I have a question about an exercise for which I already have the solution, which I do not unterstand completely.
Let $a, b \in \mathbb R$ with $0 < a < b$. Then we have \begin{align*} \|u\|^2_{L_2(a,b)} \le \frac{(b-a)^2}{2} \|u'\|^2_{L_2(a,b)} \end{align*} for all $u \in \overset{\circ}{W}_{1,2}((a,b))$. The latter space denotes the closure of the test function space with the Sobolev-norm.
Proof: By the Sobolev embedding theorem we know that $\overset{\circ}{W}_{1,2}((a,b))$ is continously embedded in the space $C([a,b])$. Thus we can write $u \in \overset{\circ}{W}_{1,2}((a,b))$ as \begin{align*} u(x) = \int_a^x u'(y) \, \mathrm dy, \quad \forall x \in (a,b). \end{align*} Why is this possible or why does that follow from the fact that $u$ is continuous? Then (summarized), we get for $x \in (a,b)$ that \begin{align*} |u(x)| \le \|u'\|_{L_2(a,b)} (x-a)^{1/2} \end{align*} and afterwards one can conclude that \begin{align*} \|u\|^2_{L_2(a,b)} \le \frac{(b-a)^2}{2} \|u'\|^2_{L_2(a,b)}. \end{align*}
I really understand the second part of the proof but not the first one.