0
$\begingroup$

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 $Y$?

Improve this question
$\endgroup$
7
  • $\begingroup$ What category does $X$ live in? Is it a Banach space, a topological vector space, or any vector space? $\endgroup$
    – Alex Becker
    Commented Sep 8, 2012 at 23:54
  • $\begingroup$ 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 $E$. $\endgroup$
    – Yemon Choi
    Commented Sep 9, 2012 at 0:17
  • $\begingroup$ That said, I think this question would have been better suited to math.stackexchange.com $\endgroup$
    – Yemon Choi
    Commented Sep 9, 2012 at 0:17
  • 1
    $\begingroup$ @Yemon I now see why I should refresh my window more often... $\endgroup$
    – Alex Becker
    Commented Sep 9, 2012 at 0:47
  • 1
    $\begingroup$ 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. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 9, 2012 at 23:24

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – student
    Commented Nov 12, 2012 at 9:30
  • $\begingroup$ @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. $\endgroup$
    – Alex Becker
    Commented Nov 23, 2012 at 20:50
  • $\begingroup$ Oops, I meant $X\times \{0\}$. $\endgroup$
    – Alex Becker
    Commented Nov 23, 2012 at 20:51

Not the answer you're looking for? Browse other questions tagged .