If $1<p<\infty$ and $\{x_n\}\subset l^p$.And $\frac{1}{p}+\frac{1}{q}=1$. Prove that $\sum\limits_{j=1}^\infty x_n(j)y(j)\rightarrow0$ for every $y\in l^q$ iff $\sup\{\|x_n\|: n\geq1\}<\infty$ and $x_n(j)\to0$ for $j\geq1$.
I think we can use the following corollary comes by the Principle of Uniform Boundedness.
So what I did is:
Let $A=\{x_n:n\geq1\}\subset l^p$.
then $\sup\{\|x_n\|: n\geq1\}<\infty$ and $x_n(j)\to0$ for $j\geq1$ implies $A$ is bounded. (The norm used for $A$ is the norm on $l^p$).
So from the corollary we have for every $f\in (l^p)^*$, $\sup\{|f(\{x_n(j)\}_{j=1}^\infty)|:n\geq1\}<\infty$.
Also I know that $(l^p)^*=l^q$.
But I'm having a hard time in understanding how to visualize an element in $l^q$ as a linear functional on $l^p$. Also how to obtain the result $\sum\limits_{j=1}^\infty x_n(j)y(j)\rightarrow0$ .
Appreciate your help
This is the Problem 2 on the exercise set of Chapter 3 section 14 of the book Functional Analysis by J.B. Conway.
One direction is by Lebesgue integrals theory (over-killed):
Note that $|x_{n}(j)|\leq\|x_{n}\|_{p}$ and hence $|x_{n}(j)y(j)|\leq\sup_{n}\|x_{n}\|_{p}|y(j)|$.
Assuming that $y\in c_{00}$, the space of sequences that are eventually zero, then $(y(j))\in l^{1}$ and hence by Lebesgue Dominated Convergence Theorem we have \begin{align*} \lim_{n\rightarrow\infty}\sum_{j=1}^{\infty}x_{n}(j)y(j)=\sum_{k=1}^{\infty}\lim_{n\rightarrow\infty}x_{n}(j)y(j)=0. \end{align*}
In fact, since $(y(j))$ is eventually zero, it is safe to interchange the limit with the sum, there is no need for Lebesgue integrals theory.
For general $y\in l^{q}$, we know that $c_{00}$ is dense in $l^{q}$ we have for $z\in c_{00}$ that \begin{align*} \left|\sum_{j=1}^{\infty}x_{n}(j)y(j)\right|&=\left|\sum_{j=1}^{\infty}x_{n}(j)(y(j)-z(j))+\sum_{j=1}^{\infty}x_{n}(j)z(j)\right|\\ &\leq\|x_{n}\|_{p}\|y-z\|_{q}+\left|\sum_{j=1}^{\infty}x_{n}(j)z(j)\right|\\ &\leq\left(\sup_{n}\|x_{n}\|_{p}\right)\cdot\|y-z\|_{q}+\left|\sum_{j=1}^{\infty}x_{n}(j)z(j)\right|, \end{align*} which can be made by arbitrarily small.
For the other direction, note that by taking $y(j)$ the $j$th coordinate sequence, that is, $(0,...,0,1,0,...)$, where $1$ is at the $j$th coordinate, one has immediately that $\displaystyle\sum_{j=1}^{\infty}x_{n}(j)y(j)=x_{n}(j)\rightarrow 0$, so part of the assertion has done.
Now we let $T_{n}(y(j))=\displaystyle\sum_{j=1}^{\infty}x_{n}(j)y(j)$, so for a fixed $(y(j))$, $(T_{n}(y(j)))$ is a bounded sequence of $n$, and hence by Uniform Boundedness Principle we deduce that $\sup_{n}\|T_{n}\|<\infty$. Now it is a standard fact by Riesz Representation Theorem that $\|T_{n}\|=\|x_{n}\|_{p}$.