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Suppose that $H$ is an Hilbert space, and $T: H \to H$ is a self-adjoint strictly positive operator (i.e. $\langle Tx,x\rangle > 0$ for all $x \neq 0$). How do I show that this operator is invertible? For example, I want to show that $\langle Tx , x\rangle$ is bounded below by some positive constant (and then I am happy, I know the rest).

Thank you, Sasha

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    $\begingroup$ $\langle Tx, x\rangle$ can never be bounded below by a positive constant because of scaling. $\endgroup$
    – geometricK
    May 27, 2017 at 1:53

1 Answer 1

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Are you sure it is true?

Consider $T:\ell_2\to\ell_2$ which maps $(a_n)$ to $(a_n/n)$. This is clearly self-adjoint, and positive: $$\langle Ta,a\rangle=\sum_{n\ge 1} \frac{|a_n|^2}n$$ and this is $>0$ whenever any $a_n\ne 0$.

On the other side, $\langle Tx,x\rangle$ is not bounded from below: for the sequence $x_n=1$ and $x_k=0$ if $k\ne n$, we have $\langle Tx,x\rangle=1/n$.

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    $\begingroup$ Thank you very much! Now I am sure, that it is not true. $\endgroup$
    – Sasha
    Jun 9, 2013 at 12:04
  • $\begingroup$ What about Id+T for any positive bounded linear T? How can I show that it is invertible? $\endgroup$ Apr 26, 2014 at 9:08
  • $\begingroup$ @ErginSuer Have you got any result about your above question? $\endgroup$
    – John
    Jun 24, 2017 at 16:24
  • $\begingroup$ @ErginSuer Look up Minty's surjectivity theorem. $\endgroup$ Jan 16, 2019 at 5:44

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