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Recall that (for $1\le p<\infty$), $\ell^p = \{\{a_n\}_{n=1}^\infty:\sum\limits_{i=1}^\infty|a_i|\lt\infty\}$, with norm $||\{a_n\}||=(\sum\limits_{i=1}^\infty|a_i|^p)^{\frac{1}{p}}$.

It is well known that $(\ell^p)^*\cong\ell^q$ where $\frac{1}{q}+\frac{1}{p}=1$, and so $$(\ell^p)^{**}=(\ell^q)^*=\ell^p=(\ell^p)\oplus(0).$$

Note that for $\ell^2$ we have $(\ell^2)^*\cong{\ell^2}$, since $\ell^2$ is a Hilbert space.

For $\ell^\infty=\{\{a_n\}: \sup |a_n| \lt \infty\}$, we have $$ (\ell^\infty)^* \cong \ell^1\oplus {\rm Null}(C_0),$$ and $$ (\ell^\infty)^{**}=({\ell}^1)^*\oplus\operatorname{Null}(C_0)^*,$$ but $(\ell^1)^\ast=\ell^\infty$, hence $$ (\ell^\infty)^{**} = \ell^\infty \oplus{\rm Null} (C_0)^*. $$

Seeing this pattern, is it true that the double dual of any space $X$ can be written in the form of $X\oplus Y$ for some other space $Y$?

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closed as too localized by Andrés Caicedo, Yemon Choi, Gerald Edgar, Bill Johnson, Alain Valette Sep 9 '12 at 20:49

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What category does $X$ live in? Is it a Banach space, a topological vector space, or any vector space? – Alex Becker Sep 8 '12 at 23:54
If you want a topological decomposition in your final question, the answer is "no" - the double dual of $c_0$ is $\ell^\infty$, but you cannot write $\ell^\infty$ as $c_0\oplus E$ for any closed subspace $E$. – Yemon Choi Sep 9 '12 at 0:17
That said, I think this question would have been better suited to – Yemon Choi Sep 9 '12 at 0:17
@Yemon I now see why I should refresh my window more often... – Alex Becker Sep 9 '12 at 0:47
Since the question has an answer with upvotes, it will likely not be deleted, so I did one final pass at editing it to fix the most painful typos. – Andrés Caicedo Sep 9 '12 at 23:24

1 Answer 1

up vote 2 down vote accepted

Note that there exists such a $Y$ iff the sequence $$0\to X\overset{\varphi}{\to} (X^\*)^\*\overset{\eta}{\to} \mathrm{coker}\:\varphi\to 0$$ splits, where $\varphi$ is the canonical injection and $\eta$ the canonical projection. This happens iff there exists an injection $j:\mathrm{coker}\:\varphi\to (X^\*)^\*$ such that $\varphi(X)\oplus j(\mathrm{coker}\:\varphi)=(X^\*)^\*$, so can only happen if $\varphi(X)$ is split in $(X^\*)^\*$.

For a counterexample, consider the space $c_0$ of sequences in $\mathbb C$ which converge to $0$. The double dual of $c_0$ is $\ell^\infty$, yet $c_0$ does not split as a subspace of $\ell^\infty$. See for reference page 46 of this PDF.

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Thanks, would you please explain more why there exists such a Y iff the sequence 0$\to$ X$\overset{\varphi}{\to}$ $(X^*)^*$$\overset{\eta}{\to}$ $\mathrm{coker}\:\varphi\to$ 0 splits? – student Nov 12 '12 at 9:30
@student If it splits, then we have $(X^*)^*\cong X\times \mathrm{coker} \phi$. If we have $(X^*)^*\cong X\times E$ then the canonical injection must take $X$ to $X\times \{1\}$ and so the sequence splits. – Alex Becker Nov 23 '12 at 20:50
Oops, I meant $X\times \{0\}$. – Alex Becker Nov 23 '12 at 20:51

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