Norms of powers of self-adjoint operator

Question

Let $X$ be a Hilbert space, $T\in L(X)$ be a linear, continuous, self-adjoint operator. Due to $$ \|T^2\| = \|TT^*\| = \|T\|^2, $$ one can prove by induction $$ \|T^{2^n}\|=\|T\|^{2^n}. $$ This can be used to deduce $\lim_{n\to\infty}\|T^n\|^{1/n}=\|T\|$.

My question is: can one prove $$ \|T^n\|=\|T\|^n $$ for all $n$ in the non-compact case ($T$ not compact) by elementary calculations like above? That is, without invoking spectral theorems or decompositions.

Answer 1

We have that

$$\lim_{n\to\infty}\|T^n\|^{1/n}= \inf_{n \in \mathbb N}\|T^n\|^{1/n}.$$

Hence, if

$$\lim_{n\to\infty}\|T^n\|^{1/n}=||T||,$$ then, for each $n \in \mathbb N,$

$$||T|| \le \|T^n\|^{1/n},$$

or

$$||T||^n \le \|T^n\|.$$

Since $$||T^n|| \le ||T||^n,$$

we get

$$\|T^n\|=\|T\|^n.$$

Answer 2

I will write a very detailed answer because this question stumped me today and it took me a while to fix the subtleties of the proof.

Step 1: Induction Hypothesis (Powers of 2)

It is clear that $\lVert T^n\rVert \leq \lVert T \rVert^n$. Furtermore, by self adjointness and Cauchy-Schwarz:

$$ \lVert T(x) \rVert^2=\langle T^2(x),x\rangle\leq \lVert T^2(x)\rVert\lVert x\rVert$$

This allows us to conclude $\lVert T\rVert^2\leq \lVert T^2 \rVert$. If this holds for $T^{2^n}$ we may argue that:

$$ \lVert T^{2^{n}}(x) \rVert^2=\langle T^{2^{n+1}}(x),x\rangle\leq \lVert T^{2^{n+1}}(x)\rVert\lVert x\rVert$$

$$\lVert T^{2^{n}}\rVert^2\leq \lVert T^{2n+1}\rVert$$

But by our induction hypothesis $\lVert T\rVert^{2^{n+1}}=\lVert T^{2^n}\rVert^{2}\leq \lVert T^{2n+1}\rVert$. Therefore, $\lVert T^{2^n}\rVert=\lVert T\rVert^{2^n}$ for all natural numbers.

Step 2: Extension to natural numbers

Suppose by way of contradiction that there is a certain $n$ such that $\lVert T^n\rVert< \lVert T\rVert^n$. Clearly, there is a certain $i_o$ such that $2^{i_o}<n<2^{i_o+1}$. In this case:

$$\lVert T^{2^{i_o+1}}\rVert=\lVert T^{2^{i_o+1}-n+n}\rVert\leq \lVert T^{2^{i_o+1}-n}\rVert \lVert T^{n}\rVert < \lVert T\rVert^{2^{i_o}-n} \lVert T\rVert^n=\lVert T\rVert^{2^{i_o}}$$

But this strict inequality contradicts the validity of the first step and allows us to conclude that in fact it hold for all natural numbers that $\lVert T^n\rVert= \lVert T\rVert^n$.