I want to prove that if $I$ is an interval in $R$, $c(x) \geq 0$ is continuous and $u \leq 0$ is $C^2$ then if for all $x$ $$u''(x) \geq c(x)u(x)$$ then the strong maximum principle holds for $u$, i.e. if there exists a point $x_0 \in I$ such that $u(x_0) = 0$ then $u$ is identically equal to 0 on the whole interval.
in the notes I'm using there's something like a proof, although it's not complete in my opinion or I just don't get it. it is said that $u$ is a solution of a differential equation $u'' - cu = g(x)$ (which is actually the definition of $g$) and so the Cauchy-Lipschitz theorem ends the proof because the zero function is also a solution. That doesn't seem to work very well because $g$ needs not be $0$ as a function of $x$. on the other hand if we take it to be a function of $(x,u)$ then it's not even well defined with respect to $u$ (because of the derivative) and we have no means of knowing if its Lipschitz with respect to $u$.
can anyone provide any suggestions please?