Let $f \colon [0, \infty] \to [0, \infty]$ be a monotonically decreasing function.

Given any two continuous non-negative real-valued functions $g$ and $h$ defined on a metric compact space $X$, I am curious that could we get the following result $$ \left|\, f \left[ \max_{x \in X} g(x) \right] - f \left[ \max_{x \in X} h(x) \right] \,\right| \leq \max_{x \in X}\, \bigg| \, f \left[ g (x) \right] - f \left[ h (x) \right] \, \bigg| \quad?$$

In fact, I tried to verify the above inequality with some concrete examples and the result indicates that this inequality does hold so far.

However, I have not figured out yet how to proof it. Thus, does this inequality really hold true?

Could anyone give me some hint or advice please? Any idea or suggestions are really appreciated! Thank you so much in advance!


1 Answer 1


$f$ must be strictly decreasing because it is bijective. Therefore it is continuous (adapt the answer of Strictly increasing continuous function) and we conclude that it is a homeomorphism such that $f([0,\infty)) = (0,\infty])$.

You have $\max g(x) = g(m)$ for some $m \in X$ because $g$ is continuous and $X$ is compact. Therefore $f(\max g(x)) = \min fg(x)$ because $g(x) \le g(m)$ implies $fg(x) \ge fg(m)$.

Hence your question is equivalent to the following: If $G,H : X \to (0,\infty]$ are continuous, do we have $\lvert \min G(x) - \min H(x) \rvert \le \max \lvert G(x) - H(x) \rvert$? This is seen by considering $g = f^{-1}G, h = f^{-1}G : X \to [0,\infty)$.

Let us first consider the case that $G(X),H(X) \subset (0,\infty)$. Then $M = \max \lvert G(x) - H(x) \rvert < \infty$.

If $M = 0$, then $G = H$ and the claim is true. Now let $M > 0$. This means $M \ge G(x) - H(x) \ge - M$ for all $x$, i.e. $H(x) + M \ge G(x)$ and $G(x) + M \ge H(x)$ for all $x$. Choose $x_G, x_H \in X$ such $G(x_G) = \min G(x)$ and $H(x_H) = \min H(x)$. Assume that $\lvert G(x_G) - H(x_H) \rvert > M$. Let $G(x_G) > H(x_H)$ (the case $G(x_G) < H(x_H)$ is treated similarly). This means $G(x_G) - H(x_H) > M$. Hence $G(x_G) > H(x_H) + M \ge G(x_H)$ which contradicts the definition of $x_G$. Therefore the assumption was wrong and we conclude $\lvert G(x_G) - H(x_H) \rvert \le M$.

If $\infty$ is in the image of $G$ or $H$, we have the problem that we get expressions like $\infty - \infty$, $\infty - t$, $t - \infty$ with $t \in (0,\infty)$. It seems to be reasonable to define $\infty - t = \infty$ and $t - \infty = -\infty$ and $\lvert \infty \rvert = \lvert -\infty \rvert = \infty$, but also other approaches are conceivable. Moreover, $\infty - \infty = 0$ could be a definition, but in this case $G - H$ will in general not be continuous (if that should be what you expect): Consider for example $X = [1,\infty], G(x) = x, H(x) = x + 1$ where we understand $\infty + 1 = \infty$.


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