If $f$ is a compactly supported smooth (infinitely differentiable) function into $[0, 1]$ such that $\int f(x)dx = 1$, $g$ is a continuous function, and $f_\epsilon(x) = \frac{1}{\epsilon}f(\frac{x}{\epsilon})$, show that $g_\epsilon(x) := \int f_\epsilon(x-y)g(y)dy$ converges to g(x) uniformly on compact sets.
I have proven that $g_\epsilon$ is also smooth, with nth derivative $g^{(n)}_\epsilon(x) = \int f^{(n)}_\epsilon(x-y)g(y)dy$, though I'm not sure how helpful that is for this part.
My first guess is to let C be any compact set and try to find some upper bound on $\sup_{x \in C} \lbrace |g(x) - \int \frac{1}{\epsilon} f((x - y)/\epsilon)g(y)dy| \rbrace$ which tends to $0$ as $\epsilon \to 0$, but I'm not really sure how to estimate this supremum since the functions and set involved seem too "general" to say anything that's both specific and helpful about them. Am I missing something obvious? Would it be better to pick some particular bunch of compact sets that behave "nicely" for this calculation and somehow extrapolate out to all of them?
Assuming you have shown $g_{\epsilon}(x)\to g(x)$ pointwise a.e. $x$, modify the estimate of $|g_{\epsilon}(x)-g(x)|$ used in that proof, noting that $g$ is uniformly continuous on compact sets (hint: you evoked the Lebesgue differentiation theorem (LDT) to prove a.e. convergence; with a uniformly continuous $g$, what can you say about the relevant limit in the LDT?).
ADDITIONAL HINTS:
I assume you mean $f$ is compactly supported on $[0,1]$, not a map into $[0,1]$.
For fixed $x\in\mathbb{R}$ note that $f_{\epsilon}$ is compactly supported on $[0,\epsilon]$.
Therefore (using the unit mass property of $f$),
$$|g_{\epsilon}(x)-g(x)|=\left|\int_{x}^{x+\epsilon}f_{\epsilon}(x-y)[g(y)-g(x)]\;dy\right|.$$
If you carry this out a bit further you will get to
$$|g_{\epsilon}(x)-g(x)|\leq\sup_{x\in[0,1]}f(x)\frac{1}{\epsilon}\int_{x}^{x+\epsilon}|g(y)-g(x)|\;dy.$$
This quantity tends to $0$ by the Lebesgue differentiation theorem as $\epsilon\to0$. This proves pointwise convergence for even an integrable $g$.
Now if the regularity of $g$ is upgraded to continuity, and we pick a compact interval $I$, then $g$ is uniformly continuous on $I$, and so the limit in the Lebesgue differentiation theorem is uniform in $x$. This implies that $g_{\epsilon}\to g$ uniformly on $I$. Now take an arbitrary compact set $K$ and use the previous argument to deduce $g_{\epsilon}\to g$ uniformly on $K$ since $K\subset I$ for some compact interval $I$.