Dtyjlui

Given $y in mathbb{K^N}$ a scalar sequence, and $1<p<infty$.

Suppose that, $forall x in l_p$, the series $sum_{n geq 1} x(n)y(n)$ is bounded. Show that $x in l_{p^*}$.

With $l_{p^*}$ I mean the dual of $l_p$, if I am not wrong, we have $(l_p)^*=l_{p^*}$.
Any suggestion? I'm not really keen in dual... what should I demonstrate?

Note that can strenghthen the assumption to
$$
sum_{n =1}^infty |x(n)||y(n)| <infty, quadforall xin l^p.
$$ by considering $widehat{x}(n) = |x(n)|e^{-itheta_n}$ where $y(n) = |y(n)|e^{itheta_n}$. Define $$
F_k ={xin l^p;|; sum_n |x(n)||y(n)|le k}.$$ We can show by Fatou's lemma(or directly) that $F_k$ is closed in $l^p$ for all $k$. Now, since $l^p = cup_k F_k$, by Baire category theorem, there exists $k$, $zin l^p$ and $delta>0$ such that
$$
z'in l^p,;|z-z'|_pleq delta Rightarrow z'in F_k.
$$ Let $xin l^p$ be $|x|_ple 1$. Then from
$$
sum_n |z(n)+delta x(n)||y(n)|le k,
$$ we have
$$
deltasum_n |x(n)||y(n)| leq sum_n |z(n)||y(n)|+k le 2k.
$$ Now, if we take $x(n) = frac{|y(n)|^{p^*-1}}{(sum_{n=1}^N |y(n)|^{p^*})^{1/p}}1_{nleq N}$ , then $|x|_ple 1$ and it follows that
$$
left(sum_{nle N}|y(n)|^{p^*}right)^{1-1/p} = sum_{nle N}|x(n)||y(n)| le sum_{n}|x(n)||y(n)| lefrac{2k}{delta}.
$$ Take $Ntoinfty$ to get
$$
|y|_{p^*}le frac{2k}{delta}<infty.
$$