Is there a continuous function $f(x)$ such that the inverse function is $1/f(x)$?

Question

A student came to me with this question and I cracked my head for one hour but I couldn't unambiguously prove that it exists or it doesn't exist. Continuous and invertible function such that $f^{-1}(x) =1/f(x)$ on its domain of definition.

Answer 1

For this answer i will assume that the function is defined on an open interval (otherwise see the answer from @Elliot G ) and real.

Because $\frac1{f(x)}$ is the inverse function, we have $x = f^{-1}(f(x)) = \frac1{f(f(x))} $ and therefore $f(f(x)) = \frac 1 x$.

Note that because $f$ is bijective, it has to be either monotone falling or increasing. if $f$ is increasing, so is $f(f(x))$, but $\frac 1 x$ is not increasing, so it is not possible.

if $f$ is decreasing, the composition $f \circ f$ is still increasing (because the composition of decreasing functions is increasing). So there is no way to do it on an open interval.

Answer 2

You might not like this, but consider $f:\{1\}\to\{1\}$ where $f(1)=1$. This satisfies the conditions since $f$ is technically continuous on the domain.

Answer 3

Given

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

Whence

$$ x = f^{-1}(f(x)) = \frac{1}{f(f(x))}. \tag 2 $$

Thus

$$ f(f(x)) = \frac{1}{x}.\tag 3 $$

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Consider the function

$$ f(x) = \frac{1 + a x}{b + c x}. \tag 4 $$

Then

$$ f(f(x)) = \frac {\displaystyle 1 + a \frac{1 + a x}{b + c x}} {\displaystyle b + c \frac{1 + a x}{b + c x}} = \frac {[b + a] + [c + a^2] x} {[b^2 + c] + [bc + ac] x}. \tag 5 $$

So we get $f(f(x)) = x^{-1}$ for

$$ \left[ \begin{array}{rcl} c + a^2 &=& 0\\ b^2 + c &=& 0\\ \displaystyle \frac{b+a}{bc+ac} &=& 1 \end{array} \tag 6 \right. $$

Therefore $c=1$, $a=b=\pm \mathbf{i}$. Whence

$$ \bbox[16px,border:2px solid #800000] { f(x) = \frac{1 \pm \mathbf{i} x}{x \pm \mathbf{i}}.} \tag 7 $$

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Consider the function

$$ f(x) = x^\alpha. \tag 8 $$

Then

$$ f(f(x)) = \big( x^\alpha \big)^\alpha = x^{\alpha^2}. \tag 9 $$

So we get $f(f(x)) = x^{-1}$ for $\alpha = \pm \mathbf{i}$. Whence

$$ \bbox[16px,border:2px solid #800000] { f(x) = x^{\displaystyle \pm \mathbf{i}}. } \tag {10} $$

I have no idea if there are other possible function.

Answer 4

I think that it is difficult to fulfill. Since $f\circ f(x)=1/x$ we have that $f\circ f \circ f \circ f(x)=x$ therefore $f\circ f$ is an involution, therefore biyective. Therefore $f$ is biyective (?). Note that we are asking for $f$ and $f^{-1}$ to be evaluated on the same $x$, therefore the domain of $f$ and of $f^{-1}$ has to be the same. Hence its derivative (assuming is sufficiently nice function) does not change sign. So we have $(f\circ f)'(x)=f'(x)f'(f(x))>0$. Which is not the case since $f\circ f(x)=1/x$. Apparently I am saying that all doubly iterated monotonous functions are increasing.

Answer 5

Consider the case with real variables.

Suppose $f$ is defined on an interval. Any continuous injection $(a,b)\to\Bbb R$ must be monotone and its square $f\circ f$ must be increasing. But the condition $f^{-1}(x)=f(x)^{-1}$ for all $x\in{\rm dom}(f)$ implies that $f(f(x))=x^{-1}$ holds on some interval, which is decreasing, a contradiction.

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Consider the case with complex variables.

On the Riemann sphere $\widehat{\Bbb C}$ there is the Mobius transformation $\displaystyle f(z)=\frac{iz+1}{z+i}$.

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In general one can define the question for topological groups. The functional condition on $f$ can be recast as a condition on its graph: $(x,y)\in{\cal G}\implies (y^{-1},x)\in{\cal G}$. I haven't been able to make much headway in this direction though. One issue is that I don't think there is any topological condition that characterizes those graphs in $X\times Y$ which are graphs of continuous functions $X\to Y$.

Answer 6

Briefly: Case i) $f$ is strictly increasing. Then $f^{-1}$ is strictly increasing, while $1/f$ is strictly decreasing. So there's no way the relation can hold. Case ii) $f$ is strictly decreasing. Same idea. Either case i) or case ii) must hold, so we're done.