I have some problem on the following exercise:
Compute the smallest constant $C>0$ such that:
\begin{equation*} \bigg(\int_{[0,1]}u(x)dx\bigg)^2\leq C\int_{[0,1]}u'(x)^2dx \end{equation*}
$\forall u\in H^1_0(0,1)$. Compute also all the function such that the inequality with the optimal constant becomes an equality.
Hint: a primitive of $1$ is $x+k$ with $k\in\mathbb{R}$, use Holder inequality and optimize $k$.
First part of the proof: I first notice that: \begin{equation*} u(x)\leq|u(x)|=|u(x)-u(0)|=\bigg\lvert\int_0^x u'(t)dt\bigg\lvert \leq \int_0^1|u'(x)|dx=\|u'\|_{L^1} \end{equation*} Now integrating both sides I obtain: \begin{equation*} \int_0^1 u(x)dx\leq \|u\|_{L^1}\leq\int_0^1\|u'\|_{L^1}dx=\|u'\|_{L^1} \end{equation*} Using Holder inequality I get: \begin{equation*} \int_0^1|u'(x)|dx\leq\bigg(\int_0^1 1^2 dx\bigg)^{1/2}\bigg(\int|u'(x)|^2 dx\bigg)^{1/2}=\bigg(\int_0^1 1^2 dx\bigg)^{1/2}\bigg(\int u'(x)^2 dx\bigg)^{1/2} \end{equation*}
where the last follows because we are in dimension $n=1$. Combining now the last two term and squaring both sides I get:
\begin{equation*} \bigg(\int_0^1 u(x)dx\bigg)^2\leq\bigg(\int_0^1 1dx\bigg)\bigg(\int_0^1 u'(x)^2dx\bigg) \end{equation*}
But from this computation I only find $C=1$.
Can someone help me?