I'm reading the proof of the uniqueness part of Cauchy-Lipschitz theorem in Brezis' Functional Analysis, i.e.,
Theorem Let $(E, |\cdot|)$ be a Banach space and $F:E \to E$ be $L$-Lipschitz for some $L>0$. Then given any $u_0 \in E$, there exists a unique solution $u \in C^1 ([0, \infty); E)$ of the problem $$ (4) \quad\begin{cases} \frac{du}{dt} (t) &= F(u(t)) \quad \text{on} \quad [0, \infty), \\ u(0) &= u_0. \end{cases} $$
Proof Uniquness Let $u, \overline u$ be two solutions of (4) and set $$ \varphi (t) := |u (t) - \overline u (t)| \quad \forall t \ge 0. $$ Then we have $$ \varphi (t) \le L \int_0^t \varphi (s) ds \quad \forall t \ge 0, \quad (*) $$ and consequently $\varphi \equiv 0$.
Could you please explain how the inequality $(*)$ implies $\varphi \equiv 0$?
Below is the solution replicated from this thread.
Let $$ \psi(t) := e^{-L t} \int_0^t \varphi(s) \, ds \quad t \ge 0. $$
Because $\varphi$ is non-negative and continuous, we get $\psi \in C^1 ([0, \infty); \mathbb R_{\ge 0})$. Furthermore, $$ \begin{align*} \psi'(t) &= -Le^{-Lt} \int_0^t \varphi(s) \, ds + e^{-Lt} \varphi (t) \\ &= -L \psi (t) + e^{-Lt} \varphi (t) \\ &\le -L \psi (t) + Le^{-Lt} \int_0^t \varphi (s) \, ds \\ &= -L \psi (t) + L \psi (t) \\ &= 0. \end{align*} $$
So $\psi$ is non-increasing. On the other hand, $\psi (0)=0$. Then $\psi \le 0$. On the other hand, $\psi$ is non-negative. Then $\psi \equiv 0$.