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Hello,

This problem bothers me for some time. Suppose that

  1. $\mu$ is a non-negative Radon measure (or positive linear functional of the space of continuous functions with compact support);
  2. $\psi$ is a continuous function, vanishing at infinity and integrable, i.e., $\psi\in C^0_0(R)\cap L^1(R)$;
  3. $\sup_{x \in R}|(\psi*\mu)(x)|<+\infty$.

Then, we would like to prove that the function $x\mapsto (\psi*\mu)(x)$ is continuous.

Thank you very much for your help and any hints!

Anand


Version 2. If we add an additional property,

  1. $\sup_{x\in R} |(G*\mu)(x)|<+\infty$, where $G(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$.

Then, is it possible to prove that the function $x\mapsto (\psi*\mu)(x)$ is continuous?

Thanks

Anand

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1  
My first guess (having recently forgotten, and then remembered, how to do something which seems similar): maybe the dominated convergence theorem would be enough. –  Yemon Choi Sep 22 '11 at 8:24
    
@ Yemon Choi, I have tried the dominated convergence theorem. It seems not obvious. The local behavior of $\psi$ can be as irregular as a Brownian motion locally. –  Anand Sep 22 '11 at 8:27
    
Ah, I misunderstood the question, because I thought $\mu$ had to be a finite measure. Apologies. –  Yemon Choi Sep 22 '11 at 10:08
    
Actually... How do you define $\psi*\mu$?? Because this seems to be, formally, $x\mapsto \int \psi(x-y) \ d\mu(y)$. But to make sense of this integral, don't we need that $y\mapsto \psi(x-y)$ is in $L^1(\mu)$? Or is this sort of implicit in (3)? –  Matthew Daws Sep 22 '11 at 10:52
    
@Mathew Daws, yes, the well-posedness of the convolution is implicit in Condition (3). Or to put in other words, we only consider those Radon measures that the convolution with the given $\psi$ is well defined. –  Anand Sep 22 '11 at 11:24

2 Answers 2

Here's a counter-example.

Let $\mu$ be counting measure supported on $\mathbb Z$; so $\int f(x) \ d\mu(x) = \sum_{m\in\mathbb Z} f(m)$ for $f$ continuous with compact support.

Choose a very rapidly decreasing sequence of positive reals $\delta_n$. Let $\psi$ be the piecewise linear function which is $1$ at $n+1/n$ for $n\geq 10$ (say), and is $0$ at $n+1/n \pm \delta_n$ (and is zero at all $x<10+1/10-\delta_{10}$). By definition, $\psi$ is continuous (and could be made smooth by the use of a bump function) and $\psi$ is Lebesgue integrable (if $\delta_n$ decreases fast enough).

Then set \[ \alpha(x) = (\psi*\mu)(x) = \int \psi(x-y) \ d\mu(y) = \sum_{m\in\mathbb Z} \psi(x+m). \] A priori, this sum might diverge, but only to $+\infty$. Clearly $\alpha$ is periodic in that $\alpha(x+k)=\alpha(x)$ for any $k\in\mathbb Z$.

Fix $x\in[-1/2,1/2)$, and consider which $m\in\mathbb Z$ are such that $\psi(x+m)>0$. This occurs iff there is $n\geq 10$ with $1/n-\delta_n < x+m-n < 1/n+\delta_n$. As $m-n\in\mathbb Z$ and $n\geq 10$ and $|x|\leq 1/2$, this can only occur if $m=n$. So $1/n-\delta_n < x < 1/n+\delta_n$. Choosing $(\delta_n)$ suitably, we can arrange that $1/(k+1)+\delta_{k+1} < 1/k-\delta_k$ for all $k$, and then $n$ is unique for any given $x$.

We conclude that for every $x$, there is at most one $m\in\mathbb Z$ with $\psi(x+m)>0$. In particular, $\alpha(x)\in[0,1]$ for all $x$.

However, clearly $\alpha(1/n)\geq 1$ for all $n\geq 10$, but $\alpha(0) = 0$ as if $\psi(m)>0$ for some $m\in\mathbb Z$, then there is $n\geq 10$ with $-\delta_n < m-n-1/n < \delta_n$ which forces $m=n$, but then $-\delta_n <-1/n<\delta_n$, which we can avoid by a suitable choice of $(\delta_n)$. So $\alpha = \psi*\mu$ is not continuous at $0$.

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Thanks Matthew Daws for your very nice counter-example. I agree with you. –  Anand Sep 22 '11 at 12:39
    
Dear Matthew Daws, I realized that I forgot to put one condition. In the condition (2), we actually need that $\psi(x)$ vanishes at $x=\pm\infty$, i.e., $\psi in C^0_0(R)$. I am sorry for such mistake. Your example is perfect if we only require $\psi\in C^0(R)\cap L^1(R)$. –  Anand Sep 22 '11 at 13:12
1  
The counter-example can be slightly modified to fit the new condition (2). We can let the values of bump decrease to zero while put increase mass on the delta measures to cancel that decrease. –  Anand Sep 22 '11 at 13:25
    
@Anand-- Yes, I agree! –  Matthew Daws Sep 22 '11 at 19:08

This is a theorem of I. Glicksberg.See http://dl.dropbox.com/u/5188175/glickrev.pdf for the math review.

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1  
Any chance you could give more details-- I thought that this paper only dealt with bounded measures, and Anand's problem seems to be that his measure is not necessarily bounded... –  Matthew Daws Sep 22 '11 at 9:02
    
Matt: what definition of Radon measure are we using here? –  Yemon Choi Sep 22 '11 at 9:31
1  
Well, Anand says "or positive linear functional of the space of continuous functions with compact support" which, if you follow e.g. Rudin, certainly allows for unbounded measures (e.g. one way to construct Lebesgue measure on $\mathbb R$ starts with Riemann integration of compact supported functions, and then runs the machinery...) –  Matthew Daws Sep 22 '11 at 9:41
1  
Thanks Matthew Daws and Igor Rivin and Yemon Choi, the Radon measure can be unbounded. This is the one difficulty of the problem. –  Anand Sep 22 '11 at 10:00
    
Ah, OK, I missed the unbounded part... –  Igor Rivin Sep 22 '11 at 10:27

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