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.

I've had two ideas:

  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?

  • $\begingroup$ Pick 2 point $\alpha>\beta$. Then use MVT. $\endgroup$ May 30, 2022 at 11:26
  • 1
    $\begingroup$ @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) $\endgroup$ May 30, 2022 at 11:28
  • $\begingroup$ See math.stackexchange.com/a/2533349/42969 $\endgroup$
    – Martin R
    May 30, 2022 at 12:35

1 Answer 1


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 $(a,b)$.


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