If $f\colon \mathbb{R} \to \mathbb{R}$ is one-to-one and differentiable at $a$ with $f'(a) \ne 0$, must $f^{-1}$ be differentiable at $f(a)$?

Question

I'd like proofs and/or counterexamples, or citation of same, or even just a good hint (I should be able to figure this out!).

I know that the answer is Yes if $f^{-1}$ is at least continuous at $f(a)$ (that's the Inverse Function Theorem in the form cited in Differentiable bijection $f:\mathbb{R} \to \mathbb{R}$ with nonzero derivative whose inverse is not differentiable) or if $f$ is differentiable (or even continuous, or even strictly monotone) on a neighbourhood of $a$ (since this proves continuity of $f^{-1}$, for example by the argument in Continuity of an inverse function. [answer], reducing to the previous case). And I know that the answer is No if $f$ is not assumed to be one-to-one (counterexample: $f(x) = x + x^2 \sin(1/x)$ for $x \ne 0$, $f(0) = 0$, $a = 0$) or if $f'(a)$ is allowed to be zero (counterexample: $f(x) = x^3$, $a = 0$). But in between, there is this case that I do not know.

(Note that this question is different from the one at Differentiable bijection $f:\mathbb{R} \to \mathbb{R}$ with nonzero derivative whose inverse is not differentiable, which asks about what happens when $f$ is differentiable on a neighbourhood of $a$, in which case the answer is Yes as remarked above. But I'm only assuming that $f$ is differentiable at $a$ itself, in which case the answer turns out to be No, as seen in the accepted answer below. The other question did, however, receive an answer to my question in the addendum to its accepted answer.)

You need $f$ to be monotone also in some interval containing $f$. Just being one one won't do. — Paramanand Singh, Oct 14 '17 at 05:46
If $f(x)$ is $x + x^2$ when $x$ is algebraic but $x - x^2$ when $x$ is transcendental (at least for $x$ near $0$), then $f$ is not monotone near $0$ but the hypotheses of the IFT are still met (in particular since $f^{-1}$ is still continuous at $f(0) = 0$). So $f$ does not have to be monotone near $a$ for $f^{-1}$ to be differentiable at $f(a)$, if that's what you mean to say. — Toby Bartels, Oct 14 '17 at 06:11
Well all I am trying to do is to figure out how to guarantee continuity of $f^{-1}$ at $f(a) $. Using monotone $f$ allows us to get this easily. I don't know if just being one one would ensure it. — Paramanand Singh, Oct 14 '17 at 06:46
The question you linked to actually has a counterexample in the second half of the accepted answer. Differentiable bijection $f:\mathbb{R} \to \mathbb{R}$ with nonzero derivative whose inverse is not differentiable — Dap, Oct 14 '17 at 07:53
Thanks, Dap, I didn't appreciate that the comments below that answer had looked at a different question (my question), and that the addendum to the answer had answered it. — Toby Bartels, Oct 17 '17 at 17:34
Although the answer to my question may be found at https://math.stackexchange.com/questions/2336977/differentiable-bijection-f-mathbbr-to-mathbbr-with-nonzero-derivative-w, I don't think that my question should be marked as a duplicate, because they are not the same question. Indeed, the answers are different! — Toby Bartels, Oct 17 '17 at 17:36

Dap · Accepted Answer · 2017-10-14T07:16:57.107

$f^{-1}$ need not be differentiable. I'll use a "just do it" proof.

Define $f(x)=x$ for irrational $x$ and for $x=0.$ Pick an enumeration of the non-zero rationals $r_1,r_2,\dots$ and inductively, assuming we have chosen rationals $f(r_1),f(r_2),\dots,f(r_{i-1}),$ pick $f(x)$ with $x=r_i$ according to the following rules.

if $-1<x<1$ then $f(x)$ is the first unchosen rational with $|f(x)-x|\leq x^2$
if $x$ is a positive integer $n$ then $f(x)$ is the first unchosen rational with $|f(x)|\leq 1/n$
otherwise $f(x)$ is the first unchosen rational

Note the choice always possible in each case. This procedure results in a total function $f$ with the following properties:

$f$ is bijective: each rational is chosen as some $f(x)$
$f'(0)=1,$ because $|f(x)-x|\leq x^2$ for $|x|<1$
$f(n)\to 0$ as $n$ ranges over the positive integers

The last point ensures that $f^{-1}$ is not differentiable: $f^{-1}(f(n))=n$ but $f(n)\to 0.$

Great, thanks! I tried to just do it, but I was focussing too much on $x \approx 0$ and not enough on $f(x) \approx 0$. — Toby Bartels, Oct 17 '17 at 17:30

If $f\colon \mathbb{R} \to \mathbb{R}$ is one-to-one and differentiable at $a$ with $f'(a) \ne 0$, must $f^{-1}$ be differentiable at $f(a)$?

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