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Source: FKMO 2013 #2

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#1 h Mar 24, 2013, 6:17 AM • 6 Y

Y by YanYau, Davi-8191, tenplusten, ashologe, Adventure10, WiseTigerJ1

Find all functions $f : \mathbb{R}\to\mathbb{R}$ satisfying following conditions.
(a) $f(x) \ge 0$ for all $x \in \mathbb{R}$ .
(b) For $a, b, c, d \in \mathbb{R}$ with $ab + bc + cd = 0$ , equality $f(a-b) + f(c-d) = f(a) + f(b+c) + f(d)$ holds.

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#2 h Mar 25, 2013, 11:58 AM • 3 Y

Y by Adventure10, Mango247, WiseTigerJ1

$a=b=c=d=0\implies f(0)=0$ and $a=2b=0,c=0\implies f(d)=f(-d)$ , so $f$ is odd function.
Now take $z=-a,x=b,y=c,t=-d$ . So now we've $xz+yt=xy$ and $P(x,y,z,t):f(x+z)+f(y+t)=f(z)+f(t)+f(x+y)$ .
Now $P(x,x,z,x-z)\implies g(x,z): f(2x)+f(z)+f(x-z)=f(2x-z)+f(x+z)$ .
Now combining $g(x,-x),g(x,-2x)$ we get $f(2x)=4f(x),f(3x)=9f(x)$ .
Now just by induction applying on $g(x,z)$ we get $f(nx)=n^2f(x)$ for all $x\in\mathbb R$ .
Thus $f(x)=x^2f(1)$ for all $x\in\mathbb Q$ . From $P(x,y,z,t)$ putting $t=\frac {x(y-z)}{y}$ we get $M(x,y,z):f(x+y)+f(z)+f(\frac {x(y-z)}{y})=f(x+z)+f(y+\frac {x(y-z)}{y})$ . Now putting $z=\frac {y(2x+y)}{2x}$ we get $y+2\frac {x(y-z)}{y}=0$
Thus $f(x+y)+f(z)=f(x+z)$ . Now note $z$ is dependable on $x,y$ , so we can take any two positive $x,z$ without any problem. Now note for those we’ve $f(x+z)\geq f(z)$ . Thus we conclude $(x)$ in increasing for positive $x$ .
So as it’s also of form $cx^2$ for all $x\in\mathbb Q$ hence it’s also of same form for all non negative $x$ .
Also as $f(x)=f(-x)$ so $f(x)=cx^2$ for all $x\in\mathbb R$ .

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#3 h Dec 29, 2015, 1:56 PM • 4 Y

Y by tony88, Adventure10, Mango247, WiseTigerJ1

Denote $P(a,b,c,d)$ as the statement that for $ab+bc+cd=0$ , we have $f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)$ .
Start with $P(0,0,0,0)$ . This gives us $f(0)=0$ . Follow with $P(0,0,0,x)$ . This gives us $f(x)=f(-x)$ .

Now what? We can mindlessly, plug $d=-\frac{ab}{c}-b$ .
Also, we want some stuff to cancel out as well. Therefore, we set $a-b=b+c$ .
Now we have $d=-\frac{(2b+c)b}{c}-b=-\frac{2b^2}{c}-2b$ .
Fractions are bad, so let $b=kc$ , giving $d=-2k^2c-2kc=-2kc(k+1)$ and $a=2b+c=(2k+1)c$

Plugging this in, we have $f((k+1)c)+f(c+2kc(k+1)) = f((2k+1)c)+f((k+1)c)+f(-2kc(k+1))$ or $f((2k^2+2k+1)c)=f((2k+1)c)+f((-2k^2-2k)c)=f((2k+1)c)+f((2k^2+2k)c)$ $\underline{\hspace{ 10 in}}$
Claim: For all reals $x,y$ , we can find a suitable reals $k, c$ such that $(2k+1)c=x$ and $(2k^2+2k)c=y$ .

Proof of Claim: If $x=0$ and $y=0$ , set $c=0$ and we are done.
If $x=0$ and $y \not= 0$ , set $k=-\frac{1}{2}$ and $c=-2y$ . If $x \not= 0$ and $y=0$ , let $k=0$ and $x=c$ .
Now let us examine the case $x, y \not= 0$ . We have $\frac{y}{x}= \frac{2k^2+2k}{2k+1}$ . Let this be $t$ for convenience.
Note that $k$ must be a solution of $2k^2+2k=(2k+1) t$ , which rearranges to $2k^2+(2-2t)k- t=0$ , which has determinant of $(2-2t)^2+8t = 4(t^2+1) > 0$ .
Now setting $c=\frac{x}{2k+1}=\frac{y}{2k^2+2k}$ will do the job. Clearly $2k+1$ and $2k^2+2k$ are nonzero since $x,y$ are nonzero.
$\underline{\hspace{ 10 in}}$

Now plugging this in, we have $f(x)+f(y)=f(\sqrt{x^2+y^2})$ .
Since $f(x) \ge 0$ , we have $f(\sqrt{x^2+y^2}) \ge f(x)$ , so $f$ is monotonically increasing.

