n, j = 0 and k > n. Then inequality (2.15) also holds true for k ≥ n and
all p. In both (2.14) and (2.15), equality holds for all p if K and L are dilates.
16
From inequality (2.14) and Corollary 4.1 in [45], one gets that
asnp
,i(K,L)asnp
,i(K◦,L◦) ≤
􀀀
asp(K)asp(K◦)
n−i􀀀
asp(L)asp(L◦)
i
≤ n2n(|K||K◦|)n−i(|L||L◦|)i (2.16)
holds true for all p ≥ 0 and 0 < i < n. The inequality also holds if i = 0 and i = n
[45]. We apply Blaschke-Santal´o inequality to inequality (2.16) and get
asp,i(K,L)asp,i(K◦,L◦) ≤ as2
p(Bn
2 )
for all p ≥ 0 and 0 ≤ i ≤ n. Equality holds true if K and L are ellipsoids that are
dilates of each other. Hence we have proved the following proposition, which, for
p ≥ 1, was proved in [42].
Proposition 2.9 Let K and L be convex bodies in C2+
with centroid at the origin.
If p ≥ 0 and 0 ≤ i ≤ n, then
asp,i(K,L)asp,i(K◦,L◦) ≤ as2
p(Bn
2 ),
with equality if K and L are ellipsoids that are dilates of each other.
We now establish isoperimetric inequalities for asp,i(K).
Proposition 2.10 Let K ∈ C2+
be a convex body with centroid at the origin.
(i) If p ≥ 0 and 0 ≤ i ≤ n, then
asp,i(K)
asp,i(Bn
2 ) ≤

|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, asp,i(K)asp,i(K◦) ≤ as2
p(Bn
2 ) with equality
if K is a ball.
(ii) If p ≥ 0 and i ≥ n, then
asp,i(K)
asp,i(Bn
2 ) ≥

|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, asp,i(K)asp,i(K◦) ≥ as2
p(Bn
2 ) with equality
if K is a ball.
17
(iii) If −n < p < 0 and i ≤ 0, then
asp,i(K)
asp,i(Bn
2 ) ≥

|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, asp,i(K)asp,i(K◦) ≥ cn−ias2
p(Bn
2 ) where c is
the universal constant in the inverse Santal´o inequality [6, 20].
(iv) If p < −n and i ≤ 0, then
asp,i(K)
asp,i(Bn
2 ) ≥ c
(n−i)p
n+p

|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
.
Moreover, asp,i(K)asp,i(K◦) ≥ cn−ias2
p(Bn
2 ) where c is the same constant as in
(iii).
(v) If i ≤ 0, then
as−n,i(K)
as−n,i(Bn
2 ) ≥

|K|
|Bn
2 |
n−i
n
.
Moreover, as−n,i(K)as−n,i(K◦) ≥ as2
−n,i(Bn
2 ).
Proof.
(i) For i = n, the equality holds trivially. For i = 0, the inequality was proved in
[45]. We now prove the case 0 < i < n. L = Bn
2 in inequality (2.14) gives

asp,i(K)
asp,i(Bn
2 )
n


asp(K)
asp(Bn
2 )
n−i
(2.17)
for all p 6= −n and 0 ≤ i ≤ n. We also use that asp,i(Bn
2 ) = asp(Bn
2 ). Then, as
asp(Bn
2 ) = n|Bn
2 |, we get for all p ≥ 0 and 0 ≤ i ≤ n, the following isoperimetric
inequality as a consequence of the Lp affine isoperimetric inequality in [45]
asp,i(K)
asp,i(Bn
2 ) ≤

asp(K)
asp(Bn
2 )
n−i
n


|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
,
with equality if K is a ball. The inequality asp,i(K)asp,i(K◦) ≤ as2
p,i(Bn
2 ) follows
from Proposition 2.9 with L = Bn
2 .
(ii) For i = n, the equality holds trivially. Similarly, let L = Bn
2 in inequality
(2.15), then for all p 6= −n, and i ≥ n or i ≤ 0,

asp,i(K)
asp,i(Bn
2 )
n


asp(K)
asp(Bn
2 )
n−i
. (2.18)
18
Hence for i ≥ n and p ≥ 0, the Lp affine isoperimetric inequality in [45] implies
that
asp,i(K)
asp,i(Bn
2 ) ≥

asp(K)
asp(Bn
2 )
n−i
n


|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, by Corollary 4.1 (i) in [45] and the remark
after it, one has for all i ≥ n

asp,i(K)asp,i(K◦)
as2
p,i(Bn
2 )
!n


asp(K)asp(K◦)
as2
p(Bn
2 )
n−i
≥ 1,
or equivalently, asp,i(K)asp,i(K◦) ≥ as2
p,i(Bn
2 ), with equality if K is a ball.
(iii) If i ≤ 0 and −n < p < 0, inequality (2.18) and Theorem 4.2 (ii) of [45] imply
that
asp,i(K)
asp,i(Bn
2 ) ≥

asp(K)
asp(Bn
2 )
n−i
n


|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
with equality if K is a ball. By Corollary 4.1 (ii) of [45] and the remark after it,

asp,i(K)asp,i(K◦)
as2
p,i(Bn
2 )
!n


asp(K)asp(K◦)
as2
p(Bn
2 )
n−i
≥ cn(n−i),
or equivalently, asp,i(K)asp,i(K◦) ≥ cn−ias2
p,i(Bn
2 ) where c is the constant in the
inverse Santal´o inequality [6, 20].
(iv) If i ≤ 0 and p < −n inequality (2.18) and Theorem 4.2 (iii) of [45] imply that
asp,i(K)
asp,i(Bn
2 ) ≥

asp(K)
asp(Bn
2 )
n−i
n
≥ c
(n−i)p
n+p

|K|
|Bn
2 |
(n−p)(n−i)
(n+p)n
.
The proof of asp,i(K)asp,i(K◦) ≥ cn−ias2
p,i(Bn
2 ) is same as in (iii).
(v) Inequality (2.15) implies that as−n,i(K)n ≥ as−n(K)n−i for i ≤ 0. As as−n,i(Bn
2 ) =
1 for all i,
as−n,i(K)
as−n,i(Bn
2 )
n


as−n(K)
as−n(Bn
2 )
n−i


|K|
|Bn
2 |
n−i
.
The second inequality follows from the L−n affine isoperimetric inequality in [45].
Moreover, by Corollary 4.2 in [45], for i ≤ 0,

as−n,i(K)as−n,i(K◦)
as2
−n,i(Bn
2 )
!n


as−n(K)as−n(K◦)
as2
−n(Bn
2 )
n−i
≥ 1,
19
or equivalently, as−n,i(K)as−n,i(K◦) ≥ as2
−n,i(Bn
2 ).
Remark. The example K(R, ε) mentioned in the remarks after Proposition 2.2
shows that we cannot expect to get strictly positive lower bounds in (i) of Proposition
2.10 for p > 0 and 0 ≤ i < n. In fact, by inequality (2.17), one has

asp,i(K(R, ε))
asp,i(Bn
2 )
n


asp(K(R, ε))
asp(Bn
2 )
n−i
.
As in [45], asp(K(R, ε)) → 0 for p > 0 as R → ∞ and ε → 0. 0 ≤ i < n implies
that n − i > 0, and therefore asp,i(K(R, ε)) → 0.
This example also shows that, likewise, we cannot expect finite upper bounds
in (ii) (for p > 0 and i > n), (iii) (for −n < p < 0 and i ≤ 0), and (iv) (for p < −n
and i ≤ 0), of Proposition 2.10. For instance, if i ≤ 0, by inequality (2.18), one has

asp,i(K(R, ε))
asp,i(Bn
2 )
n


asp(K(R, ε))
asp(Bn
2 )
n−i
.
For −2 < p < 0, one has asp(K(R, ε)) → ∞ as R → ∞ and ε → 0. Therefore, if
i ≤ 0, we obtain that asp,i(K(R, ε)) → ∞ as R → ∞ and ε → 0, i.e., there are no
finite upper bounds in (iii).
Remark. In (iv), if p = −∞, then for all i ≤ 0,
Z Z
Sn−1×Sn−1
􀀀
hK(u)hK◦ (v)
i−n dσ(u) dσ(v) ≥ cn−ias2
p,i(Bn
2 )
or equivalently, for all i ≤ 0,
Z Z
Sn−1×Sn−1
􀀀
ρK(u)ρK◦(v)
n−i dσ(u) dσ(v) ≥ cn−ias2
p,i(Bn
2 ).
In particular, if i = 0, this is equivalent to the inverse Santal´o inequality [6].
3 Illumination surface bodies
We now define a new family of bodies associated with a given convex body K.
These new bodies are a variant of the illumination bodies [44] (compare also [38]).
Definition 3.1 (Illumination surface body) Let s ≥ 0 and f : ∂K → R be a
nonnegative, integrable function. The illumination surface body Kf,s is defined as
Kf,s =
n
x : μf (∂K ∩ [x,K]\K) ≤ s
o
.
20
Obviously, K ⊆ Kf,s for any s ≥ 0 and any nonnegative, integrable function f.
Moreover, Kf,s ⊆ Kf,t for any 0 ≤ s ≤ t.
Notice also that Kf,s needs to be neither bounded nor convex:
Example 3.1 Let K = B2∞
= {x ∈ R2 : max1≤i≤2|xi| ≤ 1} and
f(x) =
(
1
12 , x ∈ [(−1, 1), (1, 1)] ∪ [(1, 1), (1,−1)]
1
6 , otherwise
Kf,s is calculated as follows. If s < 1
6 , then Kf,s = K.
If s ∈ [ 1
6 , 1
3 ), then
Kf,s = {(x1, x2) : x1 ≥ −1, x2 ∈ [−1, 1]; or x2 ≥ −1, x1 ∈ [−1, 1]}.
If s ∈ [ 1
3 , 1
2 ), then
Kf,s = {(x1, x2) : x1, x2 ≥ −1; or x1 ≤ −1, x2 ∈ [−1, 1]; or x2 ≤ −1, x1 ∈ [−1, 1]}.
If s ∈ [ 1
2 , 2
3 ), then Kf,s = {(x1, x2) : x1 ≥ −1 or x2 ≥ −1}.
Except for s < 1/6, all of them are neither bounded nor convex.
If s ≥ 2
3 , then Kf,s = R2
The following lemmas describe some of the properties of the bodies Kf,s.
Lemma 3.1 Let s ≥ 0 and f : ∂K → R be a nonnegative, integrable function.
Then
(i) Kf,s =
T
>0 Kf,s+.
(ii) Kf,s is star convex, i.e., for all x ∈ Kf,s: [0, x] ⊂ Kf,s.
Proof.
(i) We only need to show that Kf,s ⊇
T
>0 Kf,s+. Let x ∈
T
>0 Kf,s+. Then for
all δ > 0, μf (∂K ∩ [x,K]\K) ≤ s+δ. Thus, letting δ → 0, μf (∂K ∩ [x,K]\K) ≤ s.
(ii) Let x ∈ Kf,s. We claim that [0, x] ⊆ Kf,s. Let y ∈ [0, x]. Since y ∈ [0, x] ⊂
[x,K], we have [y,K] \K ⊆ [x,K] \K and thus ∂K ∩[y,K] \ K ⊆ ∂K ∩ [x,K]\K.
This implies that
μf (∂K ∩ [y,K] \ K) ≤ μf (∂K ∩ [x,K] \ K) ≤ s
21
and hence y ∈ Kf,s.
Remark. We can not expect Kf,s to be convex, even for K = Bn
2 and f smooth.
Indeed, let K = B2
2 and s = 1
64 . Define
f(x) =


1
4 , x is in the first and third quadrant
1
16 , x is in the fourth quadrant
23
16 , x is in the second quadrant
Then Kf,s is not convex. In fact, Kf,s contains the arc from the point (tan(
32 ), 1)
to the point (1, tan(
32 )) of the Euclidean ball centered at 0 with radius r = sec(
32 ).
Moreover, the point (sec(
20 ), 0) is on the boundary of Kf,s. The tangent line at
(1, tan(
32 )) of B2
2(0, r) is y = −x+r2
√r2−1
. This tangent line intersects the x-axis at
(r2, 0) = (sec2(
32 ), 0). Since sec2(
32 ) ∼ 1.009701 < sec(
20 ) ∼ 1.01264, Kf,s is not
convex.
We can modify f so that it becomes smooth also at the points (±1, 0) and
(0,±1) and ∂Kf,s still intersects the positive x-axis at the point (sec(
20 ), 0). Therefore,
Kf,s is not convex, even if f is smooth.
Lemma 3.2 Let s ≥ 0 and f : ∂K → R be an integrable, μK-almost everywhere
strictly positive function. Then
(i) K = Kf,0 .
(ii) There exists s0 > 0, such that for all 0 ≤ s ≤ s0, Kf,s is bounded.
Proof. (i) It is enough to prove that Kf,0 ⊆ K. Suppose this is not the case.
Then there is x ∈ Kf,0 but x /∈ K. Since 0 ∈ int(K), there is α > 0 such that
Bn
2 (0, α) ⊆ K ⊆ Bn
2 (0, 1/α). (3.19)
Let y ∈ [0, x] ∩ ∂K and Con(x, α) = [x,Bn
2 (0, α)] be the convex hull of x and
Bn
2 (0, α). H(y,NK(y)) ∩ Con(x, α) contains a (n − 1)-dimensional Euclidean ball
with radius (at least) r1 = αkx−yk
kxk
> 0.
Hence μK(∂K ∩ [x,K] \ K) ≥ |H(y,NK(y)) ∩ Con(x, α)| ≥ rn−1
1 |Bn−1
2 | > 0. Let
Ej =

z ∈ ∂K ∩ [x,K]\K : f(z) ≥
1
j

, j = 1, 2, · · · .
As μK ({z ∈ ∂K : f(z) = 0}) = 0 and Ej ⊆ Ej+1 for all j,
μK
􀀀
∂K ∩ [x,K] \ K

