I'm reading a proof in which the authors want to show that a certain function $f:[0,T]\to\Bbb{R}_+\cup \{0\}$ is Lipschitz continuous. They prove that $$e^{−C|t−s|}f(t)≤f(s)≤e^{C|t−s|}f(t)$$ for all $t,s\in [0,T]$ and certain $C>0$ and then say that "the result follows".
Why?
I don't even know that $f$ is continuous, a priori. I've tried the following: Suppose $f$ is not Lipschitz. Then there are sequences $t_n,s_n\in [0,T]$ (which I can suppose $t_n\to t\in [0,T]$ and $s_n\to s\in [0,T]$, by compactness) such that the definition of Lipschitz fails for constants, say, $e^n$, $n\in \Bbb N$, i.e. $$|f(t_n)-f(s_n)|>e^n|t_n-s_n|.$$ If $t\neq s$, then the inequality above says that $|f(t_n)-f(s_n)|\to +\infty$ (the images of $t_n,s_n$ by $f$ get arbitrarily far). But then... what? And what if $t=s$? Then I can't say if $e^n|t_n-s_n|\to 0$ or $+\infty$. Furthermore, I have the difficult of working with $f$ without knowing anything about its continuity...
Remark.: If I knew that $f>0$ (strictly positive), then applying the logarithm the proven inequality would be equivalent to $$|\ln (f(t))-\ln (f(s))|\leq C|t-s|.$$ This is not exactly the case (since $f$ can be zero) but I've used this as motivation to the title of this post.
Edit: I've just realized that if $f(t)=0$ for some $t\in [0,T]$, then the proven inequality says that $0\leq f(s)\leq 0$, for any $s\in [0,T]$ and therefore $f\equiv 0$. So we can indeed suppose that $f>0$ as stated in the title.
EDIT: I guess a more elegant approach is to consider that the composition of two Lipschitz functions (with adequate domains) is once again Lipschitz.
Since you've shown $\log \circ f$ is Lipschitz, it is in particular bounded above on compact intervals. Now, $\exp$ had bounded derivative on intervals that are bounded above, and is hence Lipschitz on such intervals. It follows that the composition $\exp\circ \big(\log \circ f\big) = f$ is Lipschitz.
Since $f>0$, we have
$$e^{-C(t-s)} \leqslant \frac{f(s)}{f(t)} \leqslant e^{C(t-s)}$$
Then, letting $s\to t$ we find by the squeeze thereom that
$$\lim_{s\to t}\frac{f(s)}{f(t)} = 1\implies \lim_{s\to t}f(s) = f(t),$$
from which the continuity of $f$ follows. Hence, by the Extreme Value Theorem $f$ atains some positive maximum $M$. Moreover, for all $t\in[0,T)$ and $h$ with $0<h\leqslant T-t$ we have
$$\underbrace{\left(\frac{e^{-Ch}-1}h\right)}_{<0} f(t)\leqslant\frac{f(t+h)-f(t)}h\leqslant \underbrace{\left(\frac{e^{Ch}-1}h\right)}_{>0} f(t)$$
Hence:
\begin{align} \frac{|f(t+h)-f(t)|}h &\leqslant \max\left\{\frac{1-e^{-Ch}}h, \frac{e^{Ch}-1}h\right\}\cdot f(t) \end{align}
Now, $f\leqslant M$, $\left(e^{Ch}-1\right)/h$ is increasing over $h>0$, and $\left(1-e^{-Ch}\right)/h$ is decreasing over $h>0$. Since
$$\lim_{h\to 0} \frac{1-e^{-Ch}}h = -\lim_{h\to 0} \frac{e^{-Ch}-1}h = -\left(\left.\frac{d}{dx}\,e^{-Cx}\right)\right|_{x=0} = C,$$
it follows that
\begin{align} \frac{|f(t+h)-f(t)|}h &\leqslant M\cdot\max\left\{C, \frac{e^{CT}-1}T\right\}, \end{align}
which conludes the proof.