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Let $M([0,1])$ be the set of finite signed measures on $[0,1]$ (with the topology generated by the sets $\left\{ \mu \in M([0,1]) : \left| \int f(x) \mu(dx)- a\right| \leq \delta\right\}$ for all $\delta>0$, $a \in R$ and $f \in C_b([0,1])$ (continuous and bounded). (hence weak-*-topology)

This implies that for each $f \in C_b([0,1]\times [0,1],\mathbb R)$ (continuous and bounded) the set $$ A=\left\{ \mu \in M([0,1]) : \int_{[0,1]} \int_{[0,1]} f(x,y) \mu(dx) \mu(dy) \leq 1 \right\} $$ is sequentially closed.

Here it is answered that $M([0,1])$ is not sequential. Therefore we can not infer from the sequentially closedness of $A$ that it is also closed.

Question: Is $A$ nevertheless closed?

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    $\begingroup$ Why $C_b$ rather than just $C$? $[0,1]$ and $[0,1]\times [0,1]$ are compact. $\endgroup$ – Robert Israel Jul 3 '15 at 18:09
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    $\begingroup$ It's true if $f(x,y)$ is of the form $g(x)h(y)$. Linear combinations of such functions are dense in $C[a,b]^2$ by Stone-Weierstrass, so it is true in general. $\endgroup$ – Bill Johnson Jul 3 '15 at 20:52
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    $\begingroup$ @BillJohnson: I had the same idea but the details were not immediately clear to me, particularly the density argument. If I can work it out, I will post an answer, otherwise I might ping you again with questions (unless you are inclined to write a more detailed answer anyway). $\endgroup$ – Nate Eldredge Jul 4 '15 at 3:06
  • $\begingroup$ @ChristianRemling: Yes, I understand that the class of functions of the form $f(x,y) = g_1(x) h_1(y) + \dots + g_n(x) h_n(y)$ is dense. And I can see that if $f$ is of this form then the set $A$ is closed. What I do not understand is how we conclude that if $f$ is a uniform limit of such functions, then $A$ is again closed. $\endgroup$ – Nate Eldredge Jul 4 '15 at 16:29
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    $\begingroup$ @BillJohnson: But the set $A$ is not convex. (For example, let $f \in C([0,1] \times [0,1])$ be a function with $f(0,0) = f(1,1) = 0$ and $f(0,1) = f(1,0) = 7$. Then the point masses $\delta_0$ and $\delta_1$ are in $A$, but $\frac{1}{2} (\delta_0 + \delta_1)$ is not.) $\endgroup$ – Nate Eldredge Jul 13 '15 at 15:08
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Unfortunately, the above solution contains a flaw. The essence of the argument approximates a given function of two variables by a combination of tensor products of functions of one variable. One then finds a suitable weak neighborhood on which these functions have small integrals and uses a perturbation argument to claim that the same conclusion holds for the original function. But the argument (the integral of a small function is small) is only valid if you have control over the size of the measure. In this situation, this not the case---weak neighbourhoods are not norm bounded. However, what does follow is that the function which maps $\mu$ to $\mu \otimes \mu$ is continuous for the bounded weak star topology, i.e., the finest locally convex topology (even finest topology) which agrees with the weak topology on bounded sets. This gives a weaker version of the original question, namely that the set in question is closed for this finer topology. As a final remark, this topology has many common properties with the weak topology (same convergent sequences, induces same topology on the probability measures) with the additional advantage that it is complete. These remarks apply when the unit interval is replaced by any compactum and there is even a version which is valid for any completely regular space but this involves more elaborate functional analytical tools.

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  • $\begingroup$ Yes, you are right. I was actually thinking of this issue when considering the question, but I managed to ignore it while studying Christian's answer. $\endgroup$ – Nate Eldredge Jul 8 '15 at 16:07
  • $\begingroup$ Readers <10K should note that "the above solution" refers to a now-deleted answer by Christian Remling; not to the answer by me, which I believe addresses the issue raised here. $\endgroup$ – Nate Eldredge Feb 2 '16 at 16:54
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The set $A$ may or may not be closed, depending on the function $f$.

For instance, if $f$ is of the form $f(x,y) = g(x) h(y)$ for continuous $g,h \in C([0,1])$, then $A = A_f$ is closed: if we let $I_f(\mu) = \int f\,d(\mu \times \mu)$ then by Fubini's theorem we have $$I_f(\mu) = \left(\int g\,d\mu \right) \left(\int h\,d\mu\right)$$ which is the product of two weak-* continuous functions. Hence $I_f$ is weak-* continuous and so $A_f = \{I_f \le 1\}$ is closed. A similar argument applies if $f$ is a finite sum of the form $f(x,y) = g_1(x) h_1(y) + \dots + g_n(x) h_n(y)$. Note that by the Stone-Weierstrass theorem, the set of functions $f$ with this form is dense in $C([0,1]^2)$.

However, there also exist $f$ for which $A_f$ is not closed; indeed, $A_f$ can be dense. I will follow a similar construction to that used in https://mathoverflow.net/a/211173/4832, based on a suggestion from Anthony Quas. For convenience, I will work on $[-1,1]$ instead of $[0,1]$.

Let $f \in C([-1,1]^2)$ be the function $$f(x,y) = \begin{cases} 1, & x < 0 \\ e^{xy}, & x \ge 0 \end{cases}$$ which is clearly continuous.

For later use, let $F \subset C([-1,1])$ be the set of all those functions which are constant on $[-1,0]$. Note that $F$ is a closed linear subspace of $C([-1,1])$ with infinite codimension, and that $f(\cdot, y) \in F$ for each $y \in [-1,1]$.

