Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?

Question

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?

In some sense, this is a follow-up to my answer to this question where the non-isomorphism between the spaces $L^r$ and $\ell^s$ for $1 \leq r,s \lt \infty$, unless $r$ and $s$ are both two was discussed (among other things).

There is the somewhat surprising fact that the Banach spaces $X = L^\infty[0,1]$ and $Y = \ell^\infty$ are isomorphic. More precisely, there are mutually inverse bounded linear maps $T: X \to Y$ and $S: Y \to X$ (see below for a proof of existence).

Is there a direct and explicit way to prove this? In other words: I'm wondering whether there is an explicit and natural expression for either $S$ or $T$.

Here's the argument I know: Using Pełczyński's decomposition technique one can prove that $X = L^\infty$ and $Y = \ell^\infty$ are isomorphic as Banach spaces:

1. Choose a countable partition $[0,1] = \bigcup_{n=0}^\infty E_n$ into disjoint sets of positive measure and send $(x_n)_{n \in \mathbb{N}} \in \ell^\infty$ to $\sum_{n=0}^\infty x_n [E_n]$ to get an isometric embedding $i: Y \hookrightarrow X$. Since $\ell^\infty$ is injective, its image is complemented, in particular, this yields a decomposition $X \cong Y \oplus \widetilde{Y}$.

2. Choose a dense sequence $(f_n)_{n \in \mathbb{N}}$ of the unit sphere of $L^1[0,1]$. For $h \in L^\infty[0,1]$ let $j(h) = \left( \int f_n \, h \right)_{n \in \mathbb{N}} \in \ell^\infty$ to get an isometric map $j: L^\infty[0,1] \to \ell^\infty$. Since $L^\infty[0,1]$ is injective, its image is complemented in $\ell^\infty$, so this yields a decomposition $Y \cong X \oplus \widetilde{X}$.

3. Observe that $X \cong X \oplus X$ since $L^\infty[0,1] = L^\infty[0,1/2] \oplus L^\infty[1/2,1] \cong L^\infty [0,1] \oplus L^\infty [0,1]$ and $Y \cong Y \oplus Y$ by decomposing $\mathbb{N}$ into the sets of even and odd numbers. Thus, Pełczyński's argument yields: $$X \cong Y \oplus \widetilde{Y} \cong (Y \oplus Y) \oplus \widetilde{Y} \cong Y \oplus (Y \oplus \widetilde{Y}) \cong Y \oplus X$$ and $$Y \cong X \oplus \widetilde{X} \cong (X \oplus X) \oplus \widetilde{X} \cong X \oplus (X \oplus \widetilde{X}) \cong X \oplus Y$$ so that $X \cong Y \oplus X \cong X \oplus Y \cong Y$.

Of course, one can trace through this argument and “construct” an isomorphism, but the resulting maps are rather messier than what I'm looking for. A further deficit of this argument is that the appeal to injectivity properties makes this inherently non-constructive.

Any simplifications of this argument or pointers to the literature would be welcome.

Answer 1

Suppose $T: L^\infty \to \ell^\infty$ is your isomorphism. Looking at this coordinatewise, this corresponds to a bounded sequence $\phi_n$ of bounded linear functionals on $L^\infty$ such that

1) there is $\epsilon > 0$ such that $\max_n |\phi_n(f)| \ge \epsilon \|f\|_\infty$ for all $f \in L^\infty$

2) For every bounded sequence $t_n$ of reals there is $f \in L^\infty$ such that $\phi_n(f) = t_n$.

As far as I know, the only really "constructible" bounded linear functionals on $L^\infty$ are of the form $f \to \int_0^1 f(x) g(x)\ dx$ where $g \in L^1$. If all $\phi_n$ were of this form, say $\phi_n(f) = \int_0^1 f(x) g_n(x)\ dx$, then $T^*$ would map $\ell^1$ (as a closed subspace of $(\ell^\infty)^*$) one-to-one to a closed subspace $V$ of $L^1[0,1]$, i.e. for any $y \in \ell^1$ and $f \in L^\infty$, $(T^* y)(f) = y(Tf) = \sum_n y_n \phi_n(f) = \int_0^1 \left(\sum_n y_n g_n(x)\right) f(x)\ dx$ where $\sum_n y_n g_n \in L^1$. Now $$V^\perp = \{f \in L^\infty: \int_0^1 f(x) g(x)\ dx = 0 \text{ for all } g \in V\} = (T^*)^{-1}(\ell^1)^\perp = \{0\}$$ and then $V = L^1$. But we know $\ell^1$ and $L^1[0,1]$ are not isomorphic, so this is impossible. That says that any isomorphism of $L^\infty$ onto $\ell^\infty$ must involve some of the "exotic" linear functionals on $L^\infty$.

