Lemma of Gronwall

Question

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

Answer 1

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?