Inequality in the proof of unique solution of an ODE I'm studying a proof from a unique solution theorem of an ODE where I encountered an inequality. I don't know if this is applied Gronwall, unfortunately I can't see it.

So $y'(s)=f(t,y(s))$ and $f$ is a continuous and Lipschitz $$\|f(t,y)-f(t,z)\|\leq L\|y-z\|
$$
  And we definie this norm $\|y\|_{C^0}=\max (e^{-2(s-t_0)}\|y(s)\|)$

Then this inequality occurs

$$\int_{t_0}^t L\|y(s)-z(s)\|\mathrm{d}s\leq \left(\int_{t_0}^t Le^{2L(s-t_0)}\mathrm{d}s \right)\|y-z\|_{C^0}$$

Is this Gronwall? It confuses me that there is an integral on the LHS.
I computed RHS of the integral but I'm not seeing the inequality:
$$\left(\int_{t_0}^t Le^{2L(s-t_0)}\mathrm{d}s \right)\|y-z\|_{C^0}=\frac{1}{2} \left(e^{2 L (t - t_0)} - 1\right)\max (e^{-2(s-t_0)}\|y(s)-z(s)\|)$$
 A: In the inequality, $y$ and $z$ are solutions of the ODE. Then
\begin{align}
y(t)&=y(t_0)+\int_{t_0}^tf(s,y(s))\,ds,\\
z(t)&=z(t_0)+\int_{t_0}^tf(s,z(s))\,ds.
\end{align}
If $y(t_0)=z(t_0)$ then
$$
\|y(t)-z(t)\|\le\int_{t_0}^t\|f(s,y(s))-f(s,z(s))\|\,ds\le L\int_{t_0}^t\|y(s)-z(s)\|\,ds.
$$
Can you finish from here?
A: This modified norm is part of an "alternative" proof of the Picard-Lindelöf theorem that provides global convergence if there is a global $y$-Lipschitz constant for $f$ on a domain $[a,b]×\Bbb R^n$. In the inequalities in the question, there is nothing more done than defining and applying this norm. Grönwall is only interesting in that it shows that the bound on the divergence of solutions is $e^{L(s-t_0)}$ so that the modified norm (with factor $e^{-2L(s-t_0)}$) will suppress "events" resp. the behavior of the solution at large values of $s$ in favor for values close to the initial point.
The last inequality should thus, using $\|x(s)-y(s)\|\le e^{2L(s-t_0)}\|x-y\|_{C^0}$, read as
\begin{align}
e^{-2L(t-t_0)}\|x_+(t)-y_+(t)\|&\le e^{-2L(t-t_0)}\int_{t_0}^tL\|x(s)-y(s)\|\,ds\\
&\le \|x-y\|_{C^0}\int_{t_0}^tLe^{2L(s-t)}\,ds\\
&=\frac{1-e^{-2L(t-t_0))}}2\|x-y\|_{C^0}
\end{align}
so that
$$
\|x_+-y_+\|_{C^0}\le\frac12\|x-y\|_{C^0}
$$
where $x_+(t)=x_0+\int_{t_0}^t f(s,x(s))\,ds$ etc.
The more common proof (for example reproduced here: Existence of solution to first order ODE. What is there to be proved?) only assumes the Lipschitz condition on a bounded cylindrical neighborhood of the initial point but then only gives you the existence and uniqueness of a solution on a small neighborhood $(t_0-\epsilon,t_0+ϵ)$ where $ϵ$ depends on the scale of $f$ and on its Lipschitz constant.
Both variants of the theorem have their pro's and con's, and in the end they both can serve to provide the background for maximal solutions and flows of vector fields.
For an alternative proof of the first variant without modified norm see for instance https://math.stackexchange.com/a/1587871/115115
A: As of now, this is a non-answer. But other answers suggest that OP has an error, so I wrote it down.
If the norm is given by $\|y\|_{C^0}=\max (e^{-2L(s-t_0)}\|y(s)\|)$, then 
$$\int_{t_0}^t L\|y(s)-z(s)\|ds=\int_{t_0}^t Le^{2L(s-t_0)}e^{-2L(s-t_0)}\Vert y(s)-z(s) \Vert ds$$
$$\leq \int_{t_0}^t L e^{2L(s-t_0)}\max\{e^{-2L(s-t_0)}\Vert y(s)-z(s)\Vert\}ds $$
$$ =\Vert y-z\Vert_{C^0}\int_{t_0}^t Le^{2L(s-t_0)}ds. $$
