I need to prove that if $A$ is a normal operator, then the spectral radius of $A$ is $||A||$.
I know that for any linear operator $T:H_1 \rightarrow H_2$, we have that
$||T^*T|| = ||T||^2$
Therefore, we can write
$||{A^2}^n||$$^2$ = $||{{A^*}^2}^n{A^2}^n||$ = $||{(A^*A)^2}^n||$( As $A$ is a normal operator)
Now I let $B= A^*A$.
Then, clearly $B$ is a self-adjoint.
Now I am trying to show that $||{B^2}^n||^\frac{1}{2^n}$ = $||B||$.
I am trying to prove the above by induction.
I can see that for $n = 1$ it will be true as $B$ is a normal operator therefore $||B^2|| = ||B||^2$
But I am not able to proceed further. Help, please.
After proving this, I will be proceeding like this
$||{A^2}^n||^2$ = ${||A^*A||^2}^n$ = ${||A||^2}^{n+1}$ ( Since $A$ is normal therefore $||A^*A|| = ||A||^2$)
Hence $||{A^2}^n|| = {||A||^2}^n$.
Hence ${||{A^2}^n||}^{\frac{1}{2^n}}$ = $||A||$
Hence ${||{A^2}^n||}^{\frac{1}{2^n}}$ $\rightarrow$ $||A||$ as $n \rightarrow \infty$.
Now I want to know why does $\lim_{n}{||A^n||}^{\frac{1}{n}}$ even exists (I know that $||A||$ is bounded but how does it guarantee the existence of this limit).
After this, I think I will be done.
Please clarify the two doubts in the intermediate steps.
Suppose that for some integer $n$ we have that $\|B^{2^{n-1}}\|=\|B\|^{2^{n-1}}$. $B^{2^{n-1}}$ is also self-adjoint (see the comment section), so,
$$\|(B^{2^{n-1}})^2\|=\|B^{2^{n-1}}\|^2$$ so $$\|B^{2^n}\|=(\|B\|^{2^{n-1}})^2=\|B\|^{2^n}$$ and this is the inductive step. The other question you have is the proof of the classical Gelfand-Beurling formula of the spectral radius (a quick search online will yield many proofs and there are plenty posts here about it).