Suppose $f$ and $\phi$ are continuous real values function on $\mathbb R$. Suppose $\phi(x)=0$ when $|x|>5$, and ST $\int_{\mathbb R}\phi(x)dx=0$

Question

Suppose $f$ and $\phi$ are continuous real values function on $\mathbb R$. Suppose $\phi(x)=0$ when $|x|>5$, and suppose that $\int_{\mathbb R}\phi(x)dx=0$. Show that $\lim_{h \rightarrow 0}\frac{1}{h}\int_{\mathbb R} f(x-y) \phi(\frac{y}{h}) dy=f(x)$ for $x \in \mathbb R$.

Can anyone help me to give some hint of this question?

Answer 1

\begin{align*} \dfrac{1}{h}\int f(x-y)\phi(y/h)dy&=\int f(x-hy)\phi(y)dy\\ &=\int f(x-hy)\phi(y)dy-f(x)\int\phi(y)dy\\ &=\int(f(x-hy)-f(x))\phi(y)dy\\ &=\int_{|y|<5}(f(x-hy)-f(x))\phi(y)dy, \end{align*} now invoke the continuity of $f$ at $x$.