# Self convolutions of singular continuous measure

Let $\mu$ be a finite measure on $\mathbb{R}$. Define the measures $(\mu_n)_{n\geq 1}$ by $\mu_{n+1}=\mu\ast \mu_n$ and $\mu_1=\mu$

Is there a singular (with respect to the Lebesgue measure) continuous measure $\mu$ on $\mathbb{R}$ such that for all $n\geq 2$, $\mu_n$ is also singular continuous ?

Here is a simple explicit example of a measure $\mu$ with all convolution powers singular: let $\mu$ be the distribution of $$\sum_{k=1}^\infty X_k 2^{-k!},$$ where the $X_k$ are IID taking values $0$ and $1$ with equal probability. The support of $\mu$ is the set $A$ of points whose binary expansion has non-zero digits only at places $k!,k\in\mathbb{N}$. Using this, it follows easily that for each $n$, the $n$-fold arithmetic sum $A^{+n}=A+\ldots+A$ has zero Lebesgue measure (in fact it has zero Hausdorff dimension), since the binary expansion of points in $A^{+n}$ must have zeros at all places not of the form $k!+j$ for some $0\le j\le n$. Since the convolution power $\mu^{*n}$ is supported on $A^{+n}$, it must be purely singular. On the other hand, $\mu$ has no atoms so neither does $\mu^{*n}$. This example doesn't use the Fourier transform at all.

Let me also give complete details of why the Cantor-Lebesgue measure $\mu$ on the middle-thirds Cantor set is an example of a measure all of whose self-convolutions are purely singular (but continuous, since $\mu$ itself already has no atoms).

Recall that $\mu$ can be realized as the distribution of the random sum $\sum_{i=1}^\infty X_i 3^{-i}$, where $X_i$ are independent random variables taking the values $0$ and $2$ with equal probability. In particular, $$\widehat{\mu}(\xi) =\prod_{i=1}^\infty \cos(2\pi 3^{-i}\xi)$$ from which it easily follows that $\widehat{\mu}(\xi)$ doesn't decay to $0$ as $\xi\to\infty$ (this also follows from the fact that $\mu$ is invariant under multiplication by $3$ on the circle). Hence, if $\mu^{*n}$ denotes the $n$-fold self-convolution, then $\widehat{\mu^{*n}}(\xi)=\widehat{\mu}(\xi)^n$ also doesn't decay to $0$, so $\mu^{*n}$ cannot be absolutely continuous by the Riemann-Lebesgue Lemma.

Now, recall that a Borel probability measure $\nu$ on the real line is called self-similar if there are contracting similarities $f_1,\ldots,f_m$ and a probability vector $(p_1,\ldots,p_m)$ such that $$\nu(A)= \sum_{j=1}^m p_i\, \nu(f_j^{-1}(A)).$$ Given the tuples $(f_1,\ldots,f_m)$ and $(p_1,\ldots,p_m)$ there is exactly one Borel probability measure satisfying the above identity; this follows from seeing the right-hand side as a contracting operator on an appropriate Banach space (actually we only care about uniqueness, so we only need to have a contraction, no completeness is required).

If $\nu$ is a self-similar measure as above, then the absolutely continuous and singular parts of $\nu$ are seen to satisfy the same self-similarity relation. By the uniqueness of $\nu$, one of these parts has to be trivial. So self-similar measures are either absolutely continuous or purely singular with respect to Lebesgue.

To conclude, I claim that $\mu^{*n}$ is a self-similar measure for every $n$. Since we already established that $\mu^{*n}$ cannot be absolutely continuous, this will show that it must be purely singular.

By definition, $\mu^{*n}$ is the distribution of the random sum $$\sum_{i=1}^\infty Y_i^{(n)} 3^{-i},$$ where the $Y_j^{(n)}$ are IID with the distribution of the sum of $n$ independent realizations of $X_i$ (defined above). In particular, the distribution of $Y_i^{(n)}$ has the form $\sum_{j=0}^{n} p_j^{(n)} \delta_{2j}$ for some probability vector $p_j$. It can then be easily checked that $\mu^{*n}$ is the self-similar measure corresponding to the contractions $(x/3+2j)_{j=0}^{n}$ an the probability vector $(p_j)_{j=0}^n$.

Let me mention that it follows from work of Garsia in the 60s that not only $\mu^{*n}$ is singular for all $n$, but in fact for each $n$ there is a Borel set $A_n$ of Hausdorff dimension strictly less than $1$ such that $\mu^{*n}(A_n)=1$ (the topological support of $\mu^{*n}$ is an interval for all $n\ge 2$). On the other hand, the Hausdorff dimension of $A_n$ must tend to $1$; this follows from recent work of M. Hochman.

Yes, indeed. You can build a one dimensional family of non-atomic singular measures on $[0,1]$ simply by pushing the Bernoully measure with parameter $p$ with the help of the dyadic decomposition of real numbers.

