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Let $\mu$ be a positive finite measure on $\mathbb R$. Is it true that $$\int_{\mathbb R} T \text{sinc}^2(Tx) d\mu(x) \sim\frac{\mu([-1/T,1/T])}{1/T}, \text{ as } T\to\infty?$$ Here $\text{sinc}(x)=\frac{\sin x}{x}$ and the notation $f(T)\sim g(T)$ as $T\to\infty$ means that $\lim_{T\to\infty}f(T)/g(T)=c<\infty$.

Of course, this holds if $\mu$ is absolutely continuous (a classical theorem on summability kernels), where the limit of both sides exists and is finite. On the other hand, if $\mu$ is not absolutely continuous near $0$, then the RHS tends to infinity (by theory of differntiation of measures). How to prove that the LHS tends to infinity with the same asymptotics?

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No, if $\mu=\delta_{x_0}$ then the RHS is just $0$. This is an example where there is a finite limit to both sides, and it is equal. Notice in this example $\mu$ is absolutely continuous near $0$ (I am not sure this is a formal term, what I mean is the RHS has a finite limit). – Naomi Jul 24 '12 at 20:35
up vote 0 down vote accepted

Answer: no. Let $$\mu=\sum_{n=1}^\infty 2^{-n}\delta_{2^{-n}}$$ and observe that at each point $T=2^n$ the one-sided limits of the RHS differ by the factor of $2$. On the other hand, the LHS is a continuous function of $T$. Since the ratio LHS/RHS jumps by the factor of $2$ at each $T=2^n$, it cannot approach a finite limit as $T\to\infty$.

Comment: you are using the term absolutely continuous incorrectly.

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Thanks! That's a neat example. I wanted to comment that LHS >= c RHS holds always. Also, thanks for pointing out my confusion with absolutely continuous, I guess the word I lacked was "density" at zero. – Naomi Jul 30 '12 at 16:48
@Naomi Yes, that's true with $c=\sin^2 1$. (I was about to post a counterexample to the inequality with $c=1$ before you edited the comment.) – user31373 Jul 30 '12 at 17:03

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