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The Wikipedia article on spectral decomposition, see here

https://en.wikipedia.org/wiki/Self-adjoint_operator

says the following:

A self-adjoint operator A on $H$ has pure point spectrum if and only if $H$ has an orthonormal basis ${e_i}_{i \in I}$ consisting of eigenvectors for A.

Why is this true? What is a reference for a proof? (Also to be sure, I guess that pure point spectrum means that the spectrum of the operator is equal to its eigenvalues.)

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  • $\begingroup$ I am not finding where the cited Wikipedia page says what is quoted. In any case, it is not true. Cluster points of eigenvalues are in the spectrum, but not necessarily in the point spectrum. Voting to close. $\endgroup$
    – Michael Renardy
    Commented Apr 6, 2018 at 18:18
  • $\begingroup$ Sorry, I linked the incorrect page. It's fixed now. $\endgroup$
    – Dale Wintermann
    Commented Apr 6, 2018 at 18:57
  • $\begingroup$ The cited lines are from the last section: "Pure point spectrum" $\endgroup$
    – Dale Wintermann
    Commented Apr 6, 2018 at 18:58
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    $\begingroup$ . . . . perhaps the question shouldn't be closed until we are sure it's not true? $\endgroup$
    – Dale Wintermann
    Commented Apr 6, 2018 at 18:59
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    $\begingroup$ On the wikipedia page this seems to be a definition of "pure point spectrum". So claiming whether it's true/false or asking for a proof is just pointless. $\endgroup$
    – YCor
    Commented Apr 6, 2018 at 22:13

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I think it’s a matter of definitions. Here is from Kreyszig (1978, p. 521) (admittedly not covering the “unbounded” in your title):

Kreyszig

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