In my numerical approximation of PDEs class, my professor wrote that $V=\{v\in H^1(\Omega): \int_{\Omega}v=0\}$ is a Hilbert space with $\Omega$ bounded, connected, open and with Lipschitz boundary. Unfortunately, I have failed to show that $V$ is closed in $H^1(\Omega)$. This is what I tried:
Let us try to show that $V$ is closed by showing that all sequences in $V$ that converge have their limit in $V$. Let $v\in H^1$ and $(v_n)\subset V$ be such that $\lim v_n = v$ in $H^1$. In particular, $\lim v_n = v$ in $L^2$. By a theorem from measure theory, this implies that there exists a subsequence of $v_n$, $v_{n_k}$ that converges to $v$ almost everywhere. I would like to conclude with the dominated convergence theorem and say $\lim \int v_n = 0 = \int v$ but I do not have an integrable function that bounds ||.
Might anyone be able to help?
You have $v_n \to v$ in $L^2$, and now this follows from Cauchy-Schwarz: for any $n$, since $\int_\Omega v_n = 0$, we can write $$\begin{align*} \left|\int v\right| &= \left| \int (v-v_n)\cdot 1 \right| \le \left(\int |v-v_n|^2 \cdot \int 1^2\right)^{1/2} = \|v_n-v\|_{L^2} \sqrt{m(\Omega)} \end{align*}$$ and as $n \to \infty$, the right side goes to 0. Note how we used $m(\Omega) < \infty$.
Alternatively, you can think that we have $1 \in L^2(\Omega)$ (again since $m(\Omega) < \infty$), and use the continuity of the $L^2$ inner product with respect to the $L^2$ norm (which is really Cauchy-Schwarz again) to write $\langle v, 1 \rangle_{L^2} = \lim_{n \to \infty} \langle v_n, 1 \rangle = 0.$