Notes on an Inequality
on the Discrete Cube
b y P i s i e r for F u n c t i o n s
R.. Wagner
School of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel
1
Introduction
We study functions from the discrete cube { - 1 , 1} '~ to a linear normed space
B. For such functions f define D i f ( z ) to be the vector 1(=)-l(z~)
where z i
2
has the same coordinates as z except at t h e / - t h place. Define
i=1
In [P] Pisier proved that for every 1 < p < oo and every f : { - 1 , 1} '~ -~ B
one has
tli(~) - EIIIL,(8) _
Clog'~lllVflp(~)llL,(~)"
In [T] Talagrand showed that for B = ~ the logarithmic factor can be removed, whereas in general it is sharp with dependence on p < c¢. We will
close here the final gap in this inequality, by showing that the logarithmic
factor can be dropped for p = o0. The argument we employ is an elementary
counting argument. We will then turn to revise Pisier's proof of his inequality above. The proof we present uses the same mechanism, but applies it
differently.
2
The Main Result
T h e o r e m 1. Let f : { - 1 , 1}" --~ B , where B is a normed linear space. Then
max
z,ye{-1,1}"
IIf(~)-f(v)llB<~2
max
ze{-1,1}"
IVflu(z).
Proof. Choose two points, z and y, where the left-hand m a x i m u m is attained.
Without loss of generality we can assume these are diametrically opposite
points in the discrete cube. Otherwise we will restrict to the subcube where
z and y are diametrically opposed, and prove the inequality in this context;
then the "unconditional" nature of IV f l u ( z ) allows to return the right-hand
t e r m to the full-cube context without decreasing it.
Denote z = ao = ( - i , - i , . . . )
and y = a~ = (i, i , . . . ) . We will be
interested in all monotone paths of neighbouring vertices a0, a l , . . . , a~, that
264
R. W a g n e r
is paths where ai+1 is achieved from ai by changing one of its -l's into a i.
Now
f ( : ~ ) - f(Y)
1
n-'~
:
~
( ~ f(o~) -- f(o~-l)) ;
{-o ..... -.,} i=1
(1)
a monotone
path
The normalising factor n! is just the number of monotone paths.
W e would like to express (i) using the following objects:
=
Z
y-
.ill.
neighbour of z
-2 s(a)
where Y a = 1 ifa has more l's than z, and y, : -1 ifa has less l's than z. It
is clear that [[~/(z)l[B < [V/[oo(z). Thus, our claim is proved if we rearrange
(1) in the form
i=l
where zi is an enumeration of the vertices in { - 1 , 1}'~ with repetitions. This
rearrangement will be obtained heuristically. Consider the sum
2
(2)
=e{-1,1) ~
W e will find conditions on the integers b= which force (1) and (2) to be equal.
Take a point ah with k coordinates equal to 1, and a point ak+x obtained
from ak by changing one of its -l's to a i. The term f(ak)- f(ah+1) appears
in (I) exactly k!(n- k - 1)! times, because there are exactly that m a n y
monotone paths going through (ah, ak+1). That same term appears ba~ +ba~+ I
times in (2), because it is part of both ~f(ak) and ~f(ak+l). W e conclude
that (1) and (2) are equal iff
ba~ + ba~+, = k ! ( n - k -
1)!.
Now a monotone path, as well as a ~f(z), both have exactly n terms of
the form f(ak)-f(ak+l). Since we are matching these latter terms bijectively
between (1) and (2), we have exactly as many paths in (1) as 6f's in (2). This
equality affirms proper normalisation in (2), namely ~ b® = nl. Adding the
constriction that the b~,'s be positive will allow the triangle inequality to
bound (2) by 2JVfloo(z ).
That integers satisfying our demands of the ba~'s do in fact exist is easily
verified; Actually, our ba~'s will depend only on k. The main observation
behind our construction is that the sequence k!(n - k - 1)! decreases until k
reaches (n - 1)/2, and then climbs back symmetrically.
Notes on an Inequality by Pisier for Functions on the Discrete Cube
265
Consider the case r~ odd (the even case is similar). Set
1 r~-i
r~-I
We can define inductively bk-t = (k - 1 ) ! ( n - k ) ! - bh. An induction going
backwards from k = - ~ will prove that
k ! ( n - k - 1)1 > b~, > O.
Indeed, suppose bk satisfies this premise. We have bk-1 :
(k
--
1)!(n-
k)!- bk, which implies
(k - 1)!(n - k)! >
bh-1 >
(k - 1)!(n
-
k)!
