# Always strictly positive function?

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!

• Nobody got an idea? Do I need to give further information? Thanks! Commented Apr 28, 2023 at 6:23

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 $$x$$).