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There does not exist $f\in (l^\infty)^*$ with $\ker f = c_0$.

$c_0$ is the closed subspace of $l^\infty$ with the property that if $x = (x_1, x_2,...) \in c_0$ then $$\lim_{n\rightarrow \infty }x_n = 0.$$

My attempt: Now suppose $f\in (l^\infty)^*$ and $\ker f = c_0$. For each $x\in c_0$, we have $$f(x) = 0.$$ Since the dual space of $c_0$ is $l^1$, we can identify $f\big|_{c_0}$ with the $0$ element in $l^1$.

Also I know that the natural map from $l^1$ to $(l^1)^{**} = (l^\infty)^*$ is isometric, can we conclude that $f$ has to be the zero functional in $(l^\infty)^*$? So that the $\ker f$ has to be the entire space.

Thank you very much!

Xiao
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    But there are nonzero continuous linear forms on $l^\infty$ that vanish on $c_0$. Try with the codimension of $\ker f$. – Daniel Fischer Aug 08 '14 at 11:19
  • @DanielFischer could you point me to some reference for codimension in infinite dimensional normed vector spaces. I have only studied it in the setting of linear algebra and abstract algebra. Thank you very much! – Xiao Aug 08 '14 at 11:24
  • @DanielFischer There is a nonzero continuous linear form on $\ell^{\infty}$ that vanish on $c_0$. – Hamou Aug 08 '14 at 11:25
  • @Hamou That's what I wrote (except I used the plural). Are you asking for an example, or did you think I wrote the opposite? – Daniel Fischer Aug 08 '14 at 11:35
  • @Xiao I can't point to a specific reference, but if $F\subset E$ is a linear subspace, the codimension of $F$ in $E$ is $\operatorname{codim}_E F = \dim (E/F)$. Since every linear map $T \colon E \to G$ with $F\subset \ker T$ induces an injective map $\hat{T} \colon (E/F) \to G$, what can you say about the codimension of $\ker f$ if $f$ is a linear functional? – Daniel Fischer Aug 08 '14 at 11:38
  • why $\hat{T} $ is injective? if $F=0$. did you mean surjective – Hamou Aug 08 '14 at 11:44
  • @Hamou Oops, I thought of $F = \ker T$, not $F\subset \ker T$. Thanks for the correction. (But it would be good to @-ping in such cases, if I hadn't happened by again, I wouldn't have noticed the mistake.) – Daniel Fischer Aug 08 '14 at 12:23

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Assume such functional exists. Kernel of any functional is ofcodimension $1$, hence $\ell_\infty=c_0\oplus\mathbb{C}$. Note that $c_0$ is separable, so does $c_0\oplus\mathbb{C}$. Thus $\ell_\infty$ is separable. Contradiction, because $\ell_\infty$ is not separable.

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Norbert
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