Prove: $T: X \rightarrow Y$ with $Tx=x$ is bounded

Question

I want to show that $T: X \rightarrow Y$ with $Tx=x$ is bounded and surjective.

where we have $||x||_X = \sum_{n=1}^{\infty}|x_n|$ and $||x||_Y = sup_{n \in \mathbb{N}} |x_n|$

$X:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$

$Y:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$

For boundedness I know that $T \in B(X,Y) \leftrightarrow^{def} \exists M s.t. ||Tx||_Y \leq M||x||_X \forall x \in X$.

That is, if $T$ is bounded, the number $||T|| = sup_{x \neq 0} \frac{||Tx||_Y}{||x||_X} \leq M < \infty$ is finite.

Now I am stuck on how to proceed. Help would be grateful. Thanks in advance.

Answer 1

Take $x \in X$ with $$\Vert x \Vert_X = \sum_{n=1}^{\infty}\vert x_n \vert$$ finite.

For all $n \in \mathbb N$ we have

$$\vert x_n \vert \le \Vert x \Vert_X = \sum_{n=1}^{\infty}\vert x_n \vert$$ hence $$\Vert x \Vert_Y \le \Vert x \Vert_X$$ proving that $M=1$ satisfies the required inequality and that $T$ is bounded.

Now the identity is clearly surjective as $X=Y$ in your question. So for any $y \in Y$, $y$ itself is an inverse image of $y$ under $T$.