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I've started taking my first course in multivariable analysis and in our notes there is an exercise about the dual space of the sequence space $\ell_1$ and I'm unsure how to prove that it is isomorphic to $\ell_{\infty}$ as in class we haven't gone over the dual space in detail.

If somebody could also give me some extra intuition about dual spaces that would be much appreciated :)

Martin Sleziak
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Matt
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  • Maybe this can help: http://math.stackexchange.com/questions/1729002/show-there-is-a-continuous-isomorphism-l-infty-rightarrow-leftl-1-right?rq=1 – User3773 Jan 26 '17 at 10:29
  • Let $\phi$ be a continuous linear operator $l^1 \to \mathbb{R}$, show that $\phi(a) = \sum_m a_m \phi_m$ where $\phi_m = \phi(\delta_m)$, then $(\phi_m) \in l^\infty \Leftrightarrow \phi$ is continuous, and the operator norm of $\phi$ is $|(\phi_m)|_\infty$ – reuns Jan 26 '17 at 10:41

1 Answers1

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The dual space consists of all continuous linear functionals on the space. What continuity buys you is that, if you can determine the functionals on a dense subspace $M$, you can bootstrap to the full space by continuity. Continuity of a linear function $F$ on the normed space assures that $F$ is determined by its values on the dense subspace $M$.

For example, because $\ell^1$ has a basis of sequences $$ e_0 = \{ 1, 0, 0, 0, \cdots \} \\ e_1 = \{ 0, 1, 0, 0, \cdots \} \\ e_2 = \{ 0, 0, 1, 0, \cdots \} \\ \vdots $$ then there is a natural way to bootstrap. Specifically, if $x=\{ \alpha_k \}\in\ell^1$, then $$ \left\| x - \sum_{n=0}^{N}\alpha_n e_n \right\|_{\ell^1} \le \sum_{k=N+1}^{\infty}|\alpha_n|\rightarrow 0 \mbox{ as } N\rightarrow\infty. $$ Therefore, by continuity, if $F$ is a continuous linear functional on $\ell^1$, then $$ F(x) = \lim_{N\rightarrow\infty}F\left(\sum_{k=0}^{N}\alpha_k e_k \right) = \lim_{N\rightarrow\infty} \sum_{k=0}^{N}\alpha_k F(e_k) = \sum_{k=0}^{\infty}\alpha_k F(e_k). $$ So, already you have a representation of any such linear functional. The sum on the right is guaranteed to converge. In fact it must converge absolutely because you can choose unimodular constants $u_k$ such that $u_k\alpha_k F(e_k)=|\alpha_k||F(e_k)|$ and apply the above to $\{u_k\alpha_k\}$ instead. Continuity of $F$ is equivalent to boundedness, meaning the existence of a smallest non-negative constant $\|F\|$ such that $|F(x)| \le \|F\|\|x\|_{\ell^1}$ for all $x\in\ell^1$. From this, you get $\{ F(e_k) \}\in\ell^{\infty}$ and $\|\{ F(e_k)\}\|_{\ell^{\infty}} \le \|F\|$. Conversely, $$|F(\{\alpha_k\})| \le \|\{F(e_k)\}\|_{\infty}\|\{\alpha_k\}\|_{\ell^1}$$ gives $\|F\|\le\|\{F(e_k)\}\|_{\infty}$ because $\|F\|$ is the smallest such constant. Hence, $$\|F\|=\|\{F(e_k)\}\|_{\ell^{\infty}}.$$ So there is an isometric linear correspondence between $(\ell^{1})^*$ and $\ell^{\infty}$.

Paolo Leonetti
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Disintegrating By Parts
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    I think it may be better if we express the isometric isomorphism explicitly. i.e. Write $\varphi : \ell^{\infty}\longrightarrow \ell^{1}$, $\varphi(g)=F$, and $g:(={g_k})$ is chosen to correspond to any fixed $F\in\ell^{1}$, where $\varphi_g({\alpha_k})=\sum_{k=0}^{\infty}\alpha_k F(e_k)=\sum_{k=0}^{\infty}\alpha_kg_k$, and ${\alpha_k},{e_k}\in\ell^1$, ${e_k}$ being the basis of $\ell^1$. – Sam Wong May 18 '18 at 04:37