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I am working on a problem of the form $$-(pu')'+qu=f,$$ $$u(0)=u(1)=0$$ where $p>0$ is $C^1$, $q\geq 0$ is continuous and $f\in L^2$. Using Lax-Milgram's theorem, I have been able to prove that the problem has a solution on the Sobolev space $H_0^1$: the above-mentioned theorem guarantees that there exists a unique $u\in H_0^1$ such that $$\int pu'v' + \int quv = \int fv$$ for all $v\in H^1_0$ from where we deduce that $-(pu')'+qu=f$ in $H^1_0$.

Now my question is: how do I prove that if $f$ is continuous, then $u\in C^2$? What I know is that since $pu'\in H^1$ then also $u'\in H^1$, hence $u\in H^2$ which can be injected in $C^1$... Thank you very much!

EDIT:

I tried the following approach now. Let $u\in H^1_0$ be a solution of the problem. Then $(pu')'=qu-f\in H^1$, so $pu'\in H^2$. Consequently, $u'\in H^2$ and so $u\in H^3\subseteq C^2$, proving the claim.

New question. If it is correct, where does this proof use continuity of $f$? Can I say $qu-f\in H^1$?

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    If $f$ is only continuous then $f' \not\in L^2$ necessarily, so it is not immediately clear to me that $qu-f\in H^1$. However, perhaps since you know that $u\in H^2$ then $u\in C^1[0,1]$. Then writing the equation as $-pu''=p'u' - qu+f$ we see that the RHS is continuous so the LHS must also be continuous. Does this seem to work? – 2017-02-08
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    Thank you for your remark, it makes perfect sense. Yes, that should be it; we get that $u''$ is continuous, hence $u$ is $C^2$. Thanks! – 2017-02-09

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You can use classical existence/uniqueness results of ODEs to establish the result you want. For example, the unique solutions $h$, $k$ of $$ -(pu')'+qu = 0 \\ h(0)=0, h'(0)=1,\;\;\; k(1)=0, k'(1)=1, $$ are $C^2$ by classical existence/uniqueness results. Assuming $h$ and $k$ are linearly-independent, which must be the case in order for your equation to have a solution for all $f$, then the solution $u$ of your problem is explicitly given by a constant multiple of $$ k(x)\int_{0}^{x}f(t)h(t)dt+h(x)\int_{x}^{1}f(t)k(t)dt, $$ which is really just a variation of parameters solution. From this you can directly see that $u$ must be $C^2$ if $f$ is in $C$.