Norm of the operator $T(x_1,x_2,\cdots , x_n,\cdots )=(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots).$

Question

Let $T$ be a linear operator from $l_1$ space to itself defined as $$T(x_1,x_2,\cdots , x_n,\cdots )=(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots).$$ Then which of the following statements are true ?

$1.$ $\|T\|=1$.

$2.$ $\|T\|>2$.

$3.$ $1<\|T\|\leq 2.$

$4.$ none .

I tired it as $$\|T(x)\|=\|T(x_1,x_2,\cdots , x_n,\cdots )\|=\|(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots)\|=\sum |x_{i+1}-x_i|\leq 2\|x\|.$$ So $\|T\|\leq 2$. Now how to process further ? If I choose $x=(1,0,0,\cdots)$ then $$\|T(x)\| =\|x\|$$ Now I am confused. Thank you .

Answer 1

You can choose $x = (0,1,0,\ldots)$ and have $$\|Tx\| = \|(1,-1,0,\ldots)\| = 2 = 2 \|x\|.$$ Thus, $\|T\| \ge 2$.

Answer 2

Well consider $x=(1,-1,0,0,0,\dots)$. Then $Tx=(-2,1,0,0,\dots)$ so $\|Tx\|=3$ while $\|x\|=2$, thus $\|T\|\geq\frac{3}{2}$. Continuing in this fashion, take $x=(1,-1,1,-1,0,0,\dots)$. Then $Tx=(-2,2,-2,1,0,0,0,\dots)$ so $\|Tx\|=7$ while $\|x\|=4$. In general, if $x=(1,-1,1,-1,\dots,1,-1,1,-1,0,0,0,\dots)$ where we have $n$ pairs of $(1,-1)$ appearing, then $Tx=(-2,2,-2,2,\dots,2,-2,1,0,0,\dots)$ so $\|Tx\|=2(2n-1)+1$ and $\|x\|=2n$, so $$\|T\|=\sup_{y\neq0}\frac{\|Ty\|}{\|y\|}\geq\frac{2(2n-1)+1}{2n}\to2.$$

So $\|T\|=2$.