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I've read this theorem multiple times, but never seen a proof:

Every selfadjoint operator is closed.

But it's always been stated without a proof. Is it somehow obvious? I can't see it immediately from the graph.

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    $\begingroup$ From wiki "Note that, when T is self-adjoint, the existence of the adjoint implies that T is dense and since T^* is necessarily closed, T is closed." See here $\endgroup$ – wspin Dec 17 '12 at 21:12
  • $\begingroup$ If $T$ is bounded, it is of course a closed operator, this is the closed graph theorem, so we only need to consider $T$ is unbounded, but this time, you can find the answer in Kadison and Ringrose's book "Fundamentals of the theory of operator algebras I" section 2.7. Be careful of the defnitions in that section. $\endgroup$ – ougao Dec 17 '12 at 21:15
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Let $T$ be a densely defined linear operator on a Hilbert space $H$. Recall that the adjoint $T^*$ is defined by the relation $$\langle T^*x,z\rangle=\langle x,Tz\rangle$$ – or more accurately $(x,y)$ are in the graph of $T^*$ if and only if $$\langle y,z\rangle=\langle x,Tz\rangle$$ for all $z$ in the domain of $T$. By the continuity of the maps $y\mapsto\langle y,z\rangle$ and $x\mapsto\langle x,Tz\rangle$, the set of all $(x,y)$ satisfying this relation is closed. Thus $T^*$ is a closed map.

If $T$ is selfadjoint then $T=T^*$, so $T$ itself is closed.

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  • $\begingroup$ typo? this should be $x \to \langle x, T z \rangle$? $\endgroup$ – Ben Jun 30 '14 at 15:52
  • $\begingroup$ @Ben: Indeed. Probably of the cut-and-paste variety. Fixed; thanks. $\endgroup$ – Harald Hanche-Olsen Jun 30 '14 at 16:17

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