Boundedness and pointwise convergence imply weak convergence in $\ell^p$

Let $p\in(1,+\infty)$ and consider the space $\ell^p$ with its usual norm. The following are equivalent:

(1) $x_n \rightharpoonup x$ (i.e. $x_n$ weakly converges to $x$);

(2) $$\exists M>0: \quad \sup_{n} \Vert x_n \Vert_p \le M \quad \text{ and } \quad \forall k\in\mathbb N: \,\,\ x^{(k)}_n\to x^{(k)}$$

I think I've proved (1) $\Rightarrow$ (2): indeed, every weak convergent sequence is bounded (hence we get $\sup_{n} \Vert x_n \Vert_p \le M$ for some positive $M$); on the other hand, we can simply observe that the "projections" $p_k \colon \ell^{p} \to \mathbb R$ definde by $x=(x_n)_{n=1}^{\infty} \mapsto x_k$ are linear and continuous. So, briefly, we have that weak convergence implies boundness and pointwise convergence.

What about (2) $\Rightarrow$ (1)? I'm really puzzled and I don't know how to begin. Thanks in advance.

Solution 1:

There is a more general result.

Theorem. Let $E$ be a normed space. Let $\{x_n:n\in\mathbb{N}\}\subset E$ and $x\in E$, then the following conditions are equivalent:

• $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x\in E$
• $\{x_n:n\in\mathbb{N}\}$ is bounded and for all $S\subset E^*$ such that $\mathrm{cl}(\mathrm{span} S)=E^*$ holds $\lim\limits_{n\to\infty} f(x_n)=f(x)$ for all $f\in S$.

Proof. Assume $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$. Then it is well knonw that $\{x_n:\in\mathbb{N}\}$ is bounded. Moreover for all $f\in X^*$ we know that $\lim\limits_{n\to\infty}f(x_n)=f(x)$. Hence for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ we have $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$.

Assume that $\{x_n:n\in\mathbb{N}\}$ is bounded by some conatant $M>0$ and for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ holds $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$. Take arbitrary $g\in\mathrm{span}S$, then $g=\sum\limits_{k=1}^m \alpha_k f_k$. Then it is easy to check that $\lim\limits_{n\to\infty}g(x_n)=g(x)$. So for all $g\in\mathrm{span}S$ we have $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Now take arbitrary $g\in X^*=\mathrm{cl}(\mathrm{span}S)$, then $g=\lim\limits_{k\to\infty} g_k$ for some $\{g_k:k\in\mathbb{N}\}\subset\mathrm{span}S$. Fix arbitrary $\varepsilon>0$. Since $g=\lim\limits_{k\to\infty} g_k$, then there exist $K\in\mathbb{N}$ such that $\Vert g-g_k\Vert\leq\varepsilon$ for all $k>K$. Then $$ \begin{align} |g(x_n)-g(x)| &\leq|g(x_n)-g_k(x_n)|+|g_k(x_n)-g_k(x)|+|g_k(x)-g(x)|\\ &\leq\Vert g-g_k\Vert\Vert x_n\Vert+|g_k(x_n)-g_k(x)|+\Vert g_k-g\Vert\Vert x\Vert\\ &\leq\varepsilon M+|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert \end{align} $$ Let's take a limit $n\to\infty$ in this inequality, then $$ \limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq \varepsilon M+\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert $$ Since $g_k\in\mathrm{span}S$, then $\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|=0$ and we get $$ \limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq \varepsilon M+\varepsilon\Vert x\Vert $$ Since $\varepsilon>0$ is arbitrary we conclude $\limsup\limits_{n\to\infty}|g(x_n)-g(x)|=0$. This is equivalent to $\lim\limits_{n\to\infty}|g(x_n)-g(x)|=0$ i.e. $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Since $g\in X^*$ is arbitrary, then $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$.

Solution 2:

Let $L\in(\ell^p)'$ a linear continuous functional. We can assume that $x_n^{(k)}\to 0$ for all $k$. $L$ can be represent by an element $v$ of $\ell^q$, where $p^{-1}+q^{-1}=1$. Fix $\delta>0$ and $v'$ of finite support such that $\lVert v-v'\rVert_{\ell^q}\leqslant \delta$. Then $$\lVert L(x_n)\rVert\leqslant\delta+|\langle v',x_n\rangle_{\ell^q,\ell^p}|,$$ giving what we want.

Solution 3:

Let $X$ be a general Banach space and $X^*$ its dual. Then the weak topology on a bounded subset of $X$ is determined by a dense subset of $X^*$ in the sense:

If $D\subset X^*$ is dense and $(x_\alpha)\subset X$ is uniformly bounded in norm, then $x_\alpha$ converges to $x$ weakly if and only if $x_\alpha(d)\to x(d)$ for all $d\in D$.

With this and assuming 2). The first says $x_n$ are uniformly bounded in norm, while the second one says $x_n$ converges pointwise on sum dense subset of $\ell^q$ (I am pretty sure you can figure out this dense subset). So the result follows.

Solution 4:

We claim that $\textrm{Ball}(\ell^p)$ is weakly compact and weakly metrizable.

Since $1<p<\infty$, we know $\ell^p$ is reflexive, so by Alaoglu's theorem, $\textrm{Ball}(\ell^p)$ is weakly compact. Also, $\textrm{Ball}(\ell^p)$ is weakly metrizable by the separability of $\ell ^q$ where $1/p +1/q =1$. Q.E.D.

Now, we can assume that all $x_n \in \textrm{Ball}(\ell^p)$. Consider a subsequence $(x_{n_{k}})_k$. By the above claim, there is a sub-subsequence $(x_{n_{k_{l}}})_l$and an element $y$ such that $x_{n_{k_{l}}} \rightharpoonup y$. It is easy to check that $y \equiv x$, which implies that $x_n \rightharpoonup x$.