Prove $||T^n||=||T||^n$ if $T\in B(H)$ is self-adjoint Let $H$ be a complex Hilbert space and let $T\in B(H)$ be self-adjoint. I have already proven that $||T^{2^k}||=||T||^{2^k}$ for $k=0,1,2,\dots$. 
Now I have to prove the more general statement that if $n\geq 1$, $||T^n||=||T||^n$. 
I can write $T^{2^k}=T^n T^{2^k-n}$ for $1\leq n<2^k$. But how do I argue from there?
 A: The norm and spectral radius of a selfadjoint operator $T$ are the same. And the same is true of $T^k$ for $k=1,2,3,\cdots$. So $\|T^n\|=r_{\sigma}(T^n)$. By the spectral mapping theorem, $\sigma(T^n)=\sigma(T)^n$, which leads to the desired result:
$$
                      \|T^{n}\|=r_{\sigma}(T^n)=r_{\sigma}(T)^n=\|T\|^n.
$$
A: Given $n$, choose $k$ so $2^k>n$ and let $2^k=n+m$.
If we had $||T^n||<||T||^n$ then $$||T^{2^k}||=||T^{n+m}||\le||T^n||\,||T^m||<||T||^n||T||^m=||T||^{2^k}.$$
A: Here your bounded operator $T$ is self-adjoint. Since for simply bounded $T$, $T^*$ must be linear and bounded. Furthermore we may take the relations $||T||$ $=$ $||T^*||$ ,  $(T^*)^*$ $=$ $T$.
And $||T^2||$ $=$ $||TT||$ $=$ $||T^*T||$ $=$ $||TT^*||$ $=$ $||T||^2$ $=$ $||T^*||^2$. 
Indeed, by relation $||ST||$ $\leq$ $||S||$ $||T||$ and $||T||$ $=$ $||T^*||$, $||T^*T||$ $\leq$ $||T^*||$ $||T||$ $=$ $||T||^2$ $=$ $||T^*||^2$. 
and the reverse inequality follows from the relation
$||Tx||^2$ $= $ $\langle{Tx,Tx}\rangle$ $=$ $\langle{T^*Tx,x}\rangle$ $\leq$ $||T^*Tx||$ $||x||$ $\leq$ $||T^*T||$ $||x||^2$ and $||T^2||$ $\leq$ $||T^*T||$
So $||T||^2$ $=$ $||T^*T||$  i.e. $||T||^2$ $=$ $||T^2||$ 
Now routine mathematical induction may lead to required result.
