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?


1 Answer 1


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 $$\tag{1*} g(t)-g(s)=\int_s^t\left( -ag(r)+be^{-cr} \right) dr \quad \text{for all $\, s<t\, $ in $\, I$.} $$ 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 $(2^*)$.


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