I am stuck with following exercise:
Show that for $u \in H^1{(0,1)} = W^{1,p}(0,1)$ denoting the Sobolev space with $p = 2$ and with $u(0) = 0$:
$$ \| u\|_{L^\infty{(0,1)}} \leq \| u' \|_{L^2{(0,1)}}$$
My ideas so far were going for Poincaré-inequality for an open ball. This would lead to:
$$ \| u - \bar{u}\|_{L^2{(0,1)}} \leq C\cdot r \cdot \| u' \|_{L^2{(0,1)}}$$
and $r = \frac{1}{2}$, $C$ being a constant with $C > 0$ and $\bar{u} = \int_{(0,1)} u dx$ being an average of $u$ over the ball $(0,1)$.
Another idea is to use $\|u \|_{L^2(0,1)} \leq \| u \|_{L^\infty(0,1)}$, because $(0,1)$ is bounded.
Can someone please give me a hint how to use the condition $u(0)=0$. Any other hints are welcome as well! Thank you.
Take $u \in W^{1,2}(0,1)$. We can consider a continuous representative of $u$ such that $u(x) = \int_0^x u'(t)dt, x \in [0,1],$ as is pointed out in one of the comments to this question, with appropriate reference.
Now $|u(x)| \le \int_0^1 |u'(t)|dt \le \sqrt{\int_0^1|u'(t)|^2dt}= ||u'||_{L^2}, \forall x \in [0,1]$
Hence the result.