Does LIpschitz condition implies boundedness by its constant?

Question

Let $g: \mathbb{R} \to \mathbb{R}$ be a locally Lipschitz function such that $\rho:[t_{0}, t_{1}] \to \mathbb{R}$ is the unique solution of the p.v.i defined by $$ \begin{cases} \rho'(t) & = g(\rho(t)) \\ \rho(t_{0}) & = \rho_{0} \end{cases}. $$ Defining $L := Lip(g\vert_{[t_{0}, M]})$ where $M:= \max_{t \in [t_{0}, t_{1}]}(\rho(t))$, can we say that $\vert \rho'(t) \vert \leq L \vert \rho(t) \vert$? Is obvious that $\rho'(t)$ is bounded, but is that bound correct?

Answer 1

If $$R=\max_{t\in[t_0,t_1]}\|ρ(t)-ρ_0\|$$ and $L$ is a Lipschitz constant on $\bar B(ρ_0,R)$, then indeed you get a bound from the Lipschitz condition, but it should look like $$ \|ρ'(t)\|\le \|f(ρ_0)\|+L\|ρ(t)-ρ_0\|\le \|f(ρ_0)\|+LR. $$ Per the Grönwall theorem a bound on the solution evolution follows, $$ \|ρ(t)-ρ_0\|\le \|f(ρ_0)\|\frac{e^{Lt}-1}L. $$