= μK
􀀀 ∞[
j=1
Ej

= lim
j→∞
μK(Ej).
22
Therefore there exists j1 such that μK(Ej1) > 0. Thus
μf (∂K ∩ [x,K]\K) ≥ μf (Ej1) ≥
μK(Ej1)
j1
> 0
which contradicts that x ∈ Kf,0.
(ii) is an immediate consequence of Lemma 3.1 (i) and Lemma (3.2) (i). Indeed,
these lemmas imply that K = Kf,0 =
T
s>0Kf,s. So, also using (3.19), there
exists s0 > 0 such that for all 0 ≤ s ≤ s0, Kf,s ⊂ 2K ⊂ Bn
2 (0, 2
). In particular,
Kf,s0 ⊂ Bn
2 (0, 2
).
Remark. The assumption that f is μK-almost everywhere strictly positive is
necessary in order that Kf,0 = K. To see that, let K = B2
2 and
f(x, y) =
(
0 x =
p
1 − y2, y ∈ [−1, 1],
1
otherwise.
Then Kf,0 = K ∪ {(x, y) : x ≥ 0, |y| ≤ 1}.
This example also shows that there is no s0 such that Kf,s is bounded for all
0 ≤ s ≤ s0 unless f is μK-almost everywhere strictly positive.
Let K be a convex body with 0 ∈ int(K). Let f : ∂K → R be an integrable,
μK-almost everywhere strictly positive function. For x /∈ K, let t0 = t0(x) be the
strictly positive real number such that t0x = ∂K ∩ [0, x]. Define hx(t) to be
hx(t) = μf (∂K ∩ [tx,K] \ K), t ≥ t0.
Clearly hx(t0) = 0. Moreover hx(t) ≤ s if tx ∈ Kf,s, and hx(t) > s if tx /∈ Kf,s.
Lemma 3.3 Let K be a convex body in Rn and f : ∂K → R be an integrable,
μK-almost everywhere strictly positive function.
(i) hx(t) is increasing and left continuous on [t0,∞).
(ii) Kf,s is closed for all s ≥ 0. In particular, it is compact for all 0 ≤ s ≤ s0.
If K is in addition strictly convex, then
(iii) hx(t) is continuous on [t0,∞).
(iv) For any 0 ≤ s ≤ s0 and x ∈ ∂Kf,s, one has μf (∂K ∩ [x,K] \ K) = s.
23
Proof.
(i) If t1 ≤ t2, then ∂K ∩ [t1x,K] \ K ⊆ ∂K ∩ [t2x,K] \ K. Thus hx(t1) ≤ hx(t2).
Let now t > t0 and (tm)m∈N be a sequence, increasing to t. Then, by monotonicity
of hx, hx(tm) ≤ hx(t) for all m and thus limmhx(tm) ≤ hx(t). We have to show
that limmhx(tm) ≥ hx(t). Let y ∈ relint@K(∂K ∩ [tx,K] \ K), where relintB(A)
is the relative (with respect to B) interior of a set A ⊆ B, i.e., relintB(A) =
{x ∈ A : there is a δ > 0, such that B(x, δ) ∩ B ⊂ A}. Then y ∈ int ([tx,K]), and
therefore there exists m0(y) ∈ N, such that y ∈ int([tm0(y)x,K]). This implies that
y ∈ [tm0(y)x,K] \ K ∩ ∂K and thus
relint@K([tx,K] \ K ∩ ∂K) ⊂
[
m≥1
[tmx,K] \ K ∩ ∂K.
By continuity of the measure μf from below, one has
hx(t) = μf
􀀀
[tx,K] \ K ∩ ∂K

= μf

relint@K
􀀀
[tx,K] \ K ∩ ∂K

≤ μf
[
m≥1
􀀀
[tmx,K] \ K ∩ ∂K

= lim
m
μf

∂K ∩ [tmx,K] \ K

= limmhx(tm)
(ii) It will follow from Lemma 3.2 (ii) that Kf,s is compact for 0 ≤ s ≤ s0, once we
have proved that Kf,s is closed.
To that end, we show that (Kf,s)c, the complement of Kf,s in Rn, is open for all
s ≥ 0. Suppose this is not the case. Then there exists x ∈ (Kf,s)c and a sequence
(xm)m∈N, such that xm → x as m → ∞ but xm ∈ Kf,s for all m. Without loss of
generality, we can assume that xm are not in the ray of {tx : t ≥ 0}. Otherwise,
if xm ∈ Kf,s are in the ray, then hx

kxmk
kxk

≤ s and by (i), limm hx

kxmk
kxk

=
hx(1) ≤ s. This contradicts with hx(1) > s.
Now we let Km = [xm,K]. For sufficiently big m, ∂Km∩[0, x] 6= ∅. Suppose not,
then x ∈ Km implies that [x,K] ⊂ Km, and hence [x,K] \ K∩∂K ⊂ Km \ K∩∂K.
Since μf ([x,K] \ K ∩ ∂K) > s, one gets μf (Km \ K ∩ ∂K) > s, a contradiction
with xm ∈ Kf,s. Let ym = ∂Km ∩ [0, x]. Thus μf ([ym,K] \ K ∩ ∂K) ≤ s. Let α
be as in (3.19). Similarly, ∂
􀀀
[xm,Bn
2 (0, α)]

∩ [0, x] 6= ∅ for sufficiently big m and
we denote zm = ∂
􀀀
[xm,Bn
2 (0, α)]

∩ [0, x].
24
It is easy to check that 0 ≤ kxk−kymk ≤ kxk−kzmk for any m. As α ≤ kzmk ≤
kxk and
kzmk ≤ kx−xmk
kxk−kzmk
, one has kxk − kzmk ≤ kxk kx−xmk . Thus zm → x, and
hence also ym → x, as m → ∞. Therefore we can choose a subsequence (ymk )k∈N
that is monotone increasing to x. By (i) with tmk = kymk k
kxk
, hx(tmk ) ր hx(1) as
k → ∞. Since for all k, hx(tmk ) ≤ s, one has hx(1) ≤ s, a contradiction.
(iii) It is enough to prove that hx(t) is right continuous on [t0,∞). To do so, let
t ≥ t0 and let (tm)m∈N be a sequence decreasing to t. By (i), hx(tm) ≥ hx(t) for all
m, thus limmhx(tm) ≥ hx(t) and we have to show that limmhx(tm) ≤ hx(t). We
claim that if K is strictly convex, then
∂K ∩ [tx,K] \ K =
∞\
m=1

∂K ∩ [tmx,K] \ K

. (3.20)
We only need to prove that
T
∞m=1

∂K ∩ [tmx,K] \ K

⊆ ∂K ∩ [tx,K] \ K. Let
z0 ∈
T
∞m=1

∂K ∩ [tmx,K] \ K

. Thus z0 ∈ ∂K. Let l(z0, tx) be the line passing
through tx and z0. We have two cases.
Case 1: l(z0, tx) is in a tangent hyperplane of K. Then l(z0, tx) ∩ ∂K = {z0} by
strict convexity of K. Therefore, {z0} = [z0, tx] \ K ∩ ∂K ⊆ [tx,K] \ K ∩ ∂K.
Case 2: l(z0, tx) ∩ ∂K consists of two points, z0 and z1. As z0 ∈
T
∞m=1

∂K ∩
[tmx,K] \ K

, we must have ktx−z0k < ktx−z1k. Therefore, {z0} = [z0, tx] \ K∩
∂K ⊆ [tx,K] \ K ∩ ∂K.
Hence by (3.20) and continuity of the measure μf from above,
hx(t) = μf
∞\
m=1
h
∂K ∩ [tmx,K] \ K
i
= lim
m
μf

∂K ∩ [tmx,K] \ K

= lim
m
hx(tm).
(iv) Let 0 ≤ s ≤ s0, and x ∈ ∂Kf,s which implies that hx(1) ≤ s. Define x(s) =
{t : hx(t) = s}. Then x(s) 6= ∅. Indeed, let tax = ∂Bn
2 (0, 3
) ∩ Tx, where α is as
in (3.19) and Tx = {tx : t ≥ t0(x) > 0}. The proof of Lemma 3.2 (ii) shows that
Kf,s0 ⊂ Bn
2 (0, 2
). It is clear that tx /∈ Kf,s0, and hence hx(t) > s0. In fact, if
tx ∈ Kf,s0, then tx ∈ Bn
2 (0, 2
), but by definition of t, tx ∈ ∂Bn
2 (0, 3
). This is
a contradiction.
By continuity of hx(·), there must exist t ∈ [t0, t], such that hx(t) = s. This
also shows that ¯t = supx(s) ≤ t. Clearly hx(¯t) = s and thus ¯tx ∈ Kf,s. This
implies that ¯t ≤ 1 because x ∈ ∂Kf,s. Suppose ¯t < 1. Then s = hx(¯t) ≤ hx(1) ≤ s
25
by monotonicity of hx(·), a contradiction with ¯t = sup(s). Thus ¯t = 1 and
hx(1) = s.
Remark. Strict convexity is needed in (iii) and (iv). Indeed, let x = (0, 2) and
K = conv
􀀀
{(1, 1), (−1, 1), (−2, 0), (2, 0)}

.
Then ∂K ∩ [x,K]\K = [(−1, 1), (1, 1)]. However for any point tx with t > 1,
[tx,K] \ K ∩ ∂K = ∂K \ [(−2, 0), (2, 0)] ) ∂K ∩ [x,K]\K.
Thus, for any function f with f > 0 on [(−2, 0), (−1, 1)] and/or [(1, 1), (2, 0)], hx(·)
is not right continuous on [1,∞).
To see that strict convexity is needed also in (iv), observe that Kf,1/12 = K in
Example 3.1. Thus, for x ∈ ∂Kf,1/12 = ∂K, we have
μf ([x,K] \ K ∩ ∂K) = 0 6=
1
12
.
4 Geometric interpretation of functionals on convex
bodies
We now give geometric interpretations of functionals on convex bodies, such as Lp
affine surface area and mixed p-affine surface area for all p 6= −n using the non
convex illumination surface bodies. While there are no geometric interpretations
for mixed p-affine surface area, many geometric interpretations of Lp affine surface
area have been discovered in the last years, all based on using convex bodies (e.g.,
[33, 37, 38, 45]). The remarkable new fact here is that now the bodies involved in
the geometric interpretation are not necessarily convex.
Theorem 4.1 Let K be a convex body in C2+
. Let c > 0 be a constant, and
f : ∂K → R be an integrable function such that f ≥ c μK-almost everywhere.
Then
lim
s→0
cn |Kf,s| − |K|
s
2
n−1
=
Z
@K
κK(x)
1
n−1
f(x)
2
n−1
dμK(x), (4.21)
where cn = 2|Bn−1
2 |
2
n−1 .
26
Remark. As dμK = fKdσ, we also have
lim
s→0
cn |Kf,s| − |K|
s
2
n−1
=
Z
Sn−1
fK(u)
n−2
n−1
f(N−1
K (u))
2
n−1
dσ(u), (4.22)
where N−1
K is the inverse of the Gauss map NK(·).
The geometric interpretation of Lp affine surface area is then a corollary to Theorem
4.1. The theorem also gives geometric interpretations of other known functionals
on convex bodies, e.g. the surface area and the mixed p-affine surface area. Notice
that these geometric interpretations can also be obtained using e.g. the (convex)
surface body [38, 45].
Define
˜ f(N−1
K (u)) = fK(u)
n−2
2 [fp(K1, u) · · · fp(Kn, u)]
1−n
2(n+p) ,
where fp(K, u) = hK(u)1−pfK(u).
Corollary 4.1 Let K and Ki, i = 1, · · · , n, be convex bodies in C2+
. Then
lim
s→0
cn |K ˜ f,s| − |K|
s
2
n−1
= asp(K1, · · · ,Kn).
In particular, if all Ki coincide with K, then asp(K1, · · · ,Kn) = asp(K) and we
get a geometric interpretation of asp(K)
cn lim
s→0
|Kgp,s| − |K|
s
2
n−1
= asp(K),
where gp : ∂K → R is defined by gp(x) = κK(x)
n+2p−np
2(n+p) hx,NK(x)i
n(n−1)(p−1)
2(n+p) .
Corollary 4.2 Let K be a convex body in C2+
and g(x) =
p
κK(x). Then
lim
s→0
cn |Kg,s| − |K|
s
2
n−1
= μK(∂K).
The proof of the corollaries follows immediately from Theorem 4.1. To prove
Theorem 4.1, we need several other concepts and lemmas.
As K is in C2+
, for any x ∈ ∂K, the indicatrix of Dupin is an ellipsoid. As in [38],
we apply an affine transform T : Rn → Rn to K so that the indicatrix of Dupin
27
is transformed into an (n − 1)-dimensional Euclidean ball. T has the following
properties:
T(x) = x T(NK(x)) = NK(x) det(T) = 1 (4.23)
and T maps a measurable subset of a hyperplane orthogonal to NK(x) onto a
subset of the same (n − 1)-dimensional measure. It was also shown in [38] that
for any ǫ > 0 there is 1 = 1(ε) > 0 such that for all measurable subsets A of
∂K ∩ H−(x − 1NK(x),NK(x))
(1 − ǫ) μK(A) ≤ |T(A)| ≤ (1 + ǫ) μK(A). (4.24)
T(K) can be approximated at x = T(x) by a n-dimensional Euclidean ball: For
any ǫ > 0 there is 2 = 2(ε) such that
Bn
2 (x − rNK(x), r) ∩ H− (x − 2NK(x),NK(x))
⊆ T(K) ∩ H− (x − 2NK(x),NK(x)) (4.25)
⊆ Bn
2 (x − RNK(x),R) ∩ H− (x − 2NK(x),NK(x)) ,
where r = r(x) = κK(x)− 1
n−1 and R = R(x) with r ≤ R ≤ (1 + ǫ)r . We put
= (ε) = min{1,2}. (4.26)
Moreover, for x ∈ ∂K, let
xs ∈ ∂Kf,s be such that x ∈ [0, xs] ∩ ∂K (4.27)
and define ˜xs to be the orthogonal projection of xs onto the ray {y : y = x +
tNK(x), t ≥ 0}. Clearly T(˜xs) = ˜xs, and the distance from T(xs) to the hyperplane
H(x,NK(x)) is the same as the distance from xs to this hyperplane.
We say that a family of sets Es ⊆ ∂K, 0 < s ≤ s0 shrinks nicely to a point
x ∈ ∂K (see [13]) if
(SN1) diamEs → 0, as s → 0.
(SN2) There is a constant β > 0 such that for all s ≤ s0 there exists ts with
μK (∂K ∩ B(x, ts)) ≥ μK(Es) ≥ β μK (∂K ∩ B(x, ts)) .
28
Lemma 4.1 Let K be a convex body in C2+
and f : ∂K → R an integrable, μKalmost
everywhere strictly positive function. Let x ∈ ∂K and let xs and ˜xs be as
above (4.27). Then
(i) The family ∂K ∩ [˜xs,K] \ K, 0 < s ≤ s0 shrinks nicely to x.
(ii) The family ∂K ∩ [xs,K] \ K, 0 < s ≤ s0 shrinks nicely to x.
(iii)
lim
s→0
μf (∂K ∩ [˜xs,K] \ K)
μK(∂K ∩ [˜xs,K] \ K)
= f(x) μK-a.e. (4.28)
(iv)
lim
s→0
μf (∂K ∩ [xs,K] \ K)
μK(∂K ∩ [xs,K] \ K)
= f(x) μK- a.e. (4.29)
Proof. Formulas (4.28) and (4.29) in (iii) and (iv) follow from the Lebesgue
differentiation theorem (see [13]) once we have proved that ∂K ∩ [ ˜ xs,K] \ K and
∂K ∩ [xs,K] \ K shrink nicely to x. Therefore it is enough to prove (i) and (ii).
(i) For x ∈ ∂K, let r = r(x) and R = R(x) be as in (4.25). We abbreviate
B(r) = Bn
2 (x − rNK(x), r) and B(R) = Bn
2 (x − RNK(x),R). Let
(x, s) =

x
kxk
,NK(x)

kxs − xk = hxs − x,NK(x)i
be the distance from xs to H
􀀀
x,NK(x)

. This is the same as the distance from ˜xs
(defined after formula (4.27)) to H
􀀀
x,NK(x)

.
Let hR = hR(s) = R (x,s)
R+(x,s) be the height of the cap of B(R) that is “illuminated”
by ˜xs. Then
H−
􀀀
x − hRNK(x),NK(x)

∩ ∂B(R) = [T(˜xs),B(R)] \ B(R) ∩ ∂B(R). (4.30)
Let be as in (4.26). Since (x, s) → 0 as s → 0, one can choose s1 ≤ s0, such
that for all 0 < s ≤ s1, h = 2hR < . Therefore (4.25) holds:
H−(x − hNK(x),NK(x)) ∩ B(r) ⊂ H−
􀀀
x − hNK(x),NK(x)

∩ T(K)
⊂ H−
􀀀
x − hNK(x),NK(x)

