I found the following question in the book Introdução à Análise Funcional, by César R. Oliveira.
Let $\mathcal{N}_p \subset \mathcal{l}^p(\Bbb{N})$, $1 \le p \le \infty,$ the subspace of all sequences with finitely many non-zero entries. Show that $\mathcal{N}^{\ast}_p=\mathcal{l}^q(\Bbb{N})$, with $(1/p)+(1/q)=1.$
I think something is strange in this exercise. The definition of $\mathcal{N}_p$ is independent of $p$, since all sequencies with finitely many non-zero entries belong to all $\mathcal{l}^p$. In other words, $\mathcal{N}_p=\mathcal{N_q}$, for all $p,q$. But that would imply, for example, that $\mathcal{l}^1=\mathcal{N}^{\ast}_{\infty}=\mathcal{N}^{\ast}_{1}=\mathcal{l}^{\infty}$.
Is my reasoning correct? Where I made a mistake?
If I am wrong and the exercise is correct, I would appreciate any hints to solve it, I am lost.
EDIT: After posting, I realized my mistake. As pointed in the answer below, the norms are different, making the $\mathcal{N}_p$ different. However, I still need help to prove the exercise.
Thank you!
The elements of $\mathcal{N}_p$ itself indeed do not depend on $p$. However, when you consider the dual you have to specify which your norm you put on the space and that's where the $p$-dependency kicks in.