The problem I'm trying to solve is:
Let $1\leq p, q\leq\infty$ such that $\frac{1}{q}+\frac{1}{p}=1$. If $$\sum_{n=1}^{\infty} |x_n||y_n|<\infty$$ for all $(x_n)\in l^p$, then $(y_n)\in l^q$.
$\bullet$ If $p=\infty$, then picking $(x_n)=(1,1,1,...)\in l^\infty$, we have $$\sum_{n=1}^{\infty} |y_n|<\infty$$ i.e. $(y_n)\in l^1$.
but how can I prove for $1\leq p<\infty$?
Let $x=(x_n)$ and define $\phi_j:l^p\to\mathbb{R}$ by $$\phi_j(x)=\sum_{n=1}^{j}x_ny_n.$$ Note that $\phi_j\in (l^p)'$ for all $j\in\mathbb{N}$.
By hypothesis, is well define the functional $\phi:l^p\to\mathbb{R}$ by $$\phi(x)=\sum_{n=1}^{\infty}x_ny_n\hspace{2cm}(1)$$
As $\lim_{j\to\infty}\phi_j(x)=\phi(x)$ for all $x\in l^p$, using the Uniform Boundedness Principle, we can conclude that $$\phi\in (l^p)'$$ Now, as $(l^p)'=l^q$, we know that there's a $(z_n)\in l^q$ such that $$\phi(x)=\sum_{n=1}^{\infty}x_n z_n\hspace{2cm}(2)$$
By $(1)$ and $(2)$, we conclude that $(y_n)=(z_n)\in l^q$.