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 $Y$?