An almost-Lipschitz vector field has a unique solution through zero but is not locally Lipschitz there

The quotient $\mathrm{\mid }f(x)-f(0)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }=1+\mathrm{\mid }\log \mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }$ is unbounded as $x\to 0$ by [L2], so $f$ is not locally Lipschitz at zero.

Define $\varphi (r)=r(1-\log r)$ for $0<r\le e^{-1}$ with $\varphi (0)=0$, and put $\rho =2\varphi$ on $[0,e^{-1}]$. Differentiation using [L2] shows that $\varphi$ is increasing and concave on this interval. On one side of zero, concavity with $\varphi (0)=0$ gives $\mathrm{\mid }\varphi (a)-\varphi (b)\mathrm{\mid }\le \varphi (\mathrm{\mid }a-b\mathrm{\mid })$; on opposite sides, $\mathrm{\mid }a-b\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$ and monotonicity gives $\varphi (\mathrm{\mid }a\mathrm{\mid })+\varphi (\mathrm{\mid }b\mathrm{\mid })\le 2\varphi (\mathrm{\mid }a-b\mathrm{\mid })$. Since $f$ is the odd extension of $\varphi$ near zero, $\rho$ is therefore a state modulus there. It has the properties in [L3], and the substitution $u=1-\log r$ gives its reciprocal divergence, so [L1] makes the zero solution unique through the origin.