# Norm of the operator on Hilbert Space $l_2$

Suppose $l_2 = \{x= (x_n) | \sum x_n^2 < \infty \}$ is a Hilbert Space and $T( (x_n))= (x_2 -x_1, x_3 - x_2, \dots , x_n-x_{n-1}, \cdots )$. Which of the followings are true

a) $||T|| =1$

b) $||T|| > 1$ and bounded.

c) $||T||$ is unbounded.

I knnow that $||T|| = sup \{ ||T(x)||_2 \ \ : \ \ ||x||_2 = 1\} =sup \{(x_2 -x_1, x_3 - x_2, \dots , x_n-x_{n-1}, \cdots ) : \sum x_n^2 = 1 \} = sup \{\sum (x_{n+1}-x_n)^2 : \sum x_n^2 =1\}$

but $\sum (x_{n+1} -x_n)^2 \leq \sum x_n^2$.

thus $sup \{\sum (x_{n+1}-x_n)^2 : \sum x_n^2 =1\} \leq sup \{\sum (x_n)^2 : \sum x_n^2 =1\} = 1$ we get $||T|| \leq 1$ and

take $x = (1,0,0,\cdots )$, then $||T(x)||_2 = 1$

thus $|| T|| = 1$

Please check my solution , if found any error , then correct me.

• I think your bound is not ok. Take the vector $x=(-1 , 2,0,\dots )$. Then $\| x\| = \sqrt 5$, But $\|Tx\| = \sqrt{3^2 + 2^2} > \sqrt5$ This operator is indeed bounded, but you need to use better bounds (use inequalities you know). – Ranc Feb 2 '16 at 10:30
• If I take $x=(1/\sqrt{2},-1/\sqrt{2},0,0,\ldots)$ I get $\|T(x)\|_2=\sqrt{2.5}>1$ so I think a) is wrong – Elsa Feb 2 '16 at 10:32

It is incorrect.

Let $S$ the left shift operator.

Then $TX=(x_2-x_1,x_3-x_2,x_4-x_3,...)=(x_2,x_3,...)-(x_1,x_2,..)=SX-IX=(S-I)X$.

Thus $T=S-I$. Since $S$ and $I$ are bounded operators, and the space of bounded operators are a vector space, then $T$ is bounded.

Now, if $X_n=(0,0,..,0,1,0,0,0)$, where the $1$ is in the $n$th-place, $n\ge 1$, then

$$TX_n=\left\{\begin{array}{lcc}(-1,0,0,...)&\mbox{if}&n=1\\(0,0,...,1,-1,0,0,...)&\mbox{if}&n\neq 1\end{array}\right.$$

Thus $\|TX_n\|=\sqrt{2}$ for $n>1$.

Then, $\|T\|=\sup\{TX:\|X\|=1\}\ge \|TX_n\|=\sqrt{2}>1$ for $n>1$.

Indeed, your error was $\sum(x_{n+1}-x_n)^2\le\sum x_n^2$. For example, taking $X_2$, $\sum(x_{n+1}-x_n)^2=2>1=\sum x_n^2$