$\ell^p$ = $\{${$\{a_n\}_1^n$:$\sum\limits_{i=1}^\infty$ $|a_n|$ $\textless$ $\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 $(\ell^p)^{**}$=$(\ell^q)^*=(\ell^p)$=$(\ell^p)\oplus\emptyset$

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 $\ell^\infty$=$\{${$\{a_n\}_1^n$: sup $|a_n|$ $\textless$ $\infty$$\}$

${{\ell}^\infty}^*$$\cong$${\ell}^1$$\oplus$Null$C_0$ ${{\ell}^\infty}^{**}$=${{\ell}^1}^*$${\oplus}(NullC_0)^*$, but ${{\ell}^1}^*$=$\ell^\infty$ hence ${{\ell}^\infty}^{**}$=${{\ell}^\infty}$${\oplus}(NullC_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?