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Find all functions $f: \Bbb{R} \to \Bbb{R} $ such that $f(f(x)) = -x$.

I think this equation has no continuous solution, because no polynomial can be the solution and we know that every continuous function is the limit of a sequence of polynomials.

giobrach
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amir bahadory
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    I don't know the answer. But your reason for thinking there is no such continuous $f$ is wrong. The same argument would show there is no continuous $f$ with $f(x)^2=|x|$. – David C. Ullrich May 05 '16 at 17:48

3 Answers3

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if f(a) = b, then f(b) = -a, f(-a) = -b, and f(-b) = a. for every ordered pair (a,b) that the function maps, there is a 4 fold symmetry (90 degree rotations), mapping 3 other ordered pairs.

Doug M
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Applying $f$ to the equation shows that $f$ is an odd function, in particular $f(0)=0$.

Assuming differentiable such $f$ existed, we can differentiate at $x=0$ and obtain the contradiction $f'(0)^2=-1$.

Now lets assume that $f$ is merely monotone locally at $0$. If it were monotonously increasing, then we would have $f(x)>0$ for small enough $x>0$ and $f(x)\to 0$ for $x\to 0$. Thus $f(f(x))>0$ for even smaller $x$ by continuitiy, giving a contradiction. If it were monotonously decreasing we would get that $f(x)<0$ for small enough $x>0$ and $f(x)\to 0$ for $x\to 0$. Together with oddness we obtain the contradiction $f(f(x))>0$ again.

Finally, let's show that continuous such $f$ existed. Then it cannot be monotone, and by the intermediate value theorem $f(x)=f(y)$ for $x\neq y$. Applying $f$ shows $-x=-y$, a contradiction.

FINAL REMARKS: a more succinct version of this answer can be found in one of the MO links in the comments of one of the duplicate links

Bananach
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The answer of Doug is useful.

If $c=f(0)$ then $f(c)=f(f(0))=0$ so $-c=f(f(c))=f(0)=c$ showing that $c=0$.

If $x\neq0$ then $f(f(x))=-x\neq x$ so there are no $2$-cykels.

Also $f(f(f(f(x))))=-f(f(x))=x$.

So the function must be bijective and has one $1$-cykel and further has $4$-cykels.

Construct a partition $\mathcal P$ of $\mathbb R$ with $\{0\}\in\mathcal P$ and such that every other element takes the form $\{u,v,-u,-w\}$ with $u,v>0$ and $u\neq v$

If $\{u,v,-u,-v\}$ is such an element of the partition then define $f(u)=v,f(v)=-u,f(-u)=-v,f(-v)=u$ or $f(u)=-v,f(-v)=-u,f(-u)=v,f(v)=u$

Then $f$ will be a function with the mentioned property and every function $f$ with that property will take this form.

drhab
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