# If $\,\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}\,$ exists for every $x$, what does this imply for $f$?

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?

• 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 $(x+h\in \Bbb Q\land x-h\in \Bbb Q)\implies 2x=(x+h)+(x-h)\in \Bbb Q.$ Mar 22, 2018 at 23:07
• 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. Mar 23, 2018 at 7:33
• 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. Mar 24, 2018 at 22:59
• @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, 2018 at 11:38
• @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. Feb 23, 2019 at 8:34

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 $x=0$.
• What's $f(x_0,h)$? I don't see why this answers the question. Apr 21, 2018 at 13:42