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
BlazingMuddy
14.04.2019 22:54
$f(x) = x$ for all $x\in \mathbb{R}$.
It is easy to check, or rather obvious, that this function satisfies the criterion.
Our first observation is that:
$$f(f(xf(y) - f(x) - y)) = f(yf(x) - f(y) - x) = xf(y) - f(x) - y\quad \forall x, y\in \mathbb{R}$$
Setting $x = 0$ and $y = - t - f(0)$ here gives $f(f(t)) = t$ for any real number $t$. Hence, $f$ is bijective. This also gives:
$$f(xy - f(x) - f(y)) = f(x)f(y) - x - y \quad \forall x, y \in \mathbb{R} \ldots (1)$$
Let $f(0) = c$. Substituting $x = 0$ into the original equality gives us $f(-y-c) = cy - f(y)$ and thus $f(x) + f(-x-c) = cx$ for all $x\in \mathbb{R}$, so we obtain:
$$0 = f(0) + f(-c) = f(-c) + f(-(-c)-c) = -c^2$$
Hence, $f(0) = c = 0$ and thus $f$ is odd. Now, substituting $y = -x$ into $(1)$ gives us $f(-x^2) = f(x)f(-x) = -f(x)^2$, so
$$f(x^2) = f(x)^2 \quad \forall x\in \mathbb{R}\ldots (2)$$
This is the core of our solution. Since $f$ is bijective, $f(0) = 0$ and $f(1)^2 = f(1)$, we have $f(1) = 1$. Also, this informs us that $f(x)\geq 0$ when $x\geq 0$, and $f(x)\leq 0$ when $x\leq 0$. Substituting $y = 1$ into $(1)$ gives us $f(x-f(x)-1) = f(x)-x-1$ for all $x\in \mathbb{R}$. If $f(x)-x > 1$ for some $x\in \mathbb{R}$, then $f(x) - x - 1 > 0 > x - f(x) - 1$, which is a contradiction. SImilarly, if $f(x) - x < -1$, then $x - f(x) -1 > 0 > f(x) - x - 1$, yielding another contradiction. Hence,
$$|f(x) - x|\leq 1\quad \forall x\in \mathbb{R}$$
Now, we prove that $f(x) = x$ for all $x\in \mathbb{R}$ such that $|x| > 1$. Since $x$ is odd, without loss of generality, we may assume $x > 1$. Observe that $|f(x)^{2^n} - x^{2^n}| = |f(x^{2^n}) - x^{2^n}|\leq 1$ for all positive integer $n$ (use $(2)$ and induction). However, it also holds that $|f(x)^{2^n} - x^{2^n}| = |f(x)^{2^{n-1}} - x^{2^{n-1}}|(f(x)^{2^{n-1}} + x^{2^{n-1}})\geq x^{2^{n-1}} |f(x)^{2^{n-1}} - x^{2^{n-1}}|$ for all positive integer $n$, and thus by induction we get $1\geq |f(x)^{2^n} - x^{2^n}|\geq x^{2^n-1} |f(x)-x|$ for all positive integer $n$. This is false for some $n$ big enough if $f(x)\neq x$, so indeed $f(x) = x$ for all $x > 1$.
Finally, we prove that $f(x) = x$ for all $x\geq -1$. To prove this, notice that by $(1)$ we have $f(xy-x-y) = xy-x-y$ for all $x, y\geq 1$, which can be re-written as $f(xy-1) = xy-1$ for all $x, y\geq 0$. But we can write every real number greater than or equal to $-1$ as $xy-1$, so indeed $f(x) = x$ for all $x\geq -1$, and this gives $f(x) = x$ for all $x\in \mathbb{R}$ as $f$ is odd.
rmtf1111
14.04.2019 23:24
BlazingMuddy wrote:
Our first observation is that:
$$f(f(xf(y) - f(x) - y)) = f(yf(x) - f(y) - x) = xf(y) - f(x) - y\quad \forall x, y\in \mathbb{R}$$
Setting $x = 0$ and $y = - t - f(0)$ here gives $f(f(t)) = t$ for any real number $t$. Hence, $f$ is bijective. This also gives:
$$f(xy - f(x) - f(y)) = f(x)f(y) - x - y \quad \forall x, y \in \mathbb{R} \ldots (1)$$
Let $f(0) = c$. Substituting $x = 0$ into the original equality gives us $f(-y-c) = cy - f(y)$ and thus $f(x) + f(-x-c) = cx$ for all $x\in \mathbb{R}$, so we obtain:
$$0 = f(0) + f(-c) = f(-c) + f(-(-c)-c) = -c^2$$
Hence, $f(0) = c = 0$ and thus $f$ is odd. Now, substituting $y = -x$ into $(1)$ gives us $f(-x^2) = f(x)f(-x) = -f(x)^2$, so
$$f(x^2) = f(x)^2 \quad \forall x\in \mathbb{R}\ldots (2)$$
This is the core of our solution. Since $f$ is bijective, $f(0) = 0$ and $f(1)^2 = f(1)$, we have $f(1) = 1$. Also, this informs us that $f(x)\geq 0$ when $x\geq 0$, and $f(x)\leq 0$ when $x\leq 0$. Substituting $y = 1$ into $(1)$ gives us $f(x-f(x)-1) = f(x)-x-1$ for all $x\in \mathbb{R}$.
