Let $\mu, \nu$ be finite complex measures with compact supports on the real line, and assume that they are singular with respect to the Lebesgue measure. Can their convolution $\mu\ast\nu$ have a nonzero absolutely continuous component?
Asked
Active
Viewed 1,297 times
2 Answers
8
The answer is yes and there is lots of literature on this issue (google "convolution of singular measures").
The earliest reference I could dig out is by Wiener and Wintner [1]. They gave a construction of a singular continuous measure $\mu$ such that $\mu*\mu$ is absolutely continuous.
More precisely, for arbitrary $\epsilon>0$, they construct a measure $\mu$ on $[-\pi, \pi]$ which is singular continuous wrt. the Lebesgue measure such that its Fourier coefficients satisfy
$$\widehat{\mu}(n):=\int_{-\pi}^\pi e^{-inx} d\mu(x) = O(n^{-\frac{1}{2}+\epsilon})$$
In particular, though this is not what they're after in that article, it follows that $\mu*\mu$ is absolutely continuous (for small enough $\epsilon$). In fact,
$$c_n:=\widehat{\mu*\mu}(n)=\widehat{\mu}^2 (n) = O(n^{-1+2\epsilon})$$
Thus, the series $\sum_{n=-\infty}^\infty |c_n|^2$ converges and by the Fischer-Riesz theorem we have that there exists a function $f\in L^2([-\pi,\pi])\subset L^1([-\pi,\pi])$ such that
$$\widehat{\mu*\mu}(n)=\int_{-\pi}^\pi e^{-inx} f(x) dx$$
Since $(e^{-inx})_{n\in\mathbb{Z}}$ forms an orthonormal basis of $L^2([-\pi,\pi])$, we also get
$$\mu*\mu(A)=\int_A d(\mu*\mu)(x)=\int_{-\pi}^\pi \mathbf{1}_A(x) f(x) dx$$
for measurable $A\subset[-\pi,\pi]$. Hence $f$ is the Radon-Nikodym derivative of $\mu*\mu$ and $\mu*\mu$ is absolutely continuous.
Note however, that in the original article they argue using monotone functions instead of measures (since that was the customary language in their time). This is equivalent by taking primitives.
[1] Norbert Wiener and Aurel Wintner. Fourier-Stieltjes Transforms and Singular Infinite Convolutions. American Journal of Mathematics, Vol. 60, No. 3 (Jul., 1938), pp. 513-522
J.R.
- 18,312
-
1Thank you so much for your answer. To my regret, I cannot accept the both at once. I expected something like your argument, but the reference is really important. Did's answer is impressing because an elegant elementary construction gives us precisely the Lebesgue measure on a segment. – limanac Sep 23 '14 at 18:46
8
The convolution $\mu\ast\nu$ itself may be absolutely continuous, here is a concrete example.
Let $(X_n)$ be i.i.d. with uniform distribution on $\{0,1\}$, thus, $P(X_n=0)=P(X_n=1)=\frac12$ for every $n$. Consider $X=\sum\limits_{n\geqslant1}4^{-n}X_n$, $\mu$ the distribution of $X$, and $\nu$ the distribution of $2X$.
Then $\mu$ and $\nu$ are purely singular (they have no atom and no absolutely continuous part), while $\mu\ast\nu$ is the restriction to $[0,1]$ of the Lebesgue measure, hence $\mu\ast\nu$ is absolutely continuous.
To see why $\mu\ast\nu$ is uniform on $[0,1]$, consider that this measure is the distribution of $Z=X+2Y$ where $Y=\sum\limits_{n\geqslant1}4^{-n}Y_n$ with $(Y_n)$ independent of $(X_n)$, and i.i.d. with uniform distribution on $\{0,1\}$. Thus $Y$ is a copy of $X$, independent of $X$, and $Z=\sum\limits_{n\geqslant1}4^{-n}Z_n$ where $(Z_n)$ is i.i.d. and, for every $n$, $Z_n=X_n+2Y_n$.
Now, as a Grand Finale of all this, simply note that every $Z_n$ is uniform on $\{0,1,2,3\}$... as should be the "digits" of the expansion of a uniformly distributed real number in base $4$.
A similar argument, based on the commonest form of the law of large numbers, shows that $\mu$ is indeed singular, hence $\nu$ is singular as well.
