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Source: albanian TST 2008 bmo

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Anni

#1 h May 24, 2009, 2:20 PM • 2 Y

Y by Adventure10, AlexCenteno2007

Find all functions $f: \mathbb R \to \mathbb R$ such that
$f(x+f(y))=y+f(x+1),$ for all $x,y \in \mathbb R$ .

This post has been edited 2 times. Last edited by Amir Hossein, May 1, 2018, 11:49 PM
Reason: LaTeXed.

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pco

#2 h May 24, 2009, 3:02 PM • 2 Y

Y by Adventure10, Mango247

Anni wrote:

find all f:R-R such that

f(x+f(y))=y+f(x+1)

Let $g(x) = f(x) - 1$ the equation becomes $g(x + g(y)) = y + g(x)$

$x = 0$ $\implies$ $g(g(y)) = y + g(0)$ and so $g(x)$ is bijective.
Then $y = 0$ $\implies$ $g(x + g(0)) = g(x)$ and so $g(0) = 0$ since $g(x)$ is bijective and we have $g(g(x)) = x$

Then $y = g(z)$ $\implies$ $g(x + z) = g(x) + g(z)$

So we have $g(x + y) = g(x) + g(y)$ and $g(g(x)) = x$

The first part is a classical Cauchy equation and we conclude :

The only continuous solutions are $f(x) = x + 1$ and $f(x) = 1 - x$
Infinitely many non continuous solutions (with AC) exist.

In fact all solutions may be presented in a unique method :
Let $\mathbb R$ considered as $\mathbb Q$ -vector space and $A$ any $\mathbb Q$ -vector subspace.
Let $B$ any supplementary (hope this is the good english word) vector subspace of $A$ ( $B$ always exists with Axiom of Choice)

Let $a(x)$ and $b(x)$ the projection functions of $x$ in $A$ and $B$ .

Then $f(x) = 1 + a(x) - b(x)$

$A = \mathbb R$ gives $f(x) = x + 1$
$A = \{0\}$ gives $f(x) = 1 - x$
any other $A$ give non continuous solutions.

edit : sorry, but there are a lot of other non continuous solutions

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#3 h Feb 16, 2024, 5:21 AM

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Fix $x$ and vary $y$ . This shows that $f$ is surjective. Next since
$y = f(x+f(y))-f(x+1) \implies f \text{ is injective}$ Take $x = x+1$ to get
$f(x-1+f(y)) = y+f(x) \implies f(x+y) = f(x)+f(y)-1$ where we took $y = f(y)-1$ since $f$ is surjective. Now let $g(x) = f(x)-1$ , so $g$ satisfies the additive Cauchy FE. Furthermore, by plugging in $g(x) = f(x)+1$ into the given assertion tells us that $g(g(y)) = y$ for all real $y$ . Now one can easily verify all
$\boxed{f(x) = g(x)+1} \text{ where $g$ is an involution that satisfies $g(x+y) = g(x)+g(y)$}$ works.

This post has been edited 1 time. Last edited by naonaoaz, Feb 23, 2024, 5:46 AM

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#4 h Feb 16, 2024, 4:25 PM

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naonaoaz wrote:

Fix $x$ and vary $y$ . This shows that $f$ is surjective. Next since
$y = f(x+f(y))-f(x+1) \implies f \text{ is injective}$ Take $x = x+1$ to get
$f(x-1+f(y)) = y+f(x) \implies f(x+y) = f(x)+f(y)-1$ where we took $y = f(y)-1$ since $f$ is surjective. Now let $g(x) = f(x)-1$ , so $g$ satisfies the additive Cauchy FE. Now one can easily verify all
$\boxed{f(x) = g(x)+1} \text{ where $g$ satisfies $g(x+y) = g(x)+g(y)$}$ works.

$f(x) = 1$ Clearly doesn't work.

Edit: If you changed it so that $g$ is any additive involution, then it is correct.

