Finding the spectral radius of operator $ T(x)= (0, x_1, 2x_2, x_3, 2x_4,…)$ on $l^{1}$

Question

My question is:.
We have $x=(x_1, x_2,…) \in l^{1}$ and $T: l^{1} \mapsto l^{1}$ is a linear operator defined as $$ T(x)= (0, x_1, 2x_2, x_3, 2x_4,…)$$ then we have to show that———
1. The sequence $(||T^{n}||^{1/n})||$ is not monotone
2. It's spectral radius is $\sqrt{2}$.
This is not a homework problem but a problem from a book I am self studying from.
My strategy is to find $||T^n||$ for $n=1,2,3,…$, then they would form a sequence in $n$ , conclude about the two questions asked from there using Gelfand formula for spectral radius but I think I am stuck on finding the norm of $T$.
Can someone tell me if my strategy is correct or give a hint to approach this problem some other way?
P.S.: It would be great if you could refer me some PDF or notes that have solved examples of this kind- related to find spectrum, spectral radius.
Thanks a lot in advance!

Answer 1

It is definitely a good idea to compute $\big\Vert T^n\big\|$. Here is a hint. Show that $$T^{2k}x=\big(\underbrace{0,0,\ldots,0}_{2k\ \text{zeros}},2^kx_1,2^kx_2,2^kx_3,2^kx_4,\ldots\big)$$ and $$T^{2k+1}x=\big(\underbrace{0,0,\ldots,0}_{2k+1\ \text{zeros}},2^kx_1,2^{k+1}x_2,2^kx_3,2^{k+1}x_4,\ldots\big).$$

This implies that $$\big\Vert T^{2k}x\big\Vert=2^{k}\Vert x\Vert\ \ \wedge\ \ 2^k\|x\|\leq \big\Vert T^{2k+1} x\big\Vert \leq 2^{k+1}\|x\|.$$ It follows that $\big\Vert T^{2k}\big\Vert =2^k$ and $\big\Vert T^{2k+1}\big\Vert=2^{k+1}$. The rest should be easy.