$F^1, F^2$ are the CDFs corresponding to two distributions corresponding to positive finite measures, supported in $[0,1]$, i.e. $F^i(x)$ is the measure of the set $[0,x]$ under the $i$-th measure. Each of $F^1$ and $F^2$ has a continuously differentiable density. $F^1_n,F^2_n$ are two sequences of distributions, each finitely supported, which weakly converge to $F_1$ and $F_2$ respectively, i.e. $\lim\limits_{n \rightarrow \infty} F^i_n(x)=F_i(x), i \in \{1,2\}$. Let $x^* \in [0,1]$ uniquely maximizes $F^1(x)-F^2(x)$. I need to show that every sequence of maximizers of $F^1_n(x)-F^2_n(x)$ converges to $x^*$. If this is not true, are there additional conditions under which this is true?
My attempt: I'm getting stuck exchanging $\sup$ and $\lim$. Under weak convergence the measure of every set converges but not necessarily their sup, and in this case the sup remains stuck at the measure of the whole set $[0,1]$ because the difference in the measures of that set under $F^i$ and $F^i_n$ always remains $F_i(1)$.
We don't even need $F^i$ to have density, it's enough for them to be continuous.
This answer (pointwise convergence of monotonic functions to a continuous function is uniform) proves that convergence of $F_n^i$ to $F^i$ is actually uniform. Now it's enough to show that if $f$ is continuous function with unique maximum at $x^*$, $f_n \rightrightarrows f$ and $x_n$ is (not necessary unique) maximum of $f_n$, then $x_n \to x^*$.
Fix some $\epsilon > 0$. As $f$ is continuous and has global maximum in $x^*$, $\exists \delta > 0 \forall x: |x - x^*| > \epsilon \rightarrow f(x) < f(x^*) - \delta$. For some large enough $N$ for all $n > N$ and all $x$ we have $|f(x) - f_n(x)| < \delta / 2$. Then, for all $x$ s.t. $|x - x^*| > \epsilon$ we have $f_n(x) < f(x) + \delta / 2 < f(x^*) - \delta / 2 < f_n(x^*)$, and so if $x_n$ is maximum of $f_n$, then $|x_n - x^*| < \epsilon$. This implies $x_n \to x^*$.