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

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?

By induction you have $$\lVert T^{2^k}\rVert = \lVert T\rVert ^{2^k}$$
So let $$2^{k-1} . If $$\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}$$, contradiction
If $$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.$$
• The OP asks to avoid the spectral mapping theorem, but basically $\sigma(T^n) = \sigma(T)^n$ is the fundamental theorem of algebra so I give this a pass. +1 Commented May 15, 2021 at 20:55