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I am working with a space of the form $E=L^{\infty}(0,\infty;X)$, where $X$ is a non reflexive Banach space.

If a sequence $(x_n)$ is bounded in $E$, how to extract a weakly converging subsequence?

I think that this is not always possible for an arbitrary Banach space $X$!

I tried to consider that $X$ has a separable dual and use Banach Alaoglu theorem, but I did not success to identify the dual of $E$.

What I am looking is to assume that $X$ satisfies a given property letting me extract a weakly converging subsequence.

Any help will be appreciated.

elmas
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1 Answers1

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As you thought, this is indeed not always possible.

Since $X$ isn't reflexive, a classical consequence of the Mackey-Arens theorem gives you the existence of bounded subsets of $X$ which aren't weakly relatively compact in $X$ and the Eberlein-Šmulian theorem then gives you the existence of a bounded sequence $(y_n)_{n\in\mathbf N}$ of elements of $X$ which has no weakly convergent subsequence.

For every $n\in\mathbf N$, let $x_n$ be (the equivalence class in $L^\infty(0,\infty;X)$ of ) the constant function on $(0,\infty)$ taking everywhere the value $y_n$. The sequence $(x_n)_{n\in\mathbf N}$ is a bounded sequence of $L^\infty(0,\infty;X)$ and I will show that it has no weakly convergent subsequence.

Let's proceed by contradiction and suppose that $(x_n)_{n\in\mathbf N}$ has a subsequence $(x_{n_k})_{k\in\mathbf N}$ which converges weakly to some $x_\infty\in L^\infty(0,\infty;X)$. By the Lebesgue differentiation theorem, $x_\infty$ has many Lebesgue points (maybe you only know this theorem for scalar functions, but the proof presented on the Wikipedia page can be adapted to the case of vector-valued functions without any major modification); let $t\in(0,\infty)$ be a Lebesgue point of $x_\infty$ and let $$y_\infty=\lim_{h\to 0^+}\frac 1{2h}\int_{t-h}^{t+h}x_\infty(s)\,ds.$$ I will show that $(y_{n_k})_{k\in\mathbf N}$ converges weakly to $y_\infty$, which will be a contradiction because $(y_n)_{n\in\mathbf N}$ has no weakly convergent subsequence.

Let $g$ be a continuous linear functional on $X$. Let $\mathscr U$ be an ultrafilter on $(0,t)$ such that $(0,h)\in\mathscr U$ for every $h\in(0,t)$. The linear functional $$f:x\longmapsto\lim_{h,\mathscr U}\frac 1{2h}\int_{t-h}^{t+h}g(x(s))\,ds$$ on $L^\infty(0,\infty;X)$ can be easily checked to be well-defined and continuous. Moreover, because $(x_{n_k})_{k\in\mathbf N}$ converges weakly to $x_\infty$ and $g(y_n)=f(x_n)$ for every $n\in\mathbf N\cup\{\infty\}$, we have $$g(y_{n_k})=f(x_{n_k})\underset{k\to\infty}{\longrightarrow}f(x_\infty)=g(y_\infty),$$ which concludes.

P. P. Tuong
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  • Thank you for the response. I would like to ask if we consider changing the notion of weak convergence to weak* convergence. As we know, there exist non-reflexive spaces, e.g., see for which any bounded sequence has a weak* converging subsequence. Is there anything we can assume on $X$ to guarantee that $E$ exhibits this property. – elmas Feb 23 '24 at 13:55
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    @elmas What comes to mind is if $X$ has the Radon-Nikodym property and is the dual of a separable Banach space $Y$, then $F=L^1(0,\infty;Y)$ is a separable Banach space whose dual can be identified with $E$ and so in this setting, the Banach-Alaoglu theorem + metrizability of bounded subset of $E$ for the weak* topology $\sigma(E,F)$ tell you that any bounded sequence of elements of $E$ has a subsequence that converges for the w* topology $\sigma(E,F)$. – P. P. Tuong Feb 23 '24 at 16:49