Always strictly positive function?

Question

Let $f$ be some nonnegative (let's say continuous) function defined on $(a,b)\subset \mathbb{R}$. Assuming we have the tool that, if $f(x)>0$ for some $x\in(a,b)$, then we have $\frac{1}{L} f(x) \leq f(y)$ ($L>>1$) for all $y\in (a,b)\cap [x-f(x),x+f(x)]$, does then the following statement hold: if $f(x)>0$ for any $x\in(a,b)$, then we have $f>0$ in the whole of $(a,b)$ (in fact even $f\geq c$ for some positive constant $c>0$)? In my opinion it should hold as we always stay away from zero with the above inequality - however since each interval $(a,b)\cap [x-f(x),x+f(x)]$ gets smaller I am unsure if we really get the statement in the whole of $(a,b)$ as $f$ might get smaller the closer we get to $a$ or $b$ due to $\frac{1}{L} f(x) \leq f(y)$. I have confused myself! Thanks for your help!

Answer 1

No, there are smooth functions that satisfy your condition but vanish inside $(a,b)$.

As an example, take $(a,b) = (-1,1)$ and $f(x) := \frac{x^2}{2}$. This is a smooth nonnegative function, which obviously vanishes at $x = 0$. Let us check that is satisfies your condition. By symmetry, we only need to verify it for $x \in (0,1)$. For $x \in (0,1)$, $x-f(x) = x - \frac{x^2}{2} > 0$. By monotony, we thus have, for every $y \in (x-f(x),x+f(x)) \cap (-1,1)$, $$f(y) \geq f\left(x- \frac{x^2}{2}\right) = \frac{x^2}{2} \left(1-\frac{x}{2}\right)^2 \geq \frac 1 4 f(x). $$ Hence your condition holds with $L = 4$ (and even, the condition holds with an $L$ which does not depend on $x$).

Hence, your fears were justified, the fact that the interval on which we have information shrinks prevents to draw a global conclusion.