Find the bounds for the norm $T:l_2 \to l_2$. 
Given $T:l_2 \to l_2$ define as $T((x_1,x_2,\ldots,x_n,\ldots))=(x_2-x_1,x_3-x_2,\ldots,x_{n+1}-x_n,\ldots)$ then which of the following is true,
  
  
*
  
*$\|T\|=1$
  
*$\|T\|\geq2$
  
*$1<\|T\|\leq2$
  
*None of above.
  

What I did-
I used $\|T\|^2=\langle T,T\rangle$, so after calculation I got $\|T(x)\| = \left( \sum_{i=1}^\infty (x_{i+1}-x_i)^2 \right)^{1/2}$. Now I got stuck. How to proceed further? Please help.
 A: First note that if $x,y\in\mathbb{R}$, then
$$ (x-y)^2=x^2-2xy+y^2\leq 2(x^2+y^2) $$
since $2|xy|\leq x^2+y^2$.
Therefore if $x\in \ell^2$ with $\|x\|_2=1$, then
$$ \|Tx\|^2=\sum_{i=1}^\infty (x_{i+1}-x_i)^2\leq 2\sum_{i=1}^\infty (x_{i+1}^2+x_i^2) = 2\sum_{i=1}^\infty x_i^2+2\sum_{i=1}^\infty x_{i+1}^2\leq 4$$
Hence $\|T\|\leq 2$. On the other hand, for each $n\geq 1$ we can define an element
$$ x=\Big(-\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}},\dots,\frac{(-1)^n}{\sqrt{n}},0,\dots\Big) $$
of $\ell^2$ such that $\|x\|_2=1$ and $\|Tx\|_2^2=\frac{4}{n}(n-1)+\frac{1}{n} =\frac{4n-3}{n}$, so it follows that $\|T\|=2$.
A: With $x=(x_n)_n$ and $Ux=(x_{n+1})_n$ we have obviously $\|Ux\|\leq\|x\|. $ So $$\|Tx\|=\|Ux-x\|\leq \|Ux\|+\|x\|\leq 2\|x\|.$$ So $\|T\|\leq 2.$ 
For $k\in N, $ let $x(k)=(x_{k,n})_n, $ where $ x_{k,n}=(-1)^n$ for  $n\leq k  $ and  $x_{k,n}=0  $ for $ n>k. $  Then $\|x(k)\|=\sqrt k$ and $\|Tx(k)\|=\sqrt {4(k-1)^2+1}. \;  $  So $\lim_{k\to \infty}\|Tx(k)\|/\|x(k)\|=2. $
So $ \|T\|\geq 2. $
