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Math Help - bounded operator

  1. #1
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    bounded operator

    Hi! I was just pondering about the following: Say we have a linear, continuous operator W on a Hilbertspace H with W^*=W and \langle Wx,x \rangle \geq 0 \, \forall x. Then aI-W is invertible if a<0 and a is not an eigenvalue: One can show that the operator aI-W has dense image in H and is injective. Hence one can define the inverse V of aI-W on the image. We have. \langle x,(aI-W)x \rangle \leq a \langle x,x \rangle since \langle Wx,x \rangle \geq 0. I want to follow now that V is bounded but i do not succeed at that. Can someone please give me a hint?

    Best regards
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  2. #2
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    Quote Originally Posted by gammafunction View Post
    Hi! I was just pondering about the following: Say we have a linear, continuous operator W on a Hilbertspace H with W^*=W and \langle Wx,x \rangle \geq 0 \, \forall x. Then aI-W is invertible if a<0 and a is not an eigenvalue: One can show that the operator aI-W has dense image in H and is injective. Hence one can define the inverse V of aI-W on the image. We have. \langle x,(aI-W)x \rangle \leq a \langle x,x \rangle since \langle Wx,x \rangle \geq 0. I want to follow now that V is bounded but i do not succeed at that. Can someone please give me a hint?
    Hint: Write the inequality the other way round: \langle x,(W-aI)x \rangle \geqslant (-a) \langle x,x \rangle = |a|\|x\|^2.
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  3. #3
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    So easy and so effective . Thank you very much once more.
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