Find all functions u:R→R for which there exists a strictly monotonic function f:R→R such that f(x+y)=f(x)u(y)+f(y) for all x,y∈R
Problem
Source: Azerbaijan Math Olympiad Training
Tags: algebra, functional equation, TST
15.12.2019 20:47
Kamran011 wrote: Find all functions u:R→R for which there exists a strictly monotonic function f:R→R such that f(x+y)=f(x)u(y)+f(y) for all x,y∈R Let P(x,y) be the assertion f(x+y)=f(x)u(y)+f(y) P(0,x) ⟹ f(0)u(x)=0 and so f(0)=0 (else we get f(x+y)=f(y) and so f(x) is not strictly monotonic. P(x,0) ⟹ f(x)(u(0)−1)=0 and so u(0)=1 (else f(x) is not strictly monotonic If u(t)=0 for some t, then P(x,t) implies f(x+t)=f(t), impossible since f(x) strictly monotonic. So u(x)≠0 ∀x Subtracting P(x,y) from P(y,x), we get New assertion Q(x,y) : f(x)(u(y)−1)=f(y)(u(x)−1) If u(t)=1 for some t≠0 (so that f(t)≠0), then Q(x,t) ⟹ u(x)=1 ∀x which indeed is a solution (choose f(x)≡x for example). If u(x)=1 ⟺ x=0 : Q(x,1) ⟹ f(x)=c(u(x)−1) for some c=f(1)u(1)−1≠0 Plugging this in original equation, we get u(x+y)=u(x)u(y) Since we previously got u(x)≠0 ∀x, this easily implies u(x)>0 ∀x (set x=y And so lnu(x) is additive. So u(x)=eg(x) where g(x) is some additive function. If g(x) is not linear, graph of g(x) is dense in R2 and so graph of f(x) is dense in R×(−c,+∞) (or (−∞,−c) if c<0, in contradiction with f(x) stricly monotonic. So g(x) is linear and we got u(x)=eax∀x which indeed is a solution, whatever is a∈R, choosing for example f(x)=eax−1 if a≠0 or f(x)=x if a=0
15.12.2019 20:58
04.12.2022 20:51
Same solution as post #2, posting anyways Let P(x,y) be the assertion of f.e as usual. Let the strictly monotonic property be shown as (♣). Clearly u(x)≠0 and f is injective. P(0,x):f(0)u(x)=0(♣)⇒f(0)=0P(x,0):f(x)=f(x)u(0)(♣)⇒u(0)=1P(−x,x):f(−x)u(x)+f(x)=0⇒u(x)=−f(x)f(−x)(♠)P(x,y)−P(y,x)=f(x)(u(y)−1)=f(y)(u(x)−1)⇒f(x)f(y)=u(x)−1u(y)−1(1)Assume u(a)=1 for some a≠0. By (♣) if x≠0 in (♠) we have f(x)≠f(−x). if −f(x)=f(−x), or that is, f is odd function, analyzing P(−x,x) once more yields f(x)(−u(x)+1)=0Since f is obviously not constant, we must have u(x)=1 which is a clear solution. Hence, for sake of finding another solutions, assume u(a)=1 only if a=0. Observe (1) now. We have f(x)=au(x)−a for some a such that a≠0. Put into the original equation and we have u(x+y)=u(x)u(y) Which is a well known problem and the only solutions are u(x)=eax for some parameter a which indeed works, hence we are done.