Let $g(x)=f(\sqrt{x})$ . We have $g(x^2)+g(y^2)=g(x^2+y^2)$ for all $x,y$ and $g(x)$ is monotonically increasing with $g(0)=0$ , so $g(x)=cx$ , giving $f(x)=cx^2$ as desired.

This post has been edited 5 times. Last edited by rkm0959, Dec 29, 2015, 2:03 PM

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efang

#4 h Apr 7, 2016, 12:35 AM • 6 Y

Y by Mathuzb, NMN12, tony88, Adventure10, Mango247, WiseTigerJ1

There is a much more natural way to get $f(x) + f(y) = f(\sqrt{x^2+y^2})$

the unsmart method

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#6 h Oct 12, 2020, 7:11 AM • 2 Y

Y by AMMKMMRIR, WiseTigerJ1

The answer is $f(x)=cx^2$ , where $c\geq 0$ , such functions obviously works, now we will show that they are the only ones. Label the equation
$f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)\hspace{2pt} \text{for all}\hspace{2pt} ab+bc+cd=0 \qquad(1)$ CLAIM 1. $f(0)=0$ and $f(a)=f(-a)$
Proof.
They can be done by simply substituting $a=b=c=d=0$ and $a=b=-c, d=0$ in $(1)$ . $\blacksquare$

CLAIM 2. $f(a)+f(b)=f(\sqrt{a^2+b^2})$
Proof.
Sub. $b=2a$ into $(1)$ . Then $(1)$ becomes
$f(c-d)=f(b+c)+f(d)$ for all $b,c,d$ with $\frac{1}{2}b^2+bc+cd=0$ . Rearranging this gives $(c-d)^2=(b+c)^2+d^2$ . Therefore, let $d=y$ , $c=y+\sqrt{a^2+b^2}$ and $b=x-y-\sqrt{a^2+b^2}$ we have
$f(x)+f(y)=f(\sqrt{x^2+y^2})\qquad(2)$ as desired.

Now by easy induction we have $f(m)=m^2f(1)$ for all $m\in\mathbb Q$ . From the first condition and $(2)$ , $f$ is increasing. For any $a\in\mathbb R$ , take a strictly increasing sequence of rational numbers which converges to $a$ . Then the image of that sequences converges to $a^2f(1)$ as well. This shows that $f(a)=a^2f(1)$ which completes the proof.

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#7 h Nov 29, 2023, 3:50 AM • 1 Y

Y by WiseTigerJ1

posting for storage

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#8 h Jan 28, 2024, 8:18 PM • 1 Y

Y by WiseTigerJ1

The answer is $f(x) = cx^2 (c \geq 0)$ , which clearly works.
Let $P(a, b, c, d)$ be the assertion.
$P(0, 0, 0, 0) \implies f(0) = 0$ .
$P(0, x, 0, 0) \implies f(-x) = f(x)$ , so $f$ is even.
$P(a, a(1 - \sqrt{2}), a\sqrt{2}, \frac{a}{\sqrt{2}}) \implies f(a\sqrt{2}) + f(\frac{a}{\sqrt{2}}) = 2f(a) + f(\frac{a}{\sqrt{2}})$ :what?:

:what?:

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Hence, $f(a\sqrt{2}) = 2f(a)$ .
Therefore, $f(2a) = 2f(a\sqrt{2})) = 4f(a)$ :-D

:-D

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Let's prove that $f$ is strictly increasing.
Take any $0 \leq u < v$ and plug in $P(u - c, u - c, c, c - v)$ where $c$
is a solution to $c^2 - (v+u)c + u^2$ = 0. (Which ensures the $ab + bc + cd = 0$ condition and exists because $D = (v+u)^2 - 4u^2 = (v-u)(v+3u) > 0$

).
This gives $f(v) = f(u - c) + f(u) + f(c - v) > f(u)$ , cause $f \geq 0$ and can't be equal since equality requires $f(u-c) = f(c - v) = 0$ which means $u = c = v$ . Absurd :read:

:read:

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First, we claim $f(ka) = k^2f(a)$ for $\forall k \in \mathbb{Z}$ .
$P(ka, a, a, -(k+1)a) \implies f((k-1)a) + f((k+2)a) = f(ka) + f((k+1)a) + 4f(a)$ .
We can plug in $k = 1$ and check that $f(3a) = 9f(a)$ .
After that, we can use simple induction for $k \geq 4$ . Also, $f$ is even, so it's the same for $k < 0$

