Let $\Omega \subset \mathbb{R}^3$ be an open, bounded set. I want to show that the unit sphere in $H_0^1(\Omega)$ with the $L^2$ norm, the set $S = \{u \in H_0^1(\Omega) : \int_\Omega u^2 = 1\}$, is weakly closed in $H_0^1(\Omega)$. This is a claim in an article I am reading. What I thought was the following:
Let $(u_n) \subset S$ be a sequence (weakly) converging to $u \in H_0^1 (\Omega)$. We must show that $u \in S$. Note that \begin{align*} \left| \int_\Omega u^2 \ dx - 1 \right| & = \left| \int_\Omega u^2 \ dx - \int_\Omega u_n u_n \ dx \right| \\ & = \left| \int_\Omega u^2 \ dx - \int_\Omega u u_n \ dx + \int_\Omega u u_n \ dx - \int_\Omega u_n u_n \ dx\right| \\ & \leq \left| \int_\Omega u^2 \ dx - \int_\Omega u u_n \ dx\right| + \left| \int_\Omega u u_n \ dx - \int_\Omega u_n u_n \ dx \right| . \end{align*} Now, consider the following functionals: $$ f_1(u_n) = \int_\Omega u u_n \ dx, \qquad f_2(u_n) = \int_\Omega u_j u_n \ dx, $$ with $u_j \in (u_n)$. By weak convergence, we have that $$ f_1(u_n) \to f_1(u), \qquad f_2(u_n) \to f_2(u). $$ Then, given $\varepsilon > 0$ we can find $n \in \mathbb{N}$ such that $$ \left| \int_\Omega u^2 \ dx - \int_\Omega u u_n \ dx \right| \leq \frac{\varepsilon}{2}. $$
Now, my intuition says that I can also say there is an $n$ such that $$ \left| \int_\Omega u u_n \ dx - \int_\Omega u_n u_n \ dx \right| \leq \frac{\varepsilon}{2}. $$ Is this true? If so, why? If not, what is the correct way of proving the claim?
Thanks in advance.