# Invertible operator norm bound

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.

• 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

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 $\|Xv+iv\| \ge \|v\|$.
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.