Every bounded sequence in $\ell_p$ has a weakly convergent subsequence

Question

I am trying to prove the following claim:

Let $p \in (1, \infty)$. Every $(x^n) \subseteq \ell_p(\mathbb{N})$ bounded has a weakly convergent subsequence.

Here's what I thought might work, but that so far hasn't:

Given a functional $\phi \in B(\ell_p, \mathbb{C}$), if we show that $\phi(x^n)$ is a Cauchy sequence in $\mathbb{C}$, then we get that it has a convergent subsequence $\phi(x^{n_k})$ and then we could take $x_{n_k}$ to be our subsequence. But I now realize that this subsequence would depend on $\phi$, which is not what we want.

Also, I thought of identifying some characteristics of such a subsequence by using the fact that for every $\phi$, there is a $b \in \ell_q$ such that: $\phi_b(a)= \sum_{n=0}^\infty a_nb_n$. The idea was to plug in a generic $x^{n_k}$ instead of $a$ to get a clue of what $x^{n_k}$ needs to satisfy, but that wasn't very fruitful.

What am I missing? (note, this is a problem from introductory level lecture notes on Functional Analysis, so it should be provable using elementary theorems like those I used above).

Answer 1

First let's recall some definitions. The sequence $(x^n)\subset \ell_p(\Bbb N)$ weakly converges to $x\in\ell_p(\Bbb N)$ iff $\phi(x^n)\to \phi(x)$ for all functional $\phi\in \ell_p(\Bbb N)^*$.

The description of the dual space $\ell_p(\Bbb N)^*$ is the following:

• Let $\phi_i$ be a functional that gets the $i$-th component of a sequence, that is if $x=(x_1,x_2,x_3,\ldots)\in \ell_p(\mathbb{N})$ then $\phi_i(x)=x_i$. These functionals can be called projections on $i$-th component.
• Now, for any functional $\phi\in \ell_p(\Bbb N)^*$ there exists a sequence $(a_n)\in\ell_q(\Bbb N)$, where $\frac 1p+\frac 1q = 1$, such that $\phi = \sum_{i=1}^\infty a_i\phi_i$.

In other words, all functionals are of the form: $\phi(x)=\sum_{i=1}^\infty a_ix_i$.

• Conversely, each sequence $(a_n)\in\ell_q(\Bbb N)$ defines a functional in $\phi\in \ell_p(\Bbb N)^*$. This observation leads us to the identification: $$ \ell_p(\Bbb N)^*\equiv \ell_q(\Bbb N), \text{ where }\frac 1p+\frac 1q = 1.$$

Since we would like to find a subsequence convergent weakly, it's natural to start to find a subsequence such that it converges componentwise (i.e. it behaves well if we apply projections). Then we should prove that the componentwise limit is in the considered space ($\ell_p$). Then we should prove that this subsequence behaves well if we apply any functional. So let's start.

First consider $\phi_1(x^n)$. It's a bounded sequence in $\Bbb R$, so it has a convergence subsequence. Therefore $\phi_1(x^{k^1_n}) \to c_1$.

Now, the sequence $\phi_2(x^{k^1_n})$ is bounded in $\Bbb R$, so it has a convergence subsequence. Therefore $\phi_2(x^{k^2_n}) \to c_2$, where $k^2_n$ is a subsequence of $k^1_n$. Of course $\phi_1(x^{k^2_n}) \to c_1$ (as a subsequence of o previously found sequence).

Having defined an increasing sequence of natural numbers $(x^{k^j_n})$ such that $\phi_i(x^{k^i_n}) \to c_i$ for $i\leq j$ we can define a subsequence $(k^{j+1}_n)$ of $(k^j_n)$ (as before) in such a way that $\phi_{j+1}(x^{k^{j+1}_n}) \to c_{j+1}$.

Therefore, inductively, we defined the infinite family of sequences $(x^{k^i_n})$ of natural numbers such that each is a subsequence of the previous sequence and that $\phi_i(x^{k^i_n}) \to c_i$ for all $i\in\Bbb N$.

Now, consider a sequence $s_n = k^n_n$. This is a subsequence that is eventually a subsequence of all the sequences $(k^i_n)$ (eventually here means that the elements of this sequence after the index $i$ form a subsequence of $(k^i_n)$). In my language this method is called more or less a diagonal method.

Now we have $\phi_i(x^{s_n}) \to c_i$ for all $i\in \Bbb N$. Let's denote $y_n = x^{s_n}$.

I'm quite sure that we are close to prove a weak convergence of our subsequence, since what we have to prove is that [1] $(c_i)\in\ell_p(\mathbb{N})$ and [2] for any $(a_i)\subset \ell_q(\mathbb{N})$ we have

$$ \sum_{i=1}^\infty a_i\cdot \phi_i(y^{n}) \to \sum_{i=1}^\infty a_i c_i.$$

[1] We assumed that $(x^n)$ is bounded therefore $\sum_{i=1}^\infty |y^n_i|^p\leq M$ for some $M>0$.

For a fixed $I\in\Bbb N$ we have $$ M\geq \sum_{i=1}^I |y^n_i|^p \to \sum_{i=1}^I |c_i|^p.$$ Taking $I\to\infty$ we get that $(c_i)$ is in $\ell_p$.

[2] We already know that $$y^n_i\to c_i\text{ as }n\to\infty \hspace{1cm} (*)$$ and need to prove that $$\sum_{i=1}^\infty a_iy^n_i\to \sum_{i=1}^\infty a_ic_i\text{ as }n\to\infty.$$ This is a play with number series.

Observe that for any $N\in\Bbb N$ we have from (*) that $$\sum_{i=1}^N a_iy^n_i\to \sum_{i=1}^N a_ic_i.$$ Because $(a_i)\in\ell_q$, $(c_i)\in\ell_p$, then the series $\sum a_ic_i$ is convergent. This finishes the proof.