# Sequences of linear combinations of measures

Let $X$ be a Polish space. Let $J\in\mathbb{N}$.

Let $\lbrace a^n_1\rbrace_n,\dots,\lbrace a^n_J\rbrace_n$ be $J$ sequences of reals.

Let $\lbrace \mu^n_1\rbrace_n,\dots,\lbrace \mu^n_J\rbrace_n$ be $J$ sequences of probability measures in $\Delta(X)$.

For each $j\leq J$, let $\mu^n_j$ weakly converge to $\mu_j \in \Delta(X)$.

Let $\sum_{j\leq J} a^n_j \mu^n_j \in \Delta(X)$ weakly converge to $\mu^* \in \Delta(X)$.

Conjecture: Does there exist a vector $(\beta_1,\dots,\beta_J)$ of reals such that $\mu^* = \sum_{j\leq J} \beta_j\mu_j$? It need not be unique.

Proof for when $J=2$:

$a^n_1\mu^n_1+a^n_2\mu^n_2$ is a probability measure. Therefore $a^n_2 = 1-a^n_1$.

$a^n_1\mu^n_1+(1-a^n_1)\mu^n_2 = a^n_1 (\mu^n_1-\mu^n_2)+\mu^n_2 \to \mu^*$. Since $\mu^n_2 \to \mu_2$, it follows that $a^n_1 (\mu^n_1-\mu^n_2)$ converges.

If $\mu_2 = \mu_1$, then $\mu^* = \mu_2$.

If $\mu_2 \neq \mu_1$, then $a^n_1$ must converge since $a_1^n\int g\ d(\mu^n_1-\mu^n_2)$ must converge for all continuous bounded functions $g: X \to \mathbb{R}$.

Some comments The proof above does not seem to generalize to $J>2$. We can also view the problem more generally as an infinite dimensional vector space problem.

That is, we can look at $\lbrace \mu^n_1,\dots,\mu^n_J \rbrace$ as the vector subspace defined by the linear span of its elements. If $\mu^n$ is a convergent sequence ($\mu^n \to \mu^*$) such that $\mu^n \in span(\mu^n_1,\dots,\mu^n_J)$ and $\mu^n_j \to \mu_j$, is it the case that $\mu^* \in span(\mu_1,\dots,\mu_J)$?

I think the original problem places additional constraints on the problem by requiring that certain objects are probability measures (as opposed to any vector in the space of finite signed measures), but these two problems seem reasonable close.

Any hints? Could the original conjecture be wrong? Strange things happen in infinite dimensional spaces...

• As it is, the generalization is false, even in $\mathbb{R}$ (as a TVS) and $J=1$. Take $\mu^n_1:=1/n\to\mu_1:=0$, and $\mu^n=\mu^*=1$, for all $n$. – Pietro Majer Oct 3 '11 at 19:27
• Thanks Pietro. You are right about the generalization. About your comment regarding linear independence: What if $\lbrace\mu_1^n,\dots,\mu_J^n\rbrace$ was always a linearly independent set for each given $n\in\mathbb{N}$? It happens that the particular research problem I am working on does have that restriction – Empty Set Oct 3 '11 at 19:52
• Actually I think what is needed is the linear independence of the limit family; see below. – Pietro Majer Oct 3 '11 at 21:39

The generalization you suggest is true in any real topological vector space $X$ (Hausdorff or not), under the further assumption that the limit family $(\mu^\infty _ 1,\dots,\mu^\infty _ J)$ be linearly independent. The natural generalization to net convergence is also true (with essentially the same proof) .

Fact. The set $D _ J$ of all linearly dependent $J$-uples of elements of $X$ is a closed subset of $X^J$. Indeed, let $\mathbb{S}$ be the unit sphere of $\mathbb{R}^J$. Then $D _ J$ is the projection on the second factor of the closed subset of $\mathbb{S}\times X^J$ $$F_J:=\Big\{ (\lambda,\mu) \in \mathbb{S}\times X^J\ : \ \sum_{j=1}^J\ \lambda_ j \mu _ j=0 \Big\} ,$$ and the projection $\mathbb{S}\times X^J\to X^J$ is a closed map because $\mathbb{S}$ is compact.

