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 $Y$?



Double duals characteristic
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](http://matematicas.univalle.edu.co/~juancamg/Banach2.pdf).


