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I am trying to prove that a self-adjoint compact operator on a Hilbert space ($T \colon \mathcal H \to \mathcal H$) cannot be surjective. The hint says that I need to use open mapping theorem and spectral theorem. Could someone help?

P. Factor
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1 Answers1

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If a self-adjoint operator is surjective, then its null space is trivial. Thus, the inverse operator is well defined. Since it is defined on the entire space, it is continuous by the Banach closed graph theorem. But the inverse of a compact operator on an (infinite-dimensional) Hilbert space cannot be continuous. This contradiction proves the desired assertion.

Vladimir
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  • It seems more related to compactness instead of self-adjointness. –  Sep 01 '16 at 01:32
  • @Arctic Char: Yes, sure, self-adjointness is not actually required. If $T$ is not self-adjoint, you may well take $\tilde T\colon H/\operatorname{Ker} T\to H$ and then argue as in my answer. – Vladimir Sep 01 '16 at 01:34
  • How do you get the null space is trivial? Thanks! – Toasted_Brain Oct 25 '20 at 22:36
  • @cciirrcclllee: Assume $Au=0$. Since $A$ is surjective, we have $u=Av$, and then $||u||^2=(u,u)=(Av,u)=(v,Au)=0$, so that $u=0$. – Vladimir Oct 26 '20 at 23:15