Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H$ be a Hilbert space and that $X$ are bounded. Suppose $X$ is self-adjoint. Show that $Y=X+iI$ is invertible and the inverse $Y^{-1}$ has the norm $\lVert Y^{-1} \rVert \le 1$.

I can prove $Y$ is invertible by showing that

$Y^*Y = YY^* = X^2+I\ge I$.

(This is from Lemma 7.2 in this link)

Currently I am trying to prove the inverse has norm $\le 1$.

I know $1=\lVert Y^{-1}Y \rVert \le \lVert Y^{-1} \rVert \cdot \lVert Y \rVert$ and

$\lVert Y \rVert =\sup_{\lVert x \rVert =1} \lVert Yx \rVert =\sup_{\lVert x \rVert =1}||(X+iI)x\rVert \le \sup_{\lVert x \rVert =1}\lVert Xx \rVert +1$.

I don't find how I can approach the result from what I had so far.

share|improve this question
2  
How about: $Y^{-1}$ is normal, so its norm and spectral radius agree; then use spectral mapping. –  user108903 Mar 14 '13 at 21:56
    
@user108903. Thanks for your comment, I have not learnt spectral mapping and I will look at it up. –  Qomo Mar 15 '13 at 17:51
add comment

1 Answer

up vote 1 down vote accepted

To obtain a bound on the norm of the inverse, pick some $v$ and obtain a lower bound on $\|Xv+iv\|^2$ by expanding using the inner product, and using the fact that $\langle Xv, v\rangle \in \mathbb{R}$ since $X$ is self-adjoint. You want to show that $\|Xv+iv\| \ge \|v\|$.

Details:

We have $\|Xv+iv\|^2 = \|Xv\|^2 + \|v\|^2 + 2 \operatorname{Re} \langle Xv, iv\rangle = \|Xv\|^2 + \|v\|^2$. (Since $X$ is self adjoint, $\langle Xv, v\rangle \in \mathbb{R}$). In particular, $\|Xv+iv\|^2 \ge \|v\|^2$, from which we have $\|Xv+iv\| \ge \|v\|$.

Now let $v = Y^{-1}w$ to get

$\|w\| = \|Y Y^{-1} w \| \ge \|Y^{-1} w \|$ from which the result follows.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.