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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

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A simple observation which might be silly because I don't know if Fubini's theorem holds for finitely additive probability measures: If we are in a situation where we can use Fubini's theorem the mapping in the question is constant since $$\int\int\phi(x+y)d\nu(x)d\mu(y) = \int\int\phi(x+y)d\mu(y)d\nu(x)$$ $$= \int\int\phi(y)d\mu(y)d\nu(x) = \int\phi(y)d\mu(y).$$ –  Tapio Rajala Sep 3 '11 at 8:34
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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
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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
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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
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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
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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.

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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
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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
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@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
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