A function from $f:\mathbb{R} \rightarrow \mathbb{R}$ such that :
a. $f$ is bijective,
b. $f’(0)=1$ (in particular, $f$ is differentiable and therefore continuous at 0), and
c. $f^{-1}$ is not continuous at 0.
I think it exists! But don’t know how to find it.
Thanks for any help!
On
You can even get $f(0)=0$ (which I suspect you intended to require but missed writing). For example, let $A=(0,1)\cap\mathbb Q$ and define $$ g : A \to A : g(p/q) = (pq+1)/q^2 $$ where $p/q$ is in lowest terms, and then $$ f(x) = \begin{cases} g(x) & x\in A \\ x-1 & x>1\text{ and }x-\lfloor x\rfloor \in A \setminus g(A) \\ x & \text{otherwise} \end{cases} $$
The key insight here is that $f$ only needs to be injective near $0$ while satisfying $f'(0)=1, f(0)=1$ -- you can fill in any missed points by letting it behave wildly far from $0$, and if there are missed points arbitrarily close to $0$, this then automatically prevents $f^{-1}$ from being continuous there.
Consider $$f(x)= \begin{cases} x+1 &x\in[-1,+\infty)\\ -x-2 &x\in(-2,-1)\\ x+1 &x\in(-\infty,-2] \end{cases} $$