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Every symmetric operator $A$ can be closed to an operator. Are this closed extensions symmetric? Are there any conditions that make this true such as $Dom(A)$ is densly defined in some Hilbert space?

yess
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2 Answers2

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The symmetric condition for $A$ may be written as the equivalent condition $$ \mathcal{G}(A)\perp_{H\times H} J\mathcal{G}(A), $$ where the graph of $A$ is defined as $\mathcal{G}(A)=\{ (x,Ax) \in H\times H : x\in\mathcal{D}(A)\}$, and where $J : H\times H \rightarrow H\times H$ is the unitary map defined by $J(x,y)=(-y,x)$. Let $\overline{\mathcal{G}(A)}$ denote the closure of $\mathcal{G}(A)$ in $H\times H$. Because of joint continuity of the inner product on $H\times H$, and because of the isometric nature of $J$, it follows that $$ \overline{\mathcal{G}(A)} \perp_{H\times H} J\overline{\mathcal{G}(A)}. $$ The closure $\overline{\mathcal{G}(A)}$ in $H\times H$ is $\mathcal{G}(\overline{A})$ where $\overline{A}$ is the closure of $A$. Hence, the above is equivalent to the symmetry of $\overline{A}$: $$ \mathcal{G}(\overline{A})\perp_{H\times H}J\mathcal{G}(\overline{A}). $$

Disintegrating By Parts
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  • Maybe I'm missing something, but isn't the condition you claim to be equivalent to the symmetry instead the equivalent condition for being self-adjoint? For a general densely-defined $A$, $\mathcal G(A^\ast)=(J^\ast\mathcal G(A))^\perp$. Now what I obtain is that $A$ is symmetric if and only if $\mathcal G(A)\subseteq\mathcal G(A^\ast)$ if and only if $$J^\ast\overline{\mathcal G(A)}=\overline{J^\ast\mathcal G(A)}=(J^\ast\mathcal G(A))^{\perp\perp}\subseteq{\mathcal G(A)}^\perp.$$ What am I missing? I've asked for that here: https://math.stackexchange.com/q/3426988/47771. – 0xbadf00d Nov 08 '19 at 20:19
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Here is a simpler more "analysis" proof.

Suppose $\psi,\phi\in$ Dom$(A^{cl})$. Then there are sequences $\psi_n\to\psi$ and $\phi_m\to\phi$ with $\psi_n,\phi_m \in$ Dom$(A)$ and $$\lim A\phi_n=A^{cl}\phi$$ $$\lim A\psi_m=A^{cl}\psi$$

Then by the continuity of the inner product, the following are equal $$<A^{cl}\phi,\psi>$$

$$<\lim A\phi_n,\psi>$$

$$\lim<A\phi_n,\psi>$$

$$\lim<A\phi_n,\lim \psi_m>$$

$$\lim_n\lim_m<A\phi_n,\psi_m>$$

$$\lim_n\lim_m<\phi_n,A\psi_m>$$

$$\lim<\phi_n,\lim A\psi_m>$$

$$\lim<\phi_n,A^{cl}\psi>$$

$$<\lim\phi_n,A^{cl}\psi>$$

$$<\phi,A^{cl}\psi>$$

So yes, the closure of a symmetric operator is symmetric.

juan arroyo
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