Let $f$ be a Lebesgue integratable function with the property that
$\int_{-\infty}^{\infty}f(x)dx=1$
Define for $\epsilon >0$
$f_{\epsilon}:=\frac{1}{\epsilon^n}f(\frac{x}{\epsilon})$
Then $\forall \phi \in \mathcal{D}(\mathbb{R})$
$\lim_{\epsilon \rightarrow 0} \int_{-\infty}^{\infty}f_{\epsilon}(x)\phi(x)dx=\phi(0)$
I want to proof this. Thus, I need to show that: $\langle f_{\epsilon},\phi \rangle \rightarrow \langle \delta,\phi \rangle:=\phi(0)$
I tried a few things, I think to use the substitution $x=\epsilon y$ works best. With this I get the Integral $\int_{\infty}^{\infty}f(y)\phi(\epsilon y) dy$
Then (I am not really sure about that) since $\phi$ has compact support one can look at the maximum of $\phi$:
(1) $m:=max${$\phi(x):x \in supp$($\phi$)}
I am not really sure if I can take the maximum. Or if I would need to replace the $max$ in (1) with $sup$. For example if one chooses a sequence of functions $\phi_n \in \mathcal{D}(\mathbb{R})$ such that the $supp(\phi_n)$ converges to a unbounded set, then there could be a problem. (This is my first question)
Having now the constant m it is obvious that $|f(y)\phi(\epsilon y)| \leq m f(y)$
If one takes now the limit $lim_{\epsilon \rightarrow 0}f(y)\phi(\epsilon y)=f(y)\phi(0)$
If I can interchange the limit and integral one would get the result $lim_{\epsilon \rightarrow 0} \int_{-\infty}^{\infty}f(y)\phi(\epsilon y) dy=\int_{-\infty}^{\infty}lim_{\epsilon \rightarrow 0}f(y)\phi(\epsilon y)dy=\int_{-\infty}^{\infty}f(y)\phi(0)dy=\langle f,\phi \rangle$
But I don't know why it should be possible to change limit and integral.(This is my second question)