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I came across this problem while I was studying for a preliminary exam and now I've devoted quite some time to it and can't figure it out. Any help would be greatly appreciated!

Let $f : \mathbb R \to \mathbb R$ be a Lipschitz function such that for all $x \in \mathbb R$, $$\lim_{n\to\infty} n \left[f\left( x + \tfrac 1 n \right) - f(x)\right] = 0.$$ Prove that $f$ is differentiable.

EDIT: As has been pointed out in the comments, simply making the substitution $h=1/n$ is not sufficient. If we know a priori that $f$ is differentiable at $x \in \mathbb R$ then certainly the above limit gives the derivative at $x$. However, to prove that $f$ is differentiable at $x$, we need to prove the stronger assertion: that the limit $$\lim_{n\to \infty} \frac{f(x + h_n) - f(x)}{h_n}$$ exists for any sequence $\{h_n\}$ with $h_n \to 0$ (not just $h_n = 1/n$).

Addressing a comment: the brackets in the problem statement are not meant to indicate the "floor" function; they are just meant to be parentheses. Sorry for the confusion.

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  • $\begingroup$ The discussion here seems to be leading to this being the definition of the derivative, which it is clearly not. However, since $f$ is Lipschitz it does have a derivative a.e. and this condition can be used to deduce that $f'=0$ almost everywhere (i.e., at those points at which you know that $f'(x)$ exists). Accordingly $f$ is constant (which says rather more than that it is differentiable). $\endgroup$ – B. S. Thomson Jan 22 '16 at 18:40
  • $\begingroup$ Thank you for your comment. I'm not sure that a function having derivative defined and zero almost everywhere is necessarily constant though (Cantor step function). $\endgroup$ – User8128 Jan 22 '16 at 18:56
  • $\begingroup$ A continuous function with a zero derivative a.e. need not be constant as you note. A Lipschitz function, however, must be constant if it has a derivative that vanishes almost everywhere. That seems to me to be the point of the exercise--I can't, for the moment, think of a direct simple proof of your problem that doesn't use this fact. $\endgroup$ – B. S. Thomson Jan 22 '16 at 19:10
  • $\begingroup$ Check below. zhw has just added a correct solution to your problem. If you don't know yet about absolute continuity (and that Lipschitz functions are absolutely continuous) then you were attempting, I would say, a review problem outside your skills. If you did already learn this stuff -- well now you know what you need to review! $\endgroup$ – B. S. Thomson Jan 22 '16 at 19:20
  • $\begingroup$ The problem asks to show $f$ is differentiable. Should that not mean on all $\mathbb{R}$ in contrast to the almost everywhere from the Lebesgue / Rademacher theorem? $\endgroup$ – mvw Jan 23 '16 at 20:14
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Since $f$ is Lipschitz on $[a,b],$ $f$ is absolutely continuous on $[a,b].$ Thus $f'(x)$ exists a.e. in $[a,b],$ $f'\in L^1[a,b],$ and

$$f(x) = f(a) + \int_a^xf'(t)\,dt, \,\,x\in [a,b].$$

Since the assumption in this problem gives $f'(x) = 0$ wherever $f'(x)$ exists, the above integral is $0$ for all $x\in [a,b].$ Thus $f$ is constant on $[a,b],$ and since $[a,b]$ is arbitrary, we have $f$ constant on $\mathbb R$ as desired.

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  • $\begingroup$ "Since the assumption in this problem gives $f'(x) = 0$ wherever $f'(x)$ exists" as far as I seem to understand it, the assumption by itself is weaker, that claim has to inferred, most likely using that given sequence limit. $\endgroup$ – mvw Jan 23 '16 at 20:27
  • $\begingroup$ If $f'(x)$ exists, then $f'(x) = \lim_{n\to\infty} n \left[f\left( x + \tfrac 1 n \right) - f(x)\right] = 0,$ right? $\endgroup$ – zhw. Jan 23 '16 at 20:47
  • $\begingroup$ We know it exists, and that limit corresponds to one of the feasible limit sequences, so uniqueness of the derivative value would nail it to zero. $\endgroup$ – mvw Jan 24 '16 at 9:50

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