I consider the functional $J(v) = \int_{a}^{b} \lvert v’(x)\rvert^{2}dx$ defined on
$$ K=\{ v \in H_{0}^{1}(a,b) : \int_{a}^{b} v^2(x)dx = 1 \} $$
First we notice that $J(v) = \lVert v\rVert_{H_{0}^{1}(a,b)}^{2}$ where $\lVert . \rVert_{H_{0}^{1}(a,b)}$ is an equivalent norm of $\lVert . \rVert_{H^{1}(a,b)}$.
I want to prove the existence of a minimum : consider $u_n$ a minimizing sequence on $K$ : $\lim_{n\to\infty} J(u_n) = \inf_{v\in K } J(v)$.
falseWe see that $ u_n$ is bounded as a sequence in $L^2(a,b)$ and thus admits a weakly convergent subsequence, denote it’s limit $\bar{u}$.false
Taking the constant function equals to $1$ we see by continuity of the $L^2$ norm that $\lVert \bar{u}\rVert_{L^2}^{2}$ equals to $1$
By the Banach Steinhaus theorem we have $\inf_{v\in K } J(v)\leq\lVert \bar{u}\rVert_{H_{0}^{1}(a,b)}^{2}\leq \lim\inf\lVert u_{\phi(n)}\rVert_{H_{0}^{1}(a,b)}^{2}= \inf_{v\in K } J(v)$
Hence the conclusion.
Is this seems correct please ?
Edit
I think we should use the Rellich theorem that ensures the convergence in $L^2$ of the subsequence and hence that $\lVert \bar{u}\rVert_{L^2} = 1$ so $\bar{u}\in K$.