3

EDIT: The question is edited after an error pointed out by gerw.

There are the following two results regarding weak convergence in $\ell^p$ spaces:

Let $((\beta_n^{(\alpha)}))_{\alpha \in I} \subseteq \ell_p (\mathbb{N})$ be a net and $(\beta_n) \in \ell_p (\mathbb{N})$, where $1 < p < \infty$. Then

(i). $\beta_n^{(\alpha)} \to \beta_n$ for each $n \in \mathbb{N}$ whenever $(\beta_n^{(\alpha)}) \xrightarrow[]{w} (\beta_n)$.

(ii). $(\beta_n^{(\alpha)}) \xrightarrow[]{w} (\beta_n)$ whenever the net $((\beta_n^{(\alpha)}))_{\alpha \in I}$ is bounded and $\beta_n^{(\alpha)} \to \beta_n$ for each $n \in \mathbb{N}$

Unfortunately I am not able to prove both implications. Any help is highly appreciated.

Community
  • 1

3 Answers3

3

As for (i) use the fact that the evaluation functionals are bounded. Weak convergence means convergence of the net when hit by any functional, so in particular you may take any of the evaluation functionals.

For (ii), suppose that your net is norm-bounded, and point-wise convergent. By subtracting the point-wise limit, we may suppose that this net converges point-wise to 0. Pick $g\in \ell_{p^*}$ (Here we identify $\ell_{p^*}$ with $(\ell_p)^*$ in the canonical way, and so $p^*$ is the Hölder conjugate exponent of $p$.) As $p^*\neq \infty$, the finitely supported sequences are dense in $\ell_{p^*}$. Fix $\delta > 0$ and choose a finitely supported sequence $g^\prime\in \ell_{p^*}$ such that $\|g-g^\prime\|<\delta$. Let $M$ be a bound for our net. Then

$$|\langle g, \beta^{(\alpha)} \rangle| \leqslant |\langle g - g^\prime, \beta^{(\alpha)} \rangle| + |\langle g^\prime, \beta^{(\alpha)} \rangle| \leqslant M\delta + |\langle g^\prime, \beta^{(\alpha)} \rangle|$$

As $g^\prime$ is finitely supported, we use our assumption of point-wise convergence to $0$, so the right-hand side tends to $\delta M$. As $\delta$ was arbitrary, we conclude that $|\langle g, \beta^{(\alpha)} \rangle|$ tends to $0$.

Tomasz Kania
  • 16,996
  • Could you think of an easy counterexample to (ii) when the net is not assumed to be bounded? I expect it might fail in this case. –  Aug 06 '16 at 09:04
  • Thanks. I will think about an example a bit longer and will otherwise ask it as a new question. –  Aug 06 '16 at 10:09
  • You really shouldn't have deleted comments here like that ! You deleted about 6 comments here having important information. (everyone might make mistake), instead of deleting users' comments pointing out important questions or your mistakes, you could simply say " you are right ". – Red shoes May 22 '17 at 18:30
  • @nonlinearthought I haven't deleted anything. – Tomasz Kania May 22 '17 at 18:56
2

Your assertion is not true, there are convergent nets, which are unbounded. Take $I = (0,\infty)$ with the usual order and define $x_i := 1/i \in \mathbb{R}$. Then, you can check that the net $\{x_i\}_I$ converges to $0$, but is unbounded.

There are also more advanced examples providing weakly convergent nets without bounded subnets.

Community
  • 1
gerw
  • 33,373
  • Your example is not in $\ell^p$, is it? –  Aug 05 '16 at 11:02
  • No, it is a net in the simple Banach space $\mathbb R$. – gerw Aug 05 '16 at 11:04
  • Thanks, but my question was about nets in $\ell^p$, where $1 < p < \infty$. –  Aug 05 '16 at 11:08
  • It is the same. Take your favorite $\beta \in \ell^p \setminus {0}$ and consider the net ${x_i \beta}$. – gerw Aug 05 '16 at 11:09
  • I see, you are right! However, I am quite sure the characterisation should hold. It is Exercise 2.53 in Megginson's An Introduction to Banach Space Theory. Are you aware that $(\beta_n) \in \ell^p$ and $\beta_n \in \mathbb{C}$ in the notation above? So, the first statement is about weak convergence of the net with elements $(\beta_n)$ and the second statement about the convergence of the coordinates $\beta_n$ of the element $(\beta_n) \in \ell^p$. –  Aug 05 '16 at 11:14
  • Yes, I think that I have understood your formulation. However, there are weakly convergent nets in $\ell^p$ which are unbounded in norm. – gerw Aug 05 '16 at 11:19
2

The question has already been answered by users. This is an example showing that in part (ii) the boundedness assumption is essential. Take $X=l^2$ and $(e_n) \in X$ whose nth' component is $1$ and zero otherwise. Take the sequence $\{x^k \}_{k=1}^{\infty} \subset X$ with $x^k = k e_k $. Fix $j$ and looking at $j-th$ component of this sequence, i.e. $x^k_j =0$ for all $k \in N \setminus\{j\}$, hence it convergences to zero as $k \rightarrow \infty $ (for all $j$), but $\{x_k\}$ is not weakly convergent to zero since it is unbounded!

Red shoes
  • 7,184