Prove that for the series $\sum_{n \in \mathbb{N}}|\zeta_n\mu_n|$ to be convergent for all $\zeta \in l^p \implies \mu \in l^q$

For my convenience, I will write $\zeta_n = x_n$, $\mu_n = y_n$.

Define for each $n$ functionals $T_n : \ell^p \to \Bbb{C}$ by $x=(x_n) \mapsto \sum_{i=1}^{n}x_iy_i$.

$|T_n(x)| \le \big(\sum_{i=1}^{n}|x_i|^p\big)^{1/p} \big(\sum_{i=1}^{n}|y_i|^q\big)^{1/q}$, so that

$\|T_n\| \le \big(\sum_{i=1}^{n}|y_i|^q\big)^{1/q}$, and choose $z=(z_i)$ such that $$ z_i = \left\{ \begin{eqnarray} \frac{y_i}{|y_i|^{2+p-pq}} & & \text{if} & i \le n & \text{and} &y_i\not=0 \\ 0 & & \text{if} & i > n & \text{or} &y_i=0. \end{eqnarray} \right. $$

We see that $\|T_n\|\ge |T_n(z)|/\|z\| = \big(\sum_{i=1}^{n}|y_i|^q\big)/\big(\sum_{i=1}^{n}|y_i|^q\big)^{1/p} = \big(\sum_{i=1}^{n}|y_i|^q\big)^{1/q}$. Hence, $$\|T_n\|=\big(\sum_{i=1}^{n}|y_i|^q\big)^{1/q}.$$

Therefore by the uniform boundedness principal,

$$\sup_n\|T_n\| = \big(\sum_{i=1}^{\infty}|y_i|^q\big)^{1/q} < \infty.$$