∩ B(R). (4.31)
(4.31) and (4.30) imply that for all small enough s ≤ s2 ≤ s1
T
􀀀
[˜xs,K] \ K ∩ ∂K

= [T(˜xs), T(K)] \ T(K) ∩ ∂T(K)
29
⊆ H−
􀀀
x − hNK(x),NK(x)

∩ ∂T(K) ⊆ B(R) ∩ H−
􀀀
x − hNK(x),NK(x)

.
Let ts = kx − zk =
q
4R2 (x,s)
R+(x,s) where z is any point in H
􀀀
x − hNK(x),NK(x)


∂B(R). As B(R) ∩ H−
􀀀
x − hNK(x),NK(x)

⊆ Bn
2 (x, ts),
T
􀀀
[˜xs,K] \ K

∩ ∂T(K) = T
􀀀
[˜xs,K] \ K ∩ ∂K

⊆ Bn
2 (x, ts) ∩ ∂T(K)
and ts → 0 as s → 0.
This shows that condition (SN1) is satisfied for T
􀀀
[˜xs,K] \ K

∩ ∂T(K) to shrink
nicely to T(x) = x.
We now show that condition (SN2) also holds true.
First, [T(˜xs),B(r)] \ B(r) ∩ H(x,NK(x)) is a (n − 1)-dimensional Euclidean ball
with radius r(x,s) √2r(x,s)+2(x,s)
. Then for any 0 < s < s2,
|T([˜xs,K] \ K ∩ ∂K)| ≥ |[T(˜xs),B(r)] \ B(r) ∩ H(x,NK(x))|
≥ |Bn−1
2 |

r(x, s) p
2r(x, s) + 2(x, s)
!n−1
.
We can choose ( a new, smaller) s2 such that (x,s)
r ≤ 2. Then
|T([˜xs,K] \ K ∩ ∂K)| ≥ 21−n |Bn−1
2 | (r(x, s))
n−1
2 . (4.32)
On the other hand, for ε small enough, there exists s3 < s2, such that, for all
0 < s ≤ s3 and for any subset A of H−(x − 2hNK(x),NK(x)) ∩ ∂T(K) [38]
|PH(x−2hNK(x),NK(x))(A)| ≤ |A| ≤ (1 + ε)|PH(x−2hNK(x),NK(x))(A)| (4.33)
where PH(A) is the orthogonal projection of A onto the hyperplane H. We apply
this to A = Bn
2 (x, ts) ∩ ∂T(K):
|Bn
2 (x, ts) ∩ ∂T(K)| ≤ (1 + ε)|PH(x−2hNK(x),NK(x))(Bn
2 (x, ts) ∩ ∂T(K))|
≤ (1 + ε) |B(R) ∩ H(x − 2hNK(x),N)|
≤ (1 + ε) |Bn−1
2 |

2√2R
p
R(x, s)
R + (x, s)
!n−1
≤ 4n |Bn−1
2 | (r(x, s))
n−1
2 (4.34)
The last inequality follows as r ≤ R < (1 + ε)r. It now follows from (4.32) and
(4.34) that also condition (SN2) holds true for e.g. β = 8−n.
30
Hence the family T
􀀀
[˜xs,K] \ K

∩∂T(K) shrinks nicely to T(x) = x and therefore,
as T−1 exists, the family [˜xs,K] \ K∩∂K = T−1
􀀀
T
􀀀
[˜xs,K] \ K

∩∂T(K)

shrinks
nicely to x.
(ii) Let v1 = xs − (x − rNK(x)) and v2 = xs − (x − RNK(x)). θ denotes the angle
between NK(x) and x and φi = φi(x, s), i = 1, 2 is the angle between NK(x) and
vi, i = 1, 2. These angles can be computed as follows
tan(φ1) =
(x, s) tan(θ)
r + (x, s)
tan(φ2) =
(x, s) tan(θ)
R + (x, s)
.
Then, for i = 1, 2, φi → 0 as s → 0. Since K is in C2+
, this means that for any
ε > 0 there is ¯s" ≤ s0 such that for all s ≤ ¯s"
1 − ε ≤
μK([xs,K] \ K ∩ ∂K)
μK([˜xs,K] \ K ∩ ∂K) ≤ 1 + ε. (4.35)
By (4.25) and as φi → 0, i = 1, 2 as s → 0, one can choose ˜hR = ˜hR(s) = 3R (x,s)
R+(x,s)
so small that
T

[xs,K] \ K ∩ ∂K

= [T(xs), T(K)] \ T(K) ∩ ∂T(K)
⊂ H−

x −˜hRNK(x),NK(x)

∩ B(R).
Let ˜ts =
q
6R2 (x,s)
R+(x,s) be the distance from x to any point in H

x −˜hRNK(x),NK(x)


∂B(R). Then
T

[xs,K] \ K ∩ ∂K

⊆ Bn
2 (x, ˜ts) ∩ ∂T(K).
(4.24), (4.35) and Lemma 4.1 (i) then give
|T

[xs,K] \ K ∩ ∂K

| ≥ (1 − ε)3|T

[˜xs,K] \ K ∩ ∂K

|
≥ (1 − ε)3β|Bn
2 (x, ts) ∩ ∂T(K)|. (4.36)
Furthermore, by (4.33), one has
|Bn
2 (x, ts) ∩ ∂T(K)| ≥ |H(x − hNK(x),NK(x)) ∩ T(K)|
≥ |H(x − hNK(x),NK(x)) ∩ B(r)|
=

4R2r(x, s) + 4Rr(x, s)2 − 4R2(x, s)2
(R + (x, s))2
n−1
2
|Bn−1
2 |.
31
Since (x, s) → 0 as s → 0, for ¯s" small enough, and any 0 < s < ¯s", one get
|Bn
2 (x, ts) ∩ ∂T(K)| ≥

2R
R + (x, s)
p
r(x, s) − (x, s)2
n−1
|Bn−1
2 |
≥ 2−n (r(x, s))
n−1
2 |Bn−1
2 |. (4.37)
A computation similar to (4.34) shows that for all 0 < s ≤ ¯s" with (a possibly new)
¯s" small enough
|Bn
2 (x, ˜ts) ∩ ∂T(K)| ≤ (1 + ε) |B(R) ∩ H(x −˜ hNK(x),N)|
= (1 + ε) |Bn−1
2 |
√6R
p
R(x, s)
R + (x, s)
!n−1
≤ 3n |Bn−1
2 | (r(x, s))
n−1
2 (4.38)
(4.36), (4.37) and (4.38) imply that
|T

[xs,K] \ K ∩ ∂K

| ≥ (48)−n−1β|Bn
2 (x, ˜ts) ∩ ∂T(K)|.
This shows that T

[xs,K] \ K ∩ ∂K

shrinks nicely to x. Therefore also [xs,K] \ K∩
∂K shrinks nicely to x.
Lemma 4.2 Let K be a convex body in C2+
and f : ∂K → R an integrable, μKalmost
everywhere strictly positive function. Then for μK-almost all x ∈ ∂K one
has
lim
s→0
cn
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
s
2
n−1
=
κK(x)
1
n−1
f(x)
2
n−1
, (4.39)
where xs ∈ ∂Kf,s is such that x ∈ [0, xs].
Proof. It is enough to consider x ∈ ∂K such that f(x) > 0. As x and xs are
collinear and as (1 + t)n ≥ 1 + tn for t ∈ [0, 1), one has for small enough s,
hx,NK(x)i
n

kxsk
kxk
n
− 1

= hx,NK(x)i
n

1 + kxs − xk
kxk
n
− 1

≥ (x, s).
Recall that (x, s) =
D
x
kxk
,NK(x)
E
kxs−xk = hxs−x,NK(x)i is the distance from
xs to H (x,NK(x)).
32
Similarly, as (1 + t)n ≤ 1 + nt + 2nt2 for t ∈ [0, 1), one has for s small enough,
hx,NK(x)i
n

kxsk
kxk
n
− 1

≤ (x, s)

1 +
2n
n

kxs − xk
kxk

. (4.40)
Hence for ε > 0 there exists s" ≤ s0 such that for all 0 < s ≤ s"
1 ≤
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n(x, s) ≤ 1 + ε.
K is strictly convex as K ∈ C2+
. Thus, μf (∂K ∩ [xs,K] \ K) = s by Lemma 3.3
(iv). Therefore
1 ≤
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
μf (∂K ∩ [xs,K] \ K)
2
n−1
n s
2
n−1 (x, s) ≤ 1 + ε.
By Lemma 4.1 (iv) and (4.35), it then follows that we can choose (a new) s" so
small that we have for all s ≤ s"
1 − c1ε ≤
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
f(x) μK(∂K ∩ [ ˜ xs,K] \ K)
2
n−1
n s
2
n−1 (x, s) ≤ 1 + c2ε.
(4.41)
with absolute constants c1 and c2.
Let T be as in (4.23) and let r = r(x) and R = R(x) be as in (4.25). We abbreviate
again B(r) = Bn
2 (x−rNK(x), r) and B(R) = Bn
2 (x−RNK(x),R). Let hr = hr(s) =
r(x,s)
r+(x,s) . As hr → 0 as s → 0, we have for all s sufficiently small that hr <
where is as in (4.26). Hence by (4.25)
H− (x − hrNK(x),NK(x)) ∩ B(r) ⊂ H− (x − hrNK(x),NK(x)) ∩ T(K).
If we denote by PH the orthogonal projection onto the hyperplane H− (x − hrNK(x),NK(x)),
this then implies that
PH
􀀀
∂K ∩ [˜xs,K] \ K

⊃ PH
􀀀
∂T−1(B(r)) ∩ [˜xs, T−1(B(r))] \ T−1(B(r))

= H (x − hrNK(x),NK(x)) ∩ T−1(B(r)).
33
Hence for s sufficiently small
μK(∂K ∩ [˜xs,K] \ K) ≥

PH

∂K ∩ [˜xs,K] \ K



H (x − hrNK(x),NK(x)) ∩ T−1(B(r))

=

T
􀀀
H (x − hrNK(x),NK(x)) ∩ T−1(B(r))

=

H (x − hrNK(x),NK(x)) ∩ B(r)

=
􀀀
1 − (x,s)
2r
n−1
2
􀀀
1 + (x,s)
r
n−1 (2r(x, s))
n−1
2 |Bn−1
2 |
≥ (1 − c3ε)
(2(x, s))
n−1
2 p
κK(x) |Bn−1
2 | (4.42)
where c3 > 0 is an absolute constant.
By (4.25), one has
PH
􀀀
∂K ∩ [˜xs,K] \ K

⊆ PH
􀀀

􀀀
T−1(B(R))

∩ [˜xs, T−1(B(R))] \ T−1(B(R))

= H (x − hRNK(x),NK(x)) ∩ T−1(B(R))
where H = H (x − hRNK(x),NK(x)). The equality follows as hR = R(x,s)
R+(x,s) .
Together with (4.33), for s" small enough, whenever 0 < s < s", one has
μK(∂K ∩ [˜xs,K] \ K) ≤ (1 + ε)|PH(∂K ∩ [˜xs,K] \ K)|
≤ (1 + ε)

H (x − hRNK(x),NK(x)) ∩ T−1(B(R))

.
A calculation similar to (4.42) then shows that with an absolute constant c4
μK(∂K ∩ [˜xs,K] \ K) ≤ (1 + c4ε)
(2(x, s))
n−1
2 p
κK(x) |Bn−1
2 |. (4.43)
Combining (4.41), (4.42) and (4.43), we prove the formula (4.39), i.e.,
lim
s→0
cn
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n s
2
n−1
=
κK(x)
1
n−1
f(x)
2
n−1
.
Lemma 4.3 Let K be a convex body in C2+
. Let c > 0 be a constant and f :
∂K → R an integrable function with f ≥ c μK-almost everywhere. Then there
34
exists s ≤ s0, such that for all s ≤ s,
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
s
2
n−1 ≤ c(K, n),
where c(K, n) is a constant (depending on K and n only), and x and xs are as in
Lemma 4.2.
Proof.
As K ∈ C2+
, by the Blaschke rolling theorem [36], there exists r0 > 0 such that
for all x ∈ ∂K, Bn
2 (x − r0NK(x), r0) ⊆ K. Let γ be such that 0 < γ ≤ min{1, r0}.
By Lemmas 3.1 (i) and 3.2 (ii), K = Kf,0 =
T
s>0Kf,s. Therefore there exists
s = s
≤ s0, such that for all s ≤ s, Kf,s ⊆ (1 + γ)K. Hence for xs ∈ ∂Kf,s and
x = [0, xs] ∩ ∂K, kxsk
kxk ≤ 1 + γ, or equivalently -as x and xs are collinear-
kxsk
kxk − 1 = kxs − xk
kxk ≤ γ ≤ 1. (4.44)
Together with (4.40), one has for all s ≤ s (with a possibly smaller s)
hx,NK(x)i

kxsk
kxk
n
− 1

≤ (x, s) [n + 2n] . (4.45)
As K ∈ C2+
, K is strictly convex. Hence, by Lemma 3.3 (iv), (4.25) and as
f ≥ c on ∂K
s = μf (∂K ∩ [xs,K] \ K) =
Z
@K[xs,K]\K)
fdμK ≥ c μK
􀀀
∂K ∩ [xs,K] \ K

≥ c

H (x,NK(x)) ∩ [xs, T−1(B(r))]

= c

H (x,NK(x)) ∩ [T(xs), B(r)]

where T is as in (4.23) and r = r(x) is as in (4.25).
As in the proof of Lemma 3.2 (i), H (x,NK(x)) ∩ [T(xs), B(r)] contains a (n−1)-
dimensional Euclidean ball of radius at least
r
p
2r(x, s) + 2(x, s)
2r + (x, s) ≥
α
1 + 2α
p
2r0(x, s).
The inequality follows as (x, s) = kxs−xk
kxk hx,NK(x)i ≤

≤ 1
, which is a direct
consequence of (3.19) and (4.44).
35
Hence s ≥ c


1+2
p
2r0(x, s)
n−1
|Bn−1
2 | and with (4.45) we get that
hx,NK(x)i
s
2
n−1

kxsk
kxk
n
− 1

≤ (n + 2n)

1 + 2α
α
2 􀀀
2r0 c
2
n−1 |Bn−1
2 |
2
n−1

−1.
Finally, we also need the following lemma. It is well known and we omit the proof.
Lemma 4.4 Let K be a convex body and L be a star-convex body in Rn.
(i) If 0 ∈ int(L) and L ⊂ K, then
|K| − |L| =
1
n
Z
@Khx,NK(x)i

