Prove that dual space of $\ell^1$ is $\ell^{\infty}$
My attempt : I got the answer Here but im not able to understand the answer
we know that the norm of $ x\in \ell^1$ is given by $||x||_1=\sum_{k=1}^{\infty}|a_k|$
norm of $ x\in \ell^{\infty}$ is given by $||x||_{\infty}=\sup_{k\in \mathbb{N}}|a_k|$
Now here my proof start :
Since $\ell^1$ is infinite dimensional because it contains the infinite sequence in the form $(0,0,\dots,1,0,\dots)$
So there exists a basis $\{e_1,e_2,\dots,e_k\dots\}$ of $\ell^1$ where $e_k=M_{jk}=\begin{cases} 1 &\text{ if } j=k \\ 0 & \text{ if } j \neq k. \end{cases}$
This implies that every $x \in \ell^1$ can be written as $x=a_1e_1+a_2e_2+\dots$
Now take a bounded linear functional $f$ of $\ell^1$
$f: \ell^1 \to \mathbb{R}$ defined by $f(x)= f(a_1e_1+a_2e_2+\dots)= a_1f(e_1)+a_2 f(e_2)+\dots=\sum_{k=1}^{\infty}a_kf(e_k)$
After that I am not able to proceed further..
Best Answer
Clearly, every element of $v\in\ell^\infty$ defines an element of the dual of $\ell^1$, since if $v=(v_j)$ and $x=(x_j)\in\ell^1$, then $$ v(x)=\sum_j v_jx_j\quad\text{and}\quad |v(x)|\le \sum_j |v_j||x_j|\le \big(\sup_j |v_j|\big)\sum_j|x_j|=\|v\|_\infty\|x\|_1 $$ Let $\varphi\in(\ell^1)^*$ and set $v_j=\varphi(e_j)$ and $v=(v_j)$. Clearly $$ |v_j|=|\varphi(e_j)|\le \|\varphi\|_*\|e_j\|_1=\|\varphi\|_* $$ and hence $v\in\ell^\infty$ and $\|v\|_\infty\le \|\varphi\|_\infty$. It remains to show that $\varphi(x)=v(x)$, for all $x\in\ell^1$ and $\|v\|_\infty= \|\varphi\|_*$.
Clearly, $\varphi(x)=v(x)$, for $x=e_j$ and for all $x$'s which are finite linear combinations of the $e_j$'s. They are also both bounded linear functionals, and they agree on a dense subset of $\ell^1$, and hence the agree everywhere, i.e., $v\equiv \varphi$.
For the final part, it remains to show that $\|v\|_\infty\ge\|\varphi\|_*$. Now, for every $\epsilon>0$, there exists a unit vector $w=(w_j)\in\ell^1$, such that $$ |\varphi(w)|>\|\varphi\|_*-\epsilon $$ and also there exists $n\in\mathbb N$, such that $\|w-w^n\|_1<\varepsilon$, where $w^n=(w_1,w_2,\ldots,w_n,0,0,\ldots)$ and $v(w^n)=\varphi(w^n)$, while $\|w^n\|_1\le 1$. So $$ \|v\|_\infty\ge |v(w^n)|=|\varphi(w^n)| \ge |\varphi(w)|- |\varphi(w-w^n)| \ge \|\varphi\|_*-\varepsilon-\|\varphi\|_*|w-w^n|_1 \\ \ge \|\varphi\|_*-\varepsilon-\varepsilon\|\varphi\|_*= \|\varphi\|_*-\epsilon(1+\|\varphi\|_*) $$ and this is true for all $\varepsilon>0$, which implies that $\|v\|_\infty\ge\|\varphi\|_*$.