1. 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

2. Originally Posted by gammafunction
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: $\langle x,(W-aI)x \rangle \geqslant (-a) \langle x,x \rangle = |a|\|x\|^2$.

3. So easy and so effective . Thank you very much once more.