# Showing that a continuous function is greater than zero

I've been working on a problem that wants me to show that given a function $f$ that is continuous at the point $c$ that, $$f(c)>0 \to \exists \delta\;\ \text{such that}\;\ f(x)>0\; \forall x \in(c-\delta,c+\delta)$$

I thought about taking $\epsilon < f(c)$ so that $0<f(c)<2f(c)$, but I'm pretty sure I've come up with a case where this wouldn't work. I know I'm probably missing some simple trick. I would appreciate any hints or advice.

• Why don't you tell us the case where you think this approach wouldn't work? – Michael Albanese Jul 22 '15 at 2:41
• Well I took $f(x)=x^2$ and let $c=2$ which yielded $0<f(x)<8$ for $x \in (2-\delta, 2+\delta)$ I think I may be put off by the fact that I don't have a way of finding an exact $\delta$ that would work , other than by inspection. – boseman Jul 22 '15 at 2:47
• You can find a $\delta$ if you really want, but that's not required. You just have to show the existence of some $\delta$ and you have done that (by using the continuity of $f$). – Michael Albanese Jul 22 '15 at 2:49
• So this proof wouldn't be considered "constructive", but still works? – boseman Jul 22 '15 at 2:52
• I suppose not, but there is no way it could be. You can't express the relationship between $\varepsilon$ and $\delta$ in a way which holds for every continuous function $f$. – Michael Albanese Jul 22 '15 at 3:00

Assume $f(c)>0$, then use $\epsilon := f(c)/2 >0$. By continuity, given any
$\epsilon>0$ , there is a $\delta$ with $$|f(c')-f(c)|< \epsilon$$ when $|c'-c|< \delta$.
Specializing here to $\epsilon =f(c)/2$, we get that $|f(c')-f(c)|< f(c)/2$ when $|c'-c|< \delta$, so that, expanding the expression in terms of $f(c')$....