Find all functions $ f: \mathbb{R} \to \mathbb{R} $ such that $$ f\left(xf\left(y\right)-f\left(x\right)-y\right) = yf\left(x\right)-f\left(y\right)-x $$holds for all $ x,y \in \mathbb{R} $
Problem
Source: 2019 Taiwan TST Round 1, Day 3, Problem 1
Tags: function, algebra, functional equation
14.04.2019 22:54
14.04.2019 23:24
12.05.2020 18:03
$P(0,0) \Longrightarrow f(-f(0)) = -f(0) $ $P(0,-f(0)) \Longrightarrow f(0)^2 = -2f(0) $ if $f(0)\neq 0\implies$ $f(0)=-2\implies f(2)=2$ $P(2,2)\implies f(0)=0$ a contradiction $\implies f(0)=0, P(0,y) \Longrightarrow -f(y) = f(-y) $ So $f$ is odd $P(x,0) \Longrightarrow ff(x) = x $ So $f$ is bijective $P(f(1),-1)$ it will give us $f(1)^2=f(1)$ by injective $f(1)=1$ $P(x,1)\implies f(-f(x)+x-1)=f(x)-x-1$ . now let $-f(x)+x=a$ now we are going to show that $a$ is a constant , Let's assume the opposite $\implies f(a-1)=-a-1$ now let $a=1$ $\implies f(0)=-2$ a contradiction $\implies a$ is a constant $\implies f(x)=x-a$ its easy to show from the original equation that : $a=0$ , $\implies$$f(x)=x,$$\forall x \in \mathbb{R}$
12.05.2020 18:06
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12.05.2020 18:09
Math-48 wrote: $P(0,0) \implies f(-f(0))=-f(0) \$. $P(0,-f(0))$ $\implies f(0)^2=-2f(0) $ if $f(0)$≠0 $\implies$ $f(0)=-2\implies f(2)=2$ $P(2,2)\implies f(0)=0$ a contradiction $\implies f(0)=0 $P(0,y)\implies -f(y)=f(-y)$ So $f$ is odd $P(x,0)\implies ff(x)=x$ So $f$ is bijective $P(f(1),-1)$ it will give us $f(1)^2=f(1)$ by injective $f(1)=1$ $P(x,1)\implies f(-f(x)+x-1)=f(x)-x-1$ now let $-f(x)+x=a$ now we are going to show that $a$ is a constant , Let's assume the opposite $/implies f(a-1)=-a-1$ now let $a=1$ $\implies f(0)=-2$ a contradiction $\implies a$ is a constant $\implies f(x)=x-a$ its easy to show from the original equation that : $a=0\implies\f(x)=x,\ \forall x\in \mathbb{R}$. من أين لك هذا
10.08.2020 13:02
We claim that the only solutions are $f(x)=x$ for all real $x$ . It is easy to see that these work . Now we prove that these are the only ones . Call the FE as $P(x,y)$ . Write $f(0)=c$ . We have by $P(x,0)$ and $P(0,x)$ $:=$ $$ f (cx-f(x))=-c-x \quad f(-c-x) = cx-f(x)$$This gives $f(f(-c-x))=-c-x \implies f(f(y))=y$ . Hence $f$ is a involution . Note that $P(0,0) \implies f(-c)=-c$ $P(1,1) \implies f(-1)=-1$ $P(-1,0) \implies f(1-c) =1-c$ $$ P(0,1-c) \implies -1 = f(-c+c-1)=(c-1)(1-c) \implies f(0)=0 \text { or } f(0)=2$$Assume $f(0)=2$ . Then $f(-2)=-2$ and $f(2)=0$ . However note that $P(-2,-2) \implies f(0)=0 (\rightarrow \leftarrow)$ So $f(0)=0$ This gives us $f(-f(x))=-x$ . Hence we have : $$f(-x) = f (f( -f(x)))=-f(x) \implies \text { f is odd} $$This also means that $f(1)=1$ . Next we have : $P(x,f(-x)) \implies f(x^2)=f(x)^2 \geq 0$ We are ready to finish. Consider $k= x-f(x)$ . WLOG , $k$ is non-negative, otherwise replace $x \mapsto -x$ We have $P(x-f(x)-1) = f(x)-x-1 \implies f(k-1)=-k-1$ Iterating this we obtain $:=$ $$ f( k\cdot 2^n -1) = - k\cdot 2^n -1$$for all positive integers $n$ , hence if $k \neq 0 $ , then for large $n$ we would have the RHS to be negative while the LHS stays positive, contradiction. We are done $\blacksquare$
11.08.2020 20:26
A quite unique finish for a functional equation Denote by $P(x,y)$ the assertion of $x$ and $y$ into the original condition. Plug in $P(y,x)$ and take $f$ of both sides to get $f(f(xf(y)-f(x)-y))=xf(y)-f(x)-y$, and notice by taking $x=0$ that $xf(y)-f(x)-y=-f(0)-y$ obtains every real value, so $f(f(x))=x$ for all $x$. This also means $f$ is bijective. $P(0,0)\implies f(-f(0))=-f(0)$, using it with $P(0,-f(0))\implies f(0)=-f(0)^2-f(-f(0)) $ gives $f(0)=0$ $P(0,x)\implies f(-x)=-f(x)$. $P(x,-x)\implies f(f(x)f(-x))=-x^2=f(f(-x^2))\implies -f(x)^2=-f(x)^2\implies f(x)^2=f(x^2)$. This together with the last result tells us that $f$ is positive on positive reals and negative on negatives. Similarly $P(1,-1)\implies f(1)^2=f(1) \implies f(1)=1$ $P(x,1)\implies f(f(x)-x-1)=x-f(x)-1$ Consider the last result, and assume that for some $x$ we have $x\leq f(x)$. Then $x-f(x)-1<0 \implies f(x)-x-1 < 0 \implies f(x)<x+1$. If $x>f(x)$ this is also obviously the case. Using the fact that $f$ is an involution, we also have $x=f(f(x))<f(x)-1\iff f(x)>x-1$. So we conclude $f(x)\in (x-1,x+1)$. Now denote $f(x)-x-1$ by $a$ for some random $x$. We have $f(a)=-a-2$. Also, if you denote by $n$ some large power of $2$, we have $f(a^n)=(a+2)^n$. Notice $\frac{f(a^n)}{a^n}\in (1-\frac{1}{a^n}, 1+\frac{1}{a^n})$, and this interval gets smaller as $n$ grows so $\frac{(a+2)^n}{a^n}\rightarrow 1$ as $n\rightarrow \infty$. This obviously implies $1+\frac{2}{a}$ is either $1$ or $-1$, so it has to be $-1$ and $a$ has to be $-1$, implying $x=f(x)$. Because choice of $x$ was arbitrary, it works for all of them, meaning $f(x)=x$ for all real $x$ $\blacksquare$ Just a little technicality, $a$ obviously can't be zero because $f(0)\neq -2$.
07.06.2021 23:06
Beautiful functional equation I have quite a different idea compared to the other solutions - it may seem like I am overcomplicating things, yet; my approach is quite easily motivated.
02.11.2023 22:02
injectivity,surjectivity,$f(0)=0$ $f(x)=-f(-x)$ and rest substitution $f(x)=x$
25.07.2024 14:46
24.12.2024 15:41
bin_sherlo wrote:
Why -f(\frac{y}{f(y)})=f(-f(\frac{y}{f(y)})) ?