When is the equality $\|A^n\| = \|A\|^n$ achieved for a bounded linear operator $A$?

Question

Let $A$ be a bounded linear operator on a Hilbert space. Then $$\|A^n\| \leq \|A\|\cdots\|A\| = \|A\|^n.$$ I am wondering under what conditions can this inequality be made into an equality? Or equivalently, when does the inequality $$\|A\|^n \leq \|A^n\|$$ hold?

Answer 1

The equality $\|A^n\|=\|A\|^n$ holds for every $n$ if and only if the spectral radius $r(A)$ is equal $\|A\|. $

The proof goes as follows. Assume $\|A^n\|=\|A\|^n$ for every $n.$ Then $$r(A):=\lim_n\|A^n\|^{1/n}=\|A\|$$ Conversely, assume $r(A)=\|A\|.$ Then $$\|A^k\|\ge r(A^k)=r(A)^k=\|A\|^k\ge\|A^k\|$$ For a normal operator $A$ the spectral radius is equal $\|A\|$, and no Gelfand-Naimark theory is needed in the proof. Namely if $B$ is self-adjoint then $$\|B^{2^{n}}\|=\|B^{2^{n-1}}B^{2^{n-1}}\|=\|B^{2^{n-1}}\|^2$$ Applying induction gives $$\|B^{2^n}\|=\|B\|^{2^n}$$ Hence $$\|B\|=\|B^{2^n}\|^{1/2^n}\longrightarrow r(B)$$ If $A$ is normal then $B:=A^*A$ is self-adjoint. Hence $$\|A^n\|^2=\|(A^n)^*A^n\|=\|(A^*A)^n\|\\ =\|B^n\|=\|B\|^n=\|A^*A\|^n=\|A\|^{2n}$$

Answer 2

I can prove it for normal operators using functional analysis.

Let $\mathcal{A}$ denote the $C^\star$ algebra generated by $A$ and $I$. If $A$ is normal, this algebra is commutative and unital.

By Gelfand theory we have that $\mathcal{A} \cong C(\sigma(\mathcal{A}))$, let $\hat{\cdot} : \mathcal{A}\rightarrow C(\sigma(\mathcal{A}))$ denote the Gelfand transform. In particular, for any $x\in \mathcal{A}$ we have that $\hat{x} : \sigma(\mathcal{A})\rightarrow \mathbb{C}$ is a continuous function.

By Gelfand-Neimark theorem: $\hat{\cdot}$ is an isometric $\star$-morphism, meaning that $\|x\| = \|\hat{x}\|_\infty$

For a proof check out Theorem 1.20 in "A course in abstract harmonic analysis" by Gerald B. Folland.

We then in particular have that $\|A^n\| = \|\hat{A}^n\|_\infty$, but for continuous functions we have $\|f^n\|_\infty = \|f\|_\infty^n$ which gives the proof.