1 −

kx′k
kxk
n
dμK(x)
where x ∈ ∂K and x′ ∈ ∂L ∩ [0, x].
(ii) If 0 ∈ int(K) and K ⊂ L, then
|L| − |K| =
1
n
Z
@Khx,NK(x)i

kx′k
kxk
n
− 1

dμK(x)
where x ∈ ∂K, x′ ∈ ∂L and x = ∂K ∩ [0, x′].
Proof of Theorem 4.1.
As K ∈ C2+
, K is strictly convex. By Lemmas 4.4, 4.2, 4.3 and Lebegue’s Dominated
Convergence theorem
cn lim
s→0
|Kf,s| − |K|
s
2
n−1
= cn lim
s→0
Z
@K
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n s
2
n−1
dμK(x)
= cn
Z
@K
lim
s→0
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n s
2
n−1
dμK(x)
=
Z
@K
κK(x)
1
n−1
f(x)
2
n−1
dμK(x).
References
[1] A. D. Aleksandrov, On the theory of mixed volumes of convex bodies. II. New
inequalities between mixed volumes and their applications, Mat. Sb. (N. S.) 2
(1937) 1205-1238. [Russian]
36
[2] S. Alesker, Continuous rotation invariant valuations on convex sets, Ann. of
Math. 149 (1999) 977-1005.
[3] S. Alesker, Description of translation invariant valuations on convex sets with a
solution of P. McMullen’s conjecture, Geom. Funct. Anal. 11 (2001) 244-272.
[4] B. Andrews, Contraction of Convex Hypersurfaces by their Affine Normal, J.
Differential Geom. 43 (1996) 207-230.
[5] B. Andrews, Gauss curvature flow: The fate of the rolling stones, Invent. Math.
138 (1999) 151-161.
[6] J. Bourgain and V. D. Milman, New volume ratio properties for convex symmetric
bodies in Rn, Invent. Math. 88 (1987) 319-340.
[7] Y. D. Burago and V. A. Zalgaller, Geometric Inequalities, Springer-Verlag,
Berlin, 1988.
[8] H. Busemann, Convex Surface, Interscience Tracts in Pure and Appl. Math.,
No. 6, Interscience, New York, 1958. MR 21 #3900.
[9] S. Campi and P. Gronchi, The Lp Busemann-Petty centroid inequality, Adv.
Math. 167 (2002) 128-141.
[10] W. Chen, Lp Minkowski problem with not necessarily positive data, Adv. Math.
201 (2006) 77-89.
[11] K. Chou and X. Wang, The Lp-Minkowski problem and the Minkowski problem
in centroaffine geometry, Adv. Math. 205 (2006) 33-83.
[12] B. Fleury, O. Gu´edon, and G. Paouris, A stability result for mean width of
Lp-centroid bodies, Adv. Math. 214 (2007) 865-877.
[13] G. B. Folland, Real Analysis, Wiley, New York, 1999.
[14] R. J. Gardner, Geometric Tomography, Cambridge Univ. Press, 1995.
[15] R. J. Gardner and G. Zhang, Affine inequalities and radial mean bodies, Amer.
J. Math. 120 (3) (1998) 505-528.
[16] P. M. Gruber, Aspects of approximation of convex bodies, Handbook of Convex
Geometry, vol. A, North- Holland, 1993, pp. 321-345.
[17] G. H. Hardy, J. E. Littlewood and G. P´olya, Inequalities, 2nd ed., Cambridge
Univ. Press, 1952.
[18] C. Hu, X. Ma and C. Shen, On the Christoffel-Minkowski problem of Fiery’s
p-sum, Calc. Var. Partial Differential Equations 21 (2) (2004) 137-155.
[19] B. Klartag, A central limit theorem for convex sets, Invent. Math. 168 (2007)
91-131.
37
[20] G. Kuperberg, From the Mahler conjecture to Gauss linking integrals, Geom.
Funct. Anal. 18 (2008) 870-892.
[21] K. Leichtweiss, Konvexe Mengen, Springer-Verlag, Berlin, 1980.
[22] M. Ludwig and M. Reitzner, A characterization of affine surface area, Adv.
Math. 147 (1999) 138-172.
[23] M. Ludwig andM. Reitzner, A classification of SL(n) invariant valuations, Ann.
of Math., in press.
[24] M. Ludwig, C. Sch¨utt and E. Werner, Approximation of the Euclidean ball by
polytopes, Studia Math. 173 (2006) 1-18.
[25] E. Lutwak, Dual mixed volumes, Pacific J. Math. 58 (1975) 531-538.
[26] E. Lutwak, Mixed affine surface area, J. Math. Anal. Appl. 125 (1987) 351-360.
[27] E. Lutwak, The Brunn-Minkowski-Firey theory I: mixed volumes and the
Minkowski problem, J. Differential Geom. 38 (1993) 131-150.
[28] E. Lutwak, The Brunn-Minkowski-Firey theory. II. Affine and geominimal surface
areas, Adv. Math. 118 (2) (1996) 244-294.
[29] E. Lutwak and V. Oliker, On the regularity of solutions to a generalization of
the Minkowski problem, J. Differential Geom. 41 (1995) 227-246.
[30] E. Lutwak, D. Yang and G. Zhang, Lp affine isoperimetric inequalities, J. Differential
Geom. 56 (2000) 111-132.
[31] E. Lutwak, D. Yang and G. Zhang, Sharp affine Lp Sobolev inequalities, J.
Differential Geom. 62 (2002) 17-38.
[32] E. Lutwak, D. Yang and G. Zhang, On the Lp-Minkowski Problem, Trans. Amer.
Math. Soc. 356 (2004) 4359-4370.
[33] M. Meyer and E. Werner, The Santal´o-regions of a convex body, Trans. Amer.
Math. Soc. 350 (11) (1998) 4569-4591.
[34] M. Meyer and E. Werner, On the p-affine surface area, Adv. Math. 152 (2000)
288-313.
[35] G. Sapiro and A. Tannenbaum, On Affine Plane Curve Evolution, J. Funct.
Anal. 119 (1994) 79-120.
[36] R. Schneider, Convex Bodies: The Brunn-Minkowski theory, Cambridge Univ.
Press, 1993.
[37] C. Sch¨utt and E. Werner, Random polytopes of points chosen from the boundary
of a convex body, in: GAFA Seminar Notes, in: Lecture Notes in Math., vol.
1807, Springer-Verlag, 2002, pp. 241-422.
38
[38] C. Sch¨utt and E. Werner, Surface bodies and p-affine surface area, Adv. Math.
187 (2004) 98-145.
[39] A. Stancu, The Discrete Planar L0-Minkowski Problem, Adv. Math. 167 (2002)
160-174.
[40] A. Stancu, On the number of solutions to the discrete two-dimensional L0-
Minkowski problem, Adv. Math. 180 (2003) 290-323.
[41] N. S. Trudinger and X. Wang, The affine Plateau problem, J. Amer. Math. Soc.
18 (2005) 253-289.
[42] W. Wang and G. Leng, Lp-mixed affine surface area, J. Math. Anal. Appl. 335
(2007) 341-354.
[43] X. Wang, Affine maximal hypersurfaces, in: Proceedings of the International
Congress of Mathematicians, vol. III, Beijing, 2002, pp. 221-231.
[44] E. Werner, Illumination bodies and affine surface area, Studia Math. 110 (1994)
257-269.
[45] E. Werner and D. Ye, New Lp-affine isoperimetric inequalities, Adv. Math. 218
(2008) 762-780.
Elisabeth Werner
Department of Mathematics Universit´e de Lille 1
Case Western Reserve University UFR de Math´ematique
Cleveland, Ohio 44106, U. S. A. 59655 Villeneuve d’Ascq, France
elisabeth.werner@case.edu
Deping Ye
Department of Mathematics
Case Western Reserve University
Cleveland, Ohio 44106, U. S. A.
dxy23@case.edu
39

jurnal MATEMatika

Acta Mathematica Academiae Paedagogicae Ny´ıregyh´aziensis
24 (2008), 391–395
www.emis.de/journals
ISSN 1786-0091
RANGE-COMMUTATIVITY MAPS ON *-ALGEBRAS
A. TAGHAVI
Abstract. Let A and B be ¤-algebras, it is shown that if ' be a linear map
from A into B such that Á(xp) commutes with Á(x¤) for all x in A then
the range of ' is commutative. The paper also studies other conditions for
commutativity range of ' on ¤-algebra A.
1. Introduction
A map ' from a Banach algebra A into a Banach algebra B is called preserving
commutativity if
(1) '(x)'(y) = '(y)'(x), whenever xy = yx,
and it is called Range-commutativity if
(2) '(x)'(y) = '(y)'(x), 8x, y 2 A.
The problem of describing commutativity preserving linear maps is one of
the most studied linear preserver problems. The usual solution is that they are
standard. The problem was first considered in the case where A = Mn(F) is the
algebra of n×n matrices over F [7], and has been afterwards studied in various
much more general algebras (see [3, 6] for references). A rather general theorem,
and also the first ring-theoretic result in this area, was obtained in [3] by Bresar.
This theorem treats the case where A is a prime algebra satisfying certain
technical conditions, and instead of the commutativity preserving assumption it
only requires a milder assumption that
(3) '(x2)'(x) = '(x)'(x2), 8x 2 A.
The paper [3], particularly the idea to consider the condition (3), initiated a
systematic study of different topics in ring theory and functional analysis; the
reader is referred to [1, 4, 6] for historical accounts and further references. In
the present paper we shall use this idea in a somewhat different way.
2000 Mathematics Subject Classification. 46J10, 47B48.
Key words and phrases. *-algebra, Operator algebra, Commutativity preserving map.
391
392 A. TAGHAVI
Bresar and Semrl [4] proved the following theorem:
Theorem 1.1 ([5]). Let F be a Field with char(F) = 0, and let A be a finite
dimensional central simple algebra over F with dimA 6= 4. If a linear map
f : A −! A satisfies (3), then f is either a standard commutativity preserving
map or its range is commutative.
Can the assumption that A is finite dimensional be omitted? This is the
question that initiated the present paper. The paper consider the case where A is a ¤-algebra.
We will, first consider a more general problem concerning certain bilinear
maps and then is devoted to the proof of the main result.
Let A be a ¤-algebra, X be a vector space and f : A×A −! X. Our question
is: does the condition
(4) f(x, y) = 0 8x, y 2 A
follow from the condition
(5) f(x, x) = f(x2, x¤) = 0, 8x 2 A.
The connection with the aforementioned question is simple. Namely, if Á is a
linear map ': A −! B such that Á(xp) commutes with Á(x¤) for all x in A,
then a map f defined by f(x, y) = '(x)'(y)−'(y)'(x) satisfies (5). Therefore,
the affirmative answer to the latter question implies the affirmative answer to
the former question.
This paper considers range-commutativity mappings on ¤-algebras. We show
that if ' be a linear or conjugate linear map from A into B satisfies
(6) '(x2)'(x¤) = '(x¤)'(x2), 8x 2 A,
then the range of ' is commutative. The paper also studies other conditions for
commutativity range of ' on ¤-algebra A.
2. Main results
Theorem 2.1. Let A be a unital ¤-algebra with identity e and X be a vector
space and f : A × A −! X be a linear map with respect to first (or second)
component and it is an additive map with respect to second (or first) component.
Then each of the following two conditions implies f(x, y) = 0 for every x, y 2 A:
(i) f(x, x) = f(x¤, x) = 0, 8x 2 A,
(ii) f(x, x) = f(x¤x, x) = 0, 8x 2 A
Proof. For x, y 2 A the assumption f(x, x) = 0 shows that
(7) f(x, y) + f(y, x) = 0,
and
f(x, y) + f(y, x) = f(x, y) + f(y, x) + f(x, x) + f(y, y) = f(x + y, x + y) = 0.
RANGE-COMMUTATIVITY MAPS ON *-ALGEBRAS 393
(i) Let x, y 2 A, by replacing x by x+y in (i) we have f((x+y)¤, (x+y)) = 0.
It implies that
(8) f(x¤, y) + f(y¤, x) = 0.
Now, assume for a moment that x is self-adjoint. By using (7) and (8) we
conclude
(9) f(x, y) + f(y¤, x) = −f(y, x) + f(y¤, x) = f(y − y¤, x) = 0.
By replacing y by iy in (9) we have
(10) f(y + y¤, x) = 0.
By combining (9) and (10) we obtain f(x, y) = f(y, x) = 0. Note that, the
first equality is held by (7). Now if x is arbitrary element in A we can write
x = x1 +ix2 which x1 and x2 are self-adjoint elements in A. Linearity of f with
respect to first component follows that f(x, y) = 0 for all x, y 2 A.
(ii) Let x, y 2 A, by replacing x by x + y in (ii) we obtain f((x + y)¤(x +
y), x + y) = 0. It follows that
(11) f(x¤x, y) + f(y¤y, x) + f(x¤y + y¤x, x + y) = 0,
By replacing y by −y in (11) we have
(12) −f(x¤x, y) + f(y¤y, x) + f(x¤y + y¤x,−x + y) = 0.
Combining (11) and (12) follows that
(13) f(x¤x, y) + f(x¤y + y¤x, x) = 0.
Now by using (7), applying (13) once by replacing x by e and once by replacing
y by e we have f(x, e) = 0 and f(x¤, x) = 0. By (i) we yield f(x, y) = 0 for
every x, y 2 A. ¤
Theorem 2.2. Let A be a unital ¤-algebra with identity e and X be a vector
space and f : A × A −! X be a bilinear map satisfies:
(iii) f(x, x) = 0, 8x 2 A+,
(iv) f(x2, x¤) = 0, 8x 2 A.
Then f(x, y) = 0 for every x, y 2 A. Where A+ is the set of all positive elements
in A.
Proof. Let x 2 A+ we have
(14) f(x, e)+f(e, x) = f(x, e)+f(e, x)+f(x, x)+f(e, e) = f(x+e, x+e) = 0.
Since every self adjoint element is difference of two positive elements it is true
for every self adjoint elements.
Let x, y 2 A, we can replace x once by x + |x¤| and once by x − |x¤| in (iv)
to derive
f((x + |x¤|)2, x¤ + |x¤|) = 0,
394 A. TAGHAVI
f((x − |x¤|)2, x¤ − |x¤|) = 0.
These imply that
(15) f(|x¤|2, x¤) + f(x|x¤| + |x¤|x, |x¤|) = 0.
Since f is bilinear by replacing x by ix in (15) we obtain f(|x|2, x) = 0. Similarly
to the proof of Theorem 2.1 (ii) and by applying (14) we conclude f(x, e) = 0
and f(x¤+x, x) = 0. Since f is a bilinear map we obtain f(x¤, x) = f(x, x) = 0.
By Theorem 2.1 (i) we yield f(x, y) = 0, for every x, y 2 A. ¤
Theorem 2.3. Let A and B be ¤-algebras and ': A −! B be a linear (or
conjugate linear) map, then each of the following three conditions implies the
range of ' is commutative:
v '(x¤)'(x) = '(x)'(x¤), 8x 2 A,
vi '(x¤x)'(x) = '(x)'(x¤x), 8x 2 A,
vii '(x2)'(x¤) = '(x¤)'(x2), 8x 2 A.
Proof. First, let A be a unital ¤-algebra with identity e. Define
f : A × A −! B
with
f(x, y) = '(x)'(y) − '(y)'(x), 8x, y 2 A.
f is a bilinear map and f(x, x) = f(x¤, x) = 0, for all x 2 A, if (v) is true,
f(x, x) = f(x¤x, x) = 0, for all x 2 A if (vi) is true and f(x, x) = f(x2, x¤) = 0,
for all x 2 A if (vii) is true. According to Theorem 2.1 and 2.2 the range of '
is commutative.
Now, assume that A is without identity. We can consider two case
1)B is without identity
2)B has identity e.
In the first case we set
e A = A
M
C
and
e B = B
M
C
which are unitization of A and B respectively. We can extent ' to
e': e A −! e B
as follows: e'(x, ¸) = ('(x), ¸). In the second case we can extent ' to e': e A −! B as follows: e'(x, ¸) = '(x) + ¸e. In both case it is easy to see that e' as defined
above satisfies (v), (vi) and (vii). Therefore, the range of e' is commutative. It
follows immediately the range of ' is commutative.
Now, assume that ' is conjugate linear map. Define ¯'(x) = '(x)¤, then
¯' is a linear map which satisfies (v), (vi) and (vii). Hence the range of ¯' is
commutative. Consequently, the range of ' is commutative. ¤
RANGE-COMMUTATIVITY MAPS ON *-ALGEBRAS 395
References
[1] P. Ara and M. Mathieu. Local multipliers of C¤-algebras. Springer Monographs in Mathematics.
Springer-Verlag London Ltd., London, 2003.
[2] G. M. Bergman. Centralizers in free associative algebras. Trans. Amer. Math. Soc.,
137:327–344, 1969.
[3] M. Breˇsar. Commuting traces of biadditive mappings, commutativity-preserving mappings
and Lie mappings. Trans. Amer. Math. Soc., 335(2):525–546, 1993.
[4] M. Breˇsar. Commuting maps: a survey. Taiwanese J. Math., 8(3):361–397, 2004.
[5] M. Breˇsar and P. ˇSemrl. Commutativity preserving linear maps on central simple algebras.
J. Algebra, 284(1):102–110, 2005.
[6] M. Omladiˇc, H. Radjavi, and P. ˇSemrl. Preserving commutativity. J. Pure Appl. Algebra,
156(2-3):309–328, 2001.
[7] W. Watkins. Linear maps that preserve commuting pairs of matrices. Linear Algebra and
Appl., 14(1):29–35, 1976.
Department of Mathematic,
Mazandaran University,
Babolsar, Iran,
E-mail address: taghavi@nit.ac.ir

