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I had an exam today, and I was asked about the inverse function theorem, and the exact conditions and statement (as stated in Mathematical Analysis by VA Zorich): Let $X, Y \subset \mathbb{R}$ be open sets and let the functions $f: X \to Y$ and $f^{-1}: Y \to X$ be mutually inverse and continuous at points $x_{0} \in X$ and $y_{0}=f(x_{0})$, respectively. If $f$ is differentiable at $x_{0}$ and $f'(x_{0}) \neq 0$, then $f^{-1}$ is also differentiable at $y_{0}$ and its derivative is $$(f^{-1})'(y_{0})=(f'(x_{0}))^{-1}=\frac{1}{f'(x_{0})}.$$

Then I was asked to come up with an example which would show that the condition that $f^{-1}$ be continuous at $y_{0}$ is not redundant, i.e. that there exists a differentiable bijection $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) \neq 0$ for all $x \in \mathbb{R}$, but whose inverse $f^{-1}$ is non-differentiable at some point $y_{0} \in \mathbb{R}$ (my professor stated that both the domain and range are $\mathbb{R}$, but I'd be open to any example whose domain/range are open subsets of $\mathbb{R}$). I couldn't come up with an answer on the spot, so I was left with a homework assignment, which I'm shamelessly asking for help with here.

Here are some examples that almost, but don't quite fit the bill:

  1. $f(x)=x^3, f:\mathbb{R} \to \mathbb{R}$, whose inverse $f^{-1}(y)=\sqrt[3]y$ is not differentiable at $0$, but the problem is that $f'(0)=0$. If $f'(x) \neq 0$ were removed as a conditions, many counterexamples could be easily found, because any bijection $f$ whose derivative is zero at $x$ implies that $f^{-1}$ is not differentiable at $f(x)$.

  2. An example which is possibly closer to what I'm looking for is the one found in the answer of Functions which are Continuous, but not Bicontinuous, which fits the bill completely except for the domain/range, because $f^{-1}$ is not continuous at 1, let alone differentiable, but its domain is not an open or connected set, so it wouldn't be easy to extend its domain to $\mathbb{R}$ and still keep all of its properties.

I'm aware that the inverse $f^{-1}$ of a continuous bijection $f$ defined on an interval ($\mathbb{R}$ in this case) is also continuous, so the "pathological inverse" that I'm looking for is continuous (unlike example 2), but not differentiable (i.e. "spiky") at a point, even though $f$ isn't "spiky" anywhere.

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  • $\begingroup$ What are $X$ and $Y$? $\endgroup$ – zhw. Jun 26 '17 at 16:31
  • $\begingroup$ The book does not specify, but I assume from context that they're open sets in $\mathbb{R}$. I'll edit it accordingly. $\endgroup$ – Matija Sreckovic Jun 26 '17 at 16:32
  • $\begingroup$ If it helps, your counterexample cannot be continuously differentiable. You can restate the theorem with a $C^1$-function and remove the condition that $f^{-1}$ be continuous at $y_0$. In particular, if $f$ is $C^1$ and $f'$ is nowhere zero, then $f$ is strictly monotone. Every strictly monotone continuous function has a continuous inverse. $\endgroup$ – Tyler Holden Jun 26 '17 at 16:54
  • $\begingroup$ I see, so I'm looking for something with a weird derivative (but which is also monotone), like $x^2 \sin{\frac{1}{x}}$ for $x \neq 0$ and $0$ for $x=0$. Maybe $x+x^2 \sin{\frac{1}{x}}$? $\endgroup$ – Matija Sreckovic Jun 26 '17 at 16:57
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    $\begingroup$ Oh yes, sorry. I was making the same mistake in my head that I had suggested we use; namely, that the derivative is not continuous. [The sum of differentiable functions is certainly differentiable] $\endgroup$ – Tyler Holden Jun 26 '17 at 17:12
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On whether "there exists a differentiable bijection $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) \neq 0$ for all $x \in \mathbb{R}$, but whose inverse $f^{-1}$ is non-differentiable at some point $y_{0} \in \mathbb{R}$": I don't think this is possible.

Proof: Recall that every continuous bijection on $\mathbb R$ has a continuous inverse. So certainly this holds for $f.$ Consider the difference quotient

$$\tag 1 \frac{f^{-1}(y) - f^{-1}(y_0)}{y-y_0} = \frac{f^{-1}(y) - f^{-1}(y_0)}{f(f^{-1}(y))- f(f^{-1}(y_0))}.$$

As $y\to y_0,$ $f^{-1}(y) \to f^{-1}(y_0).$ Thus $(1)$ converges to the familiar $\dfrac{1}{f'(f^{-1}(y_0))}$ and we're done.

