Uniform convergence and sequences of minima Let $\Omega \subset \mathbb{R}^n$,  $f \in C(\Omega)$ and $g \in C^1(\Omega)$ and assume that $f - g$ has a strict global minimum at $x \in \Omega$. 
Is the following true? 

there exists a sequence of functions $\{f_n\}$ such that $f_n \to f$ uniformly $\implies$ there is a sequence of points $x_n \to x$ such that $f_n(x_n) \to f(x)$ and  $f_n - g$ has a global minimum at $x_n$?

 A: Counterexample for $\Omega = (-\pi, \pi):$ Define $f(x) = \sin^2(x),g\equiv 0.$ Then $f$ has a strict global minimum at $0.$
For $n=1,2,\dots ,$ let $\varphi_n$ be the piecewise linear function on $[-\pi,\pi]$ whose graph connects the points $(-\pi,-1), (-\pi+1/n,1), (\pi-1/n,1), (\pi,-1).$ On $(-\pi, \pi),$ set $f_n = \varphi_nf.$ Do a little work to see $f_n \to f$ uniformly on $(-\pi,\pi).$
For each $n,$ the global minimum value of $f_n$ on $\Omega$ is negative and occurs twice, once to the left of $-\pi/2,$ once to the right of $\pi/2.$ So it is impossible to find $x_n \to 0$ such that $f_n$ has a global minimum at $x_n.$
A: If $\Omega$ is unbounded the result is not true.
Let $\Omega = \mathbb{R}$, $g=0$, $x=0$, and consider the function
$$
f(x) = x^2 e^{-x^2}
$$
which has a strict minimum point at $x=0$.
Let $h(x):= \min\{x^2-1, 0\}$ and define
$$
f_n(x) := f(x) + \frac{1}{n}\, h(x-n).
$$
It is clear that the sequence $(f_n)$ converges uniformly to $f$.
On the other hand, since $f_n(n) = n^2 e^{-n^2} - 1/n < 0$ for $n$ large enough, the function $f_n$ attains its global minimum at a point $x_n \in (n-1, n+1)$.
