Limit involving an integral Let $f$ and $\varphi$ be continuous real valued functions on $\mathbb{R}$. Suppose $\varphi(x)=0$ for $|x|>5$ and that $\int_{\mathbb{R}}\varphi(x)\mathbb{d}x=1$. Show that 
$$\lim_{h\to 0}\left[\frac{1}{h}\int_{\mathbb{R}}f(x-y)\varphi\left(\frac{y}{h}\right)\mathbb{d}y\right]=f(x).$$
I don't know how to proceed. Please help.
 A: Note that $\varphi(y/h) = 0$ for $|y/h|>5$, so we don't need to integrate over the entire real line.
$$
\lim_{h\to 0}\frac{1}{h}\int_{-5h}^{5h}f(x-y)\varphi(y/h)\mathbb{d}y
$$
define $\xi=y/h$.
$$
\lim_{h\to 0}\frac{1}{h}\int_{-5}^{5}f(x-h\xi)\varphi(\xi)\mathbb{d}(\xi h)
$$
$$
\mathbb{d}(h\xi)=h\mathbb{d}\xi
$$
$$
\lim_{h\to 0}\int_{-5}^{5}f(x-h\xi)\varphi(\xi)\mathbb{d}\xi
$$
$\xi$ is finite over the integration interval, so we can now apply the limit.
$$
\int_{-5}^{5}f(x-0\xi)\varphi(\xi)\mathbb{d}\xi=f(x)\int_{-5}^{5}\varphi(\xi)\mathbb{d}\xi
$$
but
$$
\int_{-5}^{5}\varphi(\xi)\mathbb{d}\xi=1
$$
as $\varphi(\xi)$ vanishes when $|\xi|>5$ and as stated at the beginning of the problem $\int_{\mathbb{R}}\varphi(\xi)\mathbb{d}\xi=1$
A: Hint. The given limit is equivalent to
$$\lim_{h\to 0^+}\frac{1}{h}\int_{-5h}^{5h}(f(x-y)-f(x))\varphi\left(\frac{y}{h}\right)dy=\lim_{h\to 0^+}\int_{-5}^{5}(f(x-th)-f(x))\varphi\left(t\right)dt=0$$
Then use the fact that $\varphi$ is bounded in $[-5,5]$ and $f$ is uniformly continuous in $[x-6,x+6]$.
A: If you do the substitution $y=ht$ and $dy=h\,dt$, then you get$$\frac1h\int_{\mathbb R}f(x-y)\varphi\left(\frac yh\right)\,dy=\int_{\mathbb R}f(x-ht)\varphi(t)\,dt$$On the other hand,\begin{align*}\int_{\mathbb R}f(x-ht)\varphi(t)\,dt-f(x)&=\int_{\mathbb R}\bigl(f(x-ht)-f(x)\bigr)\varphi(t)\,dt\\&=\int_{-5}^5\bigl(f(x-ht)-f(x)\bigr)\varphi(t)\,dt.\end{align*}In order to prove that the limit of this integral at $0$ is $0$, it will be enough to use the fact that $f$ is uniformly continuous on some interval larger than $[-5,5]$ (take $[-6,6]$, for instance), together with the fact that $\int_{\mathbb R}\varphi=1$.
