1. ## bounded operator

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

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