# Let $H$ be a hilbert space and $T\in\mathscr{L}(H)$. Then why is $\lVert T\rVert=\sup_{\lVert w\rVert=1}|\langle Tw,w\rangle|$?

Let $H$ be a complex Hilbert space, and $T\in\mathscr{L}(H)$ a bounded linear operator. Then I'm trying to prove that $$\lVert T\rVert_{\mathscr{L}(H)}=\sup_{\lVert w\rVert=1}|\langle Tw,w\rangle|$$ ideally without using too much advanced spectral theory. While a simple application of Cauchy-Schwartz gives us that $\mathsf{LHS}\ge\mathsf{RHS}$, I can't see an easy way to prove the converse. I'm aware of the fact that $$\sigma(T)\subset \overline{N(T)}$$ where $N(T)=\{\langle Tw,w\rangle\ |\ \lVert w\rVert=1\}$ is the numerical range, and $\sigma(T)\subset\Bbb C$ is the spectrum of $T$. Is this sufficient to prove the claim?

As far as I see It is enough that $T$ is normal. Because if $T$ is normal we know \begin{align} \sup |\sigma(T)| = \|T\| . \end{align} Therefore you get the inequality you need. Since from $$\|T\| \in \sigma(T)\subseteq \overline{N(T)}$$ we conclude $$\|T\| \leq \sup_{\lVert w\rVert=1}|\langle Tw,w\rangle|$$

## Details

For you further question in the comments: There is a theorem which says

Theorem If $\mathcal{A}$ is a unital Banach-algebra and $A\in\mathcal{A}$ then $\sigma(A) \neq \emptyset$ and \begin{align} \sup_{\lambda \in \sigma(A)} |\lambda| = \max_{\lambda \in \sigma(A)} |\lambda| = \lim_{n\in\mathbb{N}} \|A^n\|^{\frac{1}{n}} = \inf_{n\in\mathbb{N}} \|A^n\|^{\frac{1}{n}}. \end{align}

I found only a wikipedia reference so far. I will also use $\|TT^*\| = \|T\|^2$. However using these two facts and the fact that $T^2$ is also normal $$\|T^2\|^2 = \|(T^2)^*(T^2)\| = \|(T^*T)^2\| = \|(T^*T)^* (T^* T)\| = \|T^*T\|^2 = (\|T\|^2)^2$$ By induction $\|T^{2^k}\|= \|T\|^{2^k}$ therefore $$\sup_{\lambda \in \sigma(T)} |\lambda|= \lim_{n\in\mathbb{N}} \|T^n\|^{\frac{1}{n}} = \lim_{k\in\mathbb{N}} \|T^{2^k}\|^{\frac{1}{2^k}} = \|T\|$$

• okay I found it in "Functional analysis" by Walter Rudin it's theorem 10.13. Here is a URL 59clc.files.wordpress.com/2012/08/… If you understand german I can provide many other sources – Nathanael Skrepek Aug 17 '17 at 18:07
• Why do we have $\lim_{n \to \infty} \| T^n \|^{\frac{1}{n}} = \lim_{k \to \infty} \| T^{2^k} \|^{2^{-k}}$? – Ramanujan Oct 13 '19 at 13:21
• If the limit of a sequence exists, then the limit of every subsequence is the equal to that limit. And the limit of $\Vert T^n\Vert^{\frac{1}{n}}$ exists. – Nathanael Skrepek Oct 13 '19 at 16:50

Your inequality is true, however to have equality $T$ must be a normal operator at least in finite dimensional Hilbert space. Counter examples are easy and yet given.