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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.

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1 Answer 1

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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.

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  • $\begingroup$ Thank you. I'm not familiar with ultrafilters but your edit made it very clear. $\endgroup$
    – MrReese
    Commented Dec 16, 2013 at 11:17
  • $\begingroup$ Yes, sorry. That was the first thing I thought of, but after posting I thought myself "What the ..., you can do that simpler". $\endgroup$ Commented Dec 16, 2013 at 11:24

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