Find a real function f:R→Rf:\mathbb{R}\to\mathbb{R} such that f(f(x))=−xf(f(x)) = -x?

I’ve been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success:

Find a function f:RR satisfying f(f(x))=x for all xR.

I am also wondering, can we find f so that is continuous?

I was thinking of letting f be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist.

Source: http://www.halfaya.org/Casti/CalculusTheory2/challenge.pdf

Answer

An important piece of information is:

Theorem: f is not continuous.

Proof: Observe that f is invertible, because

f(f(f(f(x))))=f(f(x))=x

and so fff=f1. Any continuous invertible function on R is either strictly increasing or strictly decreasing.

If f is strictly increasing, then:

  • 1<2
  • f(1)<f(2)
  • f(f(1))<f(f(2))
  • 1<2

contradiction! Similarly, if f is strictly decreasing, then:

  • 1<2
  • f(1)>f(2)
  • f(f(1))<f(f(2))
  • 1<2

contradiction! Therefore, we conclude f is not continuous.


For the sake of completeness, the entire solution space for f consists of functions defined as follows:

  • Partition the set of all positive real numbers into ordered pairs (a,b)
  • Define f by, whenever (a,b) is one of our chosen pairs,
    • f(0)=0
    • f(a)=b
    • f(b)=a
    • f(a)=b
    • f(b)=a

To see that every solution is of this form, let f be a solution. Then we must have f(0)=0 because:

  • Let f(0)=a. Then f(a)=f(f(0))=0 but a=f(f(a))=f(0)=a, and so f(0)=0

If a0, then let f(a)=b. We have:

  • f(b)=f(f(a))=a
  • f(a)=f(f(b))=b
  • f(b)=f(f(a))=a

From here it's easy to see the set {(a,f(a))a>0,f(a)>0} partitions the positive real numbers and so is of the form I describe above.


One particular solution is

f(x)={0x=0x+1x>0x is odd1xx>0x is evenx1x<0x is odd1xx<0x is even

e.g. f(1/2)=3/2, f(3/2)=1/2, f(1/2)=3/2, and f(3/2)=1/2.

(This works out to be Jyrki Lahtonen's example)

Attribution
Source : Link , Question Author : Gamma Function , Answer Author : Community

Leave a Comment