Let $f: \mathbb {R} \to \mathbb {R}$ be a concave function and $g: \mathbb {R} \to \mathbb {R}$ be a continuous function . If $$ f (x + y) + f (x-y) -2f (x) = g (x) y^2 $$for all $x, y \in \mathbb {R}, $ prove that $f $ is a second degree polynomial.
Problem
Source: Shortlist BMO 2018, A5
Tags: functional equation, algebra, polynomial
06.05.2019 00:10
03.08.2019 01:21
This problem was proposed by Peter Gaydarov from Bulgaria.
21.09.2019 18:58
sqing wrote: Let $f: \mathbb {R} \to \mathbb {R}$ be a concave function and $g: \mathbb {R} \to \mathbb {R}$ be a continuous function . If $$ f (x + y) + f (x-y) -2f (x) = g (x) y^2 $$for all $x, y \in \mathbb {R}, $ prove that $f $ is a second degree polynomial. Let $P(x,y)$ denote the given assertion. Claim 1 : $g$ is linear Proof. $2 \cdot P(x,y) - P(x,2y) + P(x+y,y) + P(x-y,-y)$ yields $2 \cdot g(x) = g(x+y) + g(x-y) \quad \forall x,y \neq 0$ Since $g$ is continuous, we have that $g$ is linear. $\square$ Claim 2 : $g$ is constant Proof. Let $g(x) = Mx + N$. $P(x,x) : f(0) = (Mx +N)x^2 \quad \forall x \in \mathbb{R}$. Also, $f(0) \leq 0$. So, we have that $M = 0$ and so $g$ is constant. $\square$ Let $g(x) = 2B$. Note that $f(x+y) - f(x-y) - f(x) = 2By^2$. Let $h(x) = f(x) - Bx^2$. Then, $h(x+y) + h(x-y) = 2h(x)$. So, $h$ is linear. So, any function $f$ of the form $Ax + Bx^2$ satisfies the given condition. $\blacksquare$.
31.07.2021 13:53
I have adding to proves that g(x)=B needs to be negative (because F(x) is concave). And everbody wrote that if 2.h(x)=h(x+y) + h(x-y) h(x) is lineer but also it can be constant , right. Which makes f(x) either ,Bx^2 +c or Bx^2+dx.