So my question is quite simple:

If an operator (which corresponds to a sesquilinear form) is semi-bounded (from below) then the sesquilinear form is also semi-bounded (from below)?

I'm pretty sure that this is true, but I dont know the proof.

  • 1
    $\begingroup$ How do you define that an operator is semi-bounded from below? Normally this is done in terms of the associated sesquilinear form. $\endgroup$ – DisintegratingByParts Jul 19 '18 at 17:53
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    $\begingroup$ Oh of course, now I see it. Thanks! $\endgroup$ – ProShitposter Jul 19 '18 at 19:28

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