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This just came to my mind, and I wonder if there is some criterion for this.

Suppose $f$ is differentiable, strictly increasing function and $f^{-1}$ is its inverse. Then, can $f-f^{-1}$ be any differentiable function?

In other words, can we find function $f$ such that the difference between $f$ and $f^{-1}$ is exactly the given differentiable function? For example, is there an increasing function $f$ such that $f(x)-f^{-1}(x)=lnx+e^{x^2}$?

Joshua Woo
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  • @SubhamKarmakar Did you make a typo there? OP is probably only asking us to find the $f$ that satisfies $f - f^{-1} = \ln x + e^{x^2}$. How come you wrote $f$ in the RHS? – bigbang Jun 27 '21 at 06:45
  • ah! yeah, sorry – Subham Karmakar Jun 27 '21 at 06:47
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    What's the cardinality of the set of all differentiable, strictly increasing functions? Is it as great as the cardinality of the set of all functions? (and, doesn't $f-f^{-1}$ have to be differentiable?) – Gerry Myerson Jun 27 '21 at 06:49
  • @GerryMyerson I'm not sure if I got your comment right, but f increasing and $f^-{1}$ doesn't imply that $f-f^{-1}$ is increasing, I think. Also, I'll edit the question as "Can $f-f^{-1}$ be any differentiable function?" – Joshua Woo Jun 27 '21 at 06:56
  • A precise formulation might go like this: Let $I \subset \Bbb R$ be an interval (with finite or infinite endpoints), and $g:I\to \Bbb R$ be differentiable. Is there an increasing, differentiable function $f:I \to \Bbb R$ with $I \subset f(I)$ and $f - f^{-1} = g$ on $I$? – The condition $I \subset f(I)$ makes sure that both $f$ and $f^{-1}$ are defined on $I$. – Martin R Jun 27 '21 at 07:17
  • Here is that question for a concrete case https://math.stackexchange.com/q/3317149/42969. – Martin R Jun 27 '21 at 07:20
  • @MartinR Thanks for finding the post for me. I see that there is some algorithm on finding these sort of questions in one of the answers. I'll read this after I do some work. Again, thanks a lot. – Joshua Woo Jun 27 '21 at 07:27

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