# Proof that the following map $\Phi:\ell^1\to(\ell^\infty)'$ is not surjective

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

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 $$y\in\ell^1$$ 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 $$\Phi_\infty:\ell^1\to(\ell^\infty)',\qquad(\Phi_\infty y)(x)=\sum_{i\in\mathbb{N}}x^iy^i$$ 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 $$f:X\to Y$$ is a surjective linear isometry, then $$f$$ is a homeomorphism.

2. Let $$f: X\to Y$$ 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 $$|f|_{op}=\sup\{||f(x)||_Y\,\big|\,||x||_X\leq 1\}=\sup\{||x||_X\,\big|\,||x||_X\leq 1 \}=1<\infty$$, 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 $$y\in Y$$, $$||y||_Y=||f(f^{-1}(y))||_Y=||f^{-1}(y)||_X$$, hence $$f^{-1}$$ is an isometry, too. Then, $$|f^{-1}|_{op}=\sup\{||f^{-1}(y)||_X\,\big|\,||y||_Y\leq 1\}=\sup\{||y||_Y\,\big|\,||y||_Y\leq 1\}=1<\infty$$, 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 $$f:X\to Y$$ be a homeomorphism. Since $$X$$ is seperable, there exists a countable dense subset $$A\subseteq X$$. 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 $$A\cap f^{-1}(V)\neq\emptyset$$. By bijectivity, we see that $$\emptyset\neq f(A\cap f^{-1}(V))=f(A)\cap f(f^{-1}(V))=f(A)\cap V$$. 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

• The standard approach here is to prove that $\ell^1$ is separable, whereas $\ell^\infty$ and $(\ell^\infty)^*$ are not. There is a very over-the-top argument where you show that the multiplicative linear functionals on $\ell^\infty$ coming from $\ell^1$ are just evaluations at a point, but there must be more functionals because $\mathbb N$ is not compact, but the pure states of a von Neumann algebra must be compact in the weak star topology. – Ashwin Trisal Jan 14 at 7:48
• Related: Dual of $l^\infty$ is not $l^1$ – Martin Sleziak Jan 19 at 1:34

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.

• I have already proven that $X$ is seperable if $X'$ is seperable, sosince $\ell^\infty$ is not seperable $(\ell^\infty)^*$ is not seperable. – user408856 Jan 14 at 7:54
• Is an isometric isomorphism naturally a homeomorphism? – user408856 Jan 14 at 8:23
• @James Of course, it is. – Kavi Rama Murthy Jan 14 at 8:24

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

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

• I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^\infty)^*$ is not separable is much more than showing it is not equal to $l^1$. – GEdgar Jan 14 at 11:58
• Which functional is it? – user408856 Jan 14 at 12:12
• The functional is the one assumed to exist in the definition of $c$. – GEdgar Jan 15 at 15:37

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 ultrafilters and evaluating on them gives functionals that are not in the image of $$\ell^1$$.