$\ell^p = \{\{a_n\}_1^n:\sum\limits_{i=1}^\infty|a_n|\lt\infty\}$ and $||\ell^p||=(\sum\limits_{i=1}^\infty|a_n|^p)^{\frac{1}{P}}$

$(\ell^p)^*\cong\ell^q$ s.t $\frac{1}{q}+\frac{1}{p}=1$

The double dual of $\ell^p$ is $(\ell^q)^*=(\ell^p)=(\ell^p)\oplus\emptyset$

Note that for $\ell^2$, ${\ell^2}^*\cong{\ell^2}$ this is because $\ell^2$ is a Hilbert space. A Hilbert Space ,$\mathcal{H}$, is a vector space over $\mathbb{C}$ with an inner product such that $\mathcal{H}$ is complete in the metric d(x,y)=$||x-y||$=${\langle x-y,x-y\rangle}^{\frac{1}{2}}$

And for $\ell^\infty=\{\{a_n\}: \sup |a_n| \lt \infty\}$

$$ (\ell^\infty)^* \cong \ell^1\oplus {\rm Null}C_0 (\ell^\infty)^{**}=({\ell}^1)^*\oplus\operatorname{Null}C_0)^*$$

but $(\ell^1)^\ast=\ell^\infty$ hence $$ (\ell^\infty)^{**} = \ell^\infty \oplus({\rm Null} C_0)^* $$

Seeing this pattern is the double dual of any space X can be written in the form of $X\oplus Y s.t Y is any other space?