Added later: With reference to the comments below, I found the following example: On $(-1,2)$ define $f(x) = x$ on $(-1,0].$ On $(0,1)$ we do something more complicated while keeping $f(x)$ trapped between $g(x) = x$ and $h(x) =x+x^2.$ In so doing we are guaranteed $f'(0)=1.$

On the interval $I_n =(1/(n+1),1/n)$ define $f$ to equal the line through $(1/(n+1), h(1/(n+1))$ and $(1/n,g(1/n)).$ Then $f(I_n) = (h(1/(n+1),g(1/n)).$ Verify that $f$ is between $g$ and $h$ on each $I_n.$ Also notice that the intervals $f(I_n)$ have gaps betweem them. For example $f(I_1) = (3/4,1),$ $f(I_2) = (4/9,1/2).$

So we've defined $f$ on $(-1,1).$ Now there is a bijection from $[1,2)$ onto all the above-mentioned gaps, i. e., onto $(0,1)\setminus \cup_{n=1}^{\infty}f(I_n).$ Define $f$ to be this bijection on $[1,2).$

Then $f$ maps $(-1,2)$ bijectively onto $(-1,1),$ $f(0)=0,$ $f'(0) = 1,$ but $f^{-1}$ fails to be continuous at $0$ (because there are sequences $\to 0^+$ that $f^{-1}$ sends to $[1,2)$).

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  • $\begingroup$ I think, there might be an issue with your interpretation of the problem. The way I read the question, continuity of $f$ is assumed only at one point, $x_0$. Your answer assumes continuity on an open interval. With my interpretation, there are indeed examples that he was asked to construct. $\endgroup$ – Moishe Kohan Jun 27 '17 at 16:45
  • $\begingroup$ @MoisheCohen I was addressing the problem in the paragraph that begins "Then I was asked to come up with an example ..." $\endgroup$ – zhw. Jun 27 '17 at 16:50
  • $\begingroup$ Oh, I see, I did not read the question closely. $\endgroup$ – Moishe Kohan Jun 27 '17 at 16:53
  • $\begingroup$ @MoisheCohen I'm glad you brought it up, because there are a number of situations mentioned in the OP. I'll edit to make this clearer. $\endgroup$ – zhw. Jun 27 '17 at 16:56
  • $\begingroup$ @MoisheCohen Even though I got the answer to the question I asked here, I'm very interested in any example that sheds light on the necessity of the "continuous at $y_{0}$" condition. I left a reply for you in the OP comments section, but I see you deleted your comment from there, so I'll address you here too: could you share such an example? Either in this thread or in a new one if you think it's better. $\endgroup$ – Matija Sreckovic Jun 27 '17 at 17:09
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After some thoughts and discussions, I've come to the unfortunate realization that such a function ($f:\mathbb{R} \to \mathbb{R}$ a differentiable bijection, $f'>0$ WLOG (which is equivalent to $f' \neq 0$ because of Darboux's mean value theorem), $f^{-1}$ not everywhere-differentiable), does not exist, for a rather trivial reason.

If $f:\mathbb{R} \to \mathbb{R}$ were a differentiable bijection, it would be a continuous bijection, and therefore by the invariance of domain theorem, $f^{-1}$ is also continuous on all of $\mathbb{R}$, so all the conditions of the inverse funciton theorem are met, therefore $f^{-1}$ is differentiable everywhere.

PS: I'm not touting my horn by answering my own question, and zhw's answer is simpler than mine and correct, so I've accepted it. I just happened to write it out at the same time as them.

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    $\begingroup$ The invariance of domain theorem is very easy on $\mathbb R,$ and very difficult on $\mathbb R^n,n>1.$ $\endgroup$ – zhw. Jun 26 '17 at 17:53
  • $\begingroup$ Does it reduce to math.stackexchange.com/questions/672174/… for $n=1$? $\endgroup$ – Matija Sreckovic Jun 26 '17 at 17:56
  • $\begingroup$ Probably. It is a standard topology/real analysis result. $\endgroup$ – zhw. Jun 26 '17 at 18:06
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    $\begingroup$ If $U,V$ are open in $\mathbb R$ and $f:U\to V$ is a continuous bijection then for $x\in U$ let $x\in (a,b$) where $[a,b]\subset U.$ Then $f([a,b])$ is connected & compact (because $[a,b]$ is). So $f([a,b])=[c,d]$ with $c<d.$.... And $f(x)\not \in \{c,d\}$ (else by the MVT $f$ is not $1$-to-$1.)$... A continuous bijection from a compact $T_2$ space to another is a homeomorphism. So $f|_{[a,b]}$ is a homeomorphism .... So $f^{-1}|_{(c,d)}$ is continuous, with $f(x)\in (c,d)\subset V.$ Therefore $f^{-1}$ is locally continuous at every $f(x)\in V$ and therefore $f^{-1}$ is continuous. $\endgroup$ – DanielWainfleet Jun 27 '17 at 0:11

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