Semi-bounded operator implies semi-bounded sesquilinear form

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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.




functional-analysis operator-theory functional-inequalities




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    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 at 17:53






  • 1




    Oh of course, now I see it. Thanks!
    – ProShitposter
    Jul 19 at 19:28














up vote
1
down vote

favorite












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.




functional-analysis operator-theory functional-inequalities




share|cite|improve this question















  • 1




    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 at 17:53






  • 1




    Oh of course, now I see it. Thanks!
    – ProShitposter
    Jul 19 at 19:28












up vote
1
down vote

favorite









up vote
1
down vote

favorite











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.




functional-analysis operator-theory functional-inequalities




share|cite|improve this question











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.





functional-analysis operator-theory functional-inequalities





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 19 at 15:42









ProShitposter

787




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  • 1




    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 at 17:53






  • 1




    Oh of course, now I see it. Thanks!
    – ProShitposter
    Jul 19 at 19:28












  • 1




    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 at 17:53






  • 1




    Oh of course, now I see it. Thanks!
    – ProShitposter
    Jul 19 at 19:28







1




1




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 at 17:53




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 at 17:53




1




1




Oh of course, now I see it. Thanks!
– ProShitposter
Jul 19 at 19:28




Oh of course, now I see it. Thanks!
– ProShitposter
Jul 19 at 19:28















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