f(y+f(x))=f(y)+x

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Source: problem 1081. from http://genesiszero617.blogspot.com

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3 posts

janlaw

#1 h Feb 24, 2018, 10:54 AM • 1 Y

Y by Adventure10

For all $x,y$ real find continuous $f(x)$ from real numbers to real numbers satisfying the following equation:

$f(y+f(x))=f(y)+x$

EDITED

my solution is in the attachment.

Attachments:

This post has been edited 1 time. Last edited by janlaw, Feb 24, 2018, 11:15 AM
Reason: continuity added

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1600 posts

Vrangr

#2 h Feb 24, 2018, 11:12 AM • 2 Y

Y by Adventure10, Mango247

You can't assume continuity.

This is my 666th post.

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#3 h Feb 24, 2018, 11:32 AM • 2 Y

Y by Maths_Guy, Adventure10

janlaw wrote:

For all $x,y$ real find continuous $f(x)$ from real numbers to real numbers satisfying the following equation:

$f(y+f(x))=f(y)+x$

Let $P(x,y)$ be the assertion.
Claim: $f$ is injective.
Proof: Suppose $f(a)=f(b)$ , $P(a,y)$ and $P(b,y)$ give injectivity.

$P(0,0)$ gives $f(f(0))=f(0)\Rightarrow f(0)=0$ . $P(x,0)$ gives $f(f(x))=x$ . $P(f(x),y)$ gives $f(x+y)=f(x)+f(y)$ . $f$ is continuous, so $f(x)=cx$ . Plugging this in the original equation, $f(x)=x$ and $f(x)=-x$ are the only solutions.

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#4 h Feb 24, 2018, 11:36 AM • 1 Y

Y by Adventure10

@above, $f(x) = -x$ also works.

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janlaw

#5 h Feb 24, 2018, 11:40 AM • 2 Y

Y by Adventure10, Mango247

Once we know that the solution has the form $f(x)=cx$ then returning back to the original equation we obtain two solutions $f(x)=x$ or $f(x)=-x$ .

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nguyenhuybao_06

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nguyenhuybao_06

#6 h Nov 24, 2024, 3:34 PM

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For all $x,y,z\in\mathbb{R}$ we have $f(f(x)+y+z)=x+f(y+z).$ We also have $f(f(x)+y+z)=f(f(f(y)+x)+z)=f(y)+x+f(z),\forall x,y,z\in\mathbb{R}.$ Now we have $f(y+z)=f(y)+f(z),\forall x,y,z\in\mathbb{R}.$ From $f$ is continuous we claim that $f(x)=cx$ for all $x\in\mathbb{R}$ and $c$ is constant.
We have $c^2x+cy=x+cy$ so that $c=1$ or $c=-1.$
So that $f(x)\equiv x$ or $f(x)\equiv -x$ are solutions.

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#7 h Nov 24, 2024, 3:45 PM

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$f$ is bijective as $P(x, 0)$ shows.
In particular $f^2(x) = x+c$
Now replace $x$ with $f(x)$ and turn it into cauchy to see $f$ is linear, from where we can check for solutions

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