# Showing that the $T: \operatorname{dom}(T) \to \ell^{2}$ is closed

Let $$T:\operatorname{dom}(T) \to \ell^{2}$$ where $$T(x^{n})=(mx_{m}^{n})_{m \in \mathbb N}$$

Let $$\operatorname{dom}(T):=\{x^{n}\in \ell^{2}: (mx_{m}^{n})_{m \in \mathbb N} \in \ell^{²}\}$$

Determine whether $$T$$ is closed or not:

Initially I attempted to show that $$T$$ is closed, by assuming a $$(x^{n})_{n} \subseteq \ell^{2}$$ where $$x^{n}\xrightarrow{n \to \infty, \vert \vert \cdot \vert \vert_{2}} x$$ and $$T(x^{n})\xrightarrow {n \to \infty, \vert \vert \cdot \vert \vert_{2}}y$$. In order to show that $$x \in \operatorname{dom}(T)$$, note that from $$x^{n}\xrightarrow{n \to \infty, \vert \vert \cdot \vert \vert_{2}} x$$ that it follows: $$\lim\limits_{m \to \infty}x^{m}_{i}=x_{i}$$ for all $$i \in \mathbb N$$ and hence let $$N \in \mathbb N$$ arbitrary:

First question: convergence in $$\ell^{p}, 1\leq p<\infty$$ of $$(x^{n})_{n}$$ to $$x$$ does imply convergence in each respective coordinate, correct?

$$\sum\limits_{i=1}^{N} \vert x_{i} \vert^{2}=\lim\limits_{m\to \infty}\sum\limits_{i=1}^{N} \vert x_{i}^{m} \vert^{2}\leq\lim\limits_{m\to \infty}\vert\vert x^{m}\vert\vert_{2}^{2}$$, but this estimate does not help me.

So, I now believe that it may not be closed. Any ideas on showing that it is not closed.

Convergence in $$\ell^p$$ does indeed imply coordinatewise convergence. This is easy to see since if $$x^n \to x$$ in $$\ell^p$$ then $$|x_m^n - x_m|^p \leq \sum_{i \geq 1} |x_m^n - x_m|^p \to 0$$ as $$n \to \infty$$ so that $$x_m^n \to x_m$$.
The reason that you're having problems showing that $$x \in \operatorname{Dom}(T)$$ is that you're trying to do it without using the information that $$Tx^n \to y$$. This approach must fail since $$\operatorname{Dom}(T)$$ is not closed in $$\ell^2$$.
You want to check that when $$x^n \to x$$ in $$\ell^2$$ and $$Tx^n \to y$$ in $$\ell^2$$ then $$\sum_{i \geq 1} i^2 |x_i|^2 < \infty$$. Since we know that $$x^n \to x$$ and $$Tx^n \to y$$ coordinatewise, we know that $$y_i = \lim_{n \to \infty} i x_i^n = i x_i$$ and so $$\|y\|_{\ell^2}^2 = \sum_{i \geq 1} i^2 |x_i|^2 < \infty$$ since $$y \in \ell^2$$ and hence $$x \in \operatorname{Dom}(T)$$ and $$Tx = y$$.
• And just to clarify, $y \in \ell^{2}$ because $\ell^{2}$ is closed in itself? – SABOY Jul 21 at 16:37
• No, $y \in \ell^2$ by the definition of the limit. You could technically write an argument that says $\ell^2$ is closed in $\ell^2$ and closed sets contain their limit points so if $y_n \in \ell^2$ and $y_n \to y$ then $y \in \ell^2$ and it would be correct but also a poor argument because you use a lot of unnecessary facts for something that just follows from the definition immediately. – Rhys Steele Jul 21 at 20:00