jurnal MATEMatika

arXiv:0812.4284v1 [math.MG] 22 Dec 2008
A REMARK ABOUT MAHLER’S CONJECTURE AND
THE MAXIMUM VALUE OF BOX SPLINES
ZHIQIANG XU
Abstract. In this paper, we recast a special case of Mahler’c
conjecture by the maximum value of box splines. This is the case
of polytopes with at most 2n + 2 facets. An asymptotic formula
for univariate box splines is given. Based on the formula, Mahler’s
conjecture is proved in this case provided n is big enough.
1. introduction
Let K be a symmetric convex body in Rn, and let K∗ be its polar
{x : | hy, xi | ≤ 1 for all y ∈ K}. An old conjecture of Mahler is
(1) vol(K) · vol(K∗) ≥
4n
n!
.
We note that an n-dimensional parallelepiped, which has 2n facets,
gives equality in (1). So, the first non-trivial case of Mahler’s conjecture
is the symmetric convex body K with 2n + 2 facets. Such K can be
realized, up to affine invariance, as a one-codimensional section of an
(n + 1)-dimensional cube. This case has been raised as a separate
problem by Ball [1]. Ball also shows an interesting relation between the
special case of Mahler’s conjecture and solutions of a scaling equation.
For each r ∈ R, we set
(2) 'A(r) := vol ((H + rA) ∩ Qn)
where Qn := [−1
2 , 1
2 ]n, A := (a1, . . . , an) is an unit vector and H :=
hAi⊥. Throughout this paper, without loss of generality, we suppose
0 < a1 ≤ a2 ≤ · · · ≤ an. In particular, Ball shows that the special case
of Mahler’s conjecture is equivalent to
(3) 'A(0) · E

Xn
k=1
ak"k

!
≥ 1,
where "k is a sequence of mutually independent random variables with
distribution P{1} = P{−1} = 1/2. In [7], the authors prove (3) for
n ≤ 8 by reducing the problem to a search over a finite set of polytopes
for each fixed dimension.
1
2 ZHIQIANG XU
An interesting observation is that 'A(·) is a box spline, a popular
tool in approximation theory. Therefore, we can recast the special case
of Mahler’s conjecture by the maximum value of box splines. Using
the saddle point approximation in statistics, we give an asymptotic
formula of 'A(·). Based on the asymptotic formula, we obtain the
following result:
Theorem 1. Suppose that there is a constant c0 so that an/a1 < c0 for
any n. Then there exists a positive integer N0(c0) so that
'A(0) · E

Xn
k=1
ak"k

!
≥ 1
when n ≥ N0(c0).
This paper is organized as follows. In Section 2, we introduce box
splines and show the relation between box splines and the special case
of Mahler’s conjecture. In Section 3, we use the saddle point approximation
to give an asymptotic formula of univariate box splines. Section
4 presents the proof of Theorem 1.
2. Box splines
Suppose that M is a s×n matrix. The box spline B(·|M) associated
with M is the distribution given by the rule [2, 3]
(4)
Z
Rs
B(x|M)(x)dx =
Z
[0,1)n
(Mu)du, ∈ D(Rs).
By taking = exp(−i·) in (4), we obtain the Fourier transform of
B(·|M) as
(5) b B(|M) =
Yn
j=1
1 − exp(−iTmj)
iTmj
, ∈ Cs.
The following formula shows the relation between box splines and the
volume of the section of unit cube (see [4], page 2):
(6) B(x|M) =
voln(P ∩ [0, 1)n) p
| det(MMT )|
,
where P := {y : My = x, y ∈ Rn
P +}. Set A := (a1, . . . , an). Recall that
j a2j
= 1 and 0 < a1 ≤ a2 ≤ · · · ≤ an. Then combining (2) and (6)
we have that
'A(·) = B(· + (a1 + · · · + an)/2|A).
MAHLER’S CONJECTURE AND BOX SPLINES 3
Also, noting B(·|A) reaching the maximum value at (a1+a2+· · ·+an)/2,
(3) is equivalent to
max
x
B(x|A) ≥
1
E (|
Pn
k=1 ak"k|)
.
3. An asymptotic formula of univariate box splines
In this section, we shall present an asymptotic formula of B(·|A). In
[8], Unser et. al. proved that B(·|A) tends to the Gaussian function
as n increase provided a1 = a2 = · · · = an. Here, using the saddle
point approximation in statistics, for the general matrix A, we can
show the box spline B(·|A) also converges to the Gaussian function as
n increases:
Theorem 2.
lim
n→∞
B(x|A) =
p
6/ exp(−6(x −
X
j
aj/2)2),
where the limit may be taken pointwise or in Lp(R), p ∈ [2,∞).
Proof. The saddle point approximation of B(·|A) is (see Theorem 6.1
and 6.2 in [5])
(7)
1
(2)1/2|K′′(s0)|1/2 exp(K(s0) − s0x).
Here,
K(s) := ln
Y
i
exp(ais) − 1
ais
and s0 satisfies K′(s0) = x. We can consider s0 as a function of x. So,
(8) K′(s0) − x = 0
defines an implicit relationship between s0 and x. Noting K′(0) =
(a1+· · ·+an)/2, the equation (8) implies that s0((a1+· · ·+an)/2) = 0.
Also, by (8), we have
(9) K′′(s0)s′
0 − 1 = 0,
which implies that
s′
0((a1 + · · · + an)/2) = 1/K′′(0) = 12/
X
j
a2j
= 12.
Using the similar method, we have
s′′
0((a1 + · · · + an)/2) = 0,
4 ZHIQIANG XU
and
s′′′
0 ((a1 + · · · + an)/2) = 864

Xn
j=1
a4j
!
/5 = O(1/n).
Using Taylor expansion at (a1 + · · · + an)/2, one has
(10) s0(x) = 12 · (x − (a1 + · · · + an)/2) + O(1/n).
Also, by (9), we have
K′′(s0) = 1/s′
0 = 1/12 + O(1/n).
Combining Taylor expansion of K(·) at 0 and (10), one has
K(s0) = 6(
X
j
aj)(x−(a1+· · ·+an)/2)+6(x−(a1+· · ·+an)/2)2+O(1/n).
By (10), we have
s0 · x = 12(x − (a1 + · · · + an)/2)x + O(1/n)
= 12(x − (a1 + · · · + an)/2)2 + 6(
X
j
aj)(x − (a1 + · · · + an)/2) + O(1/n).
From (7), the saddle point approximation of B(x|A) is
p
6/ exp(−6(x −
X
j
aj/2)2) + O(1/n).
The properties of the saddle point approximation imply this theorem.

4. proof of the main result
To prove the main result, we firstly introduce a lemma.
Lemma 3. Put
F(s) :=
2

Z ∞
0
(1 −

cos(t/√s)

s
)
t

2dt, s
>
0.
Then
E

Xn
k=1
ak"k

!
≥ F(a−2
n ).
Proof. By Lemma 1.3 in [6], we have
E

|
Xn
k=1
ak"k|
!

Xn
k=1
a2
kF(a−2
k ).
MAHLER’S CONJECTURE AND BOX SPLINES 5
Since F is an increasing function (Lemma 1.4 [6]), we have
E

|
Xn
k=1
ak"k|
!

Xn
k=1
a2
kF(a−2
k ) ≥ F(a−2
n ).

Proof of Theorem 1. To prove the theorem, we only need to prove that
there exists a positive integer N0 so that
max
x
B(x|A) ≥
1
E (|
Pn
k=1 ak"k|)
when n ≥ N0. Theorem 2 implies that
(11) lim
n→∞
max
x
B(x|A) =
p
6/.
Since
P
j a2j
= 1 and an/a1 < c0, we have limn→∞ 1/an = ∞. By [6],
we have lims→∞ F(s) =
p
2/. We choose "0 so that
0 < "0 <
p
6/ −
p
/2
2
.
By Lemma 3, there is a positive integer N1 so that
(12)
1
E (|
Pn
k=1 ak"k|) ≤
1
F(a−2
n ) ≤
p
/2 + "0
provided n ≥ N1. The equation (11) implies that there is a positive
integer N2(c0) so that
(13) max
x
B(x|A) ≥
p
6/ − "0
provided n ≥ N2(c0). We set N0(c0) := max{N1,N2(c0)}. Noting that
p
6/ − "0 ≥
p
/2 + "0,
combining (12) and (13), we have
maxB(x|A) ≥
1
E (|
Pn
k=1 ak"k|)
when n ≥ N0(c0).
6 ZHIQIANG XU
References
1. K. Ball, Mahler’s conjecture and wavelets, Discrete and Computational Geometry,
Vol. 13, No.1(1995).
2. C. de Boor and R. Devore, Approximation by smooth multivariate splines,
Trans. Amer. Math. Soc., 276 (1983) 775-788.
3. C. de Boor and K. H¨ollig, B-splines from parallelepipeds, J. Anal. Math,
42(1982/83)99-115.
4. C. de Boor, K. H¨ollig and S. Riemenschneider, Box Splines, Springer-Verlag,
New York, 1993.
5. Daniels, H. E., Saddlepoint approximations in statistics, Annals Math.
Statist., 25(1954) 631-650
6. U. Haagerup, The best constants in the Khintchine inequality, Studia Mathematica,
T. LXX. (1982).
7. M. A. Lopez and S. Reisner, A special case of Mahler’s conjecture, Discrete
Comp. Geom. 20:163-177(1998).
8. M. Unser, A. Aldroubi and M. Eden, On the asymptotic convergence of Bspline
wavelets to Gabor functions, IEEE Transactions on informations theory,
Vol.38, No.2, 1992.

Tuesday, December 23, 2008

BRAinwaRE )(kumpulan tips dan trik windows terpanas

BRAINWARE

Secara umum sudah diketahui kalau Komputer memiliki dua komponen yaitu Hardware dan Software yang mana kedua komponen ini sama pentingnya, kalau salah satu tidak berfungsi sebagaimana mestinya maka sebuah komputer pasti tidak berfungsi sebagaimana mestinya juga. Hardware dapat diartikan sebagai perangkat keras pada sebuah PC, contohnya monitor , mouse, keyboard, Hard disk dan lain-lain. Sedangkan software dapat diartikan sebagai perangkat lunak atau sebuah program pada sebuah PC . Contohnya: Winamp, windows , M office , Anti Virus dan lain-lain. Tapi bisa juga kita artikan sebagai berikut: Hardware/perangkat keras adalah sebuah alat/benda yang bisa kita lihat, sentuh, pegang, Secara fisik ada bentuk dan wujudnya. Sedangkan perangkat lunak secara fisik tidak ada wujudnya. Maka tidak bisa kita sentuh, tidak bisa kita pegang. (yang bisa kita pegang hanya media penyimpanannya saja, seperti disket,cd,dsb). Perangkat lunak memiliki fungsi-fungsi tertentu, dan biasanya untuk mengaktifkan perangkat keras. Bisa juga dikatakan perangkat lunak bekerja di dalam perangkat keras. Namun selain kedua komponen tersebut ada satu lagi komponen yang paling penting dan sering terlupakan yaitu ‘BRAINWARE’. Tanpa komponen ini kedua komponen yang lain yaitu Software dan Hardware tidak bermanfaat. Yang dimaksud dengan ‘BRAINWARE’ ini adalah Orang yang mengoperasikan komputer atau yang lazim disebut User. Posisi ‘Brainware’ ini sangatlah penting dalam pengoperasian komputer karena sebuah sistem komputer akan bekerja sesuai dengan apa yang diperintahkan penggunanya. Jadi si pemgguna komputer harus tahu bagaimana menggunakan sebuah komputer dengan baik. maka dari itu pengetahuan mengenai pengguanan komputer sangatlah penting untuk diketahui agar dalam pengoperasian komputer menjadi lebih baik dan maksimal. Di sini saya akan berbagi sedikit tip atau pengetahuan dalam penggunaan komputer sehari-hari yang saya dapatkan dari internet.

Kumpulan Tips dan Trik Komputer

Trik Menghemat Hard Disk
Dengan bertambahnya data dan aplikasi yang disimpan, lambat laun “ruang kosong” dalam hard disk tentu akan berkurang. Dan kalau ruang tersebut makin sempit, penggunaan komputer akan dihadapkan pada beberapa pilihan:

* Menambah/mengganti hard disk dengan kapasitas yang lebih besar
* Memback-up data pada CD/media penyimpan yang lain
* Atau menghapus beberapa data atau aplikasi yang sudah ada

Namun disamping cara-cara tersebut diatas, ada cara lain yang dapat dilakukan untuk menghemat pemakaian ruang hard disk, yaitu :

1. Mengggunakan fasilitas kompresi yang kini menjadi fasilitas standar dalam sebuah komputer.
Dengan cara ini ukuran file bisa dipadatkan dan tentu menghemat hard disk. Fasilitas ini bisa didapat dengan mudah lewat internet atau dalam CD perangkat lunak.
2. Menghapus semua file yang ada dalam recycle bin.
Beberapa pengguna Komputer tidak menyadari bahwa data-data yang sudah dihapus dan masuk ke dalam recycle bin masih memiliki ukuran data yang memakan hard disk. Oleh karenanya kalau yakin data dalam recycle bin sudah tidak digunakan lagi, lebih baik dihapus. Kemudian akan lebih baik kalau alokasi untuk tempat sampah ini diatur sedemikian rupa sesuai dengan kebutuhan.
3. Kemudian seringkali komputer membuat file-file baru, misalnya file yang dibuat untuk mengcover ketika komputer hang, crash, atau sekedar memback up saat pengguna membuka sebuah data misalnya dengan extensi*.tmp.
File-file yang sudah tidak diperlukan ini lebih baik dihapus. Dalam sistem operasi Windows, langkah ini bisa dilakukan dengan fasilitas Scandisk.
4. Melakukan defragment secara teratur paling tidak sebulan atau dua bulan sekali menjadi langkah yang dapat dilakukan agar file-file dalam komputer tertata dengan rapi.
Ini memang tidak akan menambah ruang kosong hard disk, namun selain lebih efisien cara ini akan mempercepat akses data.