A key fact is that the functions $\{f(\cdot, y) : y \in [-1,1]\}$ are linearly independent in $C([-1,1])$. This follows from an observation by Daniel Fischer, which I'll repeat here. If there should exist distinct $y_1, \dots, y_m$ and constants $b_1, \dots, b_m$, not all zero, such that $\sum b_j f(x,y_j) = 0$ for all $x$, then in particular, taking $x_i = (i-1)/m \in [0,1]$ for $i = 1,\dots, m$, we would have that the $m \times m$ matrix $T$ with $T_{ij} = f(x_i, y_j)$ is singular, since $b = (b_1, \dots, b_m)^t$ is in its null space. But $T_{ij} = e^{x_i y_j} = (e^{y_j/m})^{i-1}$ so that $T^t$ is a Vandermonde matrix, hence nonsingular.

Now I will show that in fact $A_f$ is dense.

Let $\mu_0 \in M([-1,1])$ be any signed measure. An arbitrary basic open neighborhood $U$ of $\mu_0$ is of the form $$U = \bigcap_{i=1}^n \left\{\mu : \left| \int g_i\,d\mu - \int g_i\,d\mu_0 \right| < \epsilon\right\}$$ for some $\epsilon > 0$ and some $g_1, \dots, g_n \in C([-1,1])$. For brevity, set $a_i = \int g_i\,d\mu_0$. I will construct a measure $\mu$ with $\int g_i\,d\mu = a_i$ for each $i$ and $I_f(\mu) \le 1$, so that $\mu \in A_f \cap U$.

To do so, we can suppose without loss of generality that the functions $g_1, \dots, g_n$ are linearly independent in $C([-1,1])$. (If we show that $\int g_{i_j}\,d\mu = a_{i_j}$ for some $g_{i_1}, \dots, g_{i_m}$, then the same must hold for any other $g_i$ which is in their linear span, since the map $g \mapsto \int g\,d\mu_0$ is linear.) Let $E \subset C([-1,1])$ be the linear span of $g_1, \dots, g_n$.

Since, as argued above, the functions $\{f(\cdot, y) : y \in [-1,1]\}$ are linearly independent, given any finite dimensional subspace $E'$ of $C([-1,1])$ we can choose $y \in [-1,1]$, or even $y \in [-1,0]$, so that $f(\cdot, y) \notin E'$. As such, we can choose distinct $y_1, \dots, y_{n+2} \in [-1,0]$ such that $$\{g_1, \dots, g_n, f(\cdot, y_1), \dots, f(\cdot, y_{n+2})\}$$ are linearly independent.

Now consider the linear map $L : C([-1,1]) \to \mathbb{R}^{n+2}$ defined by $L(g) = (g(y_1), \dots, g(y_{n+2}))$. Note that $\dim L(E) \le n$ and $\dim L(F) = 1$ (recall that $y_1, \dots, y_{n+2} \in [-1,0]$, so that $L(F)$ is just the constant vectors). So $\dim L(E+F) \le n+1$ and thus we may choose $(c_1, \dots, c_{n+2}) \in \mathbb{R}^{n+2} \setminus L(E+F)$.

By the linear independence of the functions $g_i$ and $f(x, y_j)$, using the Hahn-Banach and Riesz representation theorems, we can find a signed measure $\mu_1 \in M([-1,1])$ such that $$\begin{align*}\int g_i(x)\, \mu_1(dx) &= a_i, && i = 1,\dots, n \\ \int f(x, y_j)\, \mu_1(dx) &= c_j, && j = 1, \dots, n+2.\end{align*}.$$ By construction $\mu_1 \in U$. If it should happen that $I_f(\mu_1) \le 1$, then $\mu_1 \in A_f$ and we are done.

Otherwise, let $h(y) = \int f(x,y) \,\mu_1(dx)$ which is a continuous function thanks to the dominated convergence theorem. Note that by construction, $h(y_j) = c_j$; in particular, $h \notin E+F$. Now $E$ is finite dimensional and $F$ is closed with infinite codimension, so $E+F$ is a proper closed subspace of $C([-1,1])$. Thus by Hahn-Banach and Riesz again, we can find a signed measure $\mu_2$ with $\int g\,d\mu_2 = 0$ for all $g \in E+F$, and $\int h\,d\mu = -I_f(\mu_1)$. I claim that $\mu = \mu_1 + \mu_2$ is the desired measure.

First, for the functions $g_i$ that define $U$, we have $g_i \in E$ and so $\int g_i\,d\mu_2 = 0$. Hence $\int g_i\,d\mu = \int g_i\,d\mu_1 = a_i$ so $\mu \in U$.

It remains to show that $I_f(\mu) \le 1$; in fact I claim that $I_f(\mu) = 0$. Expanding $I_f(\mu)$ in terms of $\mu_1, \mu_2$ and using Fubini's theorem, we have $$I_f(\mu) = I_{11} + I_{12} + I_{21} + I_{22}$$ where $$I_{ij} = \int f(x,y) \,\mu_i(dx) \mu_j(dy).$$ Now since $f(\cdot, y) \in F$ for each $y$, we have $\int f(x,y) \,\mu_2(dx) = 0$ for every $y$. Hence $I_{21} = I_{22} = 0$. We have $I_{11} = I_f(\mu_1)$ by definition, and $$I_{12} = \int f(x,y)\, \mu_1(dx) \mu_2(dy) = \int h(y)\,\mu_2(dy) = -I_f(\mu_1)$$ by construction. Hence $I_f(\mu) = 0$ as desired.

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