The problem reads: Find the norm of an operator in $l^2$ defined as:
$$ Ae_n=\frac{1}{\sqrt{n}}e_{n+1} $$
This is what I have done so far:
A acting on any $x\in l^2$ gives: $$ Ax = A\sum_{n=1}^{\infty} x_ne_n=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}x_ne_{n+1} $$ Therefore the norm would be: $$ ||A||=\sup_{||x||\le 1}||Ax|| = ||A||=\sup_{||x||\le 1}||\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}x_ne_{n+1}||=\sup_{||x||\le 1}|\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}x_n| $$ I'm not sure if I'm doing this correctly or how to interpret the supremum in this case. Does this just give the norm to be $\frac{1}{\sqrt{n}}$ or something else?