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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.

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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 $Tx = y$.

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  • $\begingroup$ And just to clarify, $y \in \ell^{2}$ because $\ell^{2}$ is closed in itself? $\endgroup$ – SABOY Jul 21 at 16:37
  • $\begingroup$ 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. $\endgroup$ – Rhys Steele Jul 21 at 20:00

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