Regularization from L1 by convolution

1 Introduction
The point of this note is to provide background for one of my favorite problems.
I offer a prize of 1000 USD for the solution of this problem. I do not think that
the problem by itself has any importance. It is just pretty.
It is well known that “convolution spreads regularity”. For example, if one
want to approximate a continuous function on R by a C ∞ function, one takes
convolution with a C ∞ function with (small) compact support. However, some
regularization is possible even when one takes convolution with a singular mea-
sure, and even when convolution is applied to L1 functions.
For simplicity we consider only convolution on the group G = {−1, 1}N
provided with its Haar measure λ.
Given a positive, finite measure µ on G we consider the operator Tµ on
L1 = L1 (T, dx) given by
Z
Tµ (f)(x) = f(x + y)dµ(y) .

2 Simple facts
It is quite a requirement that Tµ has some regularization properties on L1 .

Proposition 1. Assume that for some Orlicz function ϕ such that limu→∞ ϕ(u) =
∞, for all f in L1 we have

kTµ (f)kϕ ≤ Ckfk1 . (1)

Then for some constant A we have µ ≤ Aλ.

Suppose for contradiction that this is not the case, so that there is compact
set K with µ(K) > 0 and λ(K) = 0. Consider an open set U ⊂ G and a function
f ∈ L1 , such that x 6∈ U ⇒ f = 0, with f ≥ 0 and kfk1 = 1. Let us denote by
µK the restriction of µ to K. Then, using Fubini theorem in the first equality,
Z Z
µ(K) = TµK (f)dλ = f(x + y)dλ(x)dµK (y) .
G

1
Now, if f(x + y) 6= 0 and y ∈ K we have x ∈ U − K. Therefore we can restrict
the last integral to x ∈ U − K. That is, we have shown that
Z Z
µ(K) ≤ TµK (f)dλ(x) ≤ Tµ (f)dλ , (2)
U −K U −K
R
Meanwhile, consider a function g ≥ 0 with kgkϕ ≤ C, so that ϕ(g/C)dλ ≤
1. For each measurable set V from Jensen’s inequality we have
Z
ϕ( gdλ/Cλ(V ) ≤ 1 .
V

Since we assume that limu→∞ ϕ(u)du

R = ∞, this implies that for some constant
C 0 depending only on C we have V fdλ ≤ C 0 λ(V ).. Using this for V = U − K
and g = Tµ (f) gives
µ(K) ≤ C 0 λ(U − K) .
Since U is arbitary, this gives µ(K) ≤ C 0 λ(K), and since K is arbitrary this
proves the result. 
The moral is that (1) is a far too stringent requirement, so we shall consider
weaker conditions. We define the function
n o
ψµ (u) = sup uλ({Tµ (f) ≥ u}) ; f ≥ 0, kfk1 = 1 .

Since
kTµ (f)k1 = µ(G)kfk1 ,
from Markov inequality we see that

ψµ (u) ≤ µ(G) .

Here is another simple fact.

Proposition 2 If µ is absolutely continuous with respect to λ then

lim ψµ (u) = 0 .
u→∞

ThisPis also almost obvious. When µ = µ1 + µ2 , using that λ({f1 + f2 ≥

u}) ≤ j=1,2 λ({fj ≥ u/2}) we obtain that ψµ (u) ≤ 2ψµ1 (u/2) + 2ψµ2 (u/2).
Moreover the result is trivial if µ ≤ Cλ because then ψµ (u) = 0 for u ≥ C. 

The interesting phenomenon is that as we shall see it can happen that µ is

singular with respect to λ but that limu→∞ ψµ (0) = 0. As the following simple
fact shows, this does not happen when µ has a finite support.

Theorem 3 If µ has a finite support then lim infu→∞ ψµ ≥ µ(G).

2
 We may assume that µ is a probability. The support of µ is finite, so it
generates a finite subgroup H of G. Consider then a subgroup H 0 of G, so that
H + H 0 is a subgroup of G, which is invariant under translations by elements
of the support of µ. Thus if f is the indicator of H 0 + H it is invariant under
translations by elements of the support of µ, and thus Tµ (f) = f. Since the
measure of H + H 0 can be as small as we wish, the result should be obvious. 

Here is another simple fact showing that when µ is singular the function ψµ
cannot decrease too fast.

Porposition 4 If µ is singular then

Z ∞
ψµ (u)du = ∞ . (3)
1

For a function g ≥ 0 and a subset V of G we have, for any number A ≥ 0,

Z Z ∞ Z ∞
gdλ = ¶({g ≥ u} ∩ V )du ≤ A¶(V ) + ¶({g ≥ u})du .
V 0 A

Consequently, if f ≥ 0 and kfk1 = 1,

Z Z ∞
Tµ (f)dλ ≤ A¶(V ) + ψµ (u)du .
V A

Using (2) and taking V = T − U , we get

Z ∞
µ(K) ≤ Aλ(U − K) + ψµ (u)du .
A

Since we assume that µ is singular, we can find a > 0 such that we can find a
with µ(K) ≥Ra and λ(U − K) as small as we wish. We then see that for each A
∞
we have a ≤ A ψµ (u)du. 

3 The problem
Given 0 ≤ a ≤ 1 we consider the “biased coin” probabilty µa on G. It is the
product measure on G that on each factor gives weight (1 + a)/2 to the point 1
(and weight (1 − a)/2 to the point -1) so that
1 + a 1−a ⊗N
µa = δ1 + δ−1 .
2 2
Here is a simple fact

Proposition 5 Given
√ 0 < a ≤ 1 there exists C(a) > 0 such that for u ≥ 2 we
have ψµa ≥ C(a)/ log u.

3
 We simply look at the density of µa when we replace G by {−1, 1}n. On a
sequence with k terms equal to 1 this density g is (1 + a)k (1 − a)n−k , so that
 
n
λ(g ≥ (1 + a)k (1 − a)n−k ) ≥ 2−n .
k

We then choose k as close as possible to (1 + a)n/2. Letting c2 = (1 + a)1+a (1 −

a)1−a then (1 +√a)k (1 − a)n−k is about cn while, using Stirling’s formula, 2−n nk
is about 1/(cn n). 

Conjecture 6 Given a > 0 there exist C(a) > 0 such that for u ≥ 2 we have

C(a)
ψµa (u) ≤ √ .
log u

I do not know if it is true that limu→∞ ψµa (u) = 0. However the following
is proved in my paper “A conjecture on convolution operators, and operators
from L1 to a Banach lattice”, Israel J. of Math. 68, 1989, 82-88. In view of (3)
this is about as fast a decrease as can be expected.
R1
Theorem 6 The probability measure µ = 0
µexp(−t)dt satisfies (for u ≥ 10)

C log log u
ψµ (u) ≤ .
log u
Another noteworthy result is that in 2015 Ronen Eldan and James Lee have
solved the “Gaussian limiting case” of Conjecture 6 (with a slightly weaker
estimate).