If $A$ is a self-adjoint operator . Is $\|A^n\|=\|A\|^n,\;\forall n$?

Question

Let $E$ be an infinite-dimensional complex Hilbert space.

Let $A\in \mathcal{L}(E)$ be a self-adjoint operator i.e. $A^*=A$, is $$\|A^n\|=\|A\|^n,\;\forall n\in \mathbb{N}?$$

Note that if $A$ is self-adjoint, then $$\|A\|=\sup\left\{|\langle Ax, x\rangle|\,;\;x\in E,\;\|x\|= 1\right\}.$$

It is well known that if $T\in \mathcal{L}(E)$, then $$\|T^*T\|=\|T\|^2.$$ So since $A$ is self-adjoint, then $$\|A^2\|=\|A\|^2.$$ I try to show the result by induction.

Answer 1

It is known that all elements in the spectrum of a selfadjoint operator are approximate eigenvalues. So there exists a sequence $\{x_j\}$ with $\|x_j\|=1$ for all $j$ and $Ax_j-\|A\|\,x_j\to0$.

Now we proceed by induction on $n$. If $A^nx_j-\|A\|^nx_j\to0$, then $$ A^{n+1}x_j-\|A\|^{n+1}x_j=A(A^nx_j-\|A\|^nx_j)+\|A\|^n(Ax_j-\|A\|x_j)\to0. $$ It follows that, for each $n$, $A^nx_n-\|A^n\|x_j\to0$. This implies that $A^n-\|A\|^nI$ is not invertible, that is $\|A\|^n\in\sigma(A^n)$. But then $$ \|A\|^n\leq\|A^n\|, $$ which is the non-trivial inequality. Thus, $\|A^n\|=\|A\|^n$ for all $n$.

Answer 2

The spectral radius theorem tells you that for any self-adjoint operator $$ \lVert A \rVert = \sup_{\lambda \in \sigma(A)} \lvert \lambda \rvert, $$ where $\sigma(A)$ denotes the spectrum of $A$. Using that $f(\sigma(A)) = \sigma(f(A))$ for any continuous function $f$ on $\sigma(A)$, you get the desired result.