$ l^1$ not reflexive

Question

Define $ \phi:l^1\to (l^\infty)^* $ by $$ \phi(a_0,a_1,a_2...)(b_0,b_1,b_2...)=\sum_{n=0}^\infty a_nb_n.$$ Prove that $\phi$ is not onto.

Here, $(a_0,a_1,a_2...) \in l^1 $ and $(b_0,b_1,b_2...)\in l^\infty$

Clearly $\phi$ is bounded linear map which could be shown very easily by taking out sup of $(b_0,b_1,b_2...)$. But i don't have any idea to show it is not onto. Anyone help would be appreciated.

Answer 1

I never did see an example of a concrete element of $(l^{\infty})^{*}$ which is not in the range of the natural embedding you wrote above. However, you can prove that $l_1$ is not reflexive indirectly as follows:

1. $l^1$ is isometric to $c_0^{*}$, where $c_0$ is the Banach space of sequences tending to zero with the $\sup$ norm.

2. $c_0$ is not reflexive because it's unit ball has no extreme points, and in a reflexive Banach space the unit ball is weakly-compact, so by the Krein-Milman theorem, it is the norm-closure of the convex hull of it's extreme points.

3. A Banach space $X$ is reflexive if and only it's dual is reflexive. So by 2, $c_0^{*}$ is not reflexive.

It follows from 1,2,3 that $l_1$ is not reflexive.

Answer 2

Let $c$ denotes the space of convergent sequences, then $c$ is a closed sub-space of $\ell^\infty$.

Let $f=(f(1),f(2),\dots)\in \ell^\infty\setminus c$, then $$dist(f, c):=\inf_{g\in c}\|f-g\|_\infty>0$$

Hence by Hahn-Banach extension theorem, there exists $\psi\in(\ell^\infty)^*$ such that $$\|\psi\|=1,\quad \psi(g)=0 \quad\forall g\in c\text{ and }\psi(f)=dist(f,c)>0$$

Now we claim that $\psi$ is not in the image of $\phi$, that is, we show that there is no $a\in \ell^1$ satisfying $$\psi(b)=\phi(a)(b),\quad \forall b\in\ell^\infty$$

Assume not, then such an $a\in\ell^1$ exists. Let $(h_k)_{k=0}^\infty$ be a sequence of elements in $c$ defined as $$h_k(n)=\begin{cases}s(a(n)), &n\leq k\\ 0, &n>k\end{cases}$$ where the function $s$ is defined as $$s(z)=\begin{cases}\frac{|z|}{z} &z\in\mathbb{C}\setminus\{0\}\\ 0, &z=0\end{cases}$$ Then for each $k\in\mathbb{N}$ we have $$\sum_{n=0}^k|a_n|=\sum_{n=0}^\infty a(n) h_k(n)=\phi(a)(h_k)=\psi(h_k)=0$$ as $h_k\in c$. Thus we have $$\|a\|_1=\lim_{k\to\infty}\sum_{n=0}^k|a_n|=\lim_{k\to\infty}\psi(h_k)=0$$

Hence $a=0$,and thus $\psi=\phi(a)$ is the zero operator, which contradicts $\psi(f)=dist(f,c)>0$.

Answer 3

Another fun approach:

Let $e_n = (0, \dots, 0, 1, 0, \dots)$ be the element of $l^1$ which has 1 in the $n$th position and 0 elsewhere. If you view these as elements of $(\ell^\infty)^*$, the Banach-Alaoglu theorem says they have a weak-* cluster point; call it $g$. Now let $s_k = (0, 0, \dots, 0, 1, 1, \dots)$ be the element of $\ell^\infty$ which has 0s up to position $k$ and $1$s from then on. Since $\langle e_n, s_k \rangle = 1$ for all $n \ge k$, we must have $\langle g, s_k \rangle =1$ for all $k$. If $g$ is an element of $\ell^1$, this means that $\langle g, s_k \rangle = \sum_{n=k}^\infty g(n) = 1$ for every $k$, which is absurd since we were supposed to have $\sum_{n=1}^\infty g(n) < \infty$.

Answer 4

Show that the non-separable space $\ell_\infty$ embeds isometrically into $(\ell_1)^*$ (actually it is the whole of $(\ell_1)^*$ but you do not need it). Since the dual of $\ell_1$ is non-separable, so is the bidual. Thus, $\ell_1$ and $(\ell_1)^{**}$ are not isomorphic.

Answer 5

Yet another way to do this:

Take $y^*\in (\ell^1)^*\approx \ell^\infty$. By Hahn Banach, there is an element $\xi\in (\ell^1)^{**}$ such that $\xi(y^*)=\lVert y^*\rVert$ and $\lVert \xi \rVert=1$.

Suppose by way of contradiction that $\ell^1$ is reflexive. In this case, we have that $\xi=J(x)$ and so for every $y^*\in (\ell^1)^*$, we have that there exists $x$ such that $J(x)(f)=f(x)=\lVert f \rVert$ and $\lVert x \rVert=\lVert J(x) \rVert=1$.

We will take a certain $f\in (\ell^1)^*$ which fails to meet this property. Take $(1-1/2^j)_{j\in \mathbb{N}}\in \ell^\infty$ and the isometrically correspondent $f\in (\ell^1)^*$. We have $\lVert f \rVert=1$, but for every $x$ with $\lVert x\rVert=1$:

$$ f(x)=\sum_{j=1}^\infty x_j\left(1-\frac{1}{2^j}\right)$$

Because there is a certain non zero $x_j$, we have that $|x_j|(1-1/2^j)<|x_j|$ and thus we have:

$$|f(x)|\leq \sum_{j=1}^\infty |x_j|\left(1-\frac{1}{2^j}\right)<\sum_{j=1}^\infty |x_j|=\lVert x\rVert=1$$

But this strict inequality contradicts the reflexivity of $\ell^1$.