Let $\phi : \mathbb{R}\rightarrow \mathbb{R}$ be an integrable function with $\int \phi(x)dx = 1$.
Let us define $\phi_\delta = \delta^{ā1}\phi(\delta^{-1}x)$.
Show that for every continuous function $f : \mathbb{R}\rightarrow\mathbb{R}$ with compact support,
$f ā \phi_\delta(x)\rightarrow f(x)$ as $\delta\rightarrow 0$ for every $x\in\mathbb{R}$.
My idea was to apply the following THM:
if $\phi_n$ is an approximation of unity, then for $f:\mathbb{R}^d\rightarrow \mathbb{R}$ continuous: $\phi_nā f\rightarrow f$ uniformly on compact sets.
but our definition of approximation of unity requires the fact that $\phi$ has a compact support.
Any idea or help?
Thanks
Best Answer
Write $$f\star\phi_\delta(x)-f(x)=\int_\mathbb R f(x-t)\delta^{-1}\phi(\delta^{-1}t)\mathrm dt-\int_\mathbb Rf(x)\delta^{-1}\phi(\delta^{-1}t)\mathrm dt,$$ hence using the substitution $s=\delta^{-1}t$, we get $$|f\star\phi_\delta(x)-f(x)|\leqslant \int_{\mathbb R}|f(x-s\delta)-f(x)|\cdot |\phi(s)|\mathrm ds.$$ Now we conclude using uniform continuity of $f$ and integrability of $\phi$.