# Can I conclude that $F\in\ell^1$?

Let $\ell^\infty$ be the Banach space of real bounded sequences with its usual norm and $S\subset\ell^\infty$ be the space of convergent sequences. Define $f:S\rightarrow\mathbb{R}$ by $$f(x)=\lim_{n\rightarrow\infty}x_n$$

where $x=(x_1,x_2,...,x_n,...)$. Extend $f$ to a bounded linear functional $F:\ell^\infty\rightarrow\mathbb{R}$ by using Hahn-Banach Theorem. Does $F$ belongs to $\ell^1$?

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If $F$ was in $\ell^1$, we could write $F(x)=\sum_{n=0}^{+\infty}a_nx_n$, where $\{a_n\}\in\ell^1$ ($a_n:=F(e_n)$, where $e_n$ is the sequence whose only non-vanishing term is the $n$-th which is $1$). Since $e_n\in S$, we have $F(e_n)=f(e_n)=0$ so $F$ is identically $0$, which is absurd.

Note that it doesn't depend on the choice of the extension. You could be interested by Banach limits.

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What is $e_0$ in you notation? – Tomás Dec 28 '12 at 15:16
The sequence $(1,0\dots,0,\dots)$ – Davide Giraudo Dec 28 '12 at 15:17