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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.

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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 $\psi(f)=dist(f,c)>0$.

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  • $\begingroup$ It may be trivial but I am confused how we defined sign function for complex number. I know in case of real number. And could you show me how sign$(x)x=|x|$ when x is complex number. Thank you $\endgroup$ – Toeplitz Oct 21 '15 at 3:37
  • $\begingroup$ @Lusin For any complex number $x$, the sign function is defined as: $sgn(x)=|x|/x$ if $x\neq 0$ and $sgn(x)=0$ if $x=0$. $\endgroup$ – Frank Lu Oct 21 '15 at 3:41
  • $\begingroup$ Thus in either case we have $sgn(x)x=|x|$. $\endgroup$ – Frank Lu Oct 21 '15 at 3:44
  • $\begingroup$ en.wikipedia.org/wiki/Sign_function#Complex_signum, here i see sgn(z)=z/|z| $\endgroup$ – Toeplitz Oct 21 '15 at 3:56
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    $\begingroup$ @Lusin OK, I may remembered the definition of sign function incorrectly. But this is not a big deal, the key point here is to know that for any complex number $z$, we can always find a unique complex number $z'$ so that $z'z=|z|$. BTW, I've edit the answer, now it should look good. $\endgroup$ – Frank Lu Oct 21 '15 at 4:12
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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.

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  • $\begingroup$ You won't ever see a concrete one. You can only "construct" such elements using the axiom of choice in a serious way. If you only have dependent choice, it's consistent that $l^1$ is actually reflexive. $\endgroup$ – Nate Eldredge Oct 21 '15 at 4:02
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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 $\sum_{n=1}^\infty g(n) < \infty$.

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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.

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