Let $X$ be a Banach space and let us consider the linear subspace of $\ell_\infty(X)$ comprising all weakly Cauchy sequences.

Is this subspace closed?

It is not as immediate as in the case of weakly convergent sequences, so I'd suspect the answer is no.

Any references concerning the space of weakly convergent sequences would be appreciated as well.


Here's a proof that the space of weakly Cauchy sequences is closed in $\ell_\infty(X)$:

Let $WC(X)$ denote the subspace of $\ell_\infty(X)$ composed of weakly Cauchy sequences. Let $(x_n)$ be a sequence in $WC(X)$, with $x_n=(x_{nm})$, convergent to some $y=(y_n)\in\ell_\infty(X)$. Fix $f\in X^*$ and $\varepsilon>0$. Since $x_n\to y$, there is some $n\in\mathbb N$ such that $\|x_n-y\|_\infty<\varepsilon$. For this $n$, there is some $N\in\mathbb N$ such that $|f(x_{nm_1}-x_{nm_2})|<\varepsilon$ for all $m_1,m_2\geq N$. Thus we have \begin{align*} |f(y_{m_1}-y_{m_2})|&\leq|f(y_{m_1}-x_{nm_1})|+|f(x_{nm_1}-x_{nm_2})|+|f(x_{nm_2}-y_{m_2})|\\ &\leq\|f\|\|y_{m_1}-x_{nm_1}\|+|f(x_{nm_1}-x_{nm_2})|+\|f\|\|x_{nm_2}-y_{m_2}\|\\ &\leq2\|f\|\|y-x_n\|_\infty+|f(x_{nm_1}-x_{nm_2})|\\ &<(2\|f\|+1)\varepsilon. \end{align*} Thus $(y_n)$ is weakly Cauchy, so $y\in WC(X)$ and therefore $WC(X)$ is closed in $\ell_\infty(X)$.

As far as resources are concerned about the space of weakly convergent sequences, I'm afraid I'm not aware of any.


If $x_n$ is a weakly Cauchy sequence for each $n$, and $x_n \to x$ in $\ell_\infty(X)$, i.e. $x_n(j)$ converges uniformly to $x(j)$ as $n \to \infty$, I claim $x$ is weakly Cauchy. If not, there is $y \in X^*$ with (for convenience) $\|y\| \le 1$ such that $y(x(j))$ is not a Cauchy sequence, i.e. there is $\epsilon > 0$ such that for all $N$ there exist $j,k > N$ with $|y(x(j)) - y(x(k))| > \epsilon$. But if $n$ is large enough, $\|x - x_n\|_\infty < \epsilon/3$, thus $\|x(j) - x_n(j)\| < \epsilon/3$ for all $j$, and then $$|y(x_n(j)) -y(x_n(k))| \ge | y(x(j)) - y(x(k)| - |y(x_n(j)) - y(x(j))| - |y(x(k)) - y(x_n(k))|> \frac{\epsilon}3$$ contradicting the assumption that $x_n$ is weakly Cauchy.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.