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Describe all functions which are defined as $f:\mathbb R \to \mathbb R $ and satisfying $$f(x)=\dfrac1{f(x)}$$

My Attempt:

If $f$ is constant function then.

$$(f(x))^2=1$$

Then $f(x)=1$ or $f(x)=-1$

I cannot think non-constant function satisfying this

What about $f$ being continuous?

Jale'de jaled
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3 Answers3

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$f$ satisfies this equation iff there exists a set $A$ such that $f =2\chi_A-1$. If $f$ is continuous then it is identically $1$ or identically $-1$.

[For the proof define $A$ as $\{x: f(x)=1\}$; $ \chi_A$ is defined by $ \chi_A(x)=1$ if $ x \in A$ and $0$ if $x \notin A$].

Kavi Rama Murthy
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If you allow discontinuous functions, then the general solution is that you pick some subset $A$ of $\Bbb{R}$ and define a function $f_A : \Bbb{R} \to \Bbb{R}$ by:

$$ f_A(x) = \left\{\begin{array}{l@{\quad}l} 1 & \mbox{if $x \in A$} \\ -1 & \mbox{if $x \not\in A$}\end{array}\right. $$

Any solution is $f_A$ for some unique $A$. The two continuous solutions correspond to $A = \Bbb{R}$ and $A = \emptyset$.

Rob Arthan
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  • Why is there any functions other then these, I quite didnt understand the uniqueness – Jale'de jaled Feb 02 '21 at 22:25
  • My point about uniqueness is that $f_A = f_B$ iff $A=B$, so this is a complete description of such functions: every such function is $f_A$ for exactly one set $A$. – Rob Arthan Feb 02 '21 at 22:33
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Hint: $$\begin{align} &\qquad \quad f(x)=\dfrac1{f(x)} \\ &\implies \frac{f(x)^2-1}{f(x)}=0 \\ &\implies (f(x)-1)(f(x)+1)=0 \end{align}\tag{for all $x$}$$

VIVID
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