A linear functional on the periodic sequences in $\ell^\infty$

Question

Prove that there exists a continuous nonzero linear fucntional on $\ell^\infty$ that vanishes on every periodic sequence $ x \in \ell^\infty$. (A sequence is periodic if the exists an $N \in \mathbb N$ such that for all $ n \in \mathbb N$ we have $x_n = x_{n+N}$.)

Attempt:

The set $X$ of periodic sequences forms a subspace of $\ell^\infty$. Futher more this subspace has countable (Hamel) basis $\{e_j\}_{j \in \mathbb N}$. Since $\ell^\infty$ is Banach $X$ is meager in $\ell^\infty$ and thus proper. Then there exists a nonperiodic $y \in \ell^\infty \backslash X$ such that $Y := X \oplus \mathbb Cy$ is also proper. Now define on $Y$ the linear functional $f:Y \to \mathbb C$, that such that, (for all $j \in \mathbb N$,) $f(e_j) = 0$ and $f(y) = 1$. By Hahn-Banach Theorem this extends to a linear functional on $\ell^\infty$.

Questions:

• Is there a mistake in the above proof
• Is there an explicit construct of such functional
• How to prove it otherwise

Answer 1

Let $P$ be the space of periodic sequences. Let $e_1$ be the sequence $(1,0,...)$. Note that for $x \in P$, we have $\|x-e_1\|_\infty \ge \max(|x_1-1|, |x_k|)$ for all $k>1$, and since $x$ is periodic, there is some $k>1$ such that $x_1 = x_k$ and so we have $\|x-e_1\|_\infty \ge {1 \over 2}$.

Hence we can strictly separate $\overline{P}$ and $\{e_1\}$ with some functional $\phi$ such that $\phi(x) < \phi(e_1)$ for all $x \in P$. Since $P$ is a linear subspace, we must have $\phi(x) = 0$ for all $x \in P$ and hence $\phi(e_1) >0$.

Answer 2

Let $AP$ denote the norm-closure of the linear span of all periodic sequences in $\ell^\infty$. Define the linear functional $f:AP\to\mathbb{C}$ by $$f(x) = \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n x_k\hspace{10mm} \forall x=(x_k)_{k\in\mathbb{N}} \in AP$$ There exist $y\in\ell^\infty$ for which the limit $\displaystyle \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n y_k$ does not exist. Consequently, $y\notin AP$ so $AP\neq\ell^\infty$.

Second, by Hahn-Banach theorem, there exists $g\in(\ell^\infty)^*$ such that $g(AP)=\{0\}$ and $g(y) = 1$.