Let $1\leq p<\infty$ and $q$ be the conjugate exponent of $p$. Suppose that $\lbrace a^n \rbrace_{n=1}^{\infty} \subset \ell^q$ in a sequence in $\ell^q$ such that $f_{a^n}(x) \mapsto 0~( n \mapsto \infty)$ for all $x \in \ell^p$ where $f_{a^n}(x)=\sum_{i=1}^{\infty} a_i^{(n)}x_i$. Show that the sequence $\lbrace a^n \rbrace_{n=1}^{\infty}$ is bounded, i.e., there exists $M\geq0$ such that $\lVert a^n\rVert_q\leq M$ for all $n$.
I get across with a theorem in Bryan P. Rynne Book which shows that $f_{a^n}$ defines a linear functional $f_{a^n} \in (\ell^q)'$ with $\lVert f_{a^n}\rVert=\lVert a^n\rVert_q$. I think it will help but I am still confused how to prove the statement above
If $\lbrace a^n \rbrace_{n=1}^{\infty} \subset l^q$ and $x\in l^p$, consider $J(a^n)(x)=x(a^n)=\sum_{i=1}^{\infty} a_i^{(n)}x_i.$ Since the series converges to $0,$ each element of $\left \{ J(a^n) \right \}$ is pointwise bounded, so by Banach-Steinhaus, there is an $M<\infty $ such that $\|J(a^n)\|<M$ for all $n\in \mathbb N$. But since $y\mapsto J(y)$ is an isometry, it follows that $\|a^n\|<M$ as well.