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Let's say $f$ is differentiable at $x$, and $g(y) = x$. Therefore, we have:

$$ f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{f(g(y)+h) - f(g(y))}{h} $$

If $g$ is continuous and strictly monotonic, but not differentiable, at the point $y$, can we substitute $g(y) + h = g(k+y)$, and get the following:

$$ \lim_{k\to 0} \frac{f(g(y+k)) - f(g(y))}{g(k+y) - g(y)} = f'(x) $$

Does the equality hold when $f$ is continuous in a neighbourhood of $y$? If yes, how to prove that equality of limits? If no, how to prove that inequality?

Thanks!

S11n
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    Guess no, in the denominator anything can happen with g not differentiable. Would look into a counterexample e.g. with Weierstrass function. – Sam Ginrich Jan 12 '22 at 11:02
  • How about the case when $g$ is continuous in a neighbourhood (instead at one point)? – S11n Jan 12 '22 at 13:17
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    The Weierstrass function is everywhere continuous and nowhere differentiable S11n – FShrike Jan 12 '22 at 14:52
  • @FShrike Do you know how to prove that inequality? (I don't) – S11n Jan 13 '22 at 09:41
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    When you introduce new functions you are altering the limit process. Try differentiating $f(x)=x$ at $x=0$ and let $g(x)=|x|$. The process will produce two answers for the lefthand and righthand limits – FShrike Jan 13 '22 at 09:50
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    If $f(x) = x, g(x)=|x|$, then the equality still holds: $\lim_{k\to 0} \frac{f(g(0 + k)) - f(g(0))}{g(0 + k) - g(0)} = \lim_{k\to 0} \frac{g(k)}{g(k)} = 1 = f'(0)$. I think the problem is actually when we take $g$ to be a constant function, because the required $k$ does not exist. Therefore, if $g$ would be continuous and monotonic (that excludes $|x|$), the equality should probably hold, but I look forward to be proven wrong, if you can do that (e.g. in a full answer). – S11n Jan 14 '22 at 10:59
  • ... thought about "g injective" which is nearly the same https://math.stackexchange.com/questions/279435/prove-that-if-a-continuous-function-is-injective-then-it-is-monotonic . – Sam Ginrich Jan 19 '22 at 00:16
  • Yes, it's nearly the same thing. It seems that if $g$ is monotonic and continuous, it is also differentiable. Is that a known result? Is there a proof? – S11n Jan 21 '22 at 10:05
  • ... from continuous and monotonous .... not yet. Together with Lipschitz-continuous, there is a thread, saying ´differentiable up to a Lebesgue-0-set´... will find it again. – Sam Ginrich Jan 22 '22 at 16:28
  • Actually, I take back what I wrote. If f(x) = x for x < 1, f(x) = x^3 for x >= 1, the function is continuous and monotonic, but not differentiable at 1. – S11n Jan 24 '22 at 16:49
  • ... it's here https://math.stackexchange.com/questions/2444065/does-lipschitz-continuous-function-only-has-a-countable-non-differentiable-point – Sam Ginrich Jan 26 '22 at 18:48

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