How to prove that $u(\epsilon') = \inf\{f(s) - f(s+\epsilon'): s \in [0,t]\} > 0$ for strictly decreasing function $f$?

Question

Let $f$ be a monotone strictly decreasing function. Fix $\epsilon' > 0$. How to prove that $u(\epsilon') = \inf\{f(s) - f(s+\epsilon'): s \in [0,t]\} > 0$ ?

Answer 1

Clearly, $$ f(s)-f(s+\varepsilon')>0, $$ for all $\,s\in [0,t]$ and $\varepsilon'>0$. Hence $$ \inf\{f(s)-f(s+\varepsilon'): s\in [0,t]\}\ge 0. $$ Assume that for some $\varepsilon'>0$, $\inf\{f(s)-f(s+\varepsilon'): s\in [0,t]\}= 0$.

This implies that for every $n\in\mathbb N$, there exists an $s_n\in[0,t]$, such that $$ 0<f(s_n)-f(s_n+\varepsilon')<\frac{1}{n} \tag{1} $$ The sequence $\{s_n\}\subset [0,t]$ possesses a convergent subsequence, say $s_{k_n}\to s\in [0,t]$. The $\{s_n\}$ can be chosen to be monotone.

Case I. $\{s_n\}$ decreasing.

Then $s_n\to s^+$ and $f(s_n)\to \lim_{\sigma\to s^+}f(\sigma)$, while $s_n+\varepsilon'\to s^+$ and $f(s_n)\to \lim_{\sigma\to (s+\varepsilon')^+}f(\sigma)$, and according to $(1)$, $$ \lim_{\sigma\to s^+}f(\sigma)=\lim_{\sigma\to (s+\varepsilon')^+}f(\sigma) \tag{2} $$ However, if $s<x<s+\varepsilon'$, then for large enough $n$, $$ s_n<x<s_n+\varepsilon' $$ which implies that $$ f(s_n)>f(x)>f(s_n+\varepsilon') $$ and hence $$ \lim_{\sigma\to s^+}f(\sigma)>f(x)\ge \lim_{\sigma\to (s+\varepsilon')^+}f(\sigma) $$ This contradicts $(2)$.

Case II. $\{s_n\}$ increasing. Treated similarly.

Answer 2

Question has some implied assumptions!

Notice that $ s \to f(s) - f(s+\epsilon') $ is a continuous function that is positive at every $s$. Therefore, it has a minimum on the compact interval $[0,t]$. Now the proof finishes with $$ \inf\{f(s) - f(s+\epsilon'): s \in [0,t]\} = \min_{0\leq s \leq t} u(s) > 0 \, . $$