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...

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...