Answer 2

Robert Israel's answer gives a beautiful and convincing argument why some element of non-constructivity must be involved in an isomorphism between $\ell^\infty$ and $L^\infty$.

Since Nate Eldredge brought this up in the comments, I'm going to put the argument slightly differently by abandoning the full axiom of choice and using a theorem of Väth in order to establish:

Theorem $(\operatorname{ZF}+\operatorname{DC}+\operatorname{PM}_\omega)$​. There is no isomorphism between $\ell^\infty$ and $L^\infty$.

Here $\operatorname{DC}$ is the usual axiom of dependent choice and $\operatorname{PM}_\omega$ is the statement “There is no (finitely additive probability) measure on $\omega$ which is zero on all finite sets”.

• Loosely speaking, $\operatorname{PM}_\omega$ asserts that there is no functional on $\ell^\infty$ that doesn't come from $\ell^1$. Väth proves from it (and dependent choice) that $(L^\infty)^\ast = L^1$ holds for all $\sigma$-finite measure spaces: Martin Väth, The dual space of $L^\infty$ is $L^1$, Indag. Mathem., N.S., 9 (4), 1998, 619–625.

• Solovay's model (in which every set of real numbers is measurable) proves the relative consistency of $\operatorname{ZF}+\operatorname{DC}+\operatorname{PM}_\omega$. In that model there can thus be no isomorphism between $\ell^\infty$ and $L^\infty$. While Solovay's model relies on the existence of an inaccessible cardinal whose consistency strength is way beyond the one of $\operatorname{ZF}$, Pincus and Solovay showed that no large cardinal hypothesis is required for establishing the relative consistency of $\operatorname{ZF}+\operatorname{DC}+\operatorname{PM}_\omega$ see §1 of Väth's article for more background and detailed references.

Proof of the theorem. We assume $\operatorname{ZF}+\operatorname{DC}+\operatorname{PM}_\omega$ so that by Väth's results, $\ell^1 = (\ell^\infty)^\ast$ and $L^1 = (L^\infty)^\ast$.

2. If there were an isomorphism $T: L^\infty \to \ell^\infty$ we would get an isomorphism $T^\ast: \ell^1 \to L^1$.

   Indeed, $\|T^\ast\| \leq \|T\|$ follows from a direct application of the definition of the operator norm (however: establishing equality would require Hahn-Banach). The proofs of the fact that a bounded operator is continuous (it is Lipschitz) as well as $(ST)^\ast = T^\ast S^\ast$ and $\operatorname{id}_{X}^\ast = \operatorname{id}_{X^\ast}$ don't involve any choice whatsoever.

   The remainder of the argument will be to establish that $\ell^1$ and $L^1$ are not isomorphic and thus we will end up with a contradiction.

3. It is a theorem of $\operatorname{ZF}$ that $\ell^1$ has the Schur property: every weakly convergent sequence in $\ell^1$ is norm convergent.

   Indeed, the proof outlined in Pedersen's Analysis Now, Exercise E 2.4.7, page 68 goes through as well as the proof that $\ell^\infty$ is the dual space of $\ell^1$ (thanks to Brian M. Scott and Asaf Karagila for double-checking this). For the convenience of the reader here's the argument which is basically the same as the one given by Davide Giraudo in this thread:

   

   The point is that the sequence $a(n)$ and the subsequence of $(x_n)$ can be simultaneously and explicitly defined by induction using the well-order on $\mathbb{N}$, the hypothesis that $x_n$ converges weakly to zero, that $x \mapsto \sum_{k=1}^{a(n-1)} |x(k)|$ is weakly (sequentially) continuous and that $\|x_n\|_1 \geq \varepsilon$ infinitely often.

4. The space $L^1$ does not have the Schur property since one can simply write down examples of weakly convergent sequences that aren't norm convergent, e.g. $f_n(x) = \exp{(2\pi i \,n \,x})$.

   Dependent choice is amply sufficient for proving the Riemann-Lebesgue lemma stating that $\langle f_n, h \rangle \to 0$ as $n \to \infty$ for all $h \in L^1 \supset L^\infty$. See the remarks in Fremlin's Measure Theory Volume 5II, at the beginning of 566, especially 566B and 566C and compare the proof of the Riemann-Lebesgue lemma given in Volume 2, 282E and 282F for more details.

5. If $S: E \to F$ is an isomorphism and $E$ has the Schur property then so does $F$ (this follows from the remarks in part 1. of the proof).

Combining 2., 3. and 4., we see that there can be no isomorphism between $\ell^1$ and $L^1$. Thus, we have arrived at the contradiction announced in step 1.