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The fact that so many students confuse functional inverse notation $$f^{-1}(x)$$ with multiplicative inverse notation $$[f(x)]^{-1}$$ got me to thinking... does there exist a function whose inverse is its inverse? That is, is there a function $f:\mathbb R_+\mapsto \mathbb R_+$ whose functional inverse is also its multiplicative inverse, so that $$f^{-1}(x)=[f(x)]^{-1}, \space\space\space \forall x\in\mathbb R_+$$ Any ideas? I'll impose the restriction of continuity to deter nasty solutions.

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marked as duplicate by Hans Lundmark, MichaelChirico, Community Jan 5 '18 at 0:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Well, since $\frac1{f(f(x))}=x$, necessarily domain and range of $f$ cannot contain $0$. At best, then, $f:\Bbb R\setminus\{0\}\to\Bbb R\setminus\{0\}$. $\endgroup$ – user228113 Jan 4 '18 at 0:19
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    $\begingroup$ @G.Sassatelli But with the requirement of continuity, it can only go from half of the real line to the other half. $\endgroup$ – Frpzzd Jan 4 '18 at 0:22
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    $\begingroup$ @HansLundmark: This question imposes the restriction of continuity, whereas the other question you link to does not, so this does not seem to be a duplicate. Sammy Black's answer to the other question does consider differentiability, but even that is not (quite) a duplicate. $\endgroup$ – Rory Daulton Jan 4 '18 at 9:44
  • $\begingroup$ There were also some other similar posts: A function $f:\mathbb{R}\to\mathbb{R}$ such that $f^{-1}(x)=\frac{1}{f(x)}$ or When is $f^{-1}=1/f$? However, the domain is $\mathbb R$ and $\mathbb C$, respectivelly. Found using Approach0. $\endgroup$ – Martin Sleziak Jan 4 '18 at 12:46
  • $\begingroup$ In fact, despite the question stating that it is about functions on $\mathbb C$, the paper mentioned in this answer might be interesting in connection with functions defined in $(0,\infty)$. (Russell Euler and James Foran. "On Functions Whose Inverse Is Their Reciprocal." Mathematics Magazine Vol. 54, No. 4 (Sep., 1981), pp. 185-189. jstor.org/stable/2689629) $\endgroup$ – Martin Sleziak Jan 4 '18 at 12:54
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No, it is impossible.

If $f: (0,\infty) \to (0,\infty)$ is continuous and $f^{-1}$ exists, then $f$ is either increasing or decreasing. If $f$ is increasing, $f^{-1}$ is increasing but $1/f$ is decreasing. If $f$ is decreasing, $f^{-1}$ is decreasing but $1/f$ is increasing.

EDIT: However, for $f: \mathbb R \backslash \{0\} \to \mathbb R \backslash \{0\}$ it is possible. Take

$$ f(x) = \cases{ -x & if $x > 0$\cr -1/x & if $x < 0$\cr} $$

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  • $\begingroup$ Nicely done! $\space$ $\endgroup$ – Frpzzd Jan 4 '18 at 0:38
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    $\begingroup$ The linked thread gives a (discontinuous) example $f: (0,\infty) \to (0,\infty)$. $\endgroup$ – Jeppe Stig Nielsen Jan 4 '18 at 8:57
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Call $g=\ln\circ f\circ \exp:\Bbb R\to\Bbb R$. Then, $g^{-1}=\ln\circ f^{-1}\circ \exp=\ln\frac1{f\circ \exp}=-g$.

So, we want the homeomorphisms $g:\Bbb R\to\Bbb R$ such that $g^{-1}=-g$. But the homeomorphism $\Bbb R\to\Bbb R$ are strictly monotone continuous functions. And, if $g$ is strictly monotone, $g^{-1}$ must be monotone of the same sign. This is not consistent with $g^{-1}=-g$. Thus, there is no such $g$ and no such $f$.

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