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Let $I\subset \Bbb{R}$ be an open interval and consider a continuous function $f:I\to \Bbb{R}$ satisfying, for all $x\in I$ $$\displaystyle \lim_{h\to 0} \frac{f(x+h)+f(x-h)-2f(x)}{h}=0$$ Prove that the set of points at which $f$ is differentiable is dense in $I$.

I know that $$\displaystyle \lim_{h\to 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$$ when $f''(x)$ exists, but it seems to be no use here.

Mathronaut
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    You're right, it does not seem useful, because $f''$ is not guaranteed to exist. And since you will probably get that $f$ is only differentiable on a dense set, $f'$ is not even a priori defined on any subinterval of $I$, let alone differentiable. – Jean-Claude Arbaut Sep 04 '14 at 14:44
  • @Jean-ClaudeArbaut. Do you remember my problem of voters ? Cheers :-) – Claude Leibovici Sep 04 '14 at 14:47
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    For those who might be interested, such functions are often called Zygmund smooth because Zygumnd studied them extensively in a 1945 paper titled Smooth functions. Rajchman proved the result you're asking in 1919, and "dense" was strengthened to "$c$-dense" later by Zalcwasser. In the other direction, David C. Ullrich (yes, the sci.math guy) showed here that given any $F_{\sigma}$ measure zero set $E,$ there exists a continuous Zygmund smooth function having an infinite derivative at each point of $E.$ – Dave L. Renfro Sep 04 '14 at 18:47

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To show the set of points on which $f$ is differentiable is dense, we need to show an arbitrary open subinterval of $I$ contains a point where $f$ is differentiable. Restrict our attention to any open interval in $I$, say $(a,b)$.

We know that $f$, being continuous, attains a maximum on $[a,b]$. Hence, it either attains a maximum in $(a,b)$, say at $x_0$, or at one of the endpoints (so $f(a)$ or $f(b)$ are greater than $f(x)$ for $x \in (a,b)$). Suppose the mxamimum is at $x_0$, contained in $(a,b)$. If we look at your limit at this maximum, we see that it is non-positive ($\leq 0$), and can be $0$ only if $f$ is "constant" near $x_0$, implying it is differentiable. More rigorously:

$\forall \epsilon>0 \ \ \exists \ \ \delta>0$ such that $2f(x_0)-f(x_0+h)-f(x_0-h)<\epsilon |h|$ whenever $|h|<\delta$. This trivially gives $f(x_0)-f(x_0+h)<\epsilon |h|$ (as the other remaining part is non-negative, $x_0$ being a maximum) whenever $|h|<\delta$, implying $f$ differentiable at $x_0$ with derivative $0$.

If on the other hand, our maximum is at one of the end points, then do the following. Try and sub-restrict $f$ to a smaller interval $(c,d) \in (a,b)$. If, for all such restrictions, the maximum is at one of the end-points, then $f$ is monotone on $(a,b)$ and differentiable almost everywhere on $(a,b)$ (Monotone+continuous but not differentiable). If not, then the above argument applies, and we are done.

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ShakesBeer
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    Maximum is guaranteed for closed intervals. – Paramanand Singh Sep 04 '14 at 14:57
  • @ParamanandSingh I included the case where the maximum is one of the end points (obvious what I mean by that). Edit: changed the wording – ShakesBeer Sep 04 '14 at 14:58
  • Your interval to achieve maximum might require it to be compact. In this sense, setting $f=\frac{1}{x}$ might not work for the interval say $(0,1)$. – Novice Sep 04 '14 at 14:58
  • @Novice I'm not sure I see what you mean... – ShakesBeer Sep 04 '14 at 15:03
  • Is it regarding the last comment that you posed here? – Novice Sep 04 '14 at 15:05
  • Oh, I see. Yes, I've made the wording clearer on that issue. – ShakesBeer Sep 04 '14 at 15:07
  • But I think you need the interval to be compact to achieve maximum. – Novice Sep 04 '14 at 15:09
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    I don't follow your argument. What do you mean by the limit is non-positive? It is zero by assumption. Also, where does denseness come in in your argument? – copper.hat Sep 04 '14 at 15:16
  • non-positive I mean $\leq 0$. Also, I've showed denseness by restricting to an arbitrary subinterval. Heck I'm just going to rewrite this if people are going to be so picky – ShakesBeer Sep 04 '14 at 15:18
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    I'm not being picky. I just am unable to understand what you are trying to do. – copper.hat Sep 04 '14 at 15:23
  • @copper.hat I must admit I have an unusually lax style. It should be fine now. – ShakesBeer Sep 04 '14 at 15:38
  • Hi. Your first part is quite informative and clear. I have a small problem for the second part: You have found that the set that f is not differentiable is a set of measure 0. How does that show that the set that f is differentiable is dense? Thanks for the clarification. – Novice Sep 05 '14 at 04:00
  • @Novice "To show the set of points on which $f$ is differentiable is dense, we need to show an arbitrary open subinterval of $I$ contains a point where $f$ is differentiable". I do just this in my proof, although in the last part $f$ is differentiable almost everywhere on our arbitrary open subinterval $(a,b)$ – ShakesBeer Sep 05 '14 at 14:39
  • Oh ok. I see your point. f is differentiable almost everywhere in the arbitrary open subinterval. In other words, in any arbitrary open intervals there exists points such that f is differentiable. Therefore the set that f is differentiable is dense in the open interval I. Thanks. – Novice Sep 06 '14 at 03:09