prove that f is a second degree polynomial.

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prove that f is a second degree polynomial.

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Source: Shortlist BMO 2018, A5

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sqing

#1 h May 3, 2019, 11:59 AM • 1 Y

Y by Adventure10

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.

This post has been edited 1 time. Last edited by LauraZed, May 7, 2019, 5:05 PM
Reason: fixed source and typos

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#8 h May 5, 2019, 9:10 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

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#9 h Aug 2, 2019, 10:21 PM • 2 Y

Y by Adventure10, Mango247

This problem was proposed by Peter Gaydarov from Bulgaria.

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#10 h Sep 21, 2019, 3:58 PM • 3 Y

Y by Pluto1708, Adventure10, Mango247

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. $\blacksquare$ .

This post has been edited 2 times. Last edited by MathDelicacy12, Sep 21, 2019, 3:59 PM

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#11 h Jul 31, 2021, 10:53 AM

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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.

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#12 h Mar 26, 2025, 12:07 PM

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Let $P(x,y)$ be the given assertion.
Then,
$P(x+y,y):f(x+2y)+f(x)-2f(x+y)=g(x+y)y^2$
$P(x-y,y):f(x-2y)+f(x)-2f(x-y)=g(x-y)y^2$
$2P(x,y):2f(x+y)+2f(x-y)-4f(x)=2g(x)y^2$
$-P(x,2y):2f(x)-f(x+2y)-f(x-2y)=g(x)4y^2$
Added up gives $0=g(x+y)+g(x-y)-2g(x)$

By induction we get $g(nx)-g(0)=n(g(x)-g(0))$ for all positive integer $n$ . Note that $(g(x)-g(0))=-(g(-x)-g(0))$ , thus thus applies for all integer $n$ .
Thus, $g(\frac{m}{n})-g(0)=\frac{1}{n}(g(m)-g(0))=\frac{m}{n}(g(1)-g(0))$ for all integer $m,n$ , thus $(g(x)-g(0))=x(g(1)-g(0))$ for all $x\in Q$
Since $g(x)$ is continuous, $g(x)$ is linear for all real $x$ .

Since $f(x)$ is concave, $g(x)$ must be negative, thus it is constant. Let $g(x)=2k$

Let $h(x)=f(x)-kx^2$ . $h(x+y)+h(x-y)-2h(x)=f(x+y)+f(x-y)-2f(x)-2y^2=0$ . In a similar reasoning to above, we can show $h(x)$ is linear, thus $f(x)=kx^2+ax+b$ where $k<0$ and $a,b$ are constants.

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