Find all f:R→R such that f(x+f(x))=f(−x) and for all x≤y it satisfies f(x)≤f(y)
Problem
Source: Turkey JBMO TST 2023 Day 1 P3
Tags: function, algebra
30.04.2023 14:17
don't know how to do for all x≤y it satisfies f(x)≤f(y) means it increase in R but i just foundf(x)=−2xwhich not fits.
30.04.2023 14:23
Hangzhou14Z-ZY wrote: don't know how to do for all x≤y it satisfies f(x)≤f(y) means it increase in R but i just foundf(x)=−2xwhich not fits. How about f(x)≡0?
30.04.2023 14:27
qwedsazxc wrote: Hangzhou14Z-ZY wrote: don't know how to do for all x≤y it satisfies f(x)≤f(y) means it increase in R but i just foundf(x)=−2xwhich not fits. How about f(x)≡0? fine
30.04.2023 16:07
Just consider the longest interval (x1,x2).Such that f(x1)=f(x2)=c and c∈(x1,x2). Then try to prove −x1+f(−x1)∉(x1,x2) or −x2+f(−x2)∉(x1,x2). (And you need to prove such interval is exists) Then answer is f(x)≡cc∈R
30.04.2023 16:17
Quite a difficult problem for a JBMO TST. The answer is f≡c for any real constant c, which can easily be seen to work. We have the following Claim: Claim: f(x)=0⟺f(−x)=0, and f(x)f(−x)≥0 for all x. Proof: For the first part of the Claim, assume that f(u)=0. Then, f(−u)=f(f(u)+u)=f(u)=0, as desired. For the second part, assume that a x existed such that f(x)>0>f(−x). Then, since f is increasing, f(−x)=f(f(x)+x)≥f(x)>f(−x), a contradiction ◼ To the problem, we distinguish two cases: Case 1: f(x)≠0 for all x. Then, we claim that f has the same sign everywhere. Indeed, assume otherwise. Let A<B be such that f(A)f(B)<0. Then, if f(B)<0<f(A) we obtain that f(B)<f(A)≤f(B), a contradiction, while if f(A)<0<f(B) we obtain that f(−A)<0<f(−B)≤f(−A), a contradiction. WLOG assume that f(x)>0 for all x. Then, f(−x)=f(f(x)+x)≥f(x), and so f(−x)≥f(x)≥f(−x) for all x, that is f(x)=f(−x). Now, if A<B existed such that f(A)≠f(B), then since f is increasing f(A)<f(B), and so f(−A)<f(−B), which is a contradiction since f(−B)>f(−A)≥f(−B) Thus, f must be constant, as desired. The case of f being everywhere negative can similarly be handled. Case 2: There exists some x such that f(x)=0. Let S={x∈R:f(x)=0}. By assumption, the set S is nonempty. Note that by the above Claim, if x∈S then −x∈S. We distinguish two cases. Case 1: S is upper bounded. Since x∈S⟺−x∈S, S is lower bounded, too. Note that if x,y∈S, then [x,y]⊆S, since f is increasing. Let a=inf and b=\sup S. Then, for all \epsilon>0 there are x_1,x_2 \in S such that a \leq x_1<a+\epsilon and b-\epsilon<x_2 \leq b, and so [a+\epsilon,b-\epsilon] \subseteq [x_1,x_2] \subseteq S, for all \epsilon>0, which implies that [a,b] \subseteq S. Since obviously S \subseteq [a,b], we conclude that S is an interval. Moreover, it is easy to conclude that a=-b. Taking a x such that x>b we infer that f(-x) \leq f(a)=0=f(b) \leq f(x), and so by the initial Claim equality must hold, i.e. f(x)=0 for all x>b, and similarly we may extend this equality for all x<-b. Since for -b \leq x \leq b we also have f(x)=0, we conclude that f is zero everywhere. However, S= \mathbb{R} in this case, a contradiction. Case 2: S is not upper bounded. Then, for all u \in \mathbb{R} there exists a x \in S such that x>\max \{-u, u \}, and so f(-u) \leq f(x)=0=f(-x) \leq f(u), that is equality must hold everywhere, which implies that f(x)=0 for all x, as desired.
30.04.2023 19:33
My solution during the contest. Answer:f(x) is constant. f(x)=c \in R Let x>0. i)f(x) \geq 0 f(x)\leq f(x+f(x))=f(-x) \leq f(x) f(x)=f(-x) Let x \geq y \geq -x f(x) \geq f(y) \geq f(-x)=f(x) ii)f(x) \leq 0 f(x) \leq f(x+f(x))=f(-x) \leq f(x) So f(x)=f(-x) Again let x \geq y \geq -x f(x) \geq f(y) \geq f(-x)=f(x) We choose x really large, so that we get f(x) is constant
17.08.2023 17:41
Easy (why the function is in JBMO TST). Suppose that there are x,y such that f(-x) not equal to f(-y), suppose that x \geq y then f(-y) \geq f(-x), then f(y+f(y)) \geq f(x+f(x)), then y+f(y) \geq x+f(x) (1) (because if x+f(x) \geq y+f(y) then we have f(-x) \geq f(-y), which means that f(-x)=f(-y)). But we have x \geq y and f(x) \geq f(y) (2). (1) and (2) together gives that x and y must be equal, which is contradiction to f(-x) not equal to f(-y). Then answer is f(x) \equiv c.
19.10.2023 12:28
Nice problem, but why do such problems appear at all in JBMO TSTs?! I understand that FEs are conceptually better mathematically than almost all inequalities, but still?
19.10.2023 17:53
I think that inequalities is more appropriate for JBMO TSTs.