Preface. Here is the problem I am working on:

Suppose $I$,$J$ are intervals and a monotone onto $f:I\to J$ has an inverse $g:J\to I$. Suppose $x\in I$ and $y:=f(x)\in J$, and that $g$ is differentiable at $y$. Prove:

  1. If $g'(y)\ne 0$, then $f$ is differentiable at $x$.
  2. If $g'(y) = 0$, then $f$ is not differentiable at $x$.

I've been working on this assignment for so long now that I'm at my breaking point. I feel like all I need is a little push in the right direction, that is what kinds of things should I be thinking about to get started with this? I don't want a full solution, just some suggestions and pointers.



By hypothesis,

$$\lim_{h\to0}\frac{g(x+h)-g(x)}h=g'(x)$$ meaning that

$$\forall \epsilon>0:\exists\delta:\forall|h|<\delta\implies\left|\frac{g(x+h)-g(x)}h-g'(x)\right|<\epsilon.$$

As $g$ is invertible, $x=g^{-1}(y)$ and $x+h=g^{-1}(y+l)$ for some $l$, and the last condition writes


Now by choosing $\delta'=\min(|g(x)-g(x-h)|,|g(x+h)-g(x)|)$ (noting that $\forall \epsilon:\exists \delta'$) and by monotonicity, the inequation holds for all $|l|<\delta'$.

The proves that

$${\lim_{l\to0}}\dfrac l{g^{-1}(y+l)-g^{-1}(y)}=\frac1{{\lim_{l\to0}}\dfrac{g^{-1}(y+l)-g^{-1}(y)}l}=\frac1{{g^{-1}}'(y)}=g'(x),$$ unless $g'(x)=0$.

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  • $\begingroup$ Thank you for the hint. I'll try to use this as I think about this problem more. $\endgroup$ – MrStormy83 Sep 16 at 20:47

For the sake of argument lets say that $g$ is differentiable at $y_0 = f(x_0)$ $$ g^\prime(y_0) = \lim_{y\to y_0}\frac{g(y)-g(y_0)}{y-y_0} $$ Now we are going to do a substitution $x=g(y)$. If we remember that the inverse of a monotone function is continuous we can see that $\lim_{y \to y_0}g(y) = g(y_0)$. So according to the substitution theorem

$$ g^\prime(y_0) = \lim_{x\to x_0}\frac{x-x_0}{f(x)-f(x_0)} $$ Substitution theorem implies the existence of the limit on the RHS. If $g^\prime(y_0)\neq0$ then

$$ \frac{1}{g^\prime(y_0)} = \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} = f^\prime(x_0) $$ In case $g^\prime(y_0) = 0$ the limit in the second equality is equal to $0$, so the limit in the third equation cant converge.

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  • $\begingroup$ This is extremely helpful. I appreciate the pointer, I think I can come up with a reasonable proof based on this information. $\endgroup$ – MrStormy83 Sep 16 at 20:46
  • $\begingroup$ Glad I could be of help :) $\endgroup$ – Boxonix Sep 17 at 4:44

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