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

  • 1
    $\begingroup$ 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. $\endgroup$ – Ashwin Trisal Jan 14 at 7:48
  • $\begingroup$ Related: Dual of $l^\infty$ is not $l^1$ $\endgroup$ – 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.

  • $\begingroup$ I have already proven that $X$ is seperable if $X'$ is seperable, sosince $\ell^\infty$ is not seperable $(\ell^\infty)^*$ is not seperable. $\endgroup$ – user408856 Jan 14 at 7:54
  • $\begingroup$ Is an isometric isomorphism naturally a homeomorphism? $\endgroup$ – user408856 Jan 14 at 8:23
  • 1
    $\begingroup$ @James Of course, it is. $\endgroup$ – 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 $\Phi y$?

  • $\begingroup$ 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 $l^1$. $\endgroup$ – GEdgar Jan 14 at 11:58
  • $\begingroup$ Which functional is it? $\endgroup$ – user408856 Jan 14 at 12:12
  • $\begingroup$ The functional is the one assumed to exist in the definition of $c$. $\endgroup$ – 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 $\ell^1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy