# A simple special case of Gronwall's inequality for Dini derivatives

Let $$I=[t_0,t_1)\subset \Bbb R$$ an interval and $$a,b,c\ge0$$ with $$a>c$$. Assume that $$f\colon I\to\Bbb R$$ is a continuous function with

$$\tag{1} f(t)-f(s)\le \int_s^t\left( -af(r)+be^{-cr} \right) dr \quad \text{for all s,t\in I.}$$

If $$f$$ were differentiable, dividing by $$(t-s)$$ and letting $$s\to t$$ would imply

$$f'(t) \le -af(t)+be^{-ct} \quad \text{for all t\in I},$$

and hence one could apply the differential version of Gronwall's Lemma and arrive at

$$\tag{2} f(t) \le f(t_0)e^{-a(t-t_0)}+\frac{b}{a-c}e^{-at}\left(e^{(a-c)t}-e^{(a-c)t_0}\right) \quad \text{for all t\in I}.$$

Note that, contrary to the classical integral version of Gronwall's Lemma, (1) holds for all $$s,t\in I$$ and not just for $$s=t_0$$ and $$t\in I$$. Hence even if $$f$$ is not differentiable (but continuous!), the inequality (1) still implies

$$\tag{3} Df(t) \le -af(t)+be^{-ct} \quad \text{for all t\in I},$$

where $$D$$ denotes any Dini derivative. However, I do not understand completely how to continue from here in order to prove that (2) holds. In the case $$b=0$$, (3) implies $$D^+(\log f)\le-a$$ and so (2) follows immediately. I am not sure how to deal with the term $$be^{-ct}$$ if $$b$$ does not vanish, though.

So, does (1) (or (3)) imply (2) also for $$b\neq0$$ and if yes, how can we prove it?

Let $$g(t):= f(t_0)e^{-a(t-t_0)}+\frac{b}{a-c}e^{-at}\left(e^{(a-c)t}-e^{(a-c)t_0}\right) \quad \text{for \, t\in I} \,,$$ so $$g(t_0)=f(t_0)$$ and $$\tag{1*} g(t)-g(s)=\int_s^t\left( -ag(r)+be^{-cr} \right) dr \quad \text{for all \, s The goal is to show $$\tag{2*} f(t) \le g(t) \quad \text{for all \, t\in I}.$$ Given $$\epsilon \in (0,1)$$, let $$t_\epsilon:=\min \{t \in I: f(t) \ge g(t)+\epsilon\} \tag{3*}$$ if the set on the right is nonempty, and let $$t_\epsilon=t_1+1$$ otherwise. If $$t_\epsilon \le t_1$$, then we define $$s_\epsilon:= \max\{s \in[t_0, t_\epsilon]: f(s)\le g(s)\}\,.$$ In this case, $$f>g$$ in the interval $$(s_\epsilon,t_\epsilon)$$, so we infer from $$(1)$$ that $$\tag{4*} g(t_\epsilon)+\epsilon-g(s_\epsilon)=f(t_\epsilon)-f(s_\epsilon)\le \int_{s_\epsilon}^{t_\epsilon}\left( -af(r)+be^{-cr} \right) dr$$ $$<\int_{s_\epsilon}^{t_\epsilon}\left( -ag(r)+be^{-cr} \right) dr =g(t_\epsilon)-g(s_\epsilon) \,.$$ This contradiction means that the set on the right hand side of $$(3^*)$$ must be empty, i.e., $$f \le g+\epsilon$$ in $$I$$. Since $$\epsilon>0$$ can be arbitrarily small, this proves $$(2^*)$$.