-
k!(n
-
The argument for k > n is symmetric.
k -
1)! > O.
[]
R e m a r k 2. This method seems to be applicable to other highly symmetric
regular graphs. The constant 2 should in general be replaced by a number
close to the ratio 2 ~ g.
A n example where this holds (by an easy argument
following our scheme) is the crosspolytope.
3
Pisier's P r o o f R e v i s i t e d
W e present a proof of Pisier's inequality as stated in the introduction, our
proof repeats the original proof's mechanism. Here, however, this mechanism is applied to a different operator (the identity). This approach takes
the weight off the lower estimate, and m a y clarify the picture for the upper
estimate. The price we pay is the conversion of a convexity argument into a
more intense combinatorial consideration.
The proof requires some Walsh analysis. A function f : {-1, i}n -~ B can
be decomposed uniquely into ~-~Ac_{t.....~} fAwA(z), where fA E B, wA(z) =
I'LEA zi, and the functions {wA(Z)}AC_{1 .....,~} form an orthonormal basis in
the L2 norm. It is important to observe that the operator Di in this context
is simply an orthogonal projection on those Walsh functions w.4 (z) for which
lEA.
Proof. The argument is a duality argument. Let g : { - 1 , 1}~ --+ B* such
that
[]gllq:
1 and
(we assume, without loss of generality that f¢ = E f = 0). Our goal is to
rewrite this expression in the form
Ft
f=
~o(z, y ) . E
,¥E{--1'I}'~
i=1
yiDif(z)dzdy.
266
B. Wagner
Bounding I1~o(~,u)llq by C log n will complete the argument.
Since we assume f0 = 0, orthonormality of the Walsh system implies:
(3)
N o w note that the identity can be decomposed into
1%
.,= R D'D,,
i=1
where R is the following selfadjoint operator
at_Ix.....1%}
IAI>_I
aC_{i.....1%}
lal~x
Applying this decomposition, we turn (3) into
1%
Conjugating and transferring the operators to the left we get
i=~Iffe{_1,1}," D./R(g - go)(z) • D i f ( z ) d z ,
and finally,decoupling obtains
75
1%
/=1
./=1
Observe that all terms in the Walsh decomposition of the right-hand function have the form y./wA(z), where i E A. Orthonormality implies we can replace the left-hand function by any ~(z, y), which coincides with the left-hand
function on those Walsh terms.
Our candidate will be:
1
~,i
=
I,=i
k
(1)
. ,z.),
so_It
......}
IBl--k
s
where z ~ = z./ if i ~ B, and z~ = z./y./ if i E B (the reader interested in
motivation should check out the cases n = 2 and 3; there it is visible that
to is obtained by 'completing' the relevant terms from the original left hand
function into a sum of copies of g). Note that the sequence {zis },=1
.
is a
Notes on an Inequality by Pisier for Functions on the Discrete C u b e
267
sequence of independent uniform variables. This and the fact that the inner
sum has exactly (~) terms allow the triangle inequality to conclude
Ilwll~ ~_
~llgll~ _~ i ÷
logn.
k=1
Therefore, the proof is done, once we verify that ~ can indeed replace the
original left-hand function.
Let's compare the relevant Walsh terms in the two functions. Let i E A.
On the one hand,
J~=,,e[-1,1}", ( ~.~=1yjDjR(9
- ge)(z)) " (y~WA(z))dzdy=
On the other hand,
. 1 1
,,,~_l,~r ~,(~,y). y,~,(-.)d~d~ = ~ k (~)
k=i
9A.
sc{i ......}
ISL=h
AnS:{i}
To complete the proof, all we need to show is that
~(~1{
k=l~
i
B C-{l''"'n};[B[=k'AnB = {i}}]= ~l"
Indeed, the left-hand sum equals
which can be regrouped as
(n-
]A])!(]A I - 1)!
n!
k=l
A simple combinatorial identity yields
(n- lAI)l(IAl -1)! ( n )
n!
which in turn equals the desired I-A-T"
1
]A] '
[]
268
R. Wagner
References
[P]
IT]
Pisier G. (1986) Probabilistic methods in the geometry of Banach spaces.
In: Probability and Analysis (Varenna, Italy, 1985), Lecture Notes in Math.,
1206, Springer Verlag, 167-241
Talagrand M. (1993) Isoperimetry, logarithmic Sobo]ev inequalities on the
discrete cube, and Msrgulis' graph connectivity theorem. GAFA 3:295-314