# Prove that dual space of $\ell^1$ is $\ell^{\infty}$

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..

• Welcome to Math.SE! I have tried to improve the readability of your question by improving the $\rm \LaTeX$ code. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. Dec 23 '20 at 12:37
• This is a good start. Comments... (1) You say "every $x \in \ell^1$ can be written as $x=a_1e_1+a_2e_2+\dots$". Needs proof. (2) You say "$f(a_1e_1+a_2e_2+\dots)= a_1f(e_1)+a_2 f(e_2)+\dots$". Needs proof. (3) Now define your proposed bijection, $f$ maps to the sequence $(f(e_1),f(e_2),\dots))$. Then show its values are in $\ell^\infty$. Show it is injective. Show it is surjective. Dec 23 '20 at 12:46
• $\{e_1,e_2,\ldots,e_n,\ldots\}$ is not a basis of $\ell^1$, since for example, $$v=\left(1,\frac{1}{4},\frac{1}{9},\ldots,\frac{1}{n^2},\ldots\right)\in\ell^1$$ is not a finite linear combination of the $e_j$'s. Dec 23 '20 at 12:51
• @YiorgosS.Smyrlis Words mean different things in different contexts... Dec 23 '20 at 13:02
• $\{e_1,e_2,\ldots,e_n,\ldots\}$ is not a Hamel basis, but it is a Schauder basis. Dec 23 '20 at 13:03

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 $$\|v\|_\infty\ge\|\varphi\|_*$$.