.
Now let's prove the above statement for $\forall k \in \mathbb{Q}$ .
$P(\frac{a}{k}, ka, a, -(k+1)a) \implies f(\frac{a}{k}) = \frac{1}{k^2}f(a)$ .
This proves for $k \in \mathbb{Q}$ .
If $k$ is irrational we can "sandwich" it using infinitely close two rational numbers from below and above, considering the fact that $f$ is strictly increasing :rotfl:

:rotfl:

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So, $f(ka) = k^2f(a) \forall k \in \mathbb{R}$ . Plugging in $a = 1$ we get $f(x) = cx^2 (c \geq 0) \forall x \in \mathbb{R}$ . As desired :showoff:

:showoff:

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This post has been edited 1 time. Last edited by miiirz30, Jan 28, 2024, 8:25 PM

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#9 h Jan 27, 2025, 12:56 AM

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$P(0,0,0,0)$ : $f(0)=0$
$P(0,0,0,x)$ : $f(x)=f(-x)$
Look that the condition can be taken away if we consider $Q(a,b,d)=P(a,b,-\frac{ab}{b+d},d)$ with $b+d\neq 0$ .
Since $b+d=0$ is just the previous two results, the problem can be reduced to finding $f$ such as for $a,b,d$ real numbers : $Q(a,b,d)$
$Q(a,b,b)$ and $b\neq0$ : $f(a-b)+f(-\frac{a}{2}-b)=f(a)+f(b)+f(b-\frac{a}{2})=f(a-b)+f(\frac{a}{2}+b)$
$Q(b,\frac{a}{2},-a)$ and $a\neq 0$ : $f(\frac{a}{2}-b)+f(b+a)=f(b)+f(-a)+f(\frac{a}{2}+b)=f(b)+f(a)+f(\frac{a}{2}+b)$
Summing yields : $2f(a)+2f(b)=f(a-b)+f(b+a)$ for all $a,b$ ( it generalises pretty trivially to $0$ ).
From this we apply induction to get $f(n)=n^2f(1)$ , then $f(r)=r^2f(1)$ for all rationals.
Now we prove $f$ is strictly increasing which would be equivalent to ( after some calculations ) : $2f(b+a)\geq f(b)+2f(a) \iff 2f(b-a)\geq f(b)+2f(a)$ which is right by assuming the negation and summing.
Now we finish with this.

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#10 h Jan 28, 2025, 3:30 AM

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Classic Cauchy-type equation. The answer is $f \equiv kx^2$ for all nonnegative real numbers $k$ .

Setting $a=b=c=d=0$ yields $f(0) = 0$ . Furthermore, setting $a=c=d=0$ yields $f(b)=f(-b)$ . So we instead define a function $g:\mathbb R_{>0} \to \mathbb R_{\geq 0}$ defined the same way as $f$ . Then setting $b = \frac{a-c}2$ and $d = \frac{c^2-a^2}{2c}$ , we get $g\left(\frac{a^2+c^2}{2c}\right) = g(a) + g\left(\frac{c^2-a^2}{2c}\right).$ Let $h(x) = g\left(\sqrt x\right)$ for all $x > 0$ , such that the equation reads $h\left(\frac{c^4+a^4+2a^2c^2}{4c^2}\right) = h\left(a^2\right) + h\left(\frac{a^4+c^4-2a^2c^2}{4c^2}\right).$ In particular, for any pair of positive real numbers $(x, y)$ , there exists a pair $(a, c)$ such that $a = x$ and $\frac{c^2-a^2}{2c} = y$ . So $h$ is Cauchy, but it is bounded universally below by $0$ , so $h$ is linear.

Thus $h \equiv kx$ and $g \equiv kx^2$ for $k = f(1)$ . By our first two conditions, we get $f \equiv kx^2$ too.

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#11 h Yesterday at 5:50 PM • 1 Y

Y by imagien_bad

let $a=b=c=d=0$ which gives $f(0)=0.$ set $a=c=d=0$ and we get $f$ is even. take $b = 2a,$ and it magically becomes $f(a) + f(b) = f\sqrt{a^2 + b^2}$ for all $a, b.$
let $g\equiv f(\sqrt x)$ but defined over $x \geq 0.$
$g(a) + g(b) = g(a + b)$ but $g \geq 0.$ this implies $g \equiv cx,$ or $f\equiv cx^2$ for $c \geq 0.$

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#12 h Yesterday at 5:51 PM

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Ilikeminecraft wrote:

let $a=b=c=d=0$ which gives $f(0)=0.$ set $a=c=d=0$ and we get $f$ is even. take $b = 2a,$ and it magically becomes $f(a) + f(b) = f\sqrt{a^2 + b^2}$ for all $a, b.$
let $g\equiv f(\sqrt x)$ but defined over $x \geq 0.$
$g(a) + g(b) = g(a + b)$ but $g \geq 0.$ this implies $g \equiv cx,$ or $f\equiv cx^2$ for $c \geq 0.$

ilikeilikeminecraft

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