Absolute convergence of an infinite convolution

Imre Z. Ruzsa*

Mathematical Institute of the Hungarian Academy of Sciences, Budapest, Pf. 127, H–1364 Hungary.

1. Introduction

We consider regular probability measures on locally compact commutative groups, denoted additively. The convolution of the measures μ, ν will be denoted by μν.

In [2] I called an infinite convolution ∏ μ_{n} absolutely convergent, if every rearrangement is convergent to the same limit (in the weak topology). Here, as usual, we say that an infinite convolution ∏_{n=1}^{∞} μ_{n} is convergent, if the limit lim_{N→∞} ∏_{n=k}^{N} μ_{n} exists for every k. I posed the problem whether every convergent product can be made absolutely convergent by a suitable centering, that is, whether there exist elements g_{n} such that ∏ μ_{n}δ(−g_{n}) is absolutely convergent. I announced there without proof that this is true for compact groups and the additive group of reals. The aim of this note is to prove these statements. My reason for doing so now is that recently Mauclaire [1] proved that the general case can be reduced to these subcases. Thus, his result and mine together yield the following theorem.

Theorem 1.

If the infinite convolution ∏ μ_{n} of regular probability measures on a commutative locally compact group G is convergent, then there exist elements g_{n} ∈ G such that every rearrangement of the infinite convolution ∏(μ_{n}δ(−g_{n})) converges to a common limit.

2. The case of reals

* Supported by Hungarian National Foundation for Scientific Research, Grant No. T 017433.

Statement 2.1.

If the infinite convolution ∏ μ_{n} of regular probability measures on the real line is convergent, then there exist real numbers a_{n} such that every rearrangement of the infinite convolution ∏(μ_{n}δ(−a_{n})) converges to a common limit.

Proof. Let F_{n} denote the distribution function of μ_{n}. Write

a_{n} = ∫_{−1}^{1} x dF_{n}(x).

By Kolmogorov’s theorem, the convergence of the infinite convolution is equivalent to the convergence of the three series ∑ a_{n},

∑ F_{n}(−1) + (1 − F_{n}(1)),
∑ ∫_{−1}^{1}(x − a_{n})^{2} dF_{n}(x).

We can combine the second and third assumptions into one that asserts ∑ d_{n} < ∞, where

d_{n} = ∫_{−∞}^{∞} min(1, (x − a_{n})^{2}) dF_{n}(x).

Now consider μ_{n}^{∗} = μ_{n}δ(−a_{n}). We write F_{n}^{∗}, a_{n}^{∗}, d_{n}^{∗} for the analogous quantities corresponding to these measures. We have

F_{n}^{∗}(x) = F_{n}(x + a_{n}),

hence

a_{n}^{∗} = ∫_{−1}^{1} x dF_{n}^{∗}(x) = ∫_{−1+a_{n}}^{1+a_{n}}(x − a_{n}) dF_{n}(x) = ∫_{−1+a_{n}}^{−1} + ∫_{−1}^{1} + ∫_{1}^{1+a_{n}} .

Now, assume that |a_{n}| ≤ 1/3, which holds with finitely many exceptions by virtue of the convergence of the series. By the definition of a_{n} the second integral is equal to a_{n}(F_{n}(−1) + (1 − F_{n}(1)), and its absolute value is ≤ d_{n}/3. On the first and third we have

2/3 ≤ 1 − |a_{n}| ≤ |x − a_{n}| ≤ 1 + |a_{n}| ≤ 4/3

which yields

|x − a_{n}| ≤ \frac{3}{2} min(1, (x − a_{n})^{2}),

and consequently |a_{n}^{∗}| ≤ 2d_{n}. We conclude that the series ∑ a_{n}^{∗} is absolutely convergent. To estimate d_{n}^{∗} we observe that

d_{n}^{∗} = ∫ min(1, (x − a_{n})^{2}) dF_{n}^{∗}(x) = ∫ min(1, (x − a_{n} − a_{n}^{∗})^{2}) dF_{n}(x).

Applying the inequality (u + v)^{2} ≤ 2(u^{2} + v^{2}) we obtain d_{n}^{∗} ≤ 2d_{n} + 2a_{n}^{∗}^{2}, hence ∑ d_{n}^{∗} < ∞. Now, using Kolmogorov’s theorem in the other direction, we obtain that every rearrangement of ∏ μ_{n}^{∗} is convergent.

3. The case of compact groups

Here we prove the following.

