If $(x_n)$ is a real sequence and $\forall (y_n) \in \ell_p$ we have $\sum\limits_{n=1}^{\infty}|x_n||y_n|<\infty$,then $(x_n) \in \ell_q $

Question

So, we consider $(x_n)$ is a real sequence and $1<p<\infty$.
For every $y=(y_n) \in \ell_p$, we have $\sum\limits_{n=1}^{\infty}|x_n||y_n|<\infty$.
We need to show that $x=(x_n) \in \ell_q$, where $q$ is such, that $1/p + 1/q = 1$.
I encountered this exercice in a course of functional analysis.
I have no clue where to begin. Any help would be greatly appreciated. Thanks in advance.

Answer 1

My bad, I did not read carefully the question.
So let me give a more elementary solution to this question.

---

Let $$S_n = \sum_{k=1}^n |x_k|^p$$ then $$\sum_{n \ge 1} |z_n|^q < \infty$$ with $z_n= \left( |x_n|^p\frac{1}{S_n\ln(S_n)^2}\right)^{1/q}$

---

Indeed, by MVT, we see that $$|z_n|^q= |x_n|^p\frac{1}{S_n\ln(S_n)^2} \le \frac{1}{\ln(S_{n-1})} -\frac{1}{\ln(S_{n})}$$ Hence forth the convergence of $\sum |z_n|^q$

---

So now, because $z=(z_1,z_2,\dots) \in l^q$, we imply that $$\sum_{n \ge 1} |x_nz_n| <\infty$$ However, if $S_n$ converges to infinity, we have must have for all $N$ $$\sum_{n = 1}^N |x_nz_n| \ge \frac{1}{S_N^{1/q}\ln(S_N)^{2/q}}\underbrace{\sum_{n=1}^N |x_n||x_n|^{p/q}}_{= S_N}=\frac{S_N^{1/p}}{\ln(S_N)^{2/q}}\xrightarrow[]{N \rightarrow +\infty} \infty $$

which is contradicting.
Hence $$\sum_{n \ge 1} |x_n|^p < \infty$$ Q.E.D