Condition for a linear operator to be bounded.

Question

Here is a problem on which I'm stuck.

Suppose $x=\{x_n\}$ be a sequence.Let $X=\{(y_n)\in l^2 / \{x_ny_n\} \in l^2 \}$.Consider $T_x :X \to l^2$ defined as $ T_x (y_n) =\{x_ny_n\}$. Show that $T_x$ is bounded iff $x\in l^{\infty}$.

Its clear that if $x\in l^{\infty}$ then $T_x$ is bounded but I'm unable in proving the converse part.Any ideas?

if $x\in l^{\infty}$ then does $T_x$ attains its norm?

Answer 1

For the converse we need to show that, $T_x$ is bounded $\implies x\in l^ \infty$.

Let $x_n=(x_1,x_2,x_3 \dots)$

Since $T_x$ is bounded, $ \exists \ M \geq 0$ such that $||T_x(y)|| \leq M, \ \forall\ y \in X$

Now consider, $y=(1,0,0,0 \dots) \in l^2$. Here $y \in X$. So, $$ ||T_x(y)|| = |x_1| \implies |x_1| \leq M$$

Similarly, taking $1$ in the $i_{th}$ position and $0$ at all other places i.e. $y=(0,0,\dots,1,0,0 \dots)$ we get $|x_i| \leq M$, for each $i$.

Thus, $|x_i| \leq M, \ \forall \ i\in \mathbb{N}$ $\implies Sup_{i \in \mathbb{N}}|x_i| \leq M \implies ||x||_ \infty \leq M \implies x \in l^ \infty$