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

Question

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.

Answer 1

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 $0$.

Answer 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.

Answer 3

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!