A question about weighted forward unilateral shift operators

Question

We define

$$ B(x_{1}, x_{2},...)=(0, \frac{x_{1}}{2}, \frac{x_{2}}{3},...,x_{n})\in l^{2}(N), $$

How could be shown that that $B$ is a quasinilpotent?

Answer 1

1) By induction show that $$ B^n(x_1,x_2,\ldots)=\left(0,0,\ldots,0,\frac{x_1}{2\cdot\ldots\cdot (n+1)},\ldots,\frac{x_k}{(k+1)\cdot\ldots\cdot(n+k)},\ldots\right) $$

2) Then derive that $$ \Vert B^n\Vert\leq\frac{1}{(n+1)!} $$

3) Using Stirling's formula show that spectral radius $$ r(B)=\lim\limits_{n\to\infty}\Vert B^n\Vert^{1/n}=0 $$

4) Recall that spectrum $\sigma(A)$ of any bounded operator $A$ satisfies $$ \varnothing\neq\sigma(A)\subset\{\lambda\in\mathbb{C}:|\lambda|\leq r(A)\} $$ hence $\sigma(B)=\{0\}$. This means by definition that $B$ is quasinilpotent.

Answer 2

Note that an operator $B$ is quasi-nilpotent iff $$ \lim_{n\to \infty} \|B^n\|^{1/n} = 0 $$ Now note that $$ B^n(x) = (0,0,0,\ldots, 0, \frac{x_1}{(n+1)!}, \frac{2x_2}{(n+2)!}, \frac{6x_3}{(n+3)!}, \ldots ) $$ Now, $$ \frac{j!}{(n+j)!} = \frac{1}{(j+1)(j+2)\ldots (n+j)} \leq \frac{1}{(n+1)!} $$ and hence $$ \|B^n\| = \frac{1}{(n+1)!} \Rightarrow \|B^n\|^{1/n} = \frac{1}{(n+1)^{1/n}}\frac{1}{\sqrt[n]{n!}} $$ But $e^x \geq \frac{x^n}{n!}$, so by plugging in $x=n$, one sees that $$ \sqrt[n]{n!} \geq \left ( \frac{n}{e}\right ) $$ Putting all this in, one gets $$ \lim_{n\to \infty} \|B^n\|^{1/n} \leq \lim_{n\to \infty}\frac{1}{n^{1/n}}\frac{1}{(1+1/n)^{1/n}}\frac{e}{n} = 0 $$