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) = $ $\chi_{[-1,1]}(x)exp({-x^2 \over \epsilon^2}) \over \sqrt{2\pi\epsilon}$, 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 $$ \int_\mathbb{R} |j(x)|dx=\lVert j\rVert_1 <\infty, \;\int_\mathbb{R} j(x)dx=1. $$Let us define $j_\epsilon(x) = \frac{1}{\epsilon}j(\frac{x}{\epsilon})$ as a dilation of $j$. Then, it holds that $$ (f*j_\epsilon)(x) = \int_\mathbb{R} f(x-y)j_\epsilon(y)dy\to f(x) $$ uniformly as $\epsilon \to 0$. To see this, observe that for any fixed $\delta>0$, it holds $$\begin{eqnarray} |(f*j_\epsilon)(x)-f(x)|&= & |\int_\mathbb{R} (f(x-y)-f(x))j_\epsilon(y)dy|\\ &\leq&\int_{|y|<\delta} |f(x-y)-f(x)||j_\epsilon(y)|dy+ \int_{|y|\geq\delta} |f(x-y)-f(x)||j_\epsilon(y)|dy. \end{eqnarray}$$ The first term can be estimated as $$ \int_{|y|<\delta} |f(x-y)-f(x)||j_\epsilon(y)|dy\leq \lVert j\rVert_1\cdot\sup_{x\in\mathbb{R},|y|<\delta}|f(x)-f(x-y)|. $$ The second term can be estimated as $$ \int_{|y|\geq\delta} |f(x-y)-f(x)||j_\epsilon(y)|dy\leq 2\sup_{x\in\mathbb{R}}|f(x)|\cdot\int_{|z|>\frac{\delta}{\epsilon}}|j(z)|dz. $$ From this, by taking supremum on the LHS over all $x$, we have $$ \lVert(f*j_\epsilon)(\cdot)-f(\cdot)\rVert_\infty\leq \lVert j\rVert_1\cdot \sup_{x\in\mathbb{R},|y|<\delta}|f(x)-f(x-y)| + 2\sup_{x\in\mathbb{R}}|f(x)|\cdot\int_{|z|>\frac{\delta}{\epsilon}}|j(z)|dz. $$ 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 $$ \limsup_{\epsilon\to 0}\lVert(f*j_\epsilon)(\cdot)-f(\cdot)\rVert_\infty\leq \lVert j\rVert_1\cdot \sup_{x\in\mathbb{R},|y|<\delta}|f(x)-f(x-y)|, $$ for all $\delta>0$. Finally, by taking $\delta \to 0$, we get $$ \lim_{\epsilon\to 0}\lVert(f*j_\epsilon)(\cdot)-f(\cdot)\rVert_\infty =0, $$as desired.
$\textbf{EDIT:}$ As a corollary, $$j_\epsilon(x) = \frac{1}{\sqrt{2\pi}\epsilon}e^{-\frac{x^2}{2\epsilon^2}}$$ is suitable for your purpose. You don't need to multiply $\chi_{[-1,1]}(x)$, though it doesn't matter for your purpose.