I'm having a hard time showing that this lemma is true:
Suppose $h\geq 0$ is a continuous non-increasing function on $[0,T]$, $f$ is continuous on $[0,T]$, and $g\geq 0$ is integrable on $[0,T]$. If $$f(t) \geq h(t) - \int_0^t g(s)f(s) \, ds$$ for all $t\in [0,T]$, then $$f(t) \geq h(t)e^{-\int_0^t g(s) \, ds},$$ for all $t\in [0,T]$.
If you change "non-increasing" to "non-decreasing," the $\geq$ signs to $\leq$ signs, and the minus signs to positives (in both the hypothesis and the conclusion), then you have the standard Grönwall inequality, which I can prove using essentially the same proof that's on the Wikipedia page under "Integral Form for Continuous Functions." This initially made me try to use something like $\overline{f}=-f$ and apply the standard Grönwall inequality, but I couldn't get the result to come out.
The problem is that the standard proof strategy does not seem to work out for the above reversed Grönwall inequality since I always end up with the inequality going the wrong direction at the end. I could even get by if I replaced the function $h(t)$ with a constant, but cannot seem to prove it in that case either. Does anyone know how to prove this statement (or, know of a counterexample)?
I would be happy to add more details about what proof strategies I've tried if that would help. I haven't been able to find any reference or web page that mentions this type of inequality anywhere, so even a reference would be appreciated!
EDIT: It seems that the statement may not be true, but I have not found a counterexample. I was able to get around the issue by proving a similar differential inequality:
Suppose $f>0$ is differentiable and $g\geq 0$ is integrable, both on on $[0,T]$. If $$f'(t) \geq -g(t)f(t)$$ for all $t \in [0,T]$, then $$f(t) \geq f(0)e^{-\int_0^t g(s) \, ds}$$ for all $t \in [0,T]$.