Let $L$ be a symmetric closable operator in a Hilbert space $X$. Sow that the closure of $L$ is also symmetric.
Since $L$ is symmetric, i.e. for every $s,t$ in domain of $L$, we have $$ \langle Ls,t\rangle = \langle\ s, Lt\rangle $$ $L$ is closable, i.e. for any sequence $(u_n)$ in domain of $L$ and some $v\in X$ such that $u_n\to 0$ and $Lu_n \to v$ as $n \to \infty$, then $v=0.$
Lets denote the closure of $L$ by $\widetilde{L}$.
We need to show that $\widetilde{L} $ is symmetric. Let $x,y$ in the domain of $ {\widetilde{L}}$ we have $$ \widetilde{L}x = w, \widetilde{L}y = z $$ $\iff$ $(x_n)$ and $(y_n)$ are some sequences in the domian of $L$ such that \begin{align*} &\lim x_n = x&\\ &\lim Lx_n = w&\\ &\lim y_n = y&\\ &\lim Lx_n = z&\\ \end{align*} I need to establish the following $$ \langle \widetilde{L}x ,y \rangle = \langle x,\widetilde{L}y \rangle $$ I think the closability of $L$ has to play some role in the proof before one uses the symmetry of $L$, but I am not sure how to bring that into play.
For $\tilde{x},\tilde{y}\in\mathcal{D}(\tilde{L})$, there exists sequences $\{x_n\},\{y_n\}\in\mathcal{D}(L)$ such that $x_n\rightarrow\tilde{x},Lx_n\rightarrow L\tilde{x}$ and $y_n\rightarrow\tilde{y},Ly_n\rightarrow L\tilde{y}$. Then, the joint continuity of the inner product gives $$ \langle \tilde{L}\tilde{x},\tilde{y}\rangle=\lim_n\langle Lx_n,y_n\rangle=\lim_n\langle x_n,Ly_n\rangle=\langle \tilde{x},\tilde{L}\tilde{y}\rangle. $$