Ytukyg

I am a tutor for functional analysis and the students are supposed to show the here to be found characterization for weak convergence in $ell^p$, $1<p<infty$.

I am fully aware of a correct solution but I found the following approach and did not see whether this is fixable.

Let $varphi in (ell^p)'simeq ell^q$ be arbitrary.
Then we can write $varphi(z)=sum_{i=1}^infty z^{(i)}y^{(i)}$ for suitable $y=(y^{(i)})in ell^q$.
Then we have (EDIT:Note that $(x_n)$ is bounded in $ell^p$):
$$
lim_{nto infty} varphi(x_n)=lim_{nto infty} sum_{i=1}^infty x_n^{(i)}y^{(i)}.
$$

Now, the student is trying get the limit inside of the integral, saying that the sum converges uniformly and the limit is bounded independently of $n$.
Nevertheless, I am not aware of a theorem which would allow use, moreover, the counterexample for the case $p=1$ shows that for general $p$ this is not valid.

Now, my question is whether we can fix this for $p>1$. The first thing that came to my mind was to try to apply dominated convergence but I did not manage to do so.

I am happy for any insights also for showing that this cannot not be fixed.

If the argument would be valid, it would allow to show that the sequence of sequences $x_n = n , e_n$, where $e_n$ is the n'th unit sequence, would converge weakly to zero in $ell^p$.
Indeed, $lim_{ntoinfty} x_n^{(i) }= 0$ for all fixed $n$. Thus, the wrong argument would imply
$$lim_{ntoinfty}varphi(x_n) = 0.$$

However, $x_n$ cannot converge weakly since it is not bounded.