If $T$ is self-adjoint then $||T^n|| = ||T||^n$

Question

Let $T$ be a bounded linear operator on a Hilbert space $H$. If $T$ is self-adjoint then $||T^n|| = ||T||^n$.

It is easy to see that $||T||^n$ is an upper bound. Indeed, there exists a $C>0$ such that, $||T^nx|| \leq C||x||$. Then $||T^n|| \leq ||T||^n$.

To prove the other direction, \begin{align*}||T||^2 = \sup_{||x|| =1}\{||Tx||^2\} & = \sup_{||x|| =1}\{\langle Tx,Tx\rangle\}= \sup_{||x|| =1}\{\langle x,TTx\rangle\} \\ & \leq \sup_{||x|| =1}\{||x||^2||T^2||\} = ||T^2||.\end{align*}

But for the case $n = 3$ (the induction step), it seems that trick I've used doesn't work. Since \begin{align*} ||T||^3 = \sup_{||x|| =1}\{||Tx||^3\} & = \sup_{||x|| =1}\{\sqrt{\langle Tx,Tx\rangle}^3\}= \sup_{||x|| =1}\{\sqrt{\langle x,TTx\rangle}^3\} \\ & \leq \sup_{||x|| =1}\{\sqrt{||x||^2||T^2||}^3\} = ||T^2||^{\frac{3}{2}}. \end{align*}

How to show the induction step? I read some comments using the spectral theorem but I have not learnt it yet. Is there another proof using just properties of being self-adjoint?

Answer 1

By induction you have $\lVert T^{2^k}\rVert = \lVert T\rVert ^{2^k}$

So let $ 2^{k-1} <n <2^{k}$. If $\lVert T^{n}\rVert < \lVert T\rVert ^{n}$, then $\lVert T^{2^k}\rVert \leq \|T^n\|\cdot\|T^{2^k-n}\|<\|T\|^n\cdot\|T\|^{2^k-n}=\|T\|^{2^k}$, contradiction

Answer 2

If $T$ is self-adjoint, then $\|T\|=\sup_{\lambda\in\sigma(T)}|\lambda|$. Likewise, $$ \|T^n\|=\sup_{\lambda\in\sigma(T^n)}|\lambda|. $$ By the spectral mapping theorem, $\sigma(T^n)=\sigma(T)^n$. So $$ \|T^n\|=\sup_{\lambda\in \sigma(T)}|\lambda^n|=\sup_{\lambda\in\sigma(T)}|\lambda|^n=\|T\|^n. $$