An inequality that is similar to Gronwall but something different Let $f:[0,\infty)\to[0,\infty)$ be a continuous monotonically increasing function such that each $t>0$ has small $\delta>0$ satisfying
$$f(x)-f(x-h)\le hf(x)$$
for all $t-\delta\le h\le t$.
I guess that a function
$$g(x)=e^{-x}f(x)$$
is monotonically decreasing(not strictly).
How can I show that $g$ is monotonically decreasing?
It becomes an easy problem if we assume $f$ is differentiable.
I want to show it NOT assuming differentiability!
I tried to find an elementary direct proof and an approximation method by differentiable functions, but none of them was successful...
Thank you.
 A: Let $x > 0$. We have ($D_{-} g(x)$ is the left-hand lower Dini's derivative)
\begin{align}
D_{-} g(x) = &\liminf_{h\to 0^{+}} \frac{g(x) - g(x-h)}{h}\\
=\ & \liminf_{h\to 0^{+}} \frac{\mathrm{e}^{-x}f(x) - \mathrm{e}^{-(x-h)}f(x-h)}{h}\\
=\ & \liminf_{h\to 0^{+}} 
\left(\frac{\mathrm{e}^{-x}(f(x) - f(x-h))}{h} + \frac{(\mathrm{e}^{-x} - \mathrm{e}^{-(x-h)})f(x-h)}{h}\right)\\
=\ & \liminf_{h\to 0^{+}}
\frac{\mathrm{e}^{-x}(f(x) - f(x-h))}{h} + \lim_{h\to 0^{+}} \frac{(\mathrm{e}^{-x} - \mathrm{e}^{-(x-h)})f(x-h)}{h}\\
=\ & \mathrm{e}^{-x} \liminf_{h\to 0^{+}}
\frac{f(x) - f(x-h)}{h} - \mathrm{e}^{-x}\lim_{h\to 0^{+}} \frac{\mathrm{e}^h - 1}{h} f(x-h)\\
\le \ & \mathrm{e}^{-x} f(x) - \mathrm{e}^{-x} f(x) \\
=\ & 0.
\end{align}
Remarks: i) In $(3) \Rightarrow (4)$, we have used $\liminf A + \liminf B \le \liminf (A + B) \le \liminf A + \limsup B$.
ii) In $(5) \Rightarrow (6)$, we have used $\frac{f(x) - f(x-h)}{h} \le f(x) \Longrightarrow \liminf_{h\to 0^{+}}
\frac{f(x) - f(x-h)}{h} \le f(x)$.
From Ch. 1, Remark 2.2, page 12, in [1], we know that $g(x)$ is decreasing.
[1] J. Szarski, "Differential inequalities". http://www.nsc.ru/interval/Library/ApplBooks/Szarski.pdf
