Let $f: \mathbb{R}^{k} \mapsto \mathbb{R}$ be a bounded and continuous function, and suppose that $X \subset \mathbb{R}^{k}$ and $Y \subset \mathbb{R}^{k}$ are compact.
I am reading a paper that claims: \begin{align} \left| \sup_{x \in X} f(x) - \sup_{y \in Y} f(y) \right| \leq \sup_{||x-y|| \leq d_{H}(X,Y)} \left|f(x) - f(y) \right| \end{align} where $d_{H}(X,Y)$ is the Hausdorff distance between $X$ and $Y$: \begin{align} d_{H}(X,Y) = \max\left\{\sup_{x \in X} \inf_{y \in Y} ||x-y||, \: \sup_{y \in Y} \inf_{x \in X} ||x-y|| \right\} \end{align} The paper uses it as an intermediate step in a longer proof, and does not give a proof of this result. I have tried to prove it myself, but have not made much progress (in fact I am not even sure if it is true).
Can anyone provide a proof of (or counterexample to) this result? Much appreciated!
This isn't too hard but it takes a little work to set everything up rigorously if you start from definitions.
For any $\epsilon > d_H(X,Y)$ define $X_\epsilon = \{z \mid \mathrm{dist}(z,X) < \epsilon\}$ and $Y_\epsilon = \{z \mid \mathrm{dist}(z,Y) < \epsilon\}$. By definition of Hausdorff distance you have $X \subset Y_\epsilon$ and $Y \subset X_\epsilon$.
For each $n \ge 1$ define $\epsilon_n = d_H(X,Y) + \frac 1n$.
Fix a point $x \in X$. Then $x \in Y_{\epsilon_n}$, so there exists $y_n \in Y$ with $\|x-y_n\| < \epsilon_n$. In particular $$f(x) \le |f(x) - f(y_n)| + f(y_n) \le |f(x) - f(y_n)| + \sup_{y \in Y} f(y).$$
This works for all $n$. Since $Y$ is compact there exists a subsequence $\{y_{n_k}\}$ that converges to a point $z \in Y$. Since $f$ is continuous you may conclude $$ f(x) \le |f(x) - f(z)| + \sup_{y \in Y} f(y).$$ Moreover since $$ \|x-z\| \le \|x - y_{n_k}\| + \|y_{n_k} - z\| \le d_H(X,Y) + \frac 1{n_k} + \|y_{n_k} -z\|$$ it follows that $\|x - z\| \le d_H(X,Y)$, and thus $$|f(x) - f(z)| \le \sup_{\|x-y\| \le d_H(X,Y)} |f(x) - f(y)|.$$
Finally you get $$f(x) \le \sup_{\|x-y\| \le d_H(X,Y)} |f(x) - f(y)| + \sup_{y \in Y} f(y).$$ The '$x$' on the right-hand side is now a dummy letter - it doesn't represent the fixed '$x$' on the left. You can take the supremum over all $x \in X$ to get what you are looking for: $$\sup_{x \in X} f(x) \le \sup_{\|x-y\| \le d_H(X,Y)} |f(x) - f(y)| + \sup_{y \in Y} f(y).$$
The argument is completed by reversing the roles of $X$ and $Y$ to obtain the corresponding inequality.