Define a pair of operators $S\colon\ell^2 \rightarrow \ell^2$ and $L\colon\ell^2 \rightarrow \ell^2$ (of sequences of complex or real numbers) as follows:
$S(x_1,x_2,x_3,x_4,\ldots)=\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\frac{x_4}{4},\ldots\right)$
$L(x_1,x_2,x_3,x_4,\ldots)= (x_2,x_3,x_4,\ldots)$
Let $T=S \circ L$, so $T(x_1,x_2,x_3,x_4,\ldots)=\left(x_2,\frac{x_3}{2},\frac{x_4}{3},\frac{x_5}{4},\ldots\right)$ and $T^n(x_i)_{i\in\Bbb N}=\left(\frac{(i-1)!}{(i+n-1)!}x_{i+n}\right)_{i\in\Bbb N}$ if I'm not wrong. I know that $S$ is compact and $L$ is continuous, therefore T is compact. I know the norm of $T$ is $1$. How do I calculate the norm of $T^n$, and the spectral radius and the spectrum of $T$? Thank you in advance.
My attempt for the norm of $T^n$: I've tried to find a constant $K$ such that $\|T^n(x)\| \leq K \|x\|\ $ and an $x_0\in\ell^p$ such that $\|x_0\|=1$, $\|T^n(x_0)\|=K$, what would prove that $\|T^n\|=K$. That worked for $n=1$ but I don't know how to do it for the general case. Easier alternatives are welcome.
Best Answer
Note that $$\|T^nx\|_2^2 = \sum_{i=1}^\infty \left(\frac{(i-1)!}{(i-1+n)!}\right)^2 x_n^2 \le \sum_{i=1}^\infty \left(\frac1{n!}\right)^2 x_n^2 = \frac{1}{n!^2}\|x\|_2^2$$ so $\|T^n\| \le \frac1{n!}$. Therefore the spetral radius is $$r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n} \le \limsup_{n\to\infty} \frac{1}{\sqrt[n]{n!}} = 0$$ and hence $\sigma(T) = \{0\}$. For the norm of $T$ we got $\|T\| \le 1$ and considering $Te_2 = e_1$ gives $\|T\|=1$.