Let $T$ be an operator in $H$. We say self adjoint $T$ is positive iff $(\forall x\in H)\langle Tx,x\rangle \geq 0 $. As in the case of bounded operators, it is true that a self-adjoint operator $T$ is positive iff its spectrum $\sigma(T)\subset[0,\infty)$. The 'only if' part is proved in a similar way as for bounded operators. The 'if' part is proved by using the integral representation for self adjoint unbounded operators. Is there a way to prove it without using this spectral theory?

Similarly, the fact that every positive (unbounded) operator has a positive square root is proved using the spectral theorem. Is there a way to prove this without using it?

  • $\begingroup$ I think continuous function calculus is a bit easier to construct than the general spectral theorem, though I'm not sure if it's still true for unbounded operators. What's wrong with spectral theorem anyway? $\endgroup$
    – tomasz
    Commented Jan 21, 2014 at 2:14
  • 1
    $\begingroup$ The proof for the spectral theorem for unbounded self-adjoint operators requires a measurable(not necessarily bounded measurable) functional calculus. Of course once that is proved the above two results follow as easy corollaries. I am trying to see if they can be proved independently. $\endgroup$
    – Arundhathi
    Commented Jan 21, 2014 at 2:26
  • $\begingroup$ This question was bumped by Community because you had not accepted an answer to your question. If the answer that you posted answers your question for you, please accept the answer so that this does not happen again in the future. $\endgroup$ Commented Aug 4, 2015 at 4:49

1 Answer 1


I found the following paper that answered my question- http://journals.cambridge.org/download.php?file=%2FJAZ%2FJAZ8_01%2FS1446788700004560a.pdf&code=7abaaa421aab87a68a9e7b504bdd5319


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