Trik Mensetting Windows Biar Tambah Cepat
1.Scan dengan Antivirus
Salah satu penyebab paling umum PC jadi mengendor kerjanya adalah virus. Beberapa virus memang bisa bikin kinerja PC menurun. Apalagi virus yang menyerang aplikasi-aplikasi tertentu dan menyebabkan kerusakan. Untuk mengatasinya, cara yang paling ampuh adalah dengan menggunakan antivirus. Anda bisa membunuh virus-virus ini dengan antivirus yang bagus. Update-lah secara berkala agar virus-virus baru bias terdeteksi sekaligus dihapus sehingga kinerja PC tidak terganggu. (menurut saya antivirus yang cukup bagus saat ini adalah: Avira / Avast dan anti virus buatan Indonesia yaitu Ansav khusus Ansav terdapat fitur ‘hidden revolever’ yaitu sebuah fitur yang memungkinkan kita untuk memunculkan kembali file yang sudah dihidden oleh suatu virus )

2. Cek Hardware Anda
Anda juga bisa mengecek perangkat-perangkat yang terpasang, apakah masih bekerja secara optimal. Power supply misalnya. Setelah bekerja secara terus menerus, kemungkinan terjadi penurunan kinerja. Tentunya ini akan mempengaruhi kecepatan proses kerja sistem. Kalau kondisinya begini, ganti power supply-nya dengan yang baru. Memori juga demikian. Kalau Anda rasa aplikasi yang Anda jalankan makin banyak, Anda bias mengkalkulasikan, apakah perlu memori tambahan agar kinerja sistem bisa lebih baik. Suhu prosesor juga perlu Anda periksa. Maklum, makin panas prosesor tentu kinerjanya akan semakin berat. Agar tidak terjadi hal demikian, Anda harus perhatikan apakah heatsink fan prosesor masih bekerja dengan baik untuk menjamin panasnya prosesor bukan disebabkan oleh pendinginnya yang bekerja kurang optimal. Penggunaan thermal grease juga dianjurkan untuk mengatasi panas prosesor.

3. Cek Setting di BIOS
Langkah lain yang bisa dilakukan adalah memperhatikan beragam setting pada BIOS di motherboard Anda. Pastikan kecepatan prosesor, memori, dan lain-lain sudah sesuai. Di sini Anda juga bias mengatur beberapa setting agar kecepatan kerja bisa bertambah, semisal fitur Top Performance diatur pada posisi enable.Begitu pula untuk memori Anda bisa tingkatkan CAS Latencynya, dan lain-lain. Update BIOS dengan yang terbaru juga bias mengurangi kelambatan sistem Anda.

4. Cek Driver
Anda juga bias memperhatikan driver-driver yang terpasang untuk masingmasing perangkat. Apabila terdapat konflik driver, Anda bias uninstall dulu untuk kemudian menginstal kembali driver yang sesuai. Anda juga bisa mengupdate beberapa driver vital untuk memperbaiki performa kerja perangkat.

5. Defrag Harddisk
Banyaknya file yang sudah membebani harddisk juga akan mempengaruhi kecepatan sistem dalam beroperasi. Apalagi kalau file tersebut berada pada posisi yang terpencar-pencar di dalam harddisk.Untuk mengurangi masalah kelambatan PC yang disebabkan oleh hal ini, Anda bisa mendefrag harddisk secara reguler.

6. Hapus File-File Sampah
Harddisk yang terlalu disesaki dengan beragam file juga akan memperlambat sistem. Ini akan sangat terasa jika space kosong harddisk tinggal beberapa MB lagi. Untuk menanggulanginya, Anda bisa membuang file-file yang tidak perlu. Trik lainnya adalah menguninstall aplikasi-aplikasi yang tidak terlalu dipakai namun Anda punya software-nya. Kalau diperlukan baru diinstal kembali. Jangan lupa pula untuk menghapus secara berkala beberapa file cookies dan filefile yang tidak berguna lainnya yang biasanya tersimpan dalam folder Temp atau Temporary Internet Files.



Mempercepat Pemanggilan Program di Windows Xp
Kadangkala saat kita menjalankan sebuah program, perlu beberapa waktu untuk menunggu program tersebut siap untuk dijalankan. Hal ini tentunya selain menguji kesabaran juga menghambat pekerjaan apalagi jika kita butuh menggunakannya secepatnya. Kita dapat mempercepatnya dengan melakukan beberapa perubahan di setting Windows. Cara kerjanya Windows bisa diatur agar menjalankan informasi-informasi penting dari beberapa program agar bisa dijalankan lebih cepat dengan merubah banyaknya proses kerja dari suatu program yang disebut dengan Prefetcher.

Caranya:

1. Jalankan regedit.
2. Masuk ke key berikut: "HKEY_LOCAL_MACHINE / SYSTEM / CurrentControlSet / Control / Session Manager / Memory Management / PrefetchParameters".
3. Ketika Anda klik PrefetchParameters, editor bagian kanan akan menunjukkan nilainya.
4. Double click pada Enable Prefetcher dan rubah nilai datanya ke 5 (dalam desimal, jangan heksadesimal).
5. Klik OK dan tutup regedit


1. 150 Trik di Windows


01. Back-up dahulu registry sebelum diedit. Caranya, klik Start|Run,
ketik regedit. Setelah muncul window registry, pilih menu File|Export.
Pada bagian Export range, pilih All dan tentukan nama file, akhiri dengan
klik tombol Save.

02. Banyak program yang sebenarnya terinstalasi dalam Windows, namun
tidak aktif. Untuk mengaktifkannya, masuk ke Control Panel|Add/Remove
Windows Component dan beri tanda centang pada program yang belum aktif.

03. Sebelum menginstalasi program baru atau melakukan perubahan setting
Windows secara keseluruhan, lebih baik buat Restore Point secara manual
dahulu. Caranya, Klik Start|All Programs|Accessories|System Tool|System
Restore dan klik Create a restore point.

04. Jika Anda memutuskan untuk menginstalasi Windows Update yang
sebelumnya sudah didecline, masuk ke Control Panel|System, pilih tab
Automatic Updates dan klik Restore Declined Updates.

05. Untuk mengatur Windows update berjalan sesuai dengan kebutuhan
Anda, atur dulu Windows Update. Caranya, buka System di Control Panel dan
klik tab Automatic Updates. Atur enable atau disable option Keep my
computer up to date.

06. Bila Anda tidak memiliki CD bootable, jangan khawatir. Microsoft
sudah menyediakan
tool gratis untuk membuat disket booting di
http://support.microsoft.com/?kbid=310994.

07. Jika saat instalasi Windows tiba-tiba terhenti, matikan komputer
dan lepas card tambahan. Misalnya sound card. Instal ulang dan pasang
kembali card setelah instalasi selesai.

08. Untuk menambahkan System Administration Tools ada Start Menu, klik
kanan Start|Properties. Masuk dalam tab Start Menu dan klik Customize
kemudian masuk dalam tab Advanced. Geser ke bawah dan beri tanda centang
pada option Display on the All Programs and the Start Menu.

09. Untuk menginstal Back up Utility pada Windows XP Home Edition
carilah file ntbackup.msi di direktori\valudeadd\msft\ntbackup\ di CD
instalasi Windows XP. Jalankan file tersebut dan ikuti langkahlangkahnya.

10. Windows XP secara otomatis akan me-highlight setiap program baru
yang ter-install. Cara menghilangkannya, klik kanan Start|Properties.
Masuk ke tab Start Menu|Customize, kemudian klik tab Advanced dan
hilangkan tanda centang pada opsi Highlight newly installed program.

11. Untuk tampilan film atau game terbaik pada komputer, pastikan bahwa
DirextX terbaru sudah terinstalasi dengan baik. Lihat versi terbarunya
di www.microsoft.com/windows/directx.

12. Ada kalanya hardware yang akan dipasang belum support
Plug-and-Play. Untuk itu, gunakan Add Hardware Wizard yang ada di Control
Panel|System|Hardware untuk mendeteksinya.
–> TOP TIPS ! Mempercepat Kerja Komputer

13. Sebenarnya hanya dibutuhkan waktu tidak lebih dari 2 menit saja
untuk masuk ke Windows sejak komputer dihidupkan. Namun, kadang terasa
sangat lama. Untuk mempercepat loading Windows, ada beberapa hal yang bisa
dilakukan. Misalnya, mengurangi icon di desktop serta tidak menggunakan
wallpaper yang memakan banyak memory. Ganti wallpaper dengan background
berwarna, serta gunakan Desktop Cleanup Wizard yang ada bisa ditemui
dengan klik kanan pada desktop untuk membersihkan
icon. Jangan lupa juga, jalankan defrag secara berkala.
Jika Anda sering menambah atau mengurangi program di komputer,
bersihkan registry secara rutin. Gunakan software bantu seperti Registry
Mechanic dari situs www.winguides.com. Sayangnya, versi trial program ini
hanya bisa digunakan memperbaiki sebanyak 6 sections saja.
Langkah lain yang perlu dilakukan adalah me-remove program yang di-load
secara otomatis saat memulai Windows. Tentu saja, hanya program-program
yang tidak dibutuhkan. Caranya, dengan menghapus semua isi folder
startup dan membuka msconfig melalui Start|Run.

14. Gunakan fitur File and Transfer Setting Wizard untuk memindahkan
file dan setting ke komputer baru. Caranya, klik
Start|AllPrograms|Accessories|System Tools, kemudian jalankan File and Transfer Setting Wizard.

15. Gunakan pengecualian pada security setting di Internet Explorer,
agar proses update melalui halaman Windows Update berjalan lancar.
Caranya, buka Internet Option di menu Tools pada Internet Explorer. Klik tab
Security, pilih Trusted Site dan klik tombol Sites. Isikan nama situs
Windows Update, hilangkan tanda centang pada option Require server
verification… dan klik OK.

16. Jika Anda kehilangan serial number Windows XP, gunakan freeware
Magical Jelly Bean Keyfinder dari
http://www.magicaljellybean.com/keyfinder.shtml.

17. Jika nama yang teregister dalam Windows XP Anda tidak sesuai,
perbaiki melalui registry. Caranya, buka registry dan pilih MY Computer.
Klik menu Edit|Find dan ketik RegOwner. Jika sudah ditemukan, klik kanan,
pilih Modify dan isikan nama yang sesuai. Perubahan ini bisa juga
dilakukan di key RegCompany.Peningkatan Performa Internet dan Jaringan

18. Untuk men-share sebuah folder di komputer Anda ke jaringan, klik
kanan folder tersebut dan pilih Properties. Klik tab Sharing dan enable
option Share this folder on the network. Beri nama dan klik OK.

19. Buat sebuah icon My Network Places di desktop dengan mengklik kanan
area kosong di dekstop dan klik Properties. Pilih tab Desktop|Customize
Desktop. Kemudian buka tab General dan enable option My Network Places.

20. Ada cara mudah mengirim pesan ke komputer lain di jaringan, yakni
menggunakan Console Message. Buka Control
Panel|AdministrativeTools|Computer Management|Action|All Task|Send Console Message. Ketik teks yang
hendak dikirim, tambahkan nama komputer yang hendak dituju dan klik
Send.

21. Untuk mengatur Internet Connection Firewall (ICF), buka Network
Connection di Control Panel, klik kanan koneksi yang ada dan klik
Properties. Buka tab Advanced dan enable option Protect my computer and network
by limitting or preventing access to this computer from Internet.

22. Atur Internet Connection Firewall (ICF) untuk setiap koneksi yang
ada. Baik dial-up maupun broadband. Jika komputer Anda merupakan bagian
dari jaringan yang terhubung ke Internet, pasang ICF hanya di komputer
server.

23. Untuk mengetahui alamat IP Anda, masuk dalam DOS dengan mengetikkan
command di Run. Kemudian ketikkan ipconfig /all.

24. Jika Anda menerima pesan dari Internet melalui Messenger, segera
matikan. Caranya, masuk ke Contol Panel|Administrative Tools|Services,
dan klik ganda Messenger kemudian Stop. Untuk mencegah supaya tidak
terulang, atur supaya Messenger menjadi Disabled di bagian Startup.

25. Matikan Windows Messenger dengan melalui regedit. Buka
HKEY_LOCAL_MACHINE\Software\Policies\Microsoft, kemudian pilih menu Edit|New|Key,
dan beri nama Messenger. Kemudian buat key lagi dengan cara ini di dalam
direktori Messenger dengan nama key-nya Client. Setelah itu, klik menu
Edit|New|DWORD Value, dan beri nama Prevent-Run. Klik kanan value
PreventRun, pilih Modify, isi angka 1 pada Value data, dan klik OK.

26. Untuk mengetahui informasi mengenai koneksi di komputer Anda, klik
Start|All Programs|Accessories|System Tools|System Information. Pilih
menu Tools|Net Diagnostics. Pada window yang terbuka kemudian pilih
option Scan your system. Tunggu hingga proses selesai untuk melihat
hasilnya.

27. Lindungi privasi dengan mencegah aplikasi Windows Media Player
mengirim data mengenai komputer dan kebiasaan Anda menggunakan komputer
melalui Internet ke alamat-alamat tertentu. Caranya mudah, Pada Windows
Media Player, pilih menu Tools|Option. Buka tab Player dan disable option
Aloww internet sites to uniquely your player.

28. Untuk mengunci komputer yang berada dalam sebuah network domain,
tekan tombol Ctrl + Alt + Del bersamaan dan klik option Lock Computer.
Untuk membuka kembali, tekan tombol Ctrl + Alt + Deldan masukkan
password. Konfigurasi Windows yang Mudah dan Cepat.

29. Untuk men-disable fitur autorun, klik kanan pada icon drive CD,
pilih Properties dan masuk dalam tab AutoPlay. Kemudian disable autoplay
untuk setiap jenis file yang tertera pada daftar.

30. Gunakan program Microsoft Clear Type Tuning Control dari
http://www.microsoft.com/typography/cleartype/ untuk mengatur Clear
Type pada komputer.

31. Untuk melihat system file yang secara default di-hidden oleh
Windows XP, pilih tab View dalam menu Tool|Folder Option dalam Windows
Explorer. Enable Display the content of system folder.

32. Untuk meletakkan icon volume control di taskbar, masuk dalam
Control Panel|Sound and Audio dan klik tab Volume. Enable Place volume
control in the Taskbar dan klik OK.

33. Atur tombol Power di keyboard melalui Control Panel|Power Option di
tab Advanced. Tentukan pengaturan tombol Power ini dengan memilih
option yang tersedia.

34. Atur supaya Windows membersihkan Pagefile saat shut down demi
keamanan. Caranya, buka registry dan masuk dalam direktori
HKEY_LOCALMACHINE\SYSTEM\CurrentControlSet\Control|Session Manager. Edit value pada key
Clear-PageFileAtShutdown menjadi 1. Konsekuensinya, proses shut down
akan berlangsung sedikit lebih lama.

35. Atur supaya Windows menampilkan ekstensi setiap file. Caranya, di
Windows Explorer, pilih menu Tool|Folder Option dan tab View. Hilangkan
tanda centang di option Hide file extentions for known file types.

36. Menghapus Komponen yang Terinstal
Banyak komponen Windows yang tidak muncul di Add/Remove Windows
Component sehingga tidak bisa di-uninstall.
a. Buka Notepad dan pilih menu File|Open. Arahkan ke folder
Windows\inf. Isi nama file sysoc.inf. Klik Open untuk membuka file ini.
b. Pilih menu Edit|Replace. Ketik Hide pada kolom Find, namun kosongkan
kolom Replace With, klik Replace All. Tujuannya untuk menghapus semua
kata Hide dalam file ini. Setelah selesai, tutup dan simpan file.
c. Buka Control Panel dan pilih Add/Remove Programs. Kemudian pilih
Add/Remove Windows Component, pada windows yang keluar kemudian akan
tampak beberapa komponen yang sebelumnya tersembunyi.