Thus there exists a $u\in \mathbb{R}$ for which $f(u-1)=-u-1.$ Let $x\rightarrow u-1$ in $f(x-f(x)-1) = f(x)-x-1,$ this implies that $f(2u+1)=1-2u.$ Now let $x\rightarrow 2u+1$ in $f(x-f(x)-1) = f(x)-x-1,$ thus $f(4u+1)=1-4u$ and an easy induction gives us that $f(2^nu+1)=1-2^nu,$ but for $n$ big enough $2^nu+1$ and $1-2^nu$ will have different signs unless $u=0,$ thus $u=x-f(x)=0 \implies f(x)=x.$
Math-48
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}$
Math-48
12.05.2020 18:06
..........
Bst
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
Aryan-23
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 $\blacksquare$
The ending was unexpected ! I think all the steps till the $f(x^2)=f(x)^2$ are quite straightforward . It is from there I took a long time . Basically we realize that the only way we can use this relation is to somehow try to make f (positive) = negative , forcing something to be 0 in the process . This was because all assertions of the form $P(x^2,x)$ etc fail . A bit too hard for a P1 , I'd say .
niksic
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 $f(0)\neq -2$.
bora_olmez
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.
$f(x) = x$ for all $x \in \mathbb{R}$ is the only solution.
Let $P(x,y)$ denote the assertion in the original equation.
$P(0,0): f(-f(0)) = -f(0)$
$P(0,-f(0)): f(0) = 0$
$P(0,x): f(-x) = -f(x)$ meaning that $f$ is odd.
$P(x,0): f(f(x)) = x$ meaning that $f$ is involutive
$P(x,-f(x)):$
Using the fact that $f$ involutive and odd, $f(x^2) = f(x)^2$
Lemma 1: $f(a) = \frac{1}{a}$ for $a \in \mathbb{R} \implies a \in \{1,-1\}$
Proof)
Let $a \in \mathbb{R}$ such that $f(a) = \frac{1}{a}$.
Assuming that $a \neq 1, -1$, and using that $f(a)^2 = f(a^2)$ as well as the fact that $f$ is both involutive and odd, we get that there are arbitrarily large $a \in \mathbb{R}$ such that $f(a) = \frac{1}{a}$.
$P(a,\frac{1}{a}): f(a^2-\frac{2}{a}) = \frac{1}{a^2} - 2a$
Taking $a \geq 2$ (essentially any other constant in place of $2$ also works) in the above equation, we get that $a^2 \geq 4, \frac{2}{a} \leq 1$ and therefore $a^2 - \frac{2}{a} \geq 3$, on the other hand $\frac{1}{a^2} \leq \frac{1}{4}$ and $2a \geq 4$ and therefore $\frac{1}{a^2} - 2a \leq \frac{-15}{4}$
More importantly, we get that $f(\text{positive}) = \text{negative}$ which is a contradiction as $f(x^2) = f(x)^2$. $\square$
Lemma 2: $f(x) \in \{x,-x\}$ for all $x \in \mathbb{R}$
Proof)
Notice that $f(x) \neq 0$ for $x \neq 0$ because $f$ is involutive (meaning that it is injective) as $f(0) = 0$.
$P(\frac{x}{f(x)},x):$
$$-f(f(\frac{x}{f(x)})) = xf(\frac{x}{f(x)}) - f(x) - \frac{x}{f(x)}$$$\implies$ $$\frac{f(x)}{x} = f(\frac{x}{f(x)})$$for $x \in \mathbb{R}$, $x \neq 0$ as $f$ is involutive.
If we let $S = \{x \in \mathbb{R} \mid \frac{x}{f(x)} \}$ then for $s \in S$, $f(s) = \frac{1}{s}$. Using Lemma 1, we get that $f(x) \in \{x,-x\}$ for $x \in \mathbb{R}$ $\square$
From $f(x^2) = f(x)^2$, we get that $f(1) = 1$ as $f(1) \neq 0$. This also means that $f(-1) = -1$
Firstly notice that $f(x) = -x$ for all $x \in \mathbb{R}$ is not a solution and that $f(x) = x$ is indeed a solution.
