An FE lemma about you!

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An FE lemma about you!

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Source: Own, inspired by problem 556 in the FE marathon

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1069 posts

gghx

#1 h Jun 14, 2022, 11:53 PM • 15 Y

Y by David-Vieta, trinhquockhanh, doanquangdang, megarnie, rama1728, PRMOisTheHardestExam, navi_09220114, GuvercinciHoca, Lamboreghini, Inconsistent, ImSh95, player01, k12byda5h, Supercali, MS_asdfgzxcvb

Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is your favourite function, $g:\mathbb{R}\rightarrow \mathbb{R}$ is your mother's favourite function, and $h:\mathbb{R}\rightarrow \mathbb{R}$ is your father's favourite function. It was discovered that for any reals $x,y$ , $f(xy+g(x))=xf(y)+h(x)$ Prove that you are boring.

(Hint: you might need to use the quotable result that if someone's favourite function is a linear polynomial, they are boring)

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DottedCaculator

7326 posts

DottedCaculator

#2 h Jun 15, 2022, 12:10 AM • 4 Y

Y by centslordm, megarnie, ImSh95, michaelwenquan

Proof: Trivial

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1069 posts

gghx

#4 h Jun 15, 2022, 2:01 PM • 1 Y

Y by ImSh95

Application: solve $f(xf(y)+g(x))=xy+h(x)$

$P(1,y)$ gives $f(f(y)+g(1))=y+h(1)$ .
$P(x,f(y)+g(1)): f(xy+xh(1)+g(x))=xf(y)+xg(1)+h(x)$ , so $f$ is linear.

Waiting for a mathematical proof

This post has been edited 4 times. Last edited by gghx, Jun 15, 2022, 3:38 PM

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Zaro23

#5 h Jun 15, 2022, 2:33 PM • 4 Y

Y by ImSh95, Mango247, Mango247, Mango247

Thanks for the correcting

This post has been edited 2 times. Last edited by Zaro23, Jun 15, 2022, 3:40 PM

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gghx

#6 h Jun 15, 2022, 3:38 PM • 2 Y

Y by Zaro23, ImSh95

Zaro23 wrote:

$P(1,x) \to$ use iteration leema $f(x+C)=f(x)+C (C=g(1)+h(1))$
compare $(x,y+c)-(x,y) \to f(xc+xy+g(x))= f(xy+g(x))+xc$
so it becomes $f(xc+y+g(x))= f(y+g(x))+xc$
$y\to 0$ we get $f(xc)=f(0)+xc$

How do you get $C=g(1)+h(1)$ ? From $y=1$ we get $f(f(y)+g(1))=y+h(1)$ .

In any case, this fails because of the case $g(1)=h(1)=0$ (which is in fact a solution in problem 556, $f(x)=1-x$ ). This actually should not appear in the proof (because $g(1)=0$ gives nothing). There is something else

@below i see (and also this is a stronger condition that the original question, so try that too haha)

This post has been edited 5 times. Last edited by gghx, Jun 15, 2022, 4:13 PM

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Zaro23

#7 h Jun 15, 2022, 3:43 PM • 1 Y

Y by ImSh95

$f(f(x)+S1)=x+S2$
$f(f(f(x)+S1))=f(x+S1+S2)=f(x)+S1+S2$
That's how I used iteration lemma
But it"s not useful

This post has been edited 3 times. Last edited by Zaro23, Jun 15, 2022, 3:46 PM

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1069 posts

gghx

#8 h Jun 18, 2022, 10:09 AM • 2 Y

Y by ImSh95, Mango247

gghx wrote:

Suppose $f,g,h:\mathbb{R}\rightarrow\mathbb{R}$ are functions such that for any reals $x,y$ , $f(xy+g(x))=xf(y)+h(x)$ . Show that $f$ is linear.

Complete motivation here

Claim 1: If $f(y+c)=f(y)+d$ for constants $c\ne 0,d$ and all $y\in \mathbb{R}$ , then $f$ is linear.
Proof. Using $P(x,y+c)$ gives $f(xy+xc+g(x))=xf(y+c)+h(x)=xf(y)+xd+h(x)=f(xy+g(x))+xd$ . Hence, for any $x\ne 0$ we have $f(y+g(x)+xc)=f(y+g(x))+xd$ . This is still obviously true when $x=0$ . Hence, taking $y=-g(x)$ , we get $f(xc)=f(0)+xd$ , thus $f$ is linear.

