Regularization from L^{1} 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 measure, and even when convolution is applied to L^{1} 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 T_{μ} on L^{1} = L^{1}(T, dx) given by

T_{μ}(f)(x) = ∫ f(x + y)dμ(y) .

2 Simple facts

It is quite a requirement that T_{μ} has some regularization properties on L^{1}.

Proposition 1. Assume that for some Orlicz function φ such that lim_{u→∞} φ(u) = ∞, for all f in L^{1} we have

∥T_{μ}(f)∥_{φ} ≤ C∥f∥_{1} .  (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 ∈ L^{1}, such that x ∉ U ⇒ f = 0, with f ≥ 0 and ∥f∥_{1} = 1. Let us denote by μ_{K} the restriction of μ to K. Then, using Fubini theorem in the first equality,

μ(K) = ∫_{G} T_{μ_{K}} (f)dλ = ∫ f(x + y)dλ(x)dμ_{K}(y) .

Now, if f(x + y) ≠ 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

μ(K) ≤ ∫_{U−K} T_{μ_{K}} (f)dλ(x) ≤ ∫_{U−K} T_{μ}(f)dλ ,  (2)

Meanwhile, consider a function g ≥ 0 with ∥g∥_{φ} ≤ C, so that ∫ φ(g/C)dλ ≤ 1. For each measurable set V from Jensen’s inequality we have

φ(∫_{V} gdλ/Cλ(V) ≤ 1 .

Since we assume that lim_{u→∞} φ(u)du = ∞, this implies that for some constant C′ depending only on C we have ∫_{V} fdλ ≤ C′λ(V).. Using this for V = U − K and g = T_{μ}(f) gives

μ(K) ≤ C′λ(U − K) .

Since U is arbitary, this gives μ(K) ≤ C′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

ψ_{μ}(u) = sup {uλ({T_{μ}(f) ≥ u}) ; f ≥ 0, ∥f∥_{1} = 1} .

Since

∥T_{μ}(f)∥_{1} = μ(G)∥f∥_{1},

from Markov inequality we see that

ψ_{μ}(u) ≤ μ(G) .

Here is another simple fact.

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

lim_{u→∞} ψ_{μ}(u) = 0 .

This is also almost obvious. When μ = μ_{1} + μ_{2}, using that λ({f_{1} + f_{2} ≥ u}) ≤ ∑_{j=1,2} λ({f_{j} ≥ u/2}) we obtain that ψ_{μ}(u) ≤ 2ψ_{μ_{1}}(u/2) + 2ψ_{μ_{2}}(u/2). Moreover the result is trivial if μ ≤ Cλ because then ψ_{μ}(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 lim_{u→∞} ψ_{μ}(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 inf_{u→∞} ψ_{μ} ≥ μ(G).

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′ of G, so that H + H′ 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′ + H it is invariant under translations by elements of the support of μ, and thus T_{μ}(f) = f. Since the measure of H + H′ 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 ψ_{μ} cannot decrease too fast.

Porposition 4 If μ is singular then

∫_{1}^{∞} ψ_{μ}(u)du = ∞ .  (3)

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

∫_{V} gdλ = ∫_{0}^{∞} P({g ≥ u} ∩ V)du ≤ AP(V) + ∫_{A}^{∞} P({g ≥ u})du .

Consequently, if f ≥ 0 and ∥f∥_{1} = 1,

∫_{V} T_{μ}(f)dλ ≤ AP(V) + ∫_{A}^{∞} ψ_{μ}(u)du .

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

μ(K) ≤ Aλ(U − K) + ∫_{A}^{∞} ψ_{μ}(u)du .

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

3 The problem

Given 0 ≤ a ≤ 1 we consider the “biased coin” probabilty μ_{a} 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

μ_{a} = (\frac{1 + a}{2}δ_{1} + \frac{1 − a}{2}δ_{−1})^{⊗ℕ} .

Here is a simple fact

Proposition 5 Given 0 < a ≤ 1 there exists C(a) > 0 such that for u ≥ 2 we have ψ_{μ_{a}} ≥ C(a)/\sqrt{log u}.

We simply look at the density of μ_{a} 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

λ(g ≥ (1 + a)^{k}(1 − a)^{n−k}) ≥ 2^{−n}\binom{n}{k} .

We then choose k as close as possible to (1 + a)n/2. Letting c^{2} = (1 + a)^{1+a}(1 − a)^{1−a} then (1+a)^{k}(1−a)^{n−k} is about c^{n} while, using Stirling’s formula, 2^{−n}\binom{n}{k} is about 1/(c^{n}\sqrt{n}). □

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

ψ_{μ_{a}}(u) ≤ \frac{C(a)}{\sqrt{log u}} .

I do not know if it is true that lim_{u→∞} ψ_{μ_{a}}(u) = 0. However the following is proved in my paper “A conjecture on convolution operators, and operators from L^{1} 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.

Theorem 6 The probability measure μ = ∫_{0}^{1} μ_{exp(−t)}dt satisfies (for u ≥ 10)

ψ_{μ}(u) ≤ \frac{C log log u}{log u} .