This post has been edited 1 time. Last edited by tadpoleloop, Feb 16, 2024, 9:26 PM

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#5 h Feb 23, 2024, 5:46 AM

Y by

tadpoleloop wrote:

naonaoaz wrote:

Fix $x$ and vary $y$ . This shows that $f$ is surjective. Next since
$y = f(x+f(y))-f(x+1) \implies f \text{ is injective}$ Take $x = x+1$ to get
$f(x-1+f(y)) = y+f(x) \implies f(x+y) = f(x)+f(y)-1$ where we took $y = f(y)-1$ since $f$ is surjective. Now let $g(x) = f(x)-1$ , so $g$ satisfies the additive Cauchy FE. Now one can easily verify all
$\boxed{f(x) = g(x)+1} \text{ where $g$ satisfies $g(x+y) = g(x)+g(y)$}$ works.

$f(x) = 1$ Clearly doesn't work.

Edit: If you changed it so that $g$ is any additive involution, then it is correct.

Thanks for pointing this out! I totally missed that condition on $g$ .

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#6 h Apr 19, 2024, 5:52 PM

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Let $g(x):=f(x)-1$ . Then
$g(x+g(y)+1)+1=y+g(x+1)+1\stackrel{x+1\rightarrow x}{\Longrightarrow}g(x+g(y))=g(x)+y.$ Denote by $P$ the assertion into the above FE. Clearly $g$ is bijective. Thus $P(0,0)$ gives $g(0)=0$ . Now $P(0,y)$ gives $g$ is an involution, so
$g(x+y)=g(x)+g(y).$ Thus the answer is all Cauchy involutions plus one, which clearly all work. $\blacksquare$

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#7 h Mar 7, 2025, 5:48 PM

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Let $g\equiv f - 1.$ The equation becomes $g(g(y) + 1 + x) = y + g(x + 1).$ Take $x = 0$ to get $g(g(y) + 1) = y + g(1).$ Take $y = 0$ to get $g(1 + x + g(0)) = g(x + 1).$ Add 1 and then take $g$ of both sides to get $g(1 + g(1 + x + g(0))) = g(g(x + 1) + 1).$ However, recall $g(g(y) + 1) = y + g(1).$ Thus, $g(0) = 0.$

Take $x = -1$ to get $g(g(y)) = y.$ Take $x = g(z) - 1$ to get $g(g(y) + g(z)) = y + z.$ However, recall that $g$ is an involution, implying that $g$ is bijective. Hence, if we assume $a = g(y), b = g(z),$ we get $g(a + b) = g(a) + g(b),$ implying $g$ is additive.

To summarize, we have that $g$ is an involution and additive. Let $A, B$ be sets such that $g(x) = x$ if $x\in A, f(x) = -x$ if $x\in B.$ Clearly, $A, B$ are $\mathbb Q$ -vector spaces, and $A\cap B = \{0\}.$ Hence, $\frac{x + g(x)}2\in A, \frac{x - g(x)}2\in B.$ Thus, $A + B$ generates $\mathbb R,$ so $x$ can be uniquely written as $x = a(x) + b(x).$ Thus, $g \equiv a - b.$ This finishes.

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#8 h Mar 17, 2025, 8:37 PM

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First, fixing $x$ and varying $y$ , we see that $f$ is surjective. Hence let $u \in \mathbb R$ satisfy $f(u) = 1$ , from where setting $y = u$ yields $u = 0$ .

Now, set $x = -1$ in the equation to yield $f(f(y) - 1) = y+1$ ; thus the function $g(y) = f(y) - 1$ satisfies $g(g(x)) = x$ . Furthermore, substituting $g(y)$ for $y$ in the original equation, we get $g(x+y+1) = g(y) + g(x+1)$ so $g$ is Cauchy and an involution. In particular, it is well-known that $g(x) = a(x) - b(x)$ for $a(x)$ and $b(x)$ the projections of $x$ onto two $\mathbb Q$ -orthogonal subspaces of $\mathbb R$ , $A$ and $B$ , that generate $\mathbb R$ . This completes the characterization.

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