# Lemma of Gronwall

Let $T \in \mathbb R^+\cup \{\infty\}$, $t_o \in [0,T)$, $a,b \in L^\infty(t_0,T)$ and $\lambda \in L^1(T_0,t)$, $\lambda(t) \geq 0$ for almost all $t \in (t_o,T)$. From the inequality $$a(t) \leq b(t) + \int_{t_0}^t\lambda(s)a(s)ds \,\,\,\,\,$$ a.e. in $(t_o,T)$ it follows $$a(t) \leq b(t) + \int_{t_0}^t e^{\phi(t)-\phi(s)}\lambda(s)b(s) \,ds$$ for almost all $t\in (t_0,T)$ where $\phi(s):=\int_{t_o}^s \lambda(\tau)\,d\tau$.

Is there a counterexample for the case that $\lambda$ is negative? And what changes about the implication if $t<t_o$?

• Related – Giuseppe Negro Jan 16 '17 at 16:33
• (-1) downvote because this is directly from a homework problem and you haven't shown any effort of thinking of a solution yourself. – Viktor Glombik Jan 22 at 20:48

## 1 Answer

Here's a simple example of what happens when $\lambda$ is not non-negative.

Take $t_0 =0$, $T=\infty$, and $b(t) =1$ and $\lambda(t) =-1$ for all $t \ge 0$. Further assume we have the equality $$a(t) = b(t) + \int_0^t \lambda(s) a(s) ds = 1 - \int_0^t a(s) ds$$ for $t \ge 0$. Write $F(t) = \int_0^t a(s) ds$. Then the equality reads $$F'(t) = 1 - F(t) \Rightarrow (e^t F(t))' = e^t \Rightarrow F(t) = 1- e^{-t}.$$ Plugging back in shows that $a(t) = e^{-t}$.

Now let's consider the RHS of the Gronwall inequality. We compute $$b(t) + \int_0^t e^{\phi(t)-\phi(s)} \lambda(s) b(s) ds = 1 - e^t \int_0^t e^{-s} ds = 2 - e^{t}.$$ The inequality then holds if and only if $$e^{-t} = a(t) \le b(t) + \int_0^t e^{\phi(t)-\phi(s)} \lambda(s) b(s) ds = 2- e^{t}$$ for all $t \ge 0$, which is equivalent to $$\cosh(t) \le 1 \text{ for all }t \ge 0,$$ a contradiction.

For your second question I can't really give a good answer, as it's not clear to me what you want to assume. Are we supposed to assume that the first inequality also holds for $t < t_0$? If not, then we have no information about $a$ before $t_0$, so how could we say anything about the function?

• That is very elegant. How to get these perfect contradictions with $\cosh$etc? :D – Tesla Jan 16 '17 at 19:20
• And yes, for the second part we are assuming that the first inequality holds. The question is what one can then imply from that inequality, e.g. what changes about the second inequality. – Tesla Jan 17 '17 at 10:27
• I got \begin{align*} b(t) + \int_{t_0}^{t} \exp(\Lambda(t) - \Lambda(s)) \lambda(s) b(s) ds & = 1 + \int_{0}^{t} \exp\left( \int_{0}^{t} -1 d\tau - \int_{0}^{s} -1 d\tau \right) (-1) \cdot 1 ds \\ & = 1 - \int_{0}^{t} \exp(-(t - 0) + (s - 0)) = 1 - \int_{0}^{t} \exp(s-t) \\ & = 1 - \frac{1}{e^t} \int_{0}^{t} e^{s} ds = 1 - \frac{e^{t} - e^0}{e^t} \\ & = 1 - (1 - e^{-t}) = e^{-t}. \end{align*} What did I do wrong? – Viktor Glombik Jan 22 at 18:48
• For the first task, I'd choose $t_0 = 0$ und $T = \infty$, $a(t) \equiv -1$, $b(t) \equiv 1$ und $\lambda(t) \equiv - 10 < 0$ and for the second $a(t) \equiv 0$, $b(t) \equiv$\lambda(t) \equiv 1\$. – Viktor Glombik Jan 22 at 22:16