I want to show that $\ell^1( \mathbb N )$ enjoys the Schur property. More precisely, I have to prove the following
Theorem. Let $X = \ell^1 (\mathbb N )$, $\{ x^{(n)} \} \subset X$, $x \in X$. The following statements are equivalent:
- $x^{(n)} \overset{n \to \infty}{\rightharpoonup} x$ ;
- $x^{(n)} \overset{n \to \infty}{\to} x$.
Now, $(2) \implies (1)$ is trivial. For the $(1) \implies (2)$ side, I tried something like: let $f \in \ell^1(\mathbb N)^\prime$ be an arbitrary linear functional. Then, by definition,
\begin{equation}
x^{(n)} \overset{n \to \infty}{\rightharpoonup} x \iff \langle f, x^{(n)} \rangle \to \langle f, x \rangle \implies \lvert \langle f, x^{(n)} – x \rangle \rvert \to 0 \quad (n \to \infty) .
\end{equation}
Hence
\begin{equation}
\lvert \langle f, x^{(n)} – x \rangle \rvert \leq \lVert f \rVert \lVert x^{(n)} – x \rVert.
\end{equation}
(This reasoning is actually applied for the other side; I wonder if something useful could be extracted in a similar way.)
Now, by the Hahn-Banach theorem, we can choose $f$ such that $\lVert f \rVert = 1$ and $\langle f, x^{(n)} – x \rangle = \lVert x^{(n)} – x \rVert$.
Since the left hand side of the last equation tends to zero for every $f \in \ell^1(\mathbb N)^\prime$ by definition of weak convergence, and in particular for the $f$ given by H-B theorem, it seems that the result would hold for every Banach space. This is clearly absurd, but I can't understand where the error is. Notice that it is obvious that the proof relies on the particular norm of $\ell^1$ and my original idea was to exploit it in same manner from the point where I'm stucked.
I suspect there is no way to complete the preceding attempt and achieve a correct solution of the problem. (as one should expect in most cases he tries to copy the other part in a proof like that.) Nevertheless, I wanna detect the mistakes for many reasons. (proving the theorem, understanding the usage of the Hahn-Banach theorem, better myself, avoid similar mistakes in different situations, etc…)
So I ask for a "error-checking" and for a proof of the theorem. (or references as well.) Thank you!
Best Answer
As Giusepped pointed out in the comments, the problem is that $f$ is not a priori the same for each $n$.
Here are the ideas for a proof of the Schur property.