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?
Best Answer
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}