# If $T$ is a symmetric densely defined operator its adjoint $T^*$ is a closed extension of $T$

In this question Why is a densely defined symmetric operator $T$ extended by its adjoint $T^*$? the accepted answer proves that for a densely defined symmetric operator $$T$$ on a Hilbert space $$H$$, its adjoint $$T^*$$ is an extension of $$T$$. That is, $$D(T) \subset D(T^*)$$ and $$Tx = T^*x$$ for all $$x \in H$$. According to a textbook I am following, this extension is a closed extension. I think this is a simple consequence of $$T$$ being symmetric but I am not sure. How can one prove this?

Consider the unitary operator $$U \colon H \oplus H \to H \oplus H$$ given by $$U(x, y) = (-y, x)$$ and show that the graph of $$T^*$$ coincides with the orthogonal complement of the image of the graph of $$T$$ under $$U$$.
The straightforward proof also works. Assume $$(x_n,T^*x_n)\to (x,y)$$ where $$x_n\in D(T^*).$$ Then for any $$z\in D(T)$$ we have $$\langle x,Tz\rangle\leftarrow \langle x_n,Tz\rangle =\langle T^*x_n,z\rangle\to \langle y,z\rangle$$ Hence $$x\in D(T^*)$$ and $$T^*x=y,$$ i.e $$(x,y)$$ belongs to the graph of $$T^*.$$