# Show that an unbouned normal operator is closed

A linear operator $A$ is called nomal if $\mathcal{D}(A)=\mathcal{D}(A^{*})$ and $\lVert A\phi\rVert =\lVert A^{*}\phi\rVert$ for every $\phi\in \mathcal{D}(A)$. Show that normal operators are closed. Here I am not assuming that $A$ is bounded nor define on the whole Hilbert space $\mathcal{H}$.

Since $A^*$ is defined we know that $\mathcal{D}(A)$ is dense and since $\mathcal{D}(A)=\mathcal{D}(A^*)$ we have that $A^*$ is densely defined this implys that $A$ is closable. From the hypothesis I know that for any $\phi\in \mathcal{D}(A)$ that $\langle A\phi,A\phi\rangle = \langle A^*\phi,A^*\phi\rangle$ this gives that $\langle A^*A\phi,\phi\rangle=\langle A^{**}A^{*}\phi,\phi\rangle$ here I am assuming that $\phi\in \mathcal{D}(A^{**}A^{*})\cap \mathcal{D}(A^{*}A)$. But now I'm not sure where to go from here.

$A$ is closed if and only if $D(A)$ is complete under the graph norm $\|x\|_A = \|x\| + \|Ax\|$. But $A^*$ is always closed. Hence $D(A) = D(A^*)$ is complete under the graph norm $\|\cdot\|_A = \|\cdot\|_{A^*}$.