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Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be an $L^1(\mathbb{R}^2; \mathbb{R})$ function. Let $A \subset \mathbb{R}$ be a measurable set. Let $\eta: \mathbb{R}\rightarrow \mathbb{R} \in C_c^{\infty}(\mathbb{R}),$ with support in $[-1,1]$ such that $\int\limits_{\mathbb{R}}\eta(x)dx=1.$ Then consider the following limit \begin{eqnarray} \lim\limits_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int\limits_{\mathbb{R}} \int\limits_{A} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx. \end{eqnarray} What is the value of the limit? How to prove it.

P.S: Define $\eta_{\epsilon}(x):=\frac{1}{\epsilon}\eta(\frac{x}{\epsilon}).$ Then $\eta_{\epsilon} \rightarrow \delta_0.$ So, I think the limit is, \begin{eqnarray} \int\limits_{A} f(x,x)dx. \end{eqnarray} is it correct? How to prove it?

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    $\begingroup$ if you want $\eta_{\epsilon} \rightarrow \delta_0$, then you need to define $\eta_{\epsilon}(x):=\frac1\epsilon\eta(\frac{x}{\epsilon})$. Right now $\eta_{\epsilon} \rightarrow0$ almost everywhere, which means that the limit of the double integral equals $0$ right now. (Also, what is $A$?) $\endgroup$ – Greg Martin Jul 12 '20 at 7:24
  • $\begingroup$ @Greg Martin..yes you are right.. I have made the necessary changes...Thanks for the comment. Is the result true now? Or do we need some more regularity assumption on the set A $\endgroup$ – Sameera Jul 12 '20 at 8:07
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Change variables $\frac{x-y}{\epsilon}\rightarrow z$ then \begin{eqnarray} \frac{1}{\epsilon} \int\limits_{\mathbb{R}} \int\limits_{A} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx\\=- \int\limits_{\mathbb{R}} \int\limits_{\left(\frac{1}{\epsilon}x-\frac{1}{\epsilon}A\right)\cap [-1,1]} f(x,x-\epsilon z)\eta\left(z\right)dz dx \end{eqnarray} which converges by the dominated convergence theorem to \begin{eqnarray} \int\limits_{\mathbb{R}}f(x,x) \left( -\int\limits_{\mathbb{R}} \lim_{\epsilon\rightarrow 0}\mathbb{1}_{\left(\frac{1}{\epsilon}x-\frac{1}{\epsilon}A\right)\cap [-1,1]}(z) \eta\left(z\right)dz\right) dx \end{eqnarray} Since $\eta$ is compactly supported in $[-1,1]$, the inner integral converges to a number uniformly in $x$. If we know that $[-1,1]\subset A$ we get a nicer limit: \begin{eqnarray} \frac{1}{\epsilon} \int\limits_{\mathbb{R}} \int\limits_{A} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx= \frac{1}{\epsilon} \int\limits_{\mathbb{R}} \int\limits_{\mathbb{R}} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx \\= \int\limits_{\mathbb{R}} \int\limits_{\mathbb{R}} f(x,x-\epsilon z)\eta\left(z\right)dz dx \end{eqnarray} which converges by the dominated convergence theorem to \begin{eqnarray} \int\limits_{\mathbb{R}}f(x,x) dx\,\int\limits_{\mathbb{R}} \eta\left(z\right)dz=\int\limits_{\mathbb{R}}f(x,x) dx. \end{eqnarray}

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  • $\begingroup$ I think there is no minus, (as change of variable formula gives modulus of jacobian) and inner integral as a function of x converges to characteristic function of the set A..please correct me if I am wrong $\endgroup$ – Sameera Jul 12 '20 at 11:19
  • $\begingroup$ Yes, correct, but we then need to reorient the limit of integration in the "right" direction. $\endgroup$ – Medo Jul 12 '20 at 11:29

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