Bound and norm of operator

Question

Prove that operator defined as $T(x_1, x_2, . . .) := (x_3, x_2, x_1, x_6, x_5, x_4, x_9, x_8, x_7, . . .)$ is a bounded linear operator on complex space $l^2$.

$ ||(x_1,x_2,x_3,..)|| = \sqrt{\sum_{i=1}^\infty |x_i|^2}$?

Can someone please help with finding its bound and norm?

Answer 1

Hint: For all $x = (x_i)_i \in \ell^2$, $||x||_2 = \sum_{i=1}^\infty |x_i|^2 < \infty$, so $(x_i)_i$ is absolutely summable, and interchanging the order of terms inside the sum is OK.

$$||(\bbox[5pt, border:1pt solid black]{x_3,x_2,x_1}, x_6,x_5,x_4,\dots)|| = \sqrt{\sum_{i=1}^\infty |x_i|^2} = ||(\bbox[5pt, border:1pt solid black]{x_1,x_2,x_3},x_4,x_5,x_6,..)||$$ Therefore, $||Tx|| = ||x||$ for all $x \in \ell^2$, and $||T|| = 1$.