Statement 3.1. Let G be a compact commutative group, μ_{1}, ...., μ_{n}, ... regular probability measures

on G. There are elements g_{n} ∈ G such that the infinite convolution ∏ μ_{n}δ(−g_{n}) is absolutely convergent.

Note that the convergence is not assumed in advance, thus we get another proof (for the commutative case) of the fact that convergence can be achieved by a centering.

We denote the dual group (group of characters) of G by Γ. The Fourier transform of μ is denoted by μ̂.

Lemma 3.2. For any probability measure μ and γ_{1}, γ_{2} ∈ Γ we have

\begin{aligned}|μ̂(γ_{1}γ_{2}) − μ̂(γ_{1})μ̂(γ_{2})| & ≤ 2\sqrt{(1 − |μ̂(γ_{1})|)(1 − |μ̂(γ_{2})|)} \\ & ≤ (1 − |μ̂(γ_{1})|) + (1 − |μ̂(γ_{2})|), \end{aligned}  (3.1)
|1 − μ̂(γ_{1}γ_{2})| ≤ 2((|1 − μ̂(γ_{1})|) + (|1 − μ̂(γ_{2})|)),  (3.2)
1 − |μ̂(γ_{1}γ_{2})| ≤ 2((1 − |μ̂(γ_{1})|) + (1 − |μ̂(γ_{2})|)).  (3.3)

With arbitrary numbers c_{1}, c_{2} satisfying |c_{j}| ≤ 1 we have

|1 − c_{1}c_{2}μ̂(γ_{1}γ_{2})| ≤ 2((|1 − c_{1}μ̂(γ_{1})|) + (|1 − c_{2}μ̂(γ_{2})|)),  (3.4)

Proof. First, observe that for any γ and μ we have

∫ |γ − μ̂(γ)|^{2} dμ = 1 − |μ̂(γ)|^{2} ≤ 2(1 − |μ̂(γ)|).  (3.5)

Write m_{i} = μ̂(γ_{i}). By integrating the identity

γ_{1}γ_{2} − m_{1}m_{2} = m_{1}(γ_{2} − m_{2}) + m_{2}(γ_{1} − m_{1}) + (γ_{1} − m_{1})(γ_{2} − m_{2})

we obtain

μ̂(γ_{1}γ_{2}) − m_{1}m_{2} = ∫ (γ_{1} − m_{1})(γ_{2} − m_{2}) dμ.

Applying Schwarz’s inequality, the module of this expression is estimated from above by

(∫ |γ_{1} − m_{1}|^{2} dμ ∫ |γ_{2} − m_{2}|^{2} dμ)^{1/2} .

Now the first inequality in (3.1) follows by applying (3.5), and the second by using the inequality of arithmetic and geometric mean.

To prove (3.4) , observe that for any numbers satisfying |a_{i}| ≤ 1 we have

|1 − a_{1}a_{2}| = |(1 − a_{1}) + a_{1}(1 − a_{2})| ≤ |1 − a_{1}| + |1 − a_{2}|.

In particular, the choice a_{i} = c_{i}μ̂(γ_{i}) yields

|1 − c_{1}c_{2}μ̂(γ_{1})μ̂(γ_{2})| ≤ |1 − c_{1}μ̂(γ_{1})| + |1 − c_{2}μ̂(γ_{2})|,  (3.6)

Now (3.4) follows from (3.6), (3.1) and the observation that

|1 − cμ̂(γ)| ≥ 1 − |cμ̂(γ)| ≥ 1 − |μ̂(γ)|  (|c| ≤ 1).  (3.7)

(3.2) is the case c_{j} = 1 of (3.4). (3.3) follows by taking c_{j} so that |c_{j}| = 1 and c_{j}μ̂(γ_{j}) is nonnegative, and using (3.7) again.

Lemma 3.3. For every γ and positive integer k we have

|μ̂(γ^{k}) − μ̂(γ)^{k}| ≤ k(1 − |μ̂(γ)|).

Proof. This follows from (3.1) by an easy induction.

Lemma 3.4. Let a be a complex number, k a positive integer. There is a b such that b^{k} = a and

|1 − b| ≤ |1 − a|.

Proof. Exercise. Proof of the Statement. The convergence of a sequence of measures is equivalent

to the pointwise convergence of the Fourier transforms. Hence we need to find g_{n} such that every rearrangement of any product

∏ μ̂_{n}(γ)\overline{γ}(g_{n})  (3.8)

is convergent for all γ ∈ Γ.

Write

Γ_{0} = {γ ∈ Γ : ∑(1 − |μ̂_{n}(γ)|) < ∞}.