We can now assume that $f(a) = a$ and $f(b) = -b$ for $a,b \in \mathbb{R}$, $a,b \neq 0$.
$P(a,b): f(-ab-a-b) = ab +b - a$
Case 1) $f(-ab-a-b) = -ab-a-b$
We get that $-ab-a-b = ab+b-a$ then $2ab = 2b$ meaning that $a =1$, this is a contradiction as $a=-1$ is also possible.
Case 2) $f(-ab-a-b) = ab+a+b$
We get that $ab+a+b = ab+b-a$ then $2a = 0$ contradicting $a \neq 0$.
As we have dealt with both cases, we get that $f(x) = x$ for all $x \in \mathbb{R}$ is the only solution.
ismayilzadei1387
02.11.2023 22:02
injectivity,surjectivity,$f(0)=0$ $f(x)=-f(-x)$ and rest substitution $f(x)=x$
bin_sherlo
25.07.2024 14:46
$$ f\left(xf\left(y\right)-f\left(x\right)-y\right) = yf\left(x\right)-f\left(y\right)-x $$$P(0,0)\implies f(-f(0))=-f(0)$ and $P(0,-f(0))\implies f(0)=-f(0)^2-f(-f(0))=-f(0)^2+f(0)$ which gives that $f(0)=0$.
$P(x,0)\implies f(-f(x))=-x$ hence $f$ is bijective.
$P(-f(y),y)\implies f(-f(y)^2)=-y^2=f(-f(y^2))$ Since $f$ is injective, we get $f(y)^2=f(y^2)$.
$P(0,-f(y))\implies f(f(y))=y=f(-f(-y)) \implies f(y)=-f(-y)$ thus, $f$ is odd.
$P(x,\frac{x}{f(x)})\implies f(xf(\frac{x}{f(x)})-f(x)-\frac{x}{f(x)}))=-f(\frac{x}{f(x)})=f(-\frac{x}{f(x)})$ Since $f$ is injective, we have $f(\frac{x}{f(x)})=\frac{f(x)}{x}$
$P(\frac{y}{f(y)},y)\implies -\frac{f(y)}{y}=-f(\frac{y}{f(y)})=f(-f(\frac{y}{f(y)}))=yf(\frac{y}{f(y)})-f(y)-\frac{y}{f(y)}=-\frac{y}{f(y)}$ So $y^2=f(y)^2$ which yields $f(x)=\pm x$.
$f(x^2)=f(x)^2\geq 0$ and $f(-x^2)=-f(x^2)=-f(x)^2\leq 0$ hence $f(x)=x \ \forall \in R$ as desired.$\blacksquare$
deltapc
24.12.2024 15:41
bin_sherlo wrote:
$$ f\left(xf\left(y\right)-f\left(x\right)-y\right) = yf\left(x\right)-f\left(y\right)-x $$$P(0,0)\implies f(-f(0))=-f(0)$ and $P(0,-f(0))\implies f(0)=-f(0)^2-f(-f(0))=-f(0)^2+f(0)$ which gives that $f(0)=0$.
$P(x,0)\implies f(-f(x))=-x$ hence $f$ is bijective.
$P(-f(y),y)\implies f(-f(y)^2)=-y^2=f(-f(y^2))$ Since $f$ is injective, we get $f(y)^2=f(y^2)$.
$P(0,-f(y))\implies f(f(y))=y=f(-f(-y)) \implies f(y)=-f(-y)$ thus, $f$ is odd.
$P(x,\frac{x}{f(x)})\implies f(xf(\frac{x}{f(x)})-f(x)-\frac{x}{f(x)}))=-f(\frac{x}{f(x)})=f(-\frac{x}{f(x)})$ Since $f$ is injective, we have $f(\frac{x}{f(x)})=\frac{f(x)}{x}$
$P(\frac{y}{f(y)},y)\implies -\frac{f(y)}{y}=-f(\frac{y}{f(y)})=f(-f(\frac{y}{f(y)}))=yf(\frac{y}{f(y)})-f(y)-\frac{y}{f(y)}=-\frac{y}{f(y)}$ So $y^2=f(y)^2$ which yields $f(x)=\pm x$.
$f(x^2)=f(x)^2\geq 0$ and $f(-x^2)=-f(x^2)=-f(x)^2\leq 0$ hence $f(x)=x \ \forall \in R$ as desired.$\blacksquare$
Why -f(\frac{y}{f(y)})=f(-f(\frac{y}{f(y)})) ?