Start with $X_i$ iid random variables such that $P(X_i = 1) = p = 1 - P(X_i = 0)$ and define $$Y_p = \sum_{k = 1}^\infty {X_i\over 2^i}$$ The law of $Y_{1/2}$ is the Lebesgue measure. For other values of $p$ however, it is singular with respect to the Lebesgue measure because it is supported by the set $\{ \sum {x_i \over 2^i} \mid {1\over n} \sum_k x_k \rightarrow p\}$ by the strong law of large numbers and these sets are disjoints for different values of $p$.

The same remark shows that $Y_p + Y'_p$ is singular with respect to the Lebesgue measure if $Y_p$ and $Y'_p$ are independent with the same law (associated to $X_i$ and $X'_i$ iid Bernoulli) and $E(Y_p+Y'_p) = 2p \neq {1}$. And so on. So just take $p\neq 2$ and the law of $Y_p$ is such that all its products are non-atomic singular.

• Thank you for the answer, I was looking for something like that. However it seems to me that you can suppose that $E(Yp+Y′p)=2p≠1$ because $Y_p+Y'_p$ lives in $[0,2]$ – Chr Apr 5 '17 at 13:37
• Indeed. Corrected. – coudy Apr 5 '17 at 13:40
• So we just need to pick $p\neq \frac12$ and all the products are non-atomic singular. – Chr Apr 5 '17 at 13:44
• Yes. You can also consider the product of $Y_p$ for different $p$ and this is singular e.g. if the p's don't give a rational number. This is really something that happens on the symbolic space $\{0,1\}^{\bf N}$. We are just charging the digits with different weights. – coudy Apr 5 '17 at 13:56
• Indeed. By the way, is the Hausdorff dimension of the support of $Y_p$ easy to compute ? – Chr Apr 5 '17 at 13:59

You can also get this (the existence of such an example, that is) cheaply from some abstract nonsense, though I would certainly prefer a direct example.

If you put the weak $*$ topology on the probability measures, then the generic measure is purely singular continuous. In other words, the set of purely sc measures contains a dense $G_{\delta}$. In fact, here I need the more specific information that both $\{\mu: \mu_{ac}=0\}$ and $\{ \mu: \mu_{pp}=0\}$ are $G_{\delta}$'s (and dense, which is obvious), which is how the statement I just quoted is usually proved. See also Thms 1.1, 1.2 here.

Since $f_n: \mu \mapsto \mu * \mu * \ldots * \mu$ is continuous, the set $\{ \mu: f_n(\mu)_{xx}=0 \}$ still contains a dense $G_{\delta}$ set, for both $xx=ac$ and $xx=pp$. So $f_n(\mu)$ is purely singular continuous on at least a dense $G_{\delta}$ of $\mu$'s.

Since we have only countably many conditions, what you want also happens on at least a dense $G_{\delta}$ set.

A pisot number is an algebraic integer all of whose conjugates are < 1 in magnitude. If $X_i$ are i.i.d, symmetric, $\pm 1$ random variables and $\lambda$ is a pisot number look at $\sum \frac {X_i} {\lambda^i}$ whose characteristic function is $\prod cos(\frac {2 \pi t} {\lambda^i})$. You can show that this does not go to zero at $\infty$. The key is that for a pisot number $|||\lambda^i||| \rightarrow 0$ exponentially fast, where that norm is distance to the nearest integer, so look in a $t = \lambda^N$. If it does not go 0 no power of it does either, so no power is absolutely continuous. As to it's singularity, I think it follow from the law of pure types, but it is easily proven in case $\lambda > 2$ in which case $\sum^N \frac {X_i} {\lambda^i}$ consists of $2^N$ distinct equally weighted points.

• The standard example of a Pisot number is the golden mean. – Anthony Quas Apr 5 '17 at 14:12
• @Chr: this answer is convincing to me. In essence, it says that the Fourier transform of an absolutely continuous measure has the property that it decays to 0 as $t\to\infty$ (this is the Riemann Lebesgue lemma). This fails for this example (compute the Fourier transform at $\lambda^n$). But since the Fourier transform of the $n$-fold convolution is just the $n$th power of the Fourier transform of $\mu$, the same argument shows that $\mu^{*n}$ is singular. – Anthony Quas Apr 5 '17 at 14:21
• I think this only proves that the measures are never purely absolutely continuous. They could be purely singular (and even atomic) or the sum of an absolutely continuous measure and a singular one. – Chr Apr 5 '17 at 14:31
• @AnthonyQuas, that's correct, also in this case the convolutions are all self-similar, and self-similar measures are also known to be either purely singular or mutually absolutely continuous to Lebesgue – Pablo Shmerkin Apr 6 '17 at 2:23
• Also, the same argument (even easier) shows that the Cantor-Lebesgue measure on the middle-thirds Cantor set is an example that answers the OP's question – Pablo Shmerkin Apr 6 '17 at 2:24