Claim 2: If $xg(-1)+2g(x)\ne g(-1)$ for some $x\ne 0$ , then $f$ is linear.
Proof. $P(-1,y): f(-y+g(-1))=-f(y)+h(-1)$ . Hence $f(-xy+xg(-1)+g(x))=xf(-y+g(-1))+h(x)=-xf(y)+xh(-1)+h(x)$ The RHS is actually $-f(xy+g(x))+xh(-1)+2h(x)$ .

Simplifying, $f(-y+xg(-1)+g(x))=-f(y+g(x))+xh(-1)+2h(x)\text{ for all }x\ne0$ $\iff f(-y+xg(-1)+2g(x))=-f(y)+xh(-1)+2h(x)\text{ for all }x\ne0$ Suppose $tg(-1)+2g(t)\ne g(-1)$ for some $t\ne 0$ , then letting $tg(-1)+2g(t)=a$ and $th(-1)+2h(t)=b$ we get $f(-y+a)=-f(y)+b$ , but
$f(-y+a)=f(-(y-a+g(-1))+g(-1))=-f(y-a+g(-1))+h(-1)$ This means $f(y)-f(y-a+g(-1))-h(-1)+b$ , which means $f$ is linear from claim 1.

Else, we have $xg(-1)+2g(x)=g(-1)$ for all $x\ne 0$ , which means $xh(-1)+2h(x)=h(-1)$ for all $h\ne 0$ . Hence $g,h$ are both linear (except at $0$ ).

Now in the original equation, picking $y$ constant such that it is not the negative of the gradient of $g$ , we get that $f$ is linear everywhere except at one point.

We are almost done, suppose $f(x)=ax+b$ for all $x\ne k$ , $g(x)=c(x-1)$ and $h(x)=d(x-1)$ for all $x\ne 0$ . Then taking any $x\ne 0$ and $y$ really large such that $|xy+g(x)|,|y|>k$ gives $axy+ag(x)+b=axy+xb+h(x)$ , so $ac(x-1)+b=xb+d(x-1)$ , which forces $ac=b+d$ .

If $k\ne -c$ , then we can take $y=k$ , which gives $f(xk+c(x-1))=xf(k)+d(x-1)$ . By taking $x$ sufficiently large, we get $axk+ac(x-1)+b=xf(k)+d(x-1)$ . This simplifies to $axk+b(x-1)+b=xf(k)$ so $f(k)=ax+b$ as well, done.

If $k=-c$ , then if we take $y=k$ , we get $f(xk+(-k)(x-1))=xf(k)+d(x-1)$ , so $f(k)=xf(k)+d(x-1)$ . By taking $x$ sufficiently large, $f(k)=-d$ . But this means $f(k)=f(-c)=-d=b-ac=a(-c)+b$ as well, so we are done too!

This completes the proof.

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#9 h Jun 21, 2022, 1:12 AM • 3 Y

Y by ImSh95, Mango247, Acorn-SJ

Suppose $f(2x+m)=2f(x)+u$ . Then if $t(x)=f(x-m)+u$ then $t(2x)=2t(x)$

We can verify there exists functions $g', h'$ such that $t(xy+g'(x))=xt(y)+h'(x)$
For all $x,y\in \mathbb{R}$ . Call this $P(x,y)$

Case 1: $g'(x)\ne 0$ for a fixed value of $x\ne 1$ .

Then $t(2(xy+g'(x))=2t(xy+g'(x))=2(xt(y)+h'(x))=2xt(y)+2h'(x)$

Also, $t(2xy+2g'(x))=t(x(2y+\frac{g'(x)}{x})+g'(x))=xt(2y+\frac{g'(x)}{x})+h'(x)=2xt(y+\frac{g'(x)}{2x})+h'(x)$

It follows that there exists constants $d = \frac{g'(x)}{2x} \ne 0,e$ such that $t(y+d)-t(y)=e$ for all $x\in \mathbb{R}$ . If $x=1$ , this also holds.

From here, $P(\frac ud, y)- P(\frac ud, y+d)$
yields $t(\frac ud y + g'(\frac ud)) - t(\frac ud y +g'(\frac ud) +u)=\frac{eu}{d}$
So for all $y\in \mathbb{R}$ , $t(y+u)-t(y)=\frac{eu}{d}$ , so it follows that $f(x)\equiv f(0)+\frac{ex}{d}$ is linear.

Case 2: $g'(x)\equiv 0$ . Then $t(xy)=xt(y)+h(x)$ .

