Dtyjlui

I am working on the dual spaces of sequence spaces, and I want to show that the map $$
Phi:ell^1to(ell^infty)',qquad(Phi y)(x)=sum_{iinmathbb{N}}y_ix_i
$$

is not surjective. I have already shown it is a linear isometry. Can I use the Hahn-Banach theorem to find $y'in (ell^infty)'$ such that there is no $yinell^1$ with $Phi y=y'$.

I have written my own answer the following way (corollary 4.14 is a corollary in my lecture notes stating that $X$ is seperable if $X'$ is. I have proven earlier that $ell^infty$ is inseperable)

Proposition 2: The map $$Phi_infty:ell^1to(ell^infty)',qquad(Phi_infty y)(x)=sum_{iinmathbb{N}}x^iy^i$$ is not surjective.

The proof is based on the following lemma, which consists of two parts.
Lemma 1. Let $X$ and $Y$ be normed spaces. Then the following claims are true:

1. If $f:Xto Y$ is a surjective linear isometry, then $f$ is a homeomorphism.




2. Let $f: Xto Y$ be a homeomorphism. If $X$ is a seperable space, then $Y$ is a seperable space.

Proof of 1: We remark that every isometry is automatically injective. Since $f$ is also surjective, $f$ is a bijection. Now, we calculate
$|f|_{op}=sup{||f(x)||_Y,big|,||x||_Xleq 1}=sup{||x||_X,big|,||x||_Xleq 1 }=1<infty$, hence $f$ is bounded. Since $f$ is linear, $f$ is continuous. We will now show that $f^{-1}$ is continuous. We remark that $f^{-1}$ is linear. We also remark that for all $yin Y$, $||y||_Y=||f(f^{-1}(y))||_Y=||f^{-1}(y)||_X$, hence $f^{-1}$ is an isometry, too. Then, $|f^{-1}|_{op}=sup{||f^{-1}(y)||_X,big|,||y||_Yleq 1}=sup{||y||_Y,big|,||y||_Yleq 1}=1<infty$, hence $f^{-1}$ is bounded. It follows that $f^{-1}$ is continuous, hence $f$ is a homeomorphism.

Proof of 2: Let $X$ be a seperable normed space and let $f:Xto Y$ be a homeomorphism. Since $X$ is seperable, there exists a countable dense subset $Asubseteq X$. Then, $f(A)$ is a countable subset of $Y$. We will show that $f(A)$ is dense in $Y$.

Let $V$ be an open set in $Y$. By continuity of $f$, $f^{-1}(V)$ is open in $X$, so $Acap f^{-1}(V)neqemptyset$. By bijectivity, we see that
$emptysetneq f(Acap f^{-1}(V))=f(A)cap f(f^{-1}(V))=f(A)cap V$. This holds for all open sets in $Y$, hence $f(A)$ is dense in $Y$. It follows that $Y$ is seperable.

We can now prove proposition 2.

By theorem 4.6, $Phi_infty$ is a well-defined linear isometry. We remark that $ell^1$ is seperable. It follows by corollary 4.14 that $(ell^infty)'$ is inseperable since $ell^infty$ is inseperable.

We will give a proof by contradiction. Suppose that $Phi_infty$ is surjective. Then, $Phi_infty$ is a surjective linear isometry and by lemma 1 part textit{i}, a homeomorphism. Since $ell^{1}$ is seperable, it follows by lemma 1 part textit{ii} that $Phi_infty(ell^{1})=(ell^infty)'$ is seperable, but this is a contradiction since we know by corollary 4.14 that $(ell^infty)'$ is inseperable. Therefore, our assumption that $Phi_infty$ was surjective is false, hence $Phi_infty$ is not surjective

One way of showing this is to use the fact that If $X^{*}$ is separable then so is $X$. Take $X=ell^{infty}$. If the given map is surjective then $X^{*}$ is isometrically isomorphic to $ell^{1}$ which makes it separable. But $X=ell^{infty}$ is not separable.

Here is a different approach. Consider the subspace
$$c := { x in ell^infty mid lim_{ntoinfty}x_n text{ exists}}.$$

Can you imagine a functional on $c$ which (if extended to all of $ell^infty$ via Hahn-Banach) is not of the form $Phi y$?

Similarly to gerw's answer. Just use that, as a $C^ast$-algebra $ell^infty(mathbb N)$ is isomorphic to $C(beta mathbb{N})$, the continuous function over the compact space given by $beta mathbb{N}$, the Stone-Cech compactification of $mathbb{N}$. By Riesz theorem $(ell^infty)^ast = M(beta mathbb{N})$, the finite Radon measures over $beta mathbb{N}$. The points in $beta mathbb{N} setminus mathbb{N}$ are proper ultrafiters and evaluating on them gives functionals that are not in the image of $ell^1$.