Let $\ell^{\infty}$ be the set of bounded sequences in $\mathbb{F}$, with the supremum norm.
$c \subset \ell^{\infty}$ the sequences whose limit exists.
Then there exists a $f \in (\ell^{\infty})'$, the dual, such that $f(x) = \lim_{n \to \infty} x(n)$ for all $x \in c$.
Because we can define it on $c$ and then extend it with the Hahn-Banach theorem for Normed Spaces.
My question is if there is another $g \in (\ell^{\infty})'$, $g \not = f$, with $g(x) = f(x) = \lim_{n \to \infty} x(n)$ for all $x \in c$.
I tried using two different one-dimensional extensions first but I couldn't finish that proof.
Any ideas? Thanks.
Best Answer
You can "almost explicitly" give two different extensions.
Let $E \subset \ell^\infty$ be the subspace of sequences $x\in\ell^\infty$ such that $e_0(x) := \lim\limits_{n\to \infty} x_{2n}$ exists, and $O \subset \ell^\infty$ the subspace of sequences $x\in\ell^\infty$ such that $o_0(x) := \lim\limits_{n\to\infty} x_{2n+1}$ exists. Evidently $c \subset E\cap O$ and $e_0\lvert_c = o_0\lvert_c = f$. Let $e$ resp. $o$ be extensions of $e_0$ resp. $o_0$ to $\ell^\infty$.
Then $e$ and $o$ are continuous extensions of $f \in c'$, and as witnessed by $e((-1)^n) = 1 \neq -1 = o((-1)^n)$, they are distinct.