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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?

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2 Answers 2

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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 $U$.

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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 $T^*.$

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