# Showing that positive central difference limit implies function is strictly increasing

Suppose $$f:(a,b)\rightarrow \mathbb{R}$$ is continuous, and that $$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$$ Exists for all $$x\in(a,b)$$ and is strictly greater than zero.

How can I show that $$f$$ must be strictly increasing? I am comfortable with the usual MVT proof that a positive derivative implies a function is increasing, however, in this case, I cannot assume that the function is differentiable - so can't use the MVT in its standard form at least.

1. Go right back to foundations and show that the MVT still holds in this case - by showing that Rolle's theorem still holds (does it?).

2. Use the definition of a limit directly with an $$\epsilon - \delta$$ argument, showing that for all $$|h|<\delta$$, $$\frac{f(x+h)-f(x-h)}{2h}>0$$ so somehow the function must be increasing.

Any thoughts?

• Pick 2 point $\alpha>\beta$. Then use MVT. May 30, 2022 at 11:26
• @OğuzhanKılıç But this assumes the function is differentiable - a fact which may not necessarily be true (take $f(x) = |x|$ at 0 for example, for which the central limit exists but we do not have differentiability) May 30, 2022 at 11:28
• May 30, 2022 at 12:35

You can first show that the function $$f$$ is "locally increasing": for every $$x\in(a,b)$$, the assumption implies that $$f(x+h)-f(x-h)>0$$ for small $$h>0$$.
Now consider any compact interval $$[c,d]\subset (a,b)$$. Then $$[c,d]$$ can be covered by finitely many open intervals where $$f$$ is increasing. Since $$[c,d]$$ is arbitrary, $$f$$ must be increasing on $$(a,b)$$.