Consider Exercise 1.14 in Brezis' Functional Analysis, Sobolev Spaces and Partial Differential Equations below. My question regards item 3.. How to solve this? I am trying to show that there is no such hyperplane, by supposing that there is and arriving at a contradiction:
Let $Z = X - c = \{x - c \ : \ x \in X\} = \{z \in \ell^1 \ : \ z_{2n} = -1/2^n \}$. Suppose there is $u \in Y \cap Z$. Then $y_{2n - 1} = -1$ for all $n \geq 1$ and therefore $y \notin \ell^1$, a contradiction. Hence $Y \cap Z = \emptyset$.
Suppose now that there is a closed hyperplane separating $Y$ and $Z$. Then there is some $f \in (\ell^1)^*$ such that $$ f(y) \leq \alpha \leq f(z) \quad \forall y \in Y, z \in Z, $$ where $[f = \alpha]$ is the hyperplane. Then there is $u \in \ell^\infty$ such that $$ \sum_{n = 1}^\infty u_n y_n \leq \alpha \leq \sum_{n = 1}^\infty u_n z_n, \quad \forall y \in Y, z \in Z. $$ But $Y$ is a subspace, so $f$ can only be bounded in $Y$ if $Y \subset \ker f$.
Here I am stuck. Any hints will be the most appreciated. Thanks in advance and kind regards.
