Weak convergence in the Hilbert cube

Question

I'm very curious about the following problem:

How can I show that in the Hilbert cube defined as

$$C=\{x=(x_1,x_2,\dots) \in l^p: |x_n|\leq \frac{1}{n}\,\,\, \forall n \in \mathbb{N}\}, 1\leq p < \infty$$

the weak convergence implies strong convergence?

I just need some reference or hints on how to attack the problem. I know that for every $\varphi \in (l^p)^*$, it follows that if $(x_n)$ is a sequence in $C$ that converges weakly to some $x$, then $\varphi(x_n) \to \varphi(x).$ I'm hoping to find some specifics $\varphi's$ that is going to help me conclude strong convergence, that is: $||x_n-x||_p \to 0$.

I feel that the duality relation between $l^p$ and $l^q$, where $\frac{1}{p}+\frac{1}{q}=1$, can be helpful.

Thank you in advance, every help will be very much appreciated.

Answer 1

I don't remember who to give the credits to, but in this forum I once saw the following argument for why the weak convergence implies the strong one on a (strongly) compact subset $K$ of a Banach space $V$.

Consider the following two topologies on $K$: $$\begin{align*} \tau_1 &:= \{K \cap U : U \textrm{ is a strongly open subset of } V\}, \\ \tau_2 &:= \{K \cap U : U \textrm{ is a weakly open subset of } V\}. \end{align*}$$ Since every weakly open subset of $V$ is also strongly open, we have that $\tau_2 \subseteq \tau_1$ and the identity map $\operatorname{id}_K \colon (K,\tau_1) \to (K,\tau_2)$ is continuous. Now, as $(K,\tau_1)$ is a compact space, and $(K,\tau_2)$ is a Hausdorff one (why?), it follows that $\operatorname{id}_K \colon (K,\tau_1) \to (K,\tau_2)$ is a homeomorphism. That means $\tau_1 = \tau_2$, and then the weak and strong convergences are the same.

Answer 2

My solution: For $p=1$, this fact is a immediate consequence of an well known result in functional analysis stating that $l^1$ is a Schur space.

For $1<p<\infty$, let $x_n=(x_n^j)_{j=1}^{\infty}$ an sequence in $C$ converging weakly to $x=(x_j)_{j=1}^{\infty} \in l^p$. Let $q$ be the number such that $\frac{1}{p}+\frac{1}{q}=1$. By the duality relation between $l^p$ and $l^q$, it follows that: $$\lim_{n \to \infty} \sum_{j=1}^{\infty}a_j x_n^j =\sum_{j=1}^{\infty}a_jx_j\,\,\, \forall a=(a_j)_{j=1}^{\infty} \in l^q.$$

In particular taking $a=e_j$ for each $j$, we have that $\lim_{n \to \infty}x_n^j=x_j$, which implies that $\lim_{n \to \infty} |x_n^j-x_j|^p=0$. Consider now the sequence of functions $f_n:\mathbb{N} \to \mathbb{R}$ defined for each $j$ by $f_n(j)=|x_n^j-x_j|^p$. Then: $$|f_n(j)| \leq 2^p(|x_n^j|^p+|x_j|^p)\leq 2^p(\frac{1}{j^p}+|x_j|^p).$$ Where for $p>1$ we see that the series $2^p\sum_{j=1}^{\infty} (\frac{1}{j^p}+|x_j|^p)$ converges. We can now apply the Lebesgue dominated convergence theorem to conclude the proof: $$\lim_{n \to \infty} \sum_{j=1}^{\infty} |x_n^j-x_j|^p = \sum_{j=1}^{\infty} \lim_{n \to \infty} |x_n^j-x_j|^p = 0. $$ Which means $\lim_{n \to \infty}||x_n-x||_p = 0$.