Do invariant measures maximize the integral?

Update: The negative answer to the following question has been provided by Matthew Daws, who won, but also rejected, the bounty of 100 euro that I set over the question.

Let $\mathcal M(\mathbb Z)$ be the set of all finitely additive probability measures on the power set of $\mathbb Z$. Let $\phi:\mathbb Z\rightarrow\mathbb R$ be nonnegative and bounded. Observe that $\phi$ is integrable with respect to any $\mu\in\mathcal M(\mathbb Z)$. Let me say that $\mu$ is $\phi$-translation invariant if for all $y\in\mathbb Z$ one has

$$\int\phi(x+y)d\mu(x)=\int\phi(x)d\mu(x)$$

Let $I_\phi(\mathbb Z)$ be the class of $\phi$-invariant measures in $\mathcal M(\mathbb Z)$. Let $\mu\in I_\phi(\mathbb Z)$ be fixed.

Question: Is it true that the mapping $F:\nu\in\mathcal M(\mathbb Z)\rightarrow\int\int\phi(x+y)d\nu(x)d\mu(y)$ attains its maximum on a measure $\nu\in I_\phi(\mathbb Z)$?

Thanks in advance,

Valerio

• Unfortunately Fubini's theorem does not hold. For intance, let $\phi=\chi_{\mathbb N},$\mu$a invariant measure such that$\int\phi d\mu=1$e$\vu$an invariant measure such that$\int\phi d\vu=0$. The lackness of Fubini's theorem is, at the end, the point. – Valerio Capraro Sep 3 '11 at 8:44 • I think the title should be "Do invariant measures maximize the integral? (Bounty offered)" . That phrasing is slightly less crass, and stirs more curiosity; people will read through the post to find out what the bounty is and how much; I think the present phrasing of the title is in the grey area of acceptability on MathOverflow. Gerhard "It May Be Just Me" Paseman, 2011.11.18 – Gerhard Paseman Nov 18 '11 at 17:27 • @ Gerhard: After all, the late Paul Erdös also offered cash prizes for questions... (:-) – Alain Valette Nov 18 '11 at 18:47 • Indeed he did Alain. And how many of those were phrased in a fashion appropriate for MathOverflow? I am not opposing the occasional practice; I just think the presentation should be improved. Gerhard "It Might Be Others Too" Paseman, 2011.11.18 – Gerhard Paseman Nov 18 '11 at 18:58 • There is a meta discussion, specifically discussing the bounty, at tea.mathoverflow.net/discussion/1212/…. My suggestion would be to explicitly spell out that the bounty is entirely independent of the usual mathematical conventions regarding acknowledgement and priority. – Scott Morrison Nov 18 '11 at 21:28 1 Answer Edit: Here is what I think is a counter-example. Let$\phi$be the indication function of the even natural numbers, let$\mathcal U$be an ultrafilter supported on the even naturals, and let$\mathcal V$be an ultrafilter defined on the even negative integers. Define$\mu\in M(\mathbb Z)$by $$\int f(x) \ d\mu(x) = \lim_{x\in\mathcal V} f(x).$$ Then$\mu\in I_\phi$(as, in fact,$\int \phi(x+y) \ d\mu(y)=0$for all$x$). If$\nu\in I_\phi$then$\nu$must assign the same measure to$2\mathbb N$and$2\mathbb N+1$, say$\alpha\leq 1/2$. You also need to argue that$\nu$must assign zero measure to any finite set (else it won't be$\phi$-invariant). So for any$y\in\mathbb Z$, $$\int \phi(x+y) \ d\nu(x) = \nu(2\mathbb N-y) = \begin{cases} \nu(2\mathbb N) &: y\in 2\mathbb Z, \\ \nu(2\mathbb N+1) &: y\in 2\mathbb Z+1, \end{cases} = \alpha.$$ Thus $$\int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \alpha \ d\mu(y) = \alpha,$$ as$\mu$is a probability measure. By contrast, let$\nu$be defined by $$\int f(x) \ d\nu(x) = \lim_{y\in\mathcal U} f(y).$$ Then $$\int \phi(x+y) \ d\nu(x) = \begin{cases} 1 &: y\in 2\mathbb Z, \\ 0 &: y \in 2\mathbb Z+1, \end{cases}$$ and so $$\int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \chi_{2\mathbb Z}(y) \ d\mu(y) = 1.$$ So$F$is not maximised on$I_\phi$. In fact, by replacing$2\mathbb N$by$k\mathbb N$, I think you get that$F$has norm one, but$F(\nu)\leq 1/k$for any$\nu\in I_\phi$. But somehow, to my mind, what's wrong is that the$\mu\in I_\phi$you choose is very poor. So here's a revised conjecture: Let$\mu\in I_\phi$maximise the integral$\int \phi(x) \ d\mu(x)$. Then$F$attains its maximum on$I_\phi$. Old post: (Explains my thinking). I think of these questions using the Arens products, from abstract Banach algebra theory. So I work over the complex numbers; but this is not a problem. Consider$A=\ell^1(\mathbb Z)$with the convolution product, so$A$is commutative. Then$A^*=\ell^\infty(\mathbb Z) = C(\beta\mathbb Z)$is an$A$-module:$(a\cdot f)(b) = f(ba)$for$a,b\in A,f\in A^*$. Then$A^{**}=M(\beta\mathbb Z)$the space of finite Borel measures on the Stone-Cech compactification$\beta\mathbb Z$. Your space$M(\mathbb Z)$is just the positive measures$\mu\in A^{**}$with$\mu(1)=1$. We try to extend the product of$A$to$A^{**}$. Firstly we define a bilinear map$A^{**}\times A^*\rightarrow A^*$by $$(\mu\cdot f)(a) = \mu(a\cdot f) \qquad (\mu\in A^{**}, f\in A^*, a\in A).