Uniform convergence implies convergence of suprema?

Question

Let $f_n:M \rightarrow \mathbb{R}, M \subset \mathbb{R}, n\in \mathbb{N}$ be a sequence of continuous functions that converges uniformly on $M$ to a continuous function $f:M \rightarrow \mathbb{R}$. Moreover suppose that we have $x_n,x^\star \in M$ such that $$ f_n(x)<f_n(x_n) \quad \text{for all } x \neq x_n \qquad \text{and} \qquad f(x)<f(x^\star) \quad \text{for all } x \neq x^\star.$$ If $M$ is compact, we have $x_n \rightarrow x^\star$. Is this also true for non-compact $M$?

Answer 1

It's not true for non-compact $M$. Here's a counterexample: Define for $x\in M:={\mathbb R}$ $$f(x):=|1-2e^{-x^2}|.$$ The relevant properties of $f$ are that it is continuous with a unique maximum at $x^\star:=0$, at which $f$ takes value $1$, but also $f(x)$ increases to $1$ quickly as $x\to\infty$.

Now let $t_n$ be a triangle with base width $1$ and height $\frac1n$ centered at $x=n$, and define $f_n:=f+t_n$. Then each $f_n$ is continuous, $(f_n)$ converges uniformly to $f$, but the maximum value for $f_n$ will occur at $x_n=n$ (since $f_n(n)=1-2e^{-n^2}+\frac1n>1$ for every $n$). Thus as $n\to\infty$, $x_n$ escapes to infinity rather than converging to $x^\star$.