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Consider the function $$ f(x)=\left\{\begin{array}{rll} 1+x^2 & \text{if} & x \,\,\text{rational} \\ -x^2 & \text{if} & x \,\,\text{irrational}\end{array}\right. $$ Then, for $x=0$, the limit $\lim_{h\to 0}\dfrac{f(h)-f(-h)}{2h}$ exists, although $f$ nowhere continuous.

Consider now the function $$ f(x)=\left\{\begin{array}{rll} 1 & \text{if} & x=0 \\ 0 & \text{if} & x\ne 0\end{array}\right. $$ Then the limit $\lim_{h\to 0}\dfrac{f(x+h)-f(x-h)}{2h}$ exists, for every $x$, although $f$ is not continuous at $x=0$. This example can be generalised, and obtain an $f$ which is discontinuous in countably many points (for example all the rationals), while the central difference converges.

Suppose now that limit $\lim_{h\to 0}\dfrac{f(x+h)-f(x-h)}{2h}$ exists for every $x$ is some open interval. Does this imply that $f$ is not differentiable in at most countably many points?

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Yiorgos S. Smyrlis
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    For your first example, the limit does not exist for any $x\not \in \Bbb Q .$ Take $y>x$ with $y\in \Bbb Q$ and $y-x$ arbitrarily small and let $h=y-x.$ Then $x+h\in \Bbb Q$ but $x-h\not \in \Bbb Q$ because $(x+h\in \Bbb Q\land x-h\in \Bbb Q)\implies 2x=(x+h)+(x-h)\in \Bbb Q.$ – DanielWainfleet Mar 22 '18 at 23:07
  • It is a good Q but I don't have an A. – DanielWainfleet Mar 22 '18 at 23:11
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    There is a lot of material in the literature on functions which have symmetric derivatives at every point. You can do a google search for "symmetric derivative". You may also download an AMS article by L Larson in the google results page. – Kavi Rama Murthy Mar 23 '18 at 07:33
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    I posted some information about the set of points where $f$ has a finite symmetric derivative but not an ordinary finite derivative in this 2 October 1999 sci.math post. See also this 24 August 2006 follow-up post. Incidentally, without any assumptions (such as $f$ is measurable), a function such as this can have a symmetric derivative of $0$ at every point and yet still be discontinuous everywhere. – Dave L. Renfro Mar 24 '18 at 22:59
  • I'd like to ask for a clarification in the question: The limit existing in open interval in principle implies nothing, do you also want to have the assumption that $f$ is not continuous on $x$? – Mefitico Mar 27 '18 at 21:03
  • No initial assumption on $f$. – Yiorgos S. Smyrlis Mar 27 '18 at 22:17
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    @DaveL.Renfro: could you clarify the example of a discontinuous everywhere function with symmetric derivative $0$ everywhere? If $f$ is additive ($f(a+b)=f(a)+f(b)$) then it will have a symmetric derivative at any point if and only if $f(h)/h$ converges as $h\to 0,$ which can only happen if the graph is not dense. – Dap Apr 16 '18 at 11:38
  • @Dap: Wow, I really messed this up! Until 1982, it was not known whether an everywhere symmetrically differentiable function must be Lebesgue measurable. In 1927, Khintchine (p. 217 in this paper) proved that if the function is measurable and has a finite symmetric derivative everywhere (indeed, he only assumed the lim-sup of the symmetric difference quotient is finite everywhere), then the function has a finite ordinary derivative almost everywhere (i.e. all but a Lebesgue measure zero set). Many improvements and generalizations of this result (continued) – Dave L. Renfro Apr 16 '18 at 16:39
  • were obtained over the years but they included the assumption that the function was measurable (or more). For example, Mukhopadhyay proved in this 1964 paper that for continuous functions, the exceptional set is both measure zero and first Baire category. This measurability assumption was removed by Jaromír Uher in 1982 (paper received by editors 14 October 1982) in this paper, where he proved a result that implies that for arbitrary functions the exceptional set has measure zero. (continued) – Dave L. Renfro Apr 16 '18 at 17:24
  • In fact, Uher's result also implies (using other people's results, and I don't believe he explicitly said this in his paper) that for an arbitrary function the exceptional set is both measure zero and first Baire category, even $\sigma$-upper (i.e. lim-sup defined) porous (using previously proved results in this paper). This has been slightly sharpened by L. Zajicek in 1993 (continued) – Dave L. Renfro Apr 16 '18 at 17:37
  • and in 1999, and by others. Digging through these and other results is complicated by the fact that most don't involve just the assumption of a symmetric derivative everywhere (sometimes "finite" and sometimes "finite or infinite"), but typically they are results about the ordinary differentiability (sometimes just upper/lower, or even Dini derivates) at those points where the symmetric derivative (sometimes just upper/lower) exists (either "finitely" or "finitely or infinitely"). – Dave L. Renfro Apr 16 '18 at 17:39
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    Dave L. Renfro, why don't you move all your comments to an answer? – Gio67 Apr 21 '18 at 13:44
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    @Gio67: Yesterday I wrote a brief survey of relatively recent results (starting with mid 1970s) concerning the set of points at which a continuous function, that is everywhere finitely symmetrically differentiable, fails to have a finite ordinary two-sided derivative. See my answer to Existence of the derivative at a point is implied by a version of the symmetric derivative plus continuity. – Dave L. Renfro Feb 23 '19 at 08:34

1 Answers1

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Let $$g(x,h) = \dfrac{f(x+h)-f(x-h)}{2h},\quad g_0(h) = g(x_0,h).$$ Easy to see that $$g_0(h)=0\ \text{if}\ x_0 = 0,$$ as for any another rational $x_0.$

So the function $g_0(h)=0$ is differentiable in the first case, within $g'_0 = 0.$

In the second case, there is a removable discontinuity of the derivative in the point $x=0$.

In the third case, it can exist the arbitrary (countable) quantity of the removable discontinuities or the gaps of the derivative.

Yuri Negometyanov
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