$$ But then we have two choices for the product on$A^{**}$: $$(\mu \Box \lambda)(f) = \mu(\lambda\cdot f), \quad (\mu\diamond\lambda)(f) = \lambda(\mu\cdot f) \qquad (\mu,\lambda\in A^{**}, f\in A^*).$$ A little thought shows that$\mu\diamond\lambda = \lambda\Box\mu$. So if$\phi\in A^*$if positive then$\mu\in I_\phi$if and only if$\mu\cdot\phi = \mu(\phi) 1$. This follows, as writing$\delta_x\in A=\ell^1(\mathbb Z)$for the point mass at$x\in\mathbb Z$, we have $$(\phi\cdot\delta_x)(\delta_y) = \phi(\delta_{x+y}) \implies (\mu\cdot\phi)(\delta_x) = \mu(\phi\cdot\delta_x) = \int \phi(x+y) \ d\mu(y).$$ So the condition that$\mu\in I_\phi$becomes that$(\mu\cdot\phi)(\delta_x)$is constant in$x$, which is seen to be equivalent to$\mu\cdot\phi = \mu(\phi) 1$. Similarly, your map$F$is just$F(\nu) = (\mu\Box\nu)(\phi)$. As you allude to, it's known that$\lambda\Box\mu \not= \mu\Box\lambda$for arbitrary$\lambda,\mu$. However, we say that$f\in A^*$is "weakly almost periodic" (WAP) if$(\lambda\Box\mu)(f) = (\mu\Box\lambda)(f)$for all$\mu,\lambda\in A^{**}$. So if$\phi$is WAP and$\mu\in I_\phi$then for any$\nu\in M(\mathbb Z)$, $$F(\nu) = (\mu\Box\nu)(\phi) = (\nu\Box\mu)(\phi) = \nu(\mu\cdot\phi) = \nu(1) \mu(\phi) = \mu(\phi),$$ as$\nu$is a probability measure. So actually$F$is constant on$M(\mathbb Z)$and so certainly attains its maximum at a point of$I_\phi$. So, to be interesting, we need to ask the question for$\phi$which are not WAP. An alternative characterisation of$\phi$being in WAP is that the set of translates of$\phi$in$\ell^\infty(\mathbb Z)$forms a relatively weakly compact set. A nice characterisation of Grothendieck shows that this is equivalent to $$\lim_n \lim_m \phi(x_n+y_m) = \lim_m \lim_n \phi(x_n+y_m)$$ whenever all the limits exist, for sequences$(x_n),(y_m)$in$\mathbb Z$. If$\phi$is the indicator function of$\mathbb N$, then it's not in WAP. We may as well assume that$\|\phi\|_\infty=1$. Another "easy" case is when we can find$\nu\in I_\phi$with$\nu(\phi)=1$. Then$F(\nu) = \mu(\nu\cdot\phi) = \mu(1) \nu(\phi) = 1$; while for any$\lambda\in M(\mathbb Z)$, clearly$|F(\lambda)| = |\mu(\lambda\cdot\phi)| \leq 1$as$\mu$is a probability measure, and$\lambda\cdot\phi$is bounded by$1$(again, as$\lambda$is a probability measure and$\phi$is bounded by$1$). Notice that this case covers your example of when$\phi$is the indicator function of$\mathbb N$. So a test case is to find$\phi$not in WAP and with$\nu(\phi)<\|\phi\|_\infty$for all$\nu\in I_\phi$(notice that$I_\phi$is always non-empty, as$\mathbb Z$is amenable). Do you have an example of such a$\phi$? Actually, if$\phi$is the indicator function of the even natural numbers, then that's an example. And that leads to my (hopeful) counter-example. • Matthew, I have just seen your answer. Give me some time to go through the details. Let me understand well: are you claiming that the answer is negative and give a counterexample in the Edit? – Valerio Capraro Nov 18 '11 at 19:10 • Yes. Actually, let me make a final edit-- I'll make the counter-example clearer (I'll put it into your notation) and make a revised conjecture... – Matthew Daws Nov 18 '11 at 20:30 • I've some trouble to understand why that integral should give$\alpha$.. Indeed, my understanding is that$\int\int\phi(x+y)d\nu(x)d\mu(y)=\int_{2\mathbb Z}(\int\phi(x+y)d\nu(x))d\mu(y)+\int_{2\mathbb Z+1}(\int\phi(x+y)d\nu(x))d\mu(y)==\int\chi_{2\mathbb Z}d\nu(x)+\int\chi_{2\mathbb N+1}d\nu(x)=2\alpha\$ Am I wrong? – Valerio Capraro Nov 19 '11 at 8:20
• Hi Matthew, sorry for the delayed answer but I was mainly away this weekend. I am now sure that your counter-example is good. Please, contact me privately for the bounty. I am thinking about the missing property.. indeed, for my application, I have some stronger property and so there might be still a positive answer.. let me think about for a while. – Valerio Capraro Nov 20 '11 at 17:31
• @Valerio: I'm glad the counter-example seems okay. I must say that I don't think I have done anything like enough work here to justify taking 100 euros off you (I hope that doesn't seem churlish). But what I will accept is, if you are ever in the North of England (or we meet at a conference), then you can buy me a drink or two... (and let us hope that still costs less than 100 euros after the current financial mess!) I'll email you shortly... – Matthew Daws Nov 20 '11 at 20:07