Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying $f\left(x-f\left(y\right)\right)=1-x-y$, $x,\ y\in\mathbb{R}$

Question

I'm new in functional equations and stuck in this easy problem. Could anyone help with a clear solution?

This is what I have done so far:

Let $y=0$, then
$f\left(x-f\left(y\right)\right)=1-x-y\Rightarrow f\left(x\right)=1-x$
$f\left(x-f\left(y\right)\right)=1-x-y\Rightarrow f\left(x-\left(1-y\right)\right)=1-x-y$
$\Rightarrow f\left(x-1+y\right)=1-x-y\Rightarrow 1-\left(x-1+y\right)=1-x-y$
$\Rightarrow 1-x+1-y=1-x-y\Rightarrow 2=1$

But it isn't very helpful.

Thanks, Steve

There's no reason for assuming that $f(0)=0$. Indeed, it leads to a contradiction, so what you got is simply that $f(0)\ne0$. — egreg, Oct 23 '21 at 08:52
Plug $x = f(0)$, $y = 0$ to get an equation for $f(0)$. Then, plug $x = f(y)$ but don't fix any value for $y$. — Desura, Oct 23 '21 at 09:05
Setting $ x = y + f ( y ) $ in the original equation, you immediately get $ f ( y ) = \frac 1 2 - y $. It's straightforward to verify that it is indeed a solution. — Mohsen Shahriari, Oct 23 '21 at 14:49

score 1 · Accepted Answer · answered Oct 23 '21 at 08:56

1

Your argument only shows that $f(0)\ne0$, but the idea is good if used correctly.

Set $c=f(0)$. Then you know that $$ f(x-c)=1-x $$ for every $x$. For $x=y+c$, you obtain $$ f(y)=1-y-c $$ and the condition $f(0)=c$ implies $c=1-c$.

answered Oct 23 '21 at 08:56

egreg

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Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying $f\left(x-f\left(y\right)\right)=1-x-y$, $x,\ y\in\mathbb{R}$

1 Answers1