Consequence. Let $\mu ^ n _ 0 \in \operatorname{span} (\mu ^ n _ 1,\dots,\mu ^ n _ J)$ for all $n \in \mathbb{N}$ and assume that $\mu ^ n _ j \to \mu ^ \infty _ j$ as $n\to\infty$, for $j=0,1,\dots, J$. Then the $(J + 1)$-uple $(\mu ^n _ 0, \mu ^ n _ 1,\dots,\mu ^ n _ J)$ is linearly dependent, so by the above fact the limit $(J +1)$-uple $(\mu ^\infty _ 0, \mu^\infty _ 1,\dots,\mu^\infty _ J)$ is linearly dependent too. However, by assumption $(\mu^\infty _ 1,\dots,\mu^\infty _ J)$ is linearly independent, which implies that $\mu ^ \infty _ 0 \in \operatorname{span} (\mu^\infty _ 1,\dots,\mu^\infty _ J)$.

Rmk. Assuming the linear independence of the limit $J$-uple is necessary, otherwise the statement is false even for $J=1$ and $X=\mathbb{R}$ as shown in my comment above. Also, assuming that the limit $J$-uple are probability measures as in your first conjecture, is still not sufficient already for $J=2$, and in fact there's a problem in your proof. In your notation, if $\mu_2=\mu_1$, it is not guaranteed that $\mu^*=\mu_2$. Take e.g. $X=\{0,1\}$, a discrete two−points space. Take $\mu^n_1:=\delta_1$ and $\mu^n_2:=\big(1-\frac{1}{n}\big)\delta_1+\frac{1}{n}\delta_0$.They both converge to $\mu_1=\mu_2=\delta_1$, but a linear combination of them, namely $n\mu^n_2 - (n-1) \mu^n_1=\delta_0$ converges to $\mu^*:=\delta_0$, which is not in the span of $\delta_1$.

• I've rewritten the whole answer. – Pietro Majer Oct 6 '11 at 9:53
• Thanks! Even though this wasn't my original problem, it will be helpful. I was interested in the original problem as an intermediate result for something else, and I can see that your solution of the modified problem might help me in other ways. It also helps that my original conjecture was proven wrong in your answer. – Empty Set Oct 6 '11 at 20:34

Here's a preliminary answer.

If you consider convex combinations, so that in particular we have $a^n_j \in [0,1]$, then this is true. By the compactness of $[0,1]^J$, we can assume (passing to a subsequence) that $a^n_j$ converges for each $j$. If we call $\beta_j$ the limit, then weak continuity of addition and scalar multiplication shows that $\sum_j a^n_j \mu^n_j \to \sum \beta_j \mu_j$, so uniqueness of weak limits gives $\mu^* = \sum_j \beta_j \mu_j$.

Really, all we're using is that the space of measures with the weak topology is a Hausdorff topological vector space.

I have a suspicion that any probability measure $\mu$ that can be written as a linear combination of probability measures can also be written as a convex combination of those same probability measures, which would settle the general case. This may be a basic fact about cones in vector spaces. I'll think about it a little more.

• Thanks for the preliminary reply. I will wait for your full reply. Your ideas seem promising at least. However, one note about linear combinations of probability measures: If $X=\lbrace a, b, c\rbrace$, $p = (1/3,1/3,1/3), q = (1/4,1/4,1/2), r = (0,0,1) \in \Delta(X)$. $r = -3p + 4q$. There cannot exist an $a \in [0,1]$ such that $r= ap +(1-a)q$. – Empty Set Oct 3 '11 at 19:51
• Ok, good counterexample. – Nate Eldredge Oct 3 '11 at 21:01