Let c be a real number. For all x and y real numbers we have, f(x−f(y))=f(x−y)+c(f(x)−f(y))and f(x) is not constant. a) Find all possible values of c. b) Can f be periodic?
Problem
Source: Turkey TST 2021 D-3, P-2
Tags: algebra, functional equation, function
SerdarBozdag
21.05.2021 20:30
a)P(x,x) gives f(x−f(x))=f(0). P(0,x−f(x)) gives f(−f(0))=f(f(x)−x). P(0,0) gives f(−f(0))=f(0). Thus we have f(f(x)−x)=f(0). Plugging x=0 gives f(0)=f(f(0)). P(x,f(0)) gives c(f(x)−f(0))=0. If there exists x satisfying f(x)≠f(0), we have c=0. Otherwise f(x) is constant. Therefore c=0.
b)P(x+y,y) gives f(x+y−f(y))=f(x). If T is the period, we have y−f(y)=Tg(y) where g is a function from Z to Z. Putting f(x)=x−Tg(x) in the equation gives 2y=T(g(y)+g(x+y−g(y)T)−g(x−y)). Because g is Z to Z, we have 2y=h(x,y)T where h(x,y)=g(y)+g(x+y−g(y)T)−g(x−y) is a function from ZxZ to Z. Obviously, this cannot be true for all y. For example, 0<y<T/2. Thus f cannot be periodic.
NOTE: Solution of b is wrong. Because 2y=T(g(y)+g(x+y−g(y)T)−g(x−y)) is wrong.
Can anyone check this solution?
SerdarBozdag
21.05.2021 20:41
Not important.
BarisKoyuncu
21.05.2021 23:57
We claim that the only solution is c=0.
See that if c=0, then f(x)=x works.
Let c≠0 and let P(x,y) be the assertion.
Lemma: Let a and b be real numbers such that f(a)=f(b). Then f(x−a)=f(x−b) for each real number x.
Proof: P(x,a) and P(x,b) directly gives the desired result.
P(1,1) gives f(1−f(1))=f(0). By Lemma we can say that f(x−1+f(1))=f(x).
Replace x with x−f(1). Then f(x−1)=f(x−f(1)).
P(x,1) gives f(x−f(1))=f(x−1)+c(f(x)−f(1))⇒c(f(x)−f(1))=0. We assumed c≠0. Thus, f(x)=f(1) for all real numbers. Thus, f is constant. Contradiction. Hence c=0 is the only valid solution.
See that f(x)={x} works when c=0. Thus the function can be periodic.
rafaello
22.05.2021 00:43
For part a, we claim that the only possible value of c is 0, this obviously works.
Let P(x,y) be the assertion.
P(x,0), P(x,x) and P(x,x−f(x)) give us f(f(x))=f(x).
Now, P(x,y) and P(x,f(y)) imply f(x−y)=f(x−f(y)) and therefore c(f(x)−f(y))=0 for all x,y∈R. Since f is not a constant function, thus c=0.
For part b, one could use f(x)=x−⌊x⌋∀x∈R.