how to find the norm here? $T_n:l_2\to l_2; T_n(x_1,x_2,\dots)=(0,0,\dots,x_{n+1},x_{n+2},\dots)$ could any one help me how to find the norm here?
What is the difference between $\|T_n\|\to 0$ and $\|T_n(x)\|\to 0$?
Here both goes to $0$?
I was trying like below
$\|T_n\|=\sup_{\|x\|=1}[\sum |x_i|^2]^{1\over 2} $ 
 A: $T_n$ is of norm 1, as on one side, we have $\def\norm#1{\left\|#1\right\|}$for $x \in \ell_2\def\abs#1{\left|#1\right|}$
\begin{align*}
  \norm{T_n x}^2 &= \sum_{k=n+1}^\infty \abs{x_k}^2\\
  &\le \sum_{k=1}^\infty \abs{x_k}^2\\
  &\le \norm x.
\end{align*}
So 
$$ \norm{T_n} = \sup_{\norm x \le 1} \norm{T_n x} \le \sup_{\norm x \le 1}\norm x = 1 $$
On the other side, let $e_{n+1} = (0, \ldots, 0, 1, 0, \ldots)$ (the 1 in spot $n+1$), then $\norm{e_{n+1}} = 1$ and $T_ne_{n+1} = e_{n+1}$, so 
$$ \norm{T_n} = \sup_{\norm x \le 1} \norm{T_n x}\ge \norm{T_n e_{n+1}} = 1$$
So $\norm{T_n} = 1$ for each $n$ and $T_n \not\to 0$.
On the other side, for fixed $x \in \ell^2$, we have $\sum_{k=1}^\infty \abs{x_k}^2 < \infty$, that is $\sum_{k=n+1}^\infty \abs{x_k}^2 \to 0$ for $n \to \infty$ and hence $\norm{T_n x} \to 0$ for $n \to \infty$.
A: Clearly, $\|T_nx\|\le \|x\|$. On the other hand, it's clear that we can choose $x$ such that its first $n$ components are zero, hence in this case $\|T_nx\|=\|x\|$; itallows to say that $\|T_n\|=1$.
On the other hand, given a fixed $x$ it's possible to show that $\|T_nx\|\to 0$ as $n\to\infty$.
