# Semi-bounded operator implies semi-bounded sesquilinear form

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.

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