Double duals characteristic

2012-09-08T23:41:23Z

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 $Y$?

Answer by Alex Becker for Double duals characteristic

2012-09-09T00:46:46Z

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.

Double duals characteristic - MathOverflow [closed]

Double duals characteristic

Answer by Alex Becker for Double duals characteristic