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Functional Equation Without Real Solutions
By Martin Cohen on Friday, September 21, 2001 - 09:24 pm:

I have seen this question posed and I assume that there is an answer but I have not had any luck in trying to solve it:

Find a continuous real-valued function f(x) such that f(f(x))= - x.

By Michael Doré on Friday, September 21, 2001 - 10:27 pm:

Hi Martin,

I think there's no solution, unfortunately.

First, note f is a bijection from R to R. f is injective since f(x) = f(y) implies that f(f(x)) = f(f(y)) so -x = -y, and x = y. Also f is surjective since for any x we can find y such that f(y) = x. Simply set y = f(-x).

The continuity condition therefore means that f is strictly monotonic. Can you see why? You need the intermediate value theorem...

Now if f is strictly increasing then since f(-1) < f(1) we have f(f(-1)) < f(f(1)) and 1 < -1 which is impossible.

If f is strictly decreasing then since f(-1) > f(1) we have f(f(-1)) < f(f(1)) and again 1 < -1.

Hence no solution. It's a pity really, because otherwise we might have been able to find an operation on the reals which is isomorphic to multiplying by a complex number. Nice idea, though.

Regards,

Michael

[Editor's Comment: Most introductory texts to analysis will describe the Intermediate Value Theorem. Alternatively try The Intermediate Value Theorem .]