If γ ∉ Γ_{0}, then the product in (3.8) converges to 0. Hence we need to ensure (3.8) only for γ ∈ Γ_{0}. (3.3) shows that Γ_{0} is a subgroup of Γ.

We will use duality, and find g_{n} as characters of Γ. This motivates the following name. We say that (Δ, (ϕ_{n})_{n=1}^{∞}) is a partial centering, if Δ is a subgroup of Γ_{0}, each ϕ_{n} is a character of Δ and

∑ |1 − μ̂_{n}(γ)\overline{ϕ}_{n}(γ)| < ∞  (3.9)

for all γ ∈ Δ.

We say that (Δ, (ϕ_{n})) is an extension of (Δ′, (ϕ′_{n})) if Δ ⊃ Δ′ and each ϕ′_{n} is the restriction of ϕ_{n} to Δ′. A chain of partial centerings, that is, a family where from any two one is an extension of the other, obviously possesses a common extension. Hence by Zorn’s lemma there is a maximal partial centering. We show that this satisfies Δ = Γ_{0}.

Suppose that this is not the case, and take any χ ∈ Γ_{0}\Δ. Let Δ^{∗} be the subgroup of Γ generated by Δ and χ. We are going to extend ϕ_{n} to a character ϕ_{n}^{∗} of Δ^{∗}.

If there is a positive integer k such that χ^{k} ∈ Δ, then we must have

ϕ_{n}^{∗}(χ)^{k} = ϕ_{n}(χ^{k}).  (3.10)

If we select the smallest k, then this condition is also sufficient; if we fix b_{n} such that b_{n}^{k} = ϕ_{n}(χ^{k}), then there is exactly one extension of ϕ_{n} to Δ^{∗} with ϕ_{n}^{∗}(χ) = b_{n}.

We know that

∑ |1 − μ̂_{n}(χ^{k})\overline{ϕ}_{n}(χ^{k})| < ∞

and Lemma 3.3 yields that

∑ |μ̂_{n}(χ^{k}) − μ̂_{n}(χ)^{k}| ≤ k ∑(1 − |μ̂_{n}(χ)|) < ∞.

Hence

∑ |1 − μ̂_{n}(χ)^{k}\overline{ϕ}_{n}(χ^{k})| =< ∞.

By Lemma 3.4 we find b_{n} so that b_{n}^{k} = ϕ_{n}(χ^{k}) and

|1 − μ̂_{n}(χ)\overline{b}_{n}| ≤ |1 − μ̂_{n}(χ)^{k}\overline{ϕ}_{n}(χ^{k})|.

By putting ϕ_{n}^{∗}(χ) = b_{n} we achieve that

∑ |1 − μ̂_{n}^{∗}(γ)\overline{ϕ}_{n}(γ)| < ∞  (3.11)

holds for γ = χ; it also holds for all γ ∈ Δ. By (3.5), this implies that it holds for all γ ∈ Δ^{∗} and the extension is a partial centering.

If there is no k such that χ^{k} ∈ Δ, then we can prescribe ϕ_{n}^{∗}(χ) arbitrarily. We choose it so that μ̂_{n}^{∗}(γ)\overline{ϕ}_{n}(γ) be nonnegative. Then the convergence of (3.11) for γ = χ follows from χ ∈ Γ_{0} and the extension is again suitable.

We proved that there is a partial centering defined on the whole Γ_{0}. These characters ϕ_{n} can be extended to characters on Γ, and hence there exist g_{n} ∈ G such that ϕ_{n}(γ) = γ(g_{n}) for all γ. This means that

∑ |1 − μ̂_{n}(γ)γ(−g_{n})| < ∞

for all γ ∈ Γ_{0}, hence every rearrangement of

∏ μ̂_{n}(γ)γ(−g_{n})

is convergent.

4. Concluding remark

It would be interesting to know whether it is possible to generalize the method of the previous section to all commutative locally compact groups. For a long time I tried in vain to do this. Mauclaire uses a different approach. He applies the structure theory of locally compact groups to reduce the general case to certain subcases, which were studied in this paper.

References

[1] J-L. Mauclaire, On a problem of I. Z. Ruzsa, Acta Math. Acad. Sci. Hungar., to appear.

[2] Ruzsa, I. Z., Infinite convolution and shift-convergence of measures on topological groups, in: Probability measures on groups (Proc. Conf., Oberwolfach 1981), Lecture Notes in Math.928, Springer–Verlag, New York–Berlin (1982), 337–353.