It is well known that Cauchy's functional equation for $f:\mathbb{R}\to\mathbb{R}$ $$ f(x+y)=f(x)+f(y)\space\forall x,y\in\mathbb{R} $$ admits highly pathological solutions if no further conditions are given. Is the condition: $$ f(f(x))=x \space\forall x\in\mathbb{R} $$ i.e. $f$ is an involution, enough to ensure that no "ugly" function is a solution?


Nope. For example, we can define $f$ by $f(x) = x$ for $x \in \Bbb Q$ and $f(x) = -x$ over all other cosets of $\Bbb Q$.

  • $\begingroup$ What are cosets? $\endgroup$ – Redundant Aunt Nov 12 '15 at 20:19
  • $\begingroup$ that is not a solution of the equation... for example, if you take y to be some irrational number and x to be one minus y, you get a counterexample. $\endgroup$ – user3329719 Mar 11 '17 at 15:12
  • $\begingroup$ @user3329719 you've misunderstood how this function was defined. Also, this post is more than a year old. $\endgroup$ – Omnomnomnom Mar 11 '17 at 16:14

see Does this uniquely specify F(x)=x, on [0,1] to [0,1]; if F(x) is a function from [0,1] to [0,1]; F:[0,1]to [0,1], and satisfies F(F(x))=x is and is strictly monotonic increasing (or just monotone increasing) (possibly only increasing), then it appears that F(x)=x on that interval is the only solution; ven without any further regularity or continuity or cauchy's equation F(x+y)=F(x)+F(y), being explicitly presumed.

According to one one answer, then it appears that F(x)=x, under the presumption of only monotone increasing (not even strict increasing);see Does this uniquely specify F(x)=x, on [0,1] to [0,1]. I presume that the reasoning therein is correct, but perhaps it should be checked. There is a another solution given by commentator that it works even under F being only monotone increasing (rather than necessarily strictly increasing).

In a ranked say probability system this (the monotonicity increasing (or strict) hard-codes this rank order constraint into a functional form) this makes some kind of sense. AS it would be able to capture, transcendental values (it appears), in the same sense that it can capture any other value. An inequality would be violated (or strict monotonicity) would be violated; F(1/pi)neq1/pi for example; instantly, ie one singular, or finitary step.

This is because if, F(1/pi)neq1/pi then there is some m in [0,1[ s.t F(1/pi)=m m>1/pi, or m<[1/pi] equal in value to some m smaller or large, and F(m)=1/pi by involution, the rank (or monotone strict increasing would be violated at the first move if m<1/pi, F(m)m=F(1/pi) ie F(m)>F(1/pi) by involution equation

that an involution function (often a form of point reflection symmetry, could technically capture even transcendental values; F(x)~=x, the rank order is immediately violated, whether x, or F(x) is an algebraic irrational, transcendental or algebraic number-


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