Jtdylktuy

In class my professor claimed that that if $f: mathbb{R} rightarrow mathbb{R}$ is a continuous function on a compact set, then the convolution of $j_{epsilon}(x)$ and $f$ converges uniformly to $f$ as $epsilon rightarrow 0$, where $j_{epsilon}(x)$ is a smooth function with compact support that integrates to 1 over $mathbb{R}^n$.

I would like to verify this for $f = exp({-x^2})chi_{[-1,1]}$ and $j_{epsilon}(x) = $ $chi_{[-1,1]}(x)exp({-x^2 over epsilon^2}) over sqrt{2piepsilon}$, but I have no idea where to start. Any help is appreciated.

Edit: I accidentally left out the characteristic function on $j_{epsilon}(x)$. I've added it now.

Note that since support of $f$ is compact, $f$ is uniformly continuous and bounded. Let $j(x)$ be an arbitrary function such that
$$
int_mathbb{R} |j(x)|dx=lVert jrVert_1 <infty, ;int_mathbb{R} j(x)dx=1.
$$Let us define $j_epsilon(x) = frac{1}{epsilon}j(frac{x}{epsilon})$ as a dilation of $j$. Then, it holds that
$$
(f*j_epsilon)(x) = int_mathbb{R} f(x-y)j_epsilon(y)dyto f(x)
$$ uniformly as $epsilon to 0$. To see this, observe that for any fixed $delta>0$, it holds
$$begin{eqnarray}
|(f*j_epsilon)(x)-f(x)|&= & |int_mathbb{R} (f(x-y)-f(x))j_epsilon(y)dy|\
&leq&int_{|y|<delta} |f(x-y)-f(x)||j_epsilon(y)|dy+ int_{|y|geqdelta} |f(x-y)-f(x)||j_epsilon(y)|dy.
end{eqnarray}$$ The first term can be estimated as
$$
int_{|y|<delta} |f(x-y)-f(x)||j_epsilon(y)|dyleq lVert jrVert_1cdotsup_{xinmathbb{R},|y|<delta}|f(x)-f(x-y)|.
$$ The second term can be estimated as
$$
int_{|y|geqdelta} |f(x-y)-f(x)||j_epsilon(y)|dyleq 2sup_{xinmathbb{R}}|f(x)|cdotint_{|z|>frac{delta}{epsilon}}|j(z)|dz.
$$ From this, by taking supremum on the LHS over all $x$, we have
$$
lVert(f*j_epsilon)(cdot)-f(cdot)rVert_inftyleq lVert jrVert_1cdot sup_{xinmathbb{R},|y|<delta}|f(x)-f(x-y)| + 2sup_{xinmathbb{R}}|f(x)|cdotint_{|z|>frac{delta}{epsilon}}|j(z)|dz.
$$
As $epsilon to 0$, we have $frac{delta}{epsilon}to infty$, and hence that
$$
int_{|z|>frac{delta}{epsilon}}|j(z)|dz to 0,
$$by dominated convergence theorem. This implies
$$
limsup_{epsilonto 0}lVert(f*j_epsilon)(cdot)-f(cdot)rVert_inftyleq lVert jrVert_1cdot sup_{xinmathbb{R},|y|<delta}|f(x)-f(x-y)|,
$$ for all $delta>0$. Finally, by taking $delta to 0$, we get
$$
lim_{epsilonto 0}lVert(f*j_epsilon)(cdot)-f(cdot)rVert_infty =0,
$$as desired.

$textbf{EDIT:}$ As a corollary, $$j_epsilon(x) = frac{1}{sqrt{2pi}epsilon}e^{-frac{x^2}{2epsilon^2}}$$ is suitable for your purpose. You don't need to multiply $chi_{[-1,1]}(x)$, though it doesn't matter for your purpose.