Does every real function have this weak derivation property?

After this question : Does every real function have this weak continuity property?

Natrualy there are an other (more difficult) :

Is it true that for every real function $f:\mathbb{R}\to\mathbb{R}$, there exists a real sequence $(x_n)_n$, taking values different from $c$, converging to some real number $c$, such that the sequence $\left(\dfrac{f(c)-f(x_n)}{c-x_n}\right)_n$ converges ?

• See auburn.edu/~brownj4/tatras.pdf and for a higher dimensional generalization arxiv.org/pdf/1108.1756.pdf Aug 26 '18 at 14:09
• @Ashutosh: If I'm reading correctly, Theorem 6 in the Brown paper (which is attributed to Ceder) gives a positive answer to this question. Perhaps you'd like to post as an answer? Aug 26 '18 at 16:00
• What you're asking is whether an arbitrary function has, at some point in its domain, a finite derived number. I initially thought the answer could be no for some functions, but the example I had in mind clearly (in retrospect) doesn't work. Perhaps the theorem @Ashutosh and Nate Eldredge mentioned answers the question. However, I don't have time to look now (in a rush to get out the door) --- for related ideas, google the phrases "derived number" and "sequential derivative". Aug 26 '18 at 22:04
• Here's another way to see this. By Denjoy-Young-Saks theorem, for any real function $f$, the set of points $x$ for which $\lim_{h \to 0^+} |f(x + h) - f(x)|/h = \infty$ has measure zero. It follows that for for a.e. $x$, there exists an injective sequence $x_n \downarrow x$ such that $(f(x_n) - f(x))/(x_n - x)$ converges. Aug 27 '18 at 1:07
• @Ashutosh: I think you have enough for a decent answer. To give a bit more detail, the DYS theorem implication you mention follows from the fact that at any a.e. such point $x$ and for each $\delta>0,$ there exists a positive real number $M$ such that infinitely many of the difference quotients with $0<h<\delta$ lie within the compact set $[-M,M].$ Also, "almost all" cannot be strengthened in the sense of size, because given any measure zero set there exists a function (even a continuous and strictly increasing function) that has an infinite derivative at each point of that measure zero set. Aug 27 '18 at 10:55