TITLE: Problem of sobolev function in 1d
QUESTION [5 upvotes]: Let $1<p\leq\infty$ and $u\in L^p((0,1))$ be such that $|\int_0^1 u(x)\phi'(x)dx|\leq C\|\phi\|_{L^{p'}}$ for all $\phi\in C_c^\infty ((0,1))$, where $C$ is a constant. Show that $u\in W^{1,p}((0,1))$.
I am able to show that $u$ is Holder continuous, but that is not enough for $u$ to be a sobolev function. Can anyone give me a hint?

REPLY [2 votes]: You need to show that $u$ has a weak derivative (say $u'$) which is an element of $L^p$. What would be a reasonable candidate? Well, if $u$ were smooth then you'd have
$$
\int_0^1 u\phi'~dx = -\int_0^1 u'\phi~dx.
$$
Since you only know $u\in L^p$, the above integration by parts does not literally work, but the left-hand side still defines a linear functional:
$$
\ell(\phi) = -\int_0^1 u\phi'~dx.
$$
The estimate given in the hypotheses tells you that this is a bounded linear functional on $L^{p'}$. You can then use duality of $L^p$ spaces to find your weak derivative, and show it is in $L^p$ (essentially by definition).