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$\ell^p = \{\{a_n\}_1^n:\sum\limits_{i=1}^\infty|a_n|\lt\infty\}$ 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?

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 $Y$?

$\ell^p$ = $\{${$\{a_n\}_1^n$:$\sum\limits_{i=1}^\infty$ $|a_n|$ $\textless$ $\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 $\ell^\infty=\{\{a_n\}: \sup |a_n| \lt \infty\}$

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

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?

$\ell^p = \{\{a_n\}_1^n:\sum\limits_{i=1}^\infty|a_n|\lt\infty\}$ 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?