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Suppose $f:R->R$ is continuous, and that it has a continuous right derivative, i.e. the right-sided limit $$lim(\delta->0^+) (f(x+\delta)-f(x))/\delta$$

exists for all x $\in$ R and defines a continuous function on R.

Prove or find a counterexample: f must be C^1.

Thanks in advance,

Edit: I've tried the Mean Value Theorem (and trying to derive different upper bounds), which now I know is not valid to use. Also, moving some parts around in a difference quotient has not helped. So, I'm currently trying to use the fact that the continuous right derivative, call it g(x), is integrable, and apply the Fundamental Theorem of Calculus. Any help would be greatly appreciated. I will edit again to show my work in progress. Thanks,

Let $$F(x) = \int_a^x g(t)dt$$ where g(t) is the continuous right derivative, gotten from the original function, f.

Since g(x) is continuous, then by the Fundamental Theorem of Calculus, we have that $$F'(x) = g(x)$$ for all x $\in$ R.

I know I haven't said much yet. Not sure where to go from here...

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  • $\begingroup$ You can make your idea into a complete proof by using a sharper version of the fundamental theorem of calculus (math.stackexchange.com/questions/966282/…). The proof given there shows that if $f$ is differentiable from the right everywhere (even up to a countable set) and the derivative is integrable (and $f$ is continuous), then $f(x)-f(y)=\int_x^y f'(t)\, dt$. $\endgroup$ – PhoemueX Nov 22 '14 at 9:14
  • $\begingroup$ Using the FTC, I was able to prove it, but I had to use the fact that continuous right differentiability implies the differentiability of the function (and coupled with the fact that the differentiation operator is linear). On an exam, I may have to justify usage of that fact...- do you have another suggestion to offer? I see a proof of this fact on this website, but it is a bit difficult. Thanks so much, @PhoemueX. $\endgroup$ – User001 Nov 23 '14 at 6:56
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First, I would suggest you look at Darboux's theorem and the definition of a Darboux function in that webpage. If we did in fact have differentiability (possibly not $C^1$, but differentiable nonetheless), then you can easily construct an example of a right continuous Darboux function which is not continuous. I'll leave this construction up to you. Taking $f'$ as your Darboux function will yield your desired contradiction.

Thus, the question boils down to, is the condition you've given equivalent to a derivative?

The answer to that is yes.

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  • $\begingroup$ Thanks for the link, @Rikimaru. As my follow-up question above mentions: I'm wondering whether I could just state this fact that you mention - rather than give a full proof of it... $\endgroup$ – User001 Nov 23 '14 at 6:59
  • $\begingroup$ And, were you suggesting that the assertion is false? It is actually true that f is of class C^1. $\endgroup$ – User001 Nov 23 '14 at 7:03

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