37. Ubah gambar pada welcome screen dengan cara masuk ke User Account
di Control Panel. Buka account Anda dan klik Change my picture. Tentukan
gambar pilihan Anda dengan mengklik Browse untuk gambar di harddisk
atau memilih di antara gambar yang sudah tersedia.

38. Jika lebih menyukai tampilan Start Menu versi lama, Anda bisa
mengubahnya dengan mengklik kanan tombol Start, pilih Properties. Pilih
Classic Start Menu dan klik Customize untuk mengatur isinya.

39. Untuk menyempurnakan tampilan klasik pada Start Menu, klik kanan
desktop dan pilih Properties. Buka tab Themes, dan pilih Windows Classic
dari Theme list.

40. Tambahkan image pada sebuah folder, sehingga image tersebut yang
akan tampak saat Windows Explorer dalam tampilan thumbnails. Caranya,
klik kanan folder yang hendak diolah, pilih Properties. Klik tab Customize
dan klik Choose Picture. Pilih sebuah gambar dan klik Open|OK.

41. Sesuaikan kapasitas Recycle Bin dengan mengklik kanan icon Recycle
Bin dan memilih
Properties. Isi kapasitas yang Anda inginkan dan klik OK.

42. Pada saat View di-set Details di Windows Explorer, klik kanan
header salah satu kolom untukmengatur kolom apa saja yang ditampilkan. Klik
More bila perlu mengatur setting lainnya.

43. Untuk menambahkan program yang paling sering Anda gunakan dalam
Quick Launch, drag icon program tersebut dalam Quick Launch.

44. Tambahkan address bar pada taskbar, sehingga mempercepat akses ke
sebuah alamat di Internet. Caranya, klik kanan taskbar, pilih
Toolbar|Address. Klik ganda untuk membuka dan menutupnya.

45. Jadikan tampilan Windows Explorer seperti tampilan pada window My
Computer. Caranya, klik kanan icon Window Explorer dan pilih Properties.
Pada Target area, setelah %SystemRoot%\explorer.exe tambahkan /n, /e,
/select, C:\ dan klik OK.

46. Untuk menambahkan sebuah shortcut program di baris paling atas
Start Menu, klik kanan icon-nya di Start Menu kemudian klik Pin to Start
Menu.

47. Supaya sebuah drive atau folder dapat masuk dalam menu Send To,
drag shortcut-nya ke folder \Documen Anda Setting\\SendTo.

48. Mencari folder SendTo? Klik saja Start|Run dan ketik SendTo
kemudian klik OK.

49. Untuk mengosongkan daftar dokumen dalam folder My Recent Document
di Start Menu, klik kanan Start, pilih Properties. Klik Customize dan
buka tab Advanced kemudian klik tombol Clear list. Supaya tidak ada lagi
yang muncul di My Recent Documents, disable option List my most
recently opened documents.

50. Fast User Switching
Dengan Fast User Switching, seorang user tidak perlu logoff sementara
user lain login.
a. Untuk meng-enable Fast User Switching, masuk dalam Control Panel dan
pilih User Accounts. Klik option Change the way user log on or off, dan
enable Use Fast User Switching.
b. Supaya koneksi dial-up tetap berjalan meski Fast User Switching
di-enable, masuk ke registry di direktori
HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\WindowsNT\CurrentVersion\Winlogon.
c. Klik kanan pada panel sebelah kanan dan pilih New |String Value.
Beri nama KeepRasConnections dan beri nilai 1. Restart komputer.

51. Untuk melihat isi sesungguhnya folder My Recent Documents, klik
Start|Run, kemudian ketikkan %UserProfile%\Recent.
Pengaturan Multiple User

52. Untuk berpindah antar user account, tekan tombol Ctrl+Alt+Del,
kemudian pada window Task Manager yang terbuka, pilih tab Users. Klik kanan
nama user yang hendak dipakai dan klik tombol Connect.

53. Tambahkan account Guest supaya orang lain bisa menggunakan komputer
Anda. Caranya, buka User Accounts di Control Panel, klik Guest|Turn On
the Guest Account.

54. Ingatlah untuk selalu login sebagai System Administrator sebelum
melakukan perubahan yang berimbas pada performa komputer.

55. Untuk meng-copy user profile, masuk dalam Control Panel|System.
Masuk dalam tab Advance dan klik tombol Setting di User Profiles. Sorot
profile yang hendak dicopy, kemudian klik Copy to dan tentukan lokasi
penyimpanan profile. Untuk mengubah permission, klik tombol Change.

56. Ganti Welcome screen dengan login dialog untuk menambah tingkat
keamanan. Caranya, masuk dalam User Accounts di Control Panel, pilih
option Change the way user log on or off serta disable Welcome screen.

57. Gunakan folder Shared Documents untuk menyimpan filefile yang bisa
dibuka orang lain dalam jaringan. Folder ini bisa ditemui di My
Documents\Other Places area.

58. Pada Windows XP Pro Edition, password bisa kadaluarsa jika lama
tidak digunakan login. Untuk menghindarinya, atur supaya Windows tidak
melakukan hal tersebut. Caranya, klik Start|Run, ketikkan userpasswords2
dan Enter. Buka tab Advanced dan pilih Advanced user management, klik
tombol Advanced dan pilih Local Users and Groups. Klik kanan nama user
dan pilih Properties. Buka tab General dan enable option Password never
expires.

59. Beri nama setiap partisi atau drive yang ada dengan nama yang
berbeda. Hal ini penting untuk mempermudah pencarian file.

60. Buat sebuah password reset disk sebagai disket darurat saat
kehilangan password. Caranya, masuk ke Control Panel dan buka User Account.
Pilih account Anda dan pilih Prevent a forgotten password untuk memulai
wizard.

Pengaturan File dan Folder

61. Buat sebuah compressed folder dengan cara mengklik kanan area
kosong pada desktop, dan pilih New|Compressed (zipped) Folder. Beri nama dan
drag and drop file yang hendak dikompres dalam folder tersebut.

62. Sebuah compressed folder bisa dilindungi dengan password. Caranya,
buka menu File|Add a Password. Isikan password Anda dan isi sekali lagi
untuk konfirmasi.

63. Sebuah compressed folder tetap bisa ditambah isinya dengan cara
drag and drop file yang hendak dikompres ke dalamnya.

64. Saat menjalankan Disk Cleanup, pilih option Compress old files
untuk mengompresi file-file yang sudah tidak dibutuhkan, sehingga kapasitas
harddisk bisa dihemat.

65. Enkrip atau acak file bisa dilakukan di Windows XP Pro Edition
dengan mengklik kanan My Computer, pilih Properties. Buka tab General dan
klik Advanced. Aktifkan option Encrypt contents to secure data.

66. Cara termudah mengubah nama file atau folder adalah dengan
memilihnya, kemudian menekan tombol F2.

67. Supaya Anda bisa berpindah antar folder dengan cepat, terutama pada
folder-folder yang sering dibuka, buatlah shortcut untuk setiap folder
yang sering diakses.

68. Reset file association dengan cara menhakan tombol Shift saat
mengklik kanan sebuah file. Pilih Open With… Pilih program yang hendak Anda
gunakan untuk membuka file tersebut dan enable option Open use the
selected program to open this kind of file. Kemudian klik OK.

69. Sesuaikan kapasitas penyimpanan file musik dengan cara membuka
Windows Media Player dan memilih menu Tools|Options. Buka tab Copy Music
dan sesuaikan ukuran kapasitas dengan menggeser slider.

70. Jika Anda menyimpan semua dokumen dalam folder My Documents, maka
semua file tersebut lebih mudah di-back-up serta tidak terpengaruh
system restore.

71. Jika sebuah file hasil copy dari CD tidak dapat diakses, maka
hilangkan atribut Read only. Caranya, klik kanan file tersebut dan pilih
Properties. Hilangkan tanda centang pada option Read only.

72. Aturlah tampilan setiap folder dengan mengklik kanan folder
tersebut pada My Computer. Pilih Properties dan klik Customize. Pilih template
yang tersedia atau pilih gambar sesuai dengan keinginan Anda. Klik OK
untuk menyimpan hasil setting.

73. Anda bisa me-rename beberapa file sekaligus yang memiliki karakter
sama dalam folder yang sama juga. Misalnya kumpulan foto atau image.
Caranya, pilih semua file yang hendak di-rename, klik kanan file pertama
dan klik Rename. Beri nama,misalnya Koleksi.JPG, maka semua file yang
lain akan berubah namanya menjadi Koleksi (1).JPG, Koleksi (2).JPG, dan
seterusnya.

74. Mengirim Faks
Tahukah Anda, bahwa Windows XP memiliki tool untuk mengirim dan
menerima faks?
a. Buka Control Panel, pilih Add/Remove Program dan klik Add/Remove
Windows Component. Enable Fax Services dan klik Next.
b. Klik Start|All Programs|Accessories|Communication|Fax|Fax Console
untuk menjalankan Fax Configuration Wizard. Atur semua setting, termasuk
nomor faks dan modem yang digunakan serta pengaturan incoming fax.
c. Buat semua faks dari Fax Console. Periksa semua isi sebelum dikirim,
termasuk setting dan sebagainya.

Mengatur Kualitas Cetak

75. Anda bisa mencetak beberapa foto bersamaan, sekaligus mengatur
layout serta option lainnya dengan cara mudah. Pertama, buka file-file foto
dalam folder My Pictures, kemudian pilih menu File |Print. Akan muncul
Photo Printing Wizard. Atur printer yang hendak digunakan dan layout
yang akan dipakai. Beberapa printer memungkinkan mencetak beberapa foto
dalam satu kertas.

76. Daripada harus mengatur setting printer setiap kali menggunakannya,
buatlah copy printer Anda. Atur masing-masing dengan option berbeda
sesuai yang Anda butuhkan.

77. Anda bisa menggunakan karakter yang tidak ada di keyboard, namun
bisa digunakan dalam Windows dengan fitur Character Map. Untuk
membukanya, Anda bisa menemukan di Start|All Programs|Accessories|System
Toolss|Character Map. Atau ketik “charmap” tanpa tanda petik di Start|Run.

78. Untuk meng-capture sebuah adegan di file movie menjadi sebuah
image, putarlah film tersebut dan tekan tombol Stop saat adegan yang
diinginkan. Klik Take Picture, simpan di folder Anda dan beri nama.

Memunculkan Penampakan Fitur Tersembunyi

79. Anda tertarik dengan musik pembuka Windows yang orisinal? Temukan
file-nya di \windows\oobe\system32\images\tittle.wma.
80. Ada dua edisi Windows Plus! yang beredar, yakni Microsoft Plus! for
Windows dan Digital Media Edition (DME). Simak
www.microsoft.com/windows/plus/PlusHome.asp dan coba versi trial DME.

81. Ketikkan “iexpress” tanpa tanda petik pada kotak dialog Start|Run.
Anda akan menemukan sebuah utility untuk mengompres dan membuat file
self extraction.

82. Anda akan menemukan beberapa tool yang kurang populer, namun punya
kelebihan luar biasa. Tool-tool tersebut bisa ditemui di folder
support\tools dalam CD instalasi Windows XP. Gunakan program suptools.msi
untuk menginstalasi tool-tool tersebut dalam komputer Anda.

83. Ada banyak wizard atau stepby-step di Windows. Untuk mengetahui
wizard apa saja, buka Help and Support dan ketikkan “Wizard” tanpa tanda
petik pada kolom pencarian. Windows akan memunculkan daftar wizard yang
tersedia. Anda tinggal memilihnya.

84. Untuk membuat karakter sendiri, klik Start|Run dan ketikkan
Eudcedit. Buat sesuai keinginan Anda, asalakan masih dalam ukuran 64×64 grid.
Simpan dengan memilih menu Edit|Save Character.

85. Untuk meletakkan karakter dalam Character Map ke dalam dokumen,
Anda cukup membuka Character Map dan memilih karakter tersebut. Kemudian
Anda tinggal mengcopy dan paste-nya ke dalam dokumen Anda.

Menghias Desktop

86. Anda bisa mengatur pointer mouse sesuai keinginan dengan membuka
tab Pointer dalam menu Mouse di Control Panel. Klik Browse untuk mencari
pointer yang sesuai keinginan Anda.

87. Untuk menampilkan daftar dokumen yang terakhir diakses di Start
Menu pada Windows XP Home Edition, klik kanan pada tombol Start dan pilih
Properties. Klik Customize, buka tab Advanced dan enable option List my
most recently opened documents.

88. Jika Anda ingin agar tampilan Windows Explorer terbuka penuh hingga
menutupi taskbar, tekan tombol F11.

89. Anda bisa membuat shortcut key atau penekanan tombol tertentu untuk
mengakses sebuah program. Caranya, klik kanan icon program tersebut dan
pilih Properties. Buka tab Shortcut dan tentukan tombol mana yang Anda
gunakan untuk mengakses program tersebut. Secara otomatis Windows akan
menambahkan tombol Ctrl+Alt+ pilihan Anda sebagai shortcut key.

90. Urutkan secara alfabet, semua program yang ada di Start Menu untuk
memudahkan pengaksesan. Caranya, klik Start|All Programs, klik kanan
salah satu program yang ada dan pilih Sort by Name.

91. Munculkan penampakan icon indikator koneksi broadband atau dial-up
pada system tray untuk mempermudah kontrol. Caranya, buka My Network
Places, pilih View Networks Connections, klik kanan koneksi yang Anda
gunakan dan pilih Properties. Ubah option Show icon in notofication area
when connected menjadi enable.

92. Anda bisa mengedit Start Menu sesuai dengan kebutuhan. Baik
menambah, mengurangi, atau bahkan mengubah namanya. Caranya, klik kanan Start
dan pilih Open atau Open All Users. Atur setiap shortcut yang tersedia
sesuai kebutuhan.

93. Anda bisa membuka beberapa program secara bersamaan dengan menekan
tombol Shift, sementara mengklik program yang ada dalam Start Menu.

94. Untuk menghilangkan nama icon pada desktop, klik kanan icon
tersebut, dan pilih Rename. Tekan tombol Alt dan numeric key 255. Kemudan
tekan Enter.

95. Anda bisa menghilangkan tanda panah pada icon shortcut di desktop.
Caranya, masuk ke registry editor dengan mengetik “regedit” tanpa tanda
petik di kotak dialog Start|Run. Masuk ke dalam direktori
HKEY_CLASSES_ROOT\Lnkfile. Hapus value IsShortcut. 85

–> TOP TIPS !

96. Daftar Shortcut
Berikut ini beberapa shortcut yang paling banyak digunakan saat bekerja
sehari-hari menggunakan Windows.
[Windows] + [L] Lock komputer
[Windows] + Menampilkan Utility Manager
[Windows] + [R] Menampilkan Run
[Windows] + [F] Menampilkan window pencarian
[Windows] + [E] Membuka My Computer
[Windows] + [D] Mematikan atau mengaktifkan Toggle
Desktop
[Windows] + [M] Minimize semua window
[Windows] + [Shift] + [M] Restore semua window yang di-minimize
[Windows] + [Ctrl] + [F] Mencari komputer dalam jaringan
[Windows] + [F1] Menampilkan halaman Help
[Windows] + [Break] Memunculkan window System
Properties
[Windows] + [Tab] Scroll tombol di Taskbar
Jika Anda menggunakan keyboard lama yang tidak dilengkapi dengan tombol
Windows, gunakan paduan tombol Ctrl + Esc.

