1
$\begingroup$

Recently I posted a question about the following result on weak convergence:

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

I try to find a counterexample showing that the assumption that the net $((\beta_n^{(\alpha)}))_{\alpha \in I}$ should be bounded cannot be omitted, that is, the conclusion is not true if it is only assumed that $\beta_n^{(\alpha)} \to \beta_n$ for each $n \in \mathbb{N}$.

After thinking about an example for a considerable amount of time, I did not succeed yet.

Any help and/or comment is highly appreciated.

$\endgroup$
2
$\begingroup$

Set $\beta_k(n)= \delta_k(n) \exp(k)$. Then $\lim_{k \to \infty}\beta_k(n) \to 0$, i.e. the net converges pointwise to the $0$ element in $\ell^p$, but tested against the $\ell^q$ element $(1/n)_{n\in \mathbb{N}}$, does not converge as $<\beta_k(n)_{n \in \mathbb{N}}, (1/n)_{n \in \mathbb{N}}>= \frac{\exp{k}}{k}$ does not converge.

If the $\beta_k$ are bounded in $\ell^p$, then they are particularily bounded in $\ell^\infty$, so this can not happen, as each element in $\ell^p$ is converging to $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy