On the space $l_2$ we define an operator $T$ by $Tx=(x_1, {x_2\over2}, {x_3\over3}, . . . )$. Show that $T$ is bounded, and find its adjoint.

Question

On the space $l_2$ we define an operator $T$ by $Tx=(x_1, {x_2\over2}, {x_3\over3}, . . . )$. Show that $T$ is bounded

I know that $||T||\leq 1$, but I don't know how to show this. Any solutions or hints are appreciated.

Answer 1

We have to show that $T$ maps $l^2$-sequences to $l^2$ sequences. If $x \in l^2$, then $||x||=(\sum_{i=1}^{\infty} x_i^2)^{1/2} < \infty$. But $(x_i/i)^2 \leq x_i^2$ all $i \in \mathbb{N}$ and hence $||Tx||=(\sum_{i=0}^{\infty} (x_i/i)^2)^{1/2} \leq (\sum_{i=1}^{\infty} x_i^2 )^{1/2}< \infty$