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,

  1. $\|T\|=1$
  2. $\|T\|\geq2$
  3. $1<\|T\|\leq2$
  4. 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.

  • $\begingroup$ Note that $\langle T,T\rangle$ is standard notation; $<T,T>$ is not. $\qquad$ $\endgroup$ Commented Jul 5, 2016 at 2:45
  • $\begingroup$ Ok will keep in mind for future. thanks @MichaelHardy $\endgroup$ Commented Jul 5, 2016 at 2:54

2 Answers 2


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 $\|T\|=2$.


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. $


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