Claim: Given ODE $\dot x = f(x)$, $f$ is locally lipschitz, then $x$ must be locally lipschitz as well

Question

Can someone prove or disproof the claim:

Given a locally lipschitz vector field $f$ with associated ODE $\dot x = f(x)$, then the solution $x$ must be locally lipschitz

Note: local lipschitz condition:

$\exists r, L$ s.t. $\forall x,y, |x - x_o | < r, |y - y_o| < r \Rightarrow |f(x) - f(y)| \leq L |x-y|$

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I tried some examples:

$\dot x = 1 \Rightarrow x = t$. Then $f$ is locally lipschitz (in fact globally), and $x$ is locally lipschitz

$\dot x = x \Rightarrow x = x_o \exp(t)$. $f$ is locally lipschitz (but not globally), and $x$ is locally lipschitz (since$ x \in C^1$)

$\dot x = x^2$ and higher ups are a bit too involved...

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If the claim holds, does it also hold for global lipschitz vector fields?

Answer 1

$x$ is differentiable, therefore continuous; $f$ is locally Lipschitz, therefore continuous. Therefore $\dot{x} = f(x)$ is continuous, i.e. $x$ is continuously differentiable. Continuously differentiable functions are locally Lipschitz.