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Suppose that we have that $f(x)$ is a continuous function over the support of $x$. Assume also that $f(x)$ has a unique maximizer, $x^*$. I would like to show that maximizer is well-separated. In other words:

For every $\epsilon >0$, there exists a $k > 0$ such that:

$$ \sup_{x \ : \ d(x,x^*) \geq \epsilon} f(x) < f(x^*)-k $$

My idea is to construct sequences using a proof by contradiction, but am running into problems there. Is there a more straightforward way to prove this? Thanks!

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    support of $x$? – 2017-02-08
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    It is not true in general. Consider $f(x)=-1/|x| (|x|>1), f(x)=-|x| (-1\le x\le 1)$ – 2017-02-08

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Consider $f:[0,\infty) \to \mathbb R$ defined by $$ f(x) = \begin{cases} 1 - x, & x \le 1 \\ 1 - \frac{1}{x}, & x > 1. \end{cases} $$ Then, for every $\epsilon > 0$ we have $$ \sup_{x \ge \epsilon} f(x) = 1 = f(0). $$

Notes:

It is true, if you assume a compact domain, say $X$. As $X_{\epsilon} = \{ x\in X \mid d(x,x^*) \ge \epsilon \}$ is closed and thus compact, there exists some $x_\epsilon \in X_{\epsilon}$ which maximizes $f$ on $X_\epsilon$ and by uniqueness we have $$ \sup_{x\in X_\epsilon} f(x) = f(x_\epsilon) < f(x^*). $$

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    Thanks, I had meant to assume compactness of the domain. In your last line, where did the $k$ from above go? Did I need to define some sequence on $k$? Thanks – 2017-02-08
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    @user321627: Take $k = (f(x^*) - f(x_\epsilon)) / 2$ for example. – 2017-02-08
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    How does the uniqueness give the final equation? Does it rely on the fact that for continuous $f$, $\sup_{x \in X_{\epsilon}} f(x) \leq \sup_{x \in X} f(x)$ for a subset $X_{\epsilon} \subset X$? – 2017-02-09
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    @user321627: $<$ for any proper subset. Yes. If $f(x_\epsilon) = f(x^*)$ for any $\epsilon>0$, then $x_\epsilon \ne x^*$ would be an other maximizer. – 2017-02-09