I have the operator $A:C([0,1])\to C^1([0,1])$ defined to be $Af(x)=\int_0^xf(t){\rm d}t$. I am pretty sure it is compact. I know that the image of the unit ball in $C([0,1])$ under the operator is a subset of the 2-radius ball in $C^1([0,1])$ in which every function satisfies $g(0)=0$.
But I'm lost from there. Had I been successful in proving this image is precompact, I'd be done. Any help would be appreciated.
On
This answer complements that of Kavi Rama Murthy. Here I assume that you equip $C^1([0,1])$ with the norm $\lVert f \rVert_\infty + \lVert f'\rVert_\infty$ in which case $A$ is not compact.
Take $f_n(x) = x^n$. Then $(f_n)$ is bounded in $C([0,1])$. But $(Af_n)' = f_n$ does not have any subsequence converging uniformly on $[0,1]$ (as otherwise it would converge to the discontinuous function $\mathbf{1}_{\lbrace 1\rbrace}$). Therefore no subsequence of $(Af_n)_n$ can converge in the sense of $C^1([0,1])$.
You have not specified the norm on the range. I will assume that you are using the sup norm on both the range and the domain.
This is an easy application of Arzela - Ascoli Theorem. If $(f_n)$ is sequence in the in the unit ball of $C[0,1])$ then $|Af_n(x)-Af_n(y)| \leq |\int_x ^{y} 1 dt| =|x-y|$ so the family $(Af_n)$ is equi-continuous. It is also uniformly bounded by $1$ so Arzela - Ascoli Theorem tells you that $(Af_n)$ has a subsequence which converges uniformly.
If the norm on the range is $||f||_{\infty}+\|f'\|_{\infty}$ then $A$ is not compact since $(Ax^{n})$ has no convergent subsequence..