Identification of $\ell_1^n$ $(\ell_\infty^n)$ with $\ell_\infty^{n^*}$ $(\ell_1^{n^*})$.

Let $X=\ell_1^n$ or $\ell_\infty^n$. Then any member of $\ell_1^n$ ($\ell_\infty^n$) can be identified as a functional over $\ell_\infty^n$ ($\ell_1^n$) via the canonical isometric isomorphism $\psi$. What does this identification look like? For example, if $x=(a_1,a_2,\dots,a_n)\in \ell_1^n$ (or $\ell_\infty^n$) and $\psi(x)=f\in X^{**}$, then what is the action of $f$ on a member of $\ell_\infty^n$ or ($\ell_1^n$)? In other words,

$$ f(u_1,u_2, \dots, u_n)=? \qquad \forall ~(u_1,u_2, \dots, u_n)\in \ell_\infty^n ~or~ (\ell_1^n).$$

Any help will be appreciated. Please refer some credible source!

Solution 1:

It is well-known that $(\ell^p)^* \cong \ell^{p'}$ where $\frac{1}{p} + \frac{1}{p'}=1$ if $p \in [1,\infty)$ by $T: \ell^{p'} \to (\ell^p)^*$, namely, $a=(a_n)_{n=1}^\infty \mapsto T_a$; where $T_a(x):= \sum_{n=1}^\infty a_n x_n, \quad \forall x = (x_n)_{n=1}^\infty \in \ell^p$.

Solution 2:

Your notation is somewhat confusing. Correct me if I am wrong, but I think with $\ell^n_1$ and $\ell^n_\infty$ you mean the space of sequences of length $n$ equipped with the $1$-norm and the max-norm, respectively? In other words, you consider the space $\mathbb{R}^n$ (assuming the sequences consist of real numbers) equipped either with norm $\|x\|_1=\sum_{i=1}^n|x_i|$ or with norm $\|x\|_\infty=\max\{x_1,\ldots,x_n\}$, where $x\in\mathbb{R}^n$. It is well-known that these spaces are equivalent as metric spaces (that is, they are equipped with the same topology). The notation $\ell^p(X)$ for the space of function on some set $X$ taking values in $\mathbb{R}$ (or $\mathbb{C}$) and bounded with respect to the norm $\|f\|_p:=\left(\sum_{x\in X}|f(x)|^p\right)^{1/p}$, $1\leq p <\infty$, or $\|f\|_\infty=\max\{|f(x)|,\ x\in X\}$ if $p=\infty$, is very common. In view of this, it feels more natural to use the notation $\ell^1_n$ ($\ell^\infty_n$) instead, but I will stick to your notation in this response.

Now, the canonical identification $\psi:X\to X^{**}$ for general Banach space $X$ is given by $[\psi(x)](\Lambda)=\Lambda(x)$ for any $x\in X$ and $\Lambda\in X^*$. In your case $(\ell^n_1)^*$ (or $(\ell^n_\infty)^*$) are equipped with another natural mapping; they can be identified with $\ell^n_\infty$ (or $\ell^n_1$, respectively) via the map $$ \ell^n_\infty\to (\ell^n_1)^*,\quad y\mapsto \Lambda_y, $$ where $\Lambda_y(x):=\sum_{i=1}^nx_iy_i$ for all $x\in \ell^n_1$. (The map $\ell^n_1\to (\ell^n_\infty)^*$ is defined exactly the same.) We do rely on the fact that we are working with a finite-dimensional space $X$. For example, the map $\ell^1(\mathbb{N})\to(\ell^\infty(\mathbb{N}))^*$ is not surjective and therefore cannot be an identification of spaces.

Finally, any map of Banach spaces $T:X\to Y$ induces the adjoint map $T^*:Y^{*}\to X^*$ via $[T^*(\Lambda)](x):=\Lambda(T(x))$ for any $x\in X$ and $\Lambda\in Y^*$. Returning to your case, the identification $\ell^n_\infty\to (\ell^n_1)^*$ induces the map $(\ell^n_1)^{**}\to (\ell^n_\infty)^*$. Composing this map with the inverse of the other identification map $\ell^n_1\to (\ell^n_\infty)^*$, we get the map $$ (\ell^n_1)^{**}\to \ell^n_1. $$ It should be straightforward to check that this map coincides with (the inverse of) the canonical map $\psi$ above; just plug in all the definitions.

This should answer your question as well, but let me write it down explicitly. Let $x=(a_1,\ldots,a_n)\in \ell^n_1$ and consider the image of $x$ under the canonical map $\psi$, $\psi(x)\in (\ell^n_1)^{**}$. We identify $\psi(x)\in(\ell^n_1)^{**}$ with $f\in(\ell^n_\infty)^*$ (again, via the adjoint of the identification map $\ell^n_\infty\to (\ell^n_1)^*$, $y\mapsto \Lambda_y$). Then, $$ f(u_1,\ldots,u_n)=\sum_{i=1}^n a_iu_i $$ for all $u\in \ell^n_1$.