# Weak net convergence in $\ell_p$, where $1 < p < \infty$.

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.

• Are you able to show one direction of the equivalence? – gerw Aug 5 '16 at 10:52

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 $0$.

• 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. – user342207 Aug 6 '16 at 9:04
• Thanks. I will think about an example a bit longer and will otherwise ask it as a new question. – user342207 Aug 6 '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. – Tomek Kania May 22 '17 at 18:56

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.

• Your example is not in $\ell^p$, is it? – user342207 Aug 5 '16 at 11:02
• No, it is a net in the simple Banach space $\mathbb R$. – gerw Aug 5 '16 at 11:04
• Thanks, but my question was about nets in $\ell^p$, where $1 < p < \infty$. – user342207 Aug 5 '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 5 '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 $(\beta_n) \in \ell^p$. – user342207 Aug 5 '16 at 11:14

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!