For a certain Lipschitz $f,$ must there exist $x_n \downarrow 0$ with $f(x_n),f'(x_n) > 0?$

Question

Let $f:\mathbb{R}\to [0,1]$ be Lipschitz continuous with $f(0)=0$. Assume there does NOT exist $\delta>0$ such that $f(x)=0$ for all $x\in[0,\delta]$.

Under these conditions, can I conclude there exists a stricly decreasing sequence $x_n\to 0$ such that $f(x_n)>0$ and $f'(x_n)>0?$

Answer 1

Hint: Let $U= \{x\in (0,1): f(x) > 0\}.$ Then $U=\cup(a_n,b_n)$ in the usual way. Case 1. Some $a_n =0.$ Case 2. $a_{n_k}\to 0$ along a subsequence. In either case, use the absolute continuity of $f,$ which implies $f(y)-f(x)=\int_x^yf'$ whenever $0\le x <y\le 1.$

Answer 2

Suppose the contrary. Then $ff'\le 0$ a.e. in a right neighborhood of $0$, which implies $f^2$ is non-increasing (because $(f^2)'=2ff'$ and $f^2$ is Lipschitz, hence absolutely continuous). But then $f^2\equiv 0$ in that neighborhood, a contradiction.