Proving uniform convergence of an integral-defined function on compact sets 
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?
 A: 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 $I$.
