Convergence of maximum points on compact sets Suppose $f$ and $f_n$ ($n=1,2,\dots$) are continuous functions $[0,1]\rightarrow\mathbb{R}$, achieving their maxima at unique points $x^*$ and $x^*_n$ respectively. If $f_n(x)\rightarrow f(x)$ for all $x\in [0,1]$, do we also have $x^*_n\rightarrow x^*$?
Here are a few thoughts on the problem:

*

*One can ask a similar question when $[0,1]$ is replaced by $[0,\infty).$ In that case the answer is negative, as the maximum may escape to infinity. This behaviour is prohibited on $[0,1]$ due to compactness, but of course something else might go wrong.

*If $f_n\rightarrow f$ uniformly, then the answer is positive; indeed, say (after selecting a subsequence) $x_n^*\rightarrow y \neq x^*$. Then $\vert f_n(x_n^*) - f(y) \vert \le \vert f_n(x_n^*)-f(x_n^*)\vert + \vert f(x_n^*) - f(y) \vert \le \Vert f - f_n \Vert_{L^\infty[0,1]} + \vert f(x_n^*) - f(y) \vert \rightarrow 0$ and thus taking limits in the inequality $f_n(x^*) \le f_n(x_n^*)$ yields $f(x^*) \le f(y)$. This is a contradiction, as $f$ is assumed to achieve its maximum at a unique point.


Hint
What about $f(x) = x$ and $f_n = \sup(f, 2g_n)$ where
$$g_n(x)=
\begin{cases}
nx & 0 \le x \le 1/n\\
2-nx & 1/n \le x \le 2/n\\
0 & x \ge 2/n
\end{cases}$$
