Let $f:\mathbb{R}^2\to[-2,2]$ be a continuous function that is 2-Lipschitz: $ \lvert f(t,y_1)-f(t,y_2) \rvert \leq \dfrac{1}{2}\lvert y_1 - y_2 \rvert $ for each $(t,y_1),(t,y_2)\in\mathbb{R}^2$.
Use Gronwall's lemma to prove that considering the following Cauchy's problem $y' = e^{f(t,y)}, y(1)=\alpha\in\mathbb{R}$, there is continuous dependence of solutions on initial conditions ($\alpha$).
We call $y(t,1,\alpha)$ the solution of the previous problem. We have to check if
$\forall \varepsilon > 0, \exists \delta > 0 : \lvert \alpha - \beta \rvert < \delta\Longrightarrow \lvert y(t,1,\alpha) - y(t,1,\beta)\rvert < \varepsilon$
As $y(t,1,\alpha), y(t,1,\beta)$ are solutions to the IVP:
$\lvert y(t,1,\alpha) - y(t,1,\beta)\rvert=\lvert \alpha + \int_1^t e^{f(s,y(s,1,\alpha))}ds - \beta - \int_1^t e^{f(s,y(s,1,\beta))}ds\rvert \leq \lvert \alpha-\beta\rvert + \lvert\int_1^t(e^{f(s,y(s,1,\alpha))}-e^{f(s,y(s,1,\beta))})ds\rvert$
If I prove that $e^{f(t,y)}$ is 2-Lipschitz, then I can define the function used in Gronwall's lemma as $G(t)=\rvert y(t,1,\alpha) - y(t,1,\beta) \rvert$, and conclude the exercise, as Gronwall's lemma guarantee $G(t) \leq Ae^{B\lvert t - 1\rvert}$.
Is $e^{f(t,y)}$ a 2-Lipschitz function?
The answer is yes:
$\lvert e^{f(t,y_1)} - e^{f(t,y_2)}\rvert = \lvert \int_{f(t,y_2)}^{f(t,y_1)} e^s ds \rvert = e^c \lvert f(t,y_1) - f(t,y_2) \rvert \leq e^2 \lvert f(t,y_1) - f(t,y_2) \rvert \leq \dfrac{e^2}{2} \lvert y_1 - y_2 \rvert $