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Let $E=l^p$ where $1 \le p < \infty $ we know $E'=l^q$ Where $q$ is the dual exponent of $p$, i.e. $q$ is such that $\frac{1}{p}+\frac{1}{q}=1$

Does this hold for $p=\infty $, i.e., is it true that $(l^{\infty})'= l^1? $

And what is the $(l^{\infty})'= ?$

2 Answers2

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Another proof. More elementary than Urban's proof. (But less general.)

Consider the subspace $c \subseteq l^\infty$ consisting of the real sequences that converge. So $c$ is a closed linear subspace of $l^\infty$. Let $L \;:\; c \to \mathbb R$ be the "limit" linear functional. That is, for $x = (x_1,x_2,\cdots)$, define $$ L(x) := \lim_{n\to\infty} x_n. $$
Then $L$ is a bounded linear functional on $c$, $L \in c'$. By the Hahn-Banach theorem, there is an extension $\widetilde{L}$ of $L$ with $\widetilde{L} \in (l^\infty)'$.

But we claim $\widetilde{L}$ is not the linear functional arising from any element of $l^1$. Indeed, if $y=(y_1,y_2,\dots) \in l^1$ what would it mean for $\langle y, x\rangle = \widetilde{L}(x)$ for all $x \in l^\infty$? Well, fix an $n$ and let $x=e_n$, the sequence with $1$ in the $n$th spot, zero elsewhere. We have $e_n \in c$ and $L(e_n)=0$, so $$ y_n = \langle y, e_n\rangle = \widetilde{L}(x) = L(x) = 0 $$ This holds for all $n$, so $y$ is the zero element of $l^1$.

On the other hand, for $x=u$ the sequence of all ones, $u=(1,1,1,\cdots)$, we have $u \in c$ and $L(u)=1$, so $$ 1 = L(u) = \widetilde{L}(u) = \langle y,u \rangle =\sum_{n=1}^\infty y_n \cdot 1 = 0 $$ Impossible.

GEdgar
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This is not true because of the following Theorem

$\textbf{Theorem:}$ If $X$ is a normed space such that its dual $X'$ is separable, then $X$ itself is separable.

So, if $(l^{\infty})' =l_1$, then it will follow that $l^{\infty}$ is separable which is not true.

Przemysław Scherwentke
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Urban PENDU
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  • Just this is the proof ???? –  May 01 '15 at 14:23
  • Yes!!!! and also the theorem I quoted is somewhat standard...and as far as I knw the dual space of $l^{\infty}$ is a nasty space. – Urban PENDU May 01 '15 at 14:24
  • It seems that the dual of $l^\infty$ is a nasty space if you take the axiom of choice. Without the axiom of choice, it can be more simple (kinda like every set is measurable for the Lebesgue measure without the axiom of choice, so the situation is more simple) – Tryss May 01 '15 at 14:38
  • @Tryss The axiom "every subset of the line is Lebesgue measurable" is stronger than just the negation of the axiom of choice. Our usual examples use choice, but there are more complicated examples that can be proven to exist using weaker axioms. To my understanding the situation is similar when dealing with the dual space of non-separable spaces under the negation of choice. – Ian May 01 '15 at 15:02
  • @Ian wiki said (on the Lp page) "However, Saharon Shelah proved that there are relatively consistent extensions of Zermelo-Fraenkel set theory (ZF + DC + "Every subset of the real numbers has the Baire property") in which the dual of ℓ∞ is ℓ1.[5]" [5] Schechter, Eric (1997), Handbook of Analysis and its Foundations, London: Academic Press Inc. See Sections 14.77 and 27.44--47 Yeah, things are strange without a strong AC – Tryss May 01 '15 at 15:07