Prove that for the series $\sum_{n \in \mathbb{N}}|\zeta_n\mu_n|$ to be convergent for all $\zeta \in l^p \implies \mu \in l^q$ Prove that for the series $$\sum_{n \in \mathbb{N}}|\zeta_n\mu_n|$$ to be convergent for all $$\zeta \in l^p \implies \mu \in l^q$$
I used Holder's inequality to prove that $$\sum_{n \in \mathbb{N}}|\zeta_n\mu_n| \leq \|\zeta\|_p \|\mu\|_q $$
But I cannot find a $\zeta= (\zeta_1, \zeta_2,....)$ such that $A(\zeta)=\sum_{n \in \mathbb{N}}|\zeta_n\mu_n|=(\sum_{n \in \mathbb{N}} |\mu_n|^q)^{\frac{1}{q}}$.
$ \zeta=??$
 A: 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, 
$$\sup_n\|T_n\| = \big(\sum_{i=1}^{\infty}|y_i|^q\big)^{1/q} < \infty.$$
A: Actually, the absolute convergence for all $\xi\in\ell^p$ is equivalent to a simple convergence for all $\xi\in\ell^p$ since for $\mu_n=|\mu_n|e^{i\theta_n}$ one can take $\xi_n=|\xi_n|e^{-i\theta_n}$.
It means, in particular, that 

if the functional $\phi(\xi)=\sum_{n\in\Bbb N}\xi_n\mu_n$ is defined on the whole $\ell^p$ then it must be continuous.

Let's first state a lemma. We will use $\xi(n)$ notation to address the $n$-th element of the sequence $\xi$ and $\xi_n$ notation to index a sequence of sequences.
Lemma: Let $\xi_n\in\ell^p$ and $\xi_n\to\xi$ in $\ell^p$. Then there exists a dominated subsequence $\xi_{n_j}$, i.e. $\forall k\ge 1$ $|\xi_{n_j}(k)|\le \sigma(k)$ for some $\sigma\in \ell^p$.
Proof: Since $\|\xi_n-\xi\|_p\to 0$ there exists a subsequence $\xi_{n_j}$ such that
$$
\sum_{j=1}^{+\infty}\|\xi_{n_{j+1}}-\xi_{n_j}\|_p<+\infty.
$$
Let us index the subsequence as $\xi_j$ and set $\xi_0=0$ for notational simplicity. Define
$$
\sigma(k)=\sum_{j=0}^{+\infty}|\xi_{j+1}(k)-\xi_j(k)|.
$$
The partial sums
$$
\sigma_N(k)=\sum_{j=0}^N|\xi_{j+1}(k)-\xi_j(k)|
$$
are uniformly bounded in $\ell^p$ (Minkowski inequaity) as
$$
\|\sigma_N\|_p\le\sum_{j=0}^N\|\xi_{j+1}-\xi_j\|_p\le\sum_{j=0}^{+\infty}\|\xi_{j+1}-\xi_j\|_p<+\infty,
$$
hence, by monotone convergence $\sigma\in\ell^p$. Furthermore, as$\sum_{j=0}^{+\infty}(\xi_{j+1}(k)-\xi_j(k))$ converges (absolutely) we conclude that
$$
|\xi_{N+1}(k)|=\left|\sum_{j=0}^N(\xi_{j+1}(k)-\xi_j(k))\right|\le \sigma_N(k)\le\sigma(k).
$$

Corollary: if the functional $\phi\colon \ell^p\to \Bbb C$
$$
\phi(\xi)=\sum_{k=1}^{+\infty}\xi(k)\mu(k)
$$
is defined everywhere on $\ell^p$ then it is closed.
Proof: Let $\xi_n\to \xi$ in $\ell^p$ and $\phi(\xi_n)\to y$ in $\Bbb C$. By Lemma above there exists a dominatied subsequence $\xi_j$ such that $|\xi_j(k)|\le\sigma(k)$. Strong convergence implies pointwise convergence $\xi_j(k)\to\xi(k)$ for all $k\ge 1$. Then 
$$
\xi_j(k)\mu(k)\to \xi(k)\mu(k) \quad\text{and}\quad |\xi_j(k)\mu(k)|\le\sigma(k)|\mu(k)|
$$ 
and the dominated convergence theorem gives that
$$
y=\lim_{j\to +\infty}\sum_{k=0}^{+\infty}\xi_j(k)\mu(k)=
\sum_{k=0}^{+\infty}\lim_{j\to +\infty}\xi_j(k)\mu(k)=\sum_{k=0}^{+\infty}\xi(k)\mu(k)=\phi(\xi).
$$

Now the closed graph theorem ensures that the functional $\phi$ is bounded, and therefore, $\mu\in\ell^q$.