97. Anda bisa meng-enable atau disable grouping beberapa file dalam
satu program. Caranya, klik kanan taskbar dan pilih Properties. Beri atau
hilangkan tanda centang pada option Groups similar taskbar button.

98. Anda bisa menghilangkan atau menampilkan icon di desktop dengan
mengklik kanan desktop dan memilih Arrange Icons By|Show Desktop Icons.

99. Anda bisa membuat custom toolbar dengan mengklik kanan taskbar,
memilih Toolbars|New Toolbar. Atur sesuai dengan keinginan Anda, misalnya
menjadikan My Documents sebagai toolbar di taskbar supaya mudah dan
cepat diakses.

100. Klik kanan Start Menu, pilih Properties, jika tidak menggunakan
Classic Start Menu, klik Customize dan buka tab Advanced. Ada beberapa
munu yang bisa diaktifkan dengan memberi tanda centang pada option yang
ada. Mulai dari Scroll Programs hingga memunculkan penampakan beberapa
fitur dalam Start Menu.

101. Anda bisa membuka sebuah situs tanpa membuka browser terlebih
dahulu. Caranya, ketikkan alamat lengkap situs yang hendak dibuka di kotak
dialog Start|Run. Misalnya www.pcmedia.co.id. Kemudian tekan Enter.

102. Di setiap menu dalam Windows terdapat huruf yang bergaris bawah
sebagai patokan penggunaan navigasi keyboard. Misalnya menu File bisa
dibuka dengan menekan tombol Alt+F. Anda bisa menonaktifkan atau
mengaktifkan fitur yang ditandai dengan garis bawah ini. Caranya, klik kanan
area kosong di desktop. Kemudian buka tab Appereance, dan klik tombol
effects. Atur tanda centang pada option Hide underlined letter for keyboard
navigation until I press the [Alt] key.

103. Atur supaya komputer mengeluarkan suara peringatan saat tombol
Caps Lock, Num Lock atau Scroll Lock diaktifkan. Caranya, Buka Control
Panel, masuk dalam Accessibility Option. Buka tab Keyboard dan enable
option Use Toggle-Keys.

104. Untuk meng-enable Hibernate dalam Windows XP saat menekan Turn Off
di Start Menu, tahan tombol Shift. Maka tombol Stand by pada kotak
dialog Shut Down akan berubah menjadi Hibernate.

105. Jika hardware Anda support Hibernate, aktifkan segera fitur ini.
Caranya, buka Control Panel dan buka Power Options. Klik tab Hibernate
dan beri tanda centang pada option Enable Hibernate. Jika hardware Anda
tidak support Hibernate, tab ini tidak bisa ditemui.

Memperbaiki Kinerja Komputer

106. Tambah kecepatan komputer Anda dengan menghilangkan penampakan
yang bagus namun memakan banyak waktu, yakni animasi. Caranya, buka
Control Panel, dan klik ganda System. Klik tab Advanced dan tekan tombol
Performance Settings. Kemudian enable option Adjust for the best
performance.

107. Untuk mengurangi waktu booting yang terasa lama, atur di BIOS agar
booting dimulai dari harddisk baru kemudian CD atau floppy drive pada
pilihan berikutnya.

108. Atur supaya Windows hanya akan me-load program yang dibutuhkan
saja saat mulai berjalan supaya waktu loading lebih cepat dan kerja
komputer lebih ringan. Ketikkan “msconfig” tanpa tanda petik di kotak dialog
Run dan tekan Enter. Pada tab General, pilih option Selective Startup,
kemudian buka tab Startup dan nonaktifkan semua program yang tidak
perlu.

109. Anda bisa memonitor penggunaan processor dengan menekan tombol
Ctrl+Alt+Del untuk membuka Task Manager. Kemudian minimize window
tersebut, Anda akan melihat grafik penggunaan kapasitas processor di system
tray.

110. Untuk meningkatkan kecepatan menyimpan data di USB ZIP drive, buka
My Computer dan klik kanan drive ZIP tersebut. Pilih Properties, buka
tab Hardware klik Properties, kemudian buka tab Policies dan aktifkan
option Optimize for Performance.

111. Untuk mempercepat tampilan Start Menu, buka Registry Editor
(ketikkan “regedit” tanpa tanda petik di kotak dialog Run). Masuk dalam
direktori HKEY_CURRENT_USER\Control Panel\Desktop. Klik ganda value yang
bernama MenuShowDelay, ganti angka yang ada menjadi 100. Dengan cara ini,
maka delay tampilan Start Menu semakin singkat.

112. Ada cara cepat mematikan komputer. Tekan tombol Ctrl+Alt+Del, pada
windows Task Manager yang muncul, tekan menu Shut Down dan tahan tombol
Ctrl saat mengklik Turn Off. Pastikan dulu semua dokumen telah di-save.

113. Anda bisa melihat grafis yang menampilkan performance komputer
berdasarkan beberapa indikator. Misalnya penggunaan processor, memory, dan
lain sebagainya. Caranya, buka Control Panel dan klik ganda
Administrative Tools. Buka fitur Performance.

114. Anda juga bisa menambahkan beberapa indikator lain dalam fitur
Performance (di nomer 13) dengan mengklik icon + dan memilih indikator
baru apa saja yang hendak ditampilkan.

115. Anda bisa menggabungkan sebuah file registry atau .reg ke dalam
registry Windows Anda, dengan mengklik ganda file tersebut. Untuk
mengetahui fungsinya, buka dengan notepad.

116. Jika Anda menggunakan RAM sebesar 512 MB, atur agar Windows tidak
membuat paging ke disk agar performance komputer semakin meningkat.
Caranya, buka Registry Editor dan masuk dalam direktory
HKEY_LOCAL_MACHINE\SYSTEM|Current ControlSet\Control\SessionManager\MemoryManagement.
Ubah value DisablePagingExecutive menjadi 1.

117. Gunakan Bootcfg.exe untuk mengedit boot.ini Windows XP. Caranya,
ketikkan “cmd” tanpa tanda petik dalam kotak dialog Run untuk masuk
dalam DOS Prompt, dan ketikkan bootcfg: /? setelah muncul prompt untuk
memperoleh informasi mengenai fitur dalam program ini.

118. Untuk memilih operating system dalam yang digunakan dalam sebuah
komputer yang memiliki partisi drive, buka Control Panel, masuk dalam
System kemudian buka tab Advanced. Klik tombol Startup and Recovery
Setting kemudian Edit.

119. Gunakan utility klasik chkdsk untuk men-scan harddisk dan
memperbaiki error yang ditemui. Caranya, klik Start |Run dan ketikkan “chkdsk
c:/f” tanpa tanda petik untuk men-scan dan memperbaiki drive C. Jika
Anda ingin tahu lebih banyak tentang fitur dalam chkdsk, ketikkan “chkdsk
/?”.

120. Anda bisa melewati Scanner and Camera Wizard saat memasang kamera
digital. Untuk mengaksesnya, gunakan Windows Explorer atau My Computer.
Kamera digital Anda akan tampil sebagai drive.

121. Ada utility dalam Windows XP Pro Edition yang jarang digunakan,
yakni gpedit. Jalankan utility ini dengan mengetikkan “gpedit.msc” tanpa
tanda petik di kotak dialog Run dan tekan Enter. Utility ini bisa
digunakan untuk setiap tool dan komponen yang ada di Windows. Coba satu per
satu setting yang Anda inginkan kemudian perhatikan hasilnya.

122 Mengatur Jadwal Perawatan
Daripada bersusah payah membersihkan sampah di Windows, atur agar
Windows membersihkan dirinya sendiri secara berkala dengan Schedule Task.
a. Untuk mengatur jadwal perawatan Windows secara berkala, buka Control
Panel, klik ganda Scheduled Tasks. Tambahkan Scheduled Task melalui
wizard. Klik Next untuk masuk dalam list berisi program yang bisa
dijadwalkan.
b. Jika tool yang hendak Anda jadwalkan tidak tertera pada list, klik
Browse dan arahkan ke folder tempat tool yang Anda maksud berada.
Kebanyakan tool Windows berada dalam folder Windows\system32. Pilih salah
satu dan klik Open.
c. Ketik nama task tersebut dan tentukan tingkat keseringan atau
frekuensi jadwal program tersebut. Klik Next, atur setting lain yang
diperlukan dan tambahkan username beserta password untuk mengamankan Schedule
Task.

123. Gunakan utility gratis PowerToys dari
www.microsoft.com/windowsxp/home/downloads/powertoys.asp yang sangat berguna. PowerToys sendiri
terdiri dari beberapa program, di antaranya Open Command Window Here,
Alt-Tab Replacement, Tweak UI, Power Calculator, Image Resizer, CD Slide
Show Generator, Virtual Desktop Manager,Taskbar Magnifier, HTML Slide
Show Wizard, dan Webcam Timershot.

124. Windows Anda bisa memperoleh tambahan kecepatan jika Indexing
Service dinonaktifkan. Caranya mudah, buka Administrative Tools di Control
Panel, pilih Services dan disable Indexing Service.
125. Kembangkan kapasitas harddisk dengan menggunakan Disk Cleanup.
Buka My Computer, klik kanan drive yang ada dan pilih Properties. Buka tab
General dan klik tombol Disk Cleanup. Bersihkan semua file sampah
tersebut, termasuk mengosongkan isi Recycle Bin dari semua data yang sudah
tidak digunakan lagi.

126. Hilangkan semua shortcut di folder Startup dalam Start Menu.
Sebab, program yang memiliki shortcut dalam folder ini secara otomatis akan
dieksekusi saat kali pertama Windows dijalankan.

127. Periksa setting messenger yang ada dalam komputer Anda, karena
hampir setiap messenger akan me-load dirinya sendiri secara otomatis
setiap kali Windows dijalankan. Usahakan agar option Automatically Login
atau Load at Windows Start tidak aktif.

128. Anda bisa mengakses setiap program secara langsung dengan
mengetikkan nama programnya di kotak dialog Run. Misalnya calc untuk
kalkulator, winword untuk Microsoft Word, dan lain sebagainya.

Troubleshooting Sederhana

129. Jika saat defrag tiba-tiba komputer berhenti, restart dan masuk
dalam Safe Mode dengan menekan tombol F8 sesaat sebelum Windows mulai
berjalan. Defrag lagi harddisk Anda dari mode ini.

130. Cari tahu apakah drive Anda FAT atau NTFS dengan mengklik kanan
drive tersebut dan memilih Properties kemudian masuk ke tab General. Baca
detail file system pada kotak dialog yang muncul.

131. Ubah sebuah drive dalam format FAT menjadi NTFS melalui DOS dengan
perintah convert c:/FS:NTFS. Masuklah dalam registry editor dan buka
direktori HKEY_USERS\DEFAULT\Control Panel\Desktop dan buat value dalam
AutoEndTask dengan nilai 1.

133. Jika saat menginstal sebuah driver baru komputer hang, restore
komputer ke konfigurasi sebelumnya. Caranya, restart komputer dan tekan F8
untuk masuk ke Safe Mode dan pilih option Last Known Good
Configuration.

134. Gunakan Event Viewer untuk melacak aplikasi yang error. Caranya,
klik kanan My Computer, pilih Manage dan klik event Viewer. Klik ganda
setiap aplikasi atau system yang menunjukkan error untuk melihat
informasi kesalahan.

135. Ada cara paling tepat untuk mengatur ketepatan jam di komputer.
Klik ganda jam di sebelah kanan bawah layar atau di system tray, pilih
tab Internet Time. Aktifkan option Automatically synchronize with an
Internet time server. Pilih server yang tersedia dan klik Update Now!

Tip dan Trik Spesial untuk Windows 98 dan ME

136. Back-up setting dial-up dengan cara men-drag and drop file koneksi
di folder Dial Up Networking ke sebuah floppy drive. File back-up akan
disimpan dalam ekstensi.dun.

137. Cara cepat restart, tekan Start|Shut Down|Restart, sementara klik
OK, tahan tombol Shift.

138. Gunakan Tweak UI khusus Windows 98 dari
www.microsoft.com/ntworkstation/downloads/powertoys/networking/nttweakui.asp. Setelah instalasi
selesai, buka Tweak UI melalui Control Panel.

139. Gunakan tombol F3 untuk membuka fitur Find saat berada di Windows
Explorer atau desktop.

140. Atur koneksi dial-up dengan Telephony Location Manager. Fitur
Telephony Location Manager ini akan membantu Anda mengoptimalkan koneksi
dial-up. Caranya, klik Start|Run dan ketikkan “tlocmgr” tanpa tanda
kutip.

141. Jika komputer Anda tidak bisa di-set Dalam mode standby atau
suspended, maka perbaiki dengan Pmtshoot dari
http://support.microsoft.com/?kbid=185949.

142. Selalu back-up registry sebelum melakukan perubahan di dalamnya.
Caranya, gunakan Registry Checker yang bisa ditemui di Start|Programs|
Accessories|System Tools|Tools.

143. Buat sebuah disket bootable untuk Windows 98 dengan cara
memformatnya menggunakan system files. Kemudian copy file
c:\windows\command\scanreg.exe dan c:\windows\himem.sys ke dalam disket tersebut. Jangan
lupa, edit file config.sys yang berada di disket dengan memberi tambahan
baris “device =a:\himem.sys” tanpa tanda kutip.
144. Untuk me-restore registry saat komputer tidak mau berjalan, coba
booting dengan boot disk. Kemudian masuk ke drive C:\ dan ketikkan
“scanreg\restore”, tanpa tanda kutip. Ikuti angkah-langkahnya hingga
selesai. Kemudian keluar dari DOS.

145. Tambahkan shortcut Control Panel ke Start Menu dengan mengklik
kanan tombol Start dan klik Open. Kemudian klik kanan di area kosong,
pilih New|Folder. Isikan “Control
Panel.{21EC2020-3AEA-1069-A2DD-08002B30309D}” tanpa tanda kutip dan tekan Enter.

146. Anda bisa menghilangkan kotak dialog password yang muncul kali
pertama sebelum masuk dalam Windows. Caranya, masuk ke Control Panel, buka
Password dan buka tab Change Password. Isi kolom password lama dengan
password yang Anda miliki, dan kosongkan kolom new password. Setelah
itu, klik tab User Profiles dan enable option All users of this PC use the
same preferences and desktop settings. Perubahan ini akan berjalan
setelah Windows di-restart.

147. Untuk membatalkan perintah Print, buka icon Printer di system tray
dengan mengklik ganda. Kemudian sorot file yang sedang di-print pada
Window Printer dan klik kanan kemudian Cancel.

148. Temukan tool-tool under DOS yang masih layak dipergunakan dalam
folder tools\MSDOS di CD Instalasi Windows 98.

149. Jika Windows 98 atau Me berjalan lambat, periksa memory. Caranya,
klik kanan My Computer dan pilih Properties. Buka tab Performance dan
lihat System Resource. Jika mencapai 80 persen, restart komputer. Ingat,
tutup semua program sebelum melihat System Resource.

150. Gunakan Sysedit untuk mengedit file-file system. Misalnya
Autoexec.bat dan sebagainya. Ketikkan “sysedit” di kotak dialog Run tanpa
diberi tanda petik.







OKey sekian dulu daaaaah........

tips n trik yang ku dapat dari internet......

wasalam...




by : MIFTAH FARID