Operator norm of powers of bounded normal operators and self adjoint operators on Hilbert space

Question

I saw this problem in Sheldon Axler's Measure,Integration and Real Analysis Ex. $10B$, problem $17$ and $18$.

Let's just restrict to the self adjoint case. What I am trying to show is that $||T^{n}||=||T||^{n}$ for all natural numbers. However, I am stuck with this problem.

Firstly, I can directly show that $||T^{2^{n}}||=||T||^{2^{n}}$ by a simple induction. But I don't know how to do this for odd numbers now.

If $T$ was compact, then I can say that $\pm ||T||$ is an eigen value and has the largest modulus in the spectrum. Then by the polynomial spectral mapping theorem, I can say that $||T||^{n}$ has the largest modulus in the spectrum of $T^{n}$ . Hence I can say that $||T^{n}||=||T||^{n}$. I can even apply the same logic to normal operators I guess.

But how do I solve it in general?

Answer 1

Here is a more elementary proof, which does not use the spectral radius formula:

For an arbitrary positive integer $n$, we know that $$ \|{T^n}\| \le \|{T}\|^n. $$ To prove that this inequality is an equality, suppose $n$ is a positive integer and $\|{T^n}\| < \|T\|^n$. Let $m$ be a positive integrer such that $n + m = 2^k$ for some positive integer $k$. Then \begin{align} \| T^{2^k} \| &= \| T^n \, T^m \| \\[4pt] &\le \|T^n\| \, \|T^m\| \\[4pt] &< \|T\|^n \|T\|^m \\[4pt] & = \|T\|^{2^k}. \end{align} Thus $\| T^{2^k} \| < \|T\|^{2^k}$, which contradicts the equality stated by the original proposer.

Answer 2

If you know about the Gelfand representation of abelian $C^*$-algebras, the assertion is obvious because we can regard the $C^*$-algebra generated by $1$ and $T$ as $C(X)$ for some compact Hausdorff space $X$.

For a more elementary approach, let $T$ be a self-adjoint operator. Let us recall the spectral radius formula $$r(T)=\lim_{n\to\infty}\lVert T^n\rVert^{1/n}, $$ where $r(T)$ is the spectral radius of $T$. Then we have $$r(T)=\lim_{n\to\infty}\lVert T^{2^n}\rVert^{2^{-n}}=\lVert T\rVert, $$ and now the spectral mapping theorem gives $$ \lVert T^n\rVert=r(T^n)=r(T)^n=\lVert T\rVert^n, $$ as desired.

If now $T$ is a normal operator, one can apply the preceding argument to $T^*T$ to obtain $$\lVert T^n\rVert^2=\lVert (T^*T)^n\rVert=\lVert T^*T\rVert^n=\lVert T\rVert^{2n}. $$