Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 – wspin Dec 17 '12 at 21:12
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. – ougao Dec 17 '12 at 21:15
up vote 6 down vote accepted

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.