I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:

Let $X := c_0(\mathbb{N}), \hspace{3mm}x_0 \in X^*=\ell^1(\mathbb{N}), \hspace{3mm}\{x_n\}_{n \in \mathbb{N}} \subset X^*$ bounded. Show that

\begin{equation} x_n \rightharpoonup^* x_0 \Longleftrightarrow x_n(k) \rightarrow x_0(k) \end{equation} for fixed $k \in \mathbb{N}$.


Hello everyone, I took back this exercise and I was trying to complete the proof, well:

1) for $(\Rightarrow)$ I know from HP that \begin{equation} \forall \{y_k\}_k \in \ell^1(\mathbb{N}) \quad, \sum_{k=1}^{\infty}x^{(n)}_ky_k -\sum_{k=1}^{\infty}x^{(0)}_k \rightarrow 0 \quad as \hspace{2mm}n \rightarrow \infty \end{equation}

then of course it is equivalent to say that

\begin{equation} \sum_{k=1}^{\infty}y_k(x^{(n)}_k-x^{(0)}_k) \rightarrow 0 \hspace{2mm}as \hspace{2mm} n \rightarrow \infty \end{equation}

So what I'm saying is that \begin{equation} \forall \epsilon \gt 0, k \in \mathbb{N}, \hspace{2mm}\exists n_{\epsilon,k} : \forall n>n_{\epsilon,k} \quad |x^{(n)}_k-x^{(0)}_k|\lt\epsilon \end{equation}

So I think I proved ($\Rightarrow$) this way (Tell me what you think)


What you are supposed to prove is the following: suppose $\sum_j |a_j| <\infty, \sum_j |a_{nj}|$ is bounded and $a_{nj} \to a_j$ as $n \to \infty$ for each $j$; then $\sum_j a_{nj} c_j \to \sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $\epsilon >0$ and choose $N$ such that $|c_j| <\epsilon$ for all $j \geq N$. Then $|\sum_j a_{nj} c_j - \sum_j a_jc_j| \leq |\sum_j^{N-1} a_{nj} c_j - \sum_j^{N-1} a_jc_j|+\epsilon\sum_{j=N}^{\infty} |a_{nj}|$. Can you now complete the proof?

[I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].


To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:

Theorem. Let $X$ be a topological space and $f\in C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $\mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.

The proof is straightforward if you know the definition of equicontinuity.

Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $\sup_n\lVert f_n\rVert<\infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).


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