Find all functions f:R→R such that for all real numbers x and y, f(f(x−y)−yf(x))=xf(y).
Problem
Source: Kosovo Mathematical Olympiad 2022, Grade 11, Problem 2
Tags: function, functional equation
06.03.2022 17:50
assuming this is correct, did it in 6:59 Let P(x,y) denote the given assertion. P(0,0):f(f(0))=0. P(x,0):f(f(x))=xf(0). P(f(0),0):f(0)=f(0)2⟹f(0)∈{0,1}. Case 1: f(0)=0. Then f(f(x))=0. If exist k≠0 with f(k)=0 then P(k,x):kf(x)=0⟹f≡0. If not, then f(x)=0⟹f≡0, contradiction. Case 2: f(0)=1. Then f(f(x))=x. Clearly must exist k with f(k)=0 because of P(0,x). Then P(k,x):f(f(k−x))=kf(x)⟹k−x=kf(x)⟹f(x)=k−xk so f is linear. Clearly if ax+b, then by involution a=1 or −1. But f(0)=1 so b=1. Thus, f(x)=x+1, not possible or f(x)=1−x, which works.
06.03.2022 18:53
dangerousliri wrote: Find all functions f:R→R such that for all real numbers x and y, f(f(x−y)−yf(x))=xf(y). S1 : f(x)=0∀x is a solution.So let us from now look only for non allzero solutions. Let P(x,y) be the assertion f(f(x−y)−yf(x))=xf(y) Let a=f(0) Since f is not allzero, P(x,y) implies f(x) surjective, and so f(f(x)) surjective too. P(x,0) ⟹ f(f(x))=ax and so a≠0 (else f(f(x)) would not be surjective). P(0,−x) ⟹ f(f(x)+ax)=0 P(f(x)+ax,0) ⟹ a=(f(x)+ax)a and so, since a≠0, f(x)=1−ax Setting there x=0, we get a=1 and so S2 : f(x)=1−x∀x xhich indeed fits.
09.03.2022 12:19
dangerousliri wrote: Find all functions f:R→R such that for all real numbers x and y, f(f(x−y)−yf(x))=xf(y). P(0,0):f(f(0))=0 P(f(0),0):f(f(f(0)))=f(0)2⇒f(0)=f(0)2⇒f(0)=0or1 Case1If f(0)=0 then: if there is t sucj that f(t)≠0 then from the first we have f surjective. P(x,0):f(f(x))=0f,surjective→f(x)=0 contradiction Otherwise f(x)=0 Case2If f(0)=1 then: P(x,0):f(f(x))=x⇒f:1−1 P(x,f(y)):f(f(x−f(y))−f(x)f(y))=xy=f(f(y−f(x))−f(x)f(y))f:1−1→f(x−f(y))−f(x)f(y)=f(y−f(x))−f(x)f(y)⇒f(x−f(y))=f(y−f(x))f:1−1→x−f(y)=y−f(x)⇒f(x)=c−x Move to the first to take c=1.
09.03.2022 14:00
Case 1: f is constant function so we have f(x)=0 Case 2: f not constant Claim f is injective. Proof: x1=x2→f(x1−y)=f(x2−y)→f(x1−y)−yf(x1)=f(x2−y)−yf(x2)→f(f(x1−y)−yf(x1))=f(f(x2−y)−yf(x2))→x1f(y)=x2f(y) Proven. Subs x=1 we have f(f(1−y)−yf(1))=f(y)→f(1−y)−yf(1)=y…(1) Subs y=0 to (1) we have f(1)=0 From (1) Subs f(1)=0 we have f(1−y)=y→f(y)=1−y
09.03.2022 16:46
Jufri wrote: Case 1: f is constant function so we have f(x)=0 Case 2: f not constant Claim f is injective. Proof: x1=x2→f(x1−y)=f(x2−y)→f(x1−y)−yf(x1)=f(x2−y)−yf(x2)→f(f(x1−y)−yf(x1))=f(f(x2−y)−yf(x2))→x1f(y)=x2f(y) Proven. Subs x=1 we have f(f(1−y)−yf(1))=f(y)→f(1−y)−yf(1)=y…(1) Subs y=0 to (1) we have f(1)=0 From (1) Subs f(1)=0 we have f(1−y)=y→f(y)=1−y You have the injectivity totally wrong.
10.03.2022 02:52
Did this in three minutes. P(0,0)⇒f(f(0))=0 P(f(0),0)⇒f(0)∈{0,1} If f(0)=0: P(x,0)⇒f(f(x))=0 P(f(x),x)⇒f(x)=0 If f(0)=1: P(x,0)⇒f(f(x))=x⇒f injective P(0,−x)⇒f(f(x)+x)=0⇒f(x)=c−x, testing gives f(x)=1−x. Both of these are solutions.
20.03.2022 18:28
The only constant that works is f≡0, which is our first solution. Let f be non-constant. Let (x,y) be the assertion in given equation. As f is nonzero, we directly get f is surjective from given equation. P(x,0): f(f(x))=xf(0) Hence, f(0)≠0 (otherwise combining with f surjective, we get f is constant. Contradiction.) Hence f is injective. P(0,0) : f(f(0))=0 P(0,−y): f(f(y)+yf(0))=0 Hence, f(0)=f(y)+yf(0) (as f is injective.) Hence, f(y)=f(0)(1−y)...(1) y=f(0) in (1) and f(f(0))=0 gives: f(0)=0 or 1 . As, f(0)≠0, f(0)=1. Hence from (1) we get that : f(x)=1−x which indeed works and is our second solution.
05.08.2022 13:25
Denote the assertion by P(x,y). We have f(f(0))=0 by P(0,0). And moreover f(0)∈{0,1} by P(f(0),0). If f(0)=0 then P(x,0) implies f(f(x))=0 and so P(f(x),x) implies f(x)=0 which fits. Otherwise P(x,0) implies f(f(x))=x and in particular f(1)=0 and f is injective. Finally P(1,1−x) implies f(x)=1−x which fits.
12.02.2023 01:22
18.01.2025 17:00
Let P(x,y) denote the given assertion. P(x,0) gives us that f(f(x))=x.f(0) Also P(xf(y),y) gives us that f is surjective. Now there are two cases: Case 1: f(0)=0. So f(f(x))=0 and by subjectivity, we have f(x)=0 which is a solution. Case 2: f(0)≠0. So f(f(x))=x.f(0) results in f being injective. Now, P(1,1) gives us that f(f(0)−f(1))=f(1) and by injectivity we have f(0)−f(1)=1 so f(0)=f(1)+1 Also, P(1,x) gives f(f(1−x)−x.f(1))=f(y) which by injectivity means f(1−x)−xf(1)=x which means f(1−x)=x(f(1)+1)=x.f(0). Now if we put x:=1−x we have f(x)=(1−x)f(0). This means f(1) = 0 so f(0) = 1 and so the equation becomes f(x)=1−x, which is indeed a solution. In conclusion, we proved that the only possible functions are f(x)=0 and f(x)=1−x, in which the equality holds. P. S: In the third line, I supposed that there exists an f(y) such that f(y) is not equal to 0. I forgot to clarify about it. This is kinda like proving by contradiction which helps us prove f(x) = 0 in the first case, and it's perfectly fine in the second case.