$t(2xy)=2t(xy)=2(xt(y)+h(x))$ and $t(2xy)=t(x(2y))=x(t(2y))+h(x)=2xt(y)+h(x)$ , so $2h(x)=h(x)\rightarrow h(x)\equiv 0$

It follows $t(xy)=xt(y)$ , so $\frac{t(xy)}{xy}=\frac{t(y)}{y}$ so it readily follows $t(x)\equiv cx$ , so $t$ is linear.

Since $t$ is linear, $f$ is linear, and I am boring, and I am also geo0 for failing to get a point on any geo.

One day, I will be smart and I can get points on geos, but I will forever be boring since I am trapped in GGHX's world.

This post has been edited 5 times. Last edited by CANBANKAN, Jun 21, 2022, 1:21 AM

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407420

#10 h Jun 24, 2022, 2:09 AM • 1 Y

Y by ImSh95

Where is the FE Marathon?

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ZETA_in_olympiad

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ZETA_in_olympiad

#11 h Jun 24, 2022, 4:28 AM • 2 Y

Y by ImSh95, adorefunctionalequation

GianDR wrote:

Where is the FE Marathon?

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407420

#12 h Jun 24, 2022, 8:28 PM • 1 Y

Y by ImSh95

ZETA_in_olympiad wrote:

GianDR wrote:

Where is the FE Marathon?

Thank you!

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11165 posts

#13 h Mar 30, 2025, 3:25 AM

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Not smart enough for the flavortext ig, but for me this is a more intuitive solution

gghx wrote:

Suppose $f,g,h:\mathbb{R}\rightarrow\mathbb{R}$ are functions such that for any reals $x,y$ , $f(xy+g(x))=xf(y)+h(x)$ . Show that $f$ is linear.

Suppose $f$ weren't linear. Let $P(x,y)$ be the assertion $f(xy+g(x))=xf(y)+h(x)$ .

Claim 1: if $f(x+a)=f(x)+b$ for all $x\in\mathbb R$ given some constants $a,b\in\mathbb R$ then $a=b=0$
$P(x,y+a)\Rightarrow f(xy+g(x)+ax)=xf(y)+h(x)+xb=f(xy+g(x))+xb$
Substituting $y=-\frac{g(x)}x$ gives $f(ax)=f(0)+bx$ for all $x\ne0$ , but we can see that $f(ax)=f(0)+bx$ for $x=0$ as well. If $a\ne0$ , this means that $f$ is linear, so we must have $a=0$ and $b=f(x+a)-f(x)=0$ as well.

Claim 2: $g(xy)=g(x)+xg(y)$ and $h(xy)=h(x)+xh(y)$ for $x,y\ne0$
$P(x,yz+g(y))\Rightarrow f(xyz+xg(y)+g(x))=xyf(z)+xh(y)+h(x)$
$P(xy,z)\Rightarrow f(xyz+g(xy))=xyf(z)+h(xy)$
Comparing these, we get:
$f(xyz+xg(y)+g(x))+h(xy)=f(xyz+g(xy))+xh(y)+h(x)$ which implies, for all $x,y\ne0$ :
$f(z+xg(y)+g(x))+h(xy)=f(z+g(xy))+xh(y)+h(x)$ or
$f(z+xg(y)+g(x)-g(xy))=f(z)+xh(y)+h(x)-h(xy).$ By claim 1 (fix $x$ and $y$ ) we get $xg(y)+g(x)-g(xy)=xh(y)+h(x)-h(xy)=0$ , hence the result.

Claim 3: $g(x)=c(x-1)$ and $h(x)=d(x-1)$ for all $x\ne0$ , for some $c,d\in\mathbb R$
Given $g(xy)=g(x)+xg(y)$ , setting $x=1$ and $y=1$ gives $g(1)=0$ . Now for $x,y\ne0,1$ , swapping $x,y$ gives $g(x)+xg(y)=g(y)+yg(x)$ which rearranges to $\frac{g(x)}{x-1}=\frac{g(y)}{y-1}$ , so $g(x)=c(x-1)$ for $x\ne0,1$ for some constant $c$ . This will additionally be fulfilled for $x=1$ since $g(1)=0$ . Repeat similarly with $h(x)$ to finish.

Finish
For all $x\ne-c$ we have:
$P(x+c,1-c)\Rightarrow f(x)=(f(1-c)+d)x+cf(1-c)+cd-d$ (here we are just fixing $y$ in $P(x,y)$ )
$P(2,-c)\Rightarrow f(-c)=-d$
So we get $f(x)=(f(1-c)+d)x+cf(1-c)+cd-d$ for $x=-c$ as well, hence $f$ is linear.

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