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

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.

• It would be interesting to see what the image of polynomials in $L^\infty$ looks like in $l^\infty$ under such an isomorphism. – Nick Alger Feb 17 '12 at 19:02
• The remark on page 111 in Lindenstrauss and Tzafriri's $Classical\ Banach\ Spaces\ I$ gives a slightly different argument. It relies, however on injectivity and the decomposition method too. They give a reference to the article containing the result: Pelczynski, A: On the isomorphism of the spaces $m$ and $M$. Bull. Acad. Pol. Sci. $\bf 6$ 695-696 (1958). (I surmise the argument contained theiren is the decomposition method.) I assume you know all this, though :) – David Mitra Feb 17 '12 at 19:28
• Thank you, @David I'm away from my books, so I couldn't check in LT, I reproduced the argument from memory, hence all the typos :) – t.b. Feb 17 '12 at 19:49

## 2 Answers

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 $L^\infty$.

• Nice. It would be interesting to see a result like "it is consistent with ZF or ZF+DC that no such isomorphism exists." – Nate Eldredge Feb 18 '12 at 14:36
• @Nate: It might be possible to prove such a result using this article by Väth. It is consistent with ZF+DC+(something called $PM_\omega$) that the dual spaces of $\ell^\infty$ and $L^\infty$ are $\ell^1$ and $L^1$. I'm not sure if one can still tell $\ell^1$ and $L^1$ apart in that model, though... – t.b. Feb 18 '12 at 15:43
• Thanks again, Robert for your answer! This is exactly what I was looking for. @Nate: I posted an answer with an argument along the lines of my previous comment. – t.b. Feb 19 '12 at 17:11

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 $L^1 = (L^\infty)^\ast$.

1. 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.

2. 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.

3. 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.

4. 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.

• Time permitting, I might add another answer giving a pretty explicit construction of the isomorphism, which is in line with Väth's arguments and also the arguments used in Fremlin's arguments referred to in the answer to this related thread. It'll be a while before this is actually written up... – t.b. Feb 24 '12 at 15:29
• Regarding the Pincus-Solovay proof, I think Vaeth mentions that it follows from the fact every set has the Baire property, which Shelah proved to be consistent without large cardinals (perfect set property and Lebesgue measurability everywhere do require an inaccessible). – Asaf Karagila Feb 24 '12 at 16:33