RagvaloD wrote: Find all functions $f:\mathbb{R}\to \mathbb{R},g:\mathbb{R}\to \mathbb{R}$ such that $$f(x-2f(y))= xf(y)-yf(x)+g(x)$$for all real $x,y$ Case 1: $f(0) \neq 0$ $P(0, \frac{g(0)}{f(0)})$ $\implies$ $f(t)=0$ for some $t$. $P(x, t)$ $\implies$ $g(x)=(t + 1)f(x)$ for all $x$. $P(x, y)$ becomes $f(x-2f(y))=xf(y)+(t+1-y)f(x)$. $P(0, x)$ $\implies$ $f(-2f(x))=(t+1-x)f(0)$ for all $x$, hence $f$ bijective. $P(x, t + 1)$ $\implies$ $f(x-2f(t+1))=xf(t+1)$, take $x=1$, we get $-2f(t+1)=t$, so $f(t+1)=-\frac{t}{2}$. Therefore $f(x+t)=-\frac{xt}{2}$ for all $x$ hence $f(x)=-\frac{t(x-t)}{2}$ for all $x$. Easy to check that $f:x\to \frac{t(x+t)}{2}$ and $g: x \to \frac{t(1-t)(x+t)}{2}$ is a solution, whenever $t \neq 0$. Case 2: $f(0)=0$ $P(x, 0)$ $\implies$ $f(x)=g(x)$. $P(0, x)$ $\implies$ $f(-2f(x))=0$ for all $x$. $P(x, -2f(x))$ $\implies$ $f(x)=0$ for all $x$. So $f = g\equiv 0$. Easy to check that $f=g \equiv 0$ is indeed a solution.

cazanova19921 wrote: RagvaloD wrote: Find all functions $f:\mathbb{R}\to \mathbb{R},g:\mathbb{R}\to \mathbb{R}$ such that $$f(x-2f(y))= xf(y)-yf(x)+g(x)$$for all real $x,y$ Case 1: $f(0) \neq 0$ $P(0, \frac{g(0)}{f(0)})$ $\implies$ $f(t)=0$ for some $t$. $P(x, t)$ $\implies$ $g(x)=(t + 1)f(x)$ for all $x$. $P(x, y)$ becomes $f(x-2f(y))=xf(y)+(t+1-y)f(x)$. $P(0, x)$ $\implies$ $f(-2f(x))=(t+1-x)f(0)$ for all $x$, hence $f$ bijective. $P(x, t + 1)$ $\implies$ $f(x-2f(t+1))=xf(t+1)$, take $x=1$, we get $-2f(t+1)=t$, so $f(t+1)=-\frac{t}{2}$. Therefore $f(x+t)=-\frac{xt}{2}$ for all $x$ hence $f(x)=-\frac{t(x-t)}{2}$ for all $x$. Replace in the original functional equation, we get $t(2t-ty+t-x)=0$ for all $x, y$ absurd (because $t \neq 0$). Case 2: $f(0)=0$ $P(x, 0)$ $\implies$ $f(x)=g(x)$. $P(0, x)$ $\implies$ $f(-2f(x))=0$ for all $x$. $P(x, -2f(x))$ $\implies$ $f(x)=0$ for all $x$. So $f = g\equiv 0$. Easy to check that $f=g \equiv 0$ is indeed a solution. Most of the solution seems right. Just in first case you did say that it doesn't have solution. But as I see $f(x)=\frac{x+1}{2}$ and $g(x)=0$ for all real numbers $x$ works. So please check again I think on last part of case 1 $t=-1$ work and get what I just did said. Oh and also it seems $f(x)=\frac{1-x}{2}$ and $g(x)=1-x$ works too which is by taking $t=1$.

dangerousliri wrote: Most of the solution seems right. Just in first case you did say that it doesn't have solution. But as I see $f(x)=\frac{x+1}{2}$ and $g(x)=0$ for all real numbers $x$ works. So please check again I think on last part of case 1 $t=-1$ work and get what I just did said. Oh and also it seems $f(x)=\frac{1-x}{2}$ and $g(x)=1-x$ works too which is by taking $t=1$. Thanks, I messed up the verification part. Actually, it works for any $t$.

cazanova19921 wrote: dangerousliri wrote: Most of the solution seems right. Just in first case you did say that it doesn't have solution. But as I see $f(x)=\frac{x+1}{2}$ and $g(x)=0$ for all real numbers $x$ works. So please check again I think on last part of case 1 $t=-1$ work and get what I just did said. Oh and also it seems $f(x)=\frac{1-x}{2}$ and $g(x)=1-x$ works too which is by taking $t=1$. Thanks, I messed up the verification part. Actually, it works for any $t$. Yeah I just did realised that since I did see that it works for $t=1$ and $t=-1$ without the paper and after I did get the paper I did realised that it is true for any $t$. Also an advice which solution can be shorten a little bit. You don't have necessarily to do with two cases. You do it like this: It does exist a real number $t$ such that $f(t)=0$ and the proof is on this way: If $f(0)=0$ it is done and if $f(0)\neq 0$ then you can do it as you did with $P(0,g(0)/f(0))$. So from that point you can suppose exist real number $t$ such that $f(t)=0$ and proceed same way.

dangerousliri wrote: You don't have necessarily to do with two cases. You do it like this: It does exist a real number $t$ such that $f(t)=0$ and the proof is on this way: If $f(0)=0$ it is done and if $f(0)\neq 0$ then you can do it as you did with $P(0,g(0)/f(0))$. So from that point you can suppose exist real number $t$ such that $f(t)=0$ and proceed same way. Yeah you are right, that will shoten the solution by three rows. But Let's just leave it this way, you may actually write this as alternative, it's always great to have different solutions in a thread .

cazanova19921 wrote: dangerousliri wrote: You don't have necessarily to do with two cases. You do it like this: It does exist a real number $t$ such that $f(t)=0$ and the proof is on this way: If $f(0)=0$ it is done and if $f(0)\neq 0$ then you can do it as you did with $P(0,g(0)/f(0))$. So from that point you can suppose exist real number $t$ such that $f(t)=0$ and proceed same way. Yeah you are right, that will shoten the solution by three rows. But Let's just leave it this way, you may actually write this as alternative, it's always great to have different solutions in a thread . Yeah I'm lazy to do that. It is everything okay with your solution too, there is nothing wrong. Just the solution would look more cleaner.

This problem was actually quite similar to this. https://artofproblemsolving.com/community/u454758h1441128p17262732 The ans is $f(x)=-\frac{t}{2}x+\frac{t^2}{2}, g(x)=(1+t)\left(-\frac{t}{2}x+\frac{t^2}{2}\right)$, where $t\in \mathbb{R}$. For the sake of completeness, let $P(x,y)$ be the given assertion, if $f(x)=0$, then we have $g(x)=0$, otherwise, choose $j\in \mathbb{R}$ such that $f(j)\ne 0$ and assume there exists $a,b \in \mathbb{R}$, \[P(j,a), P(j,b)\implies a=b\]so $f$ is injective. Now, if $f(0)=0$, then we have \[P(0,0)\implies g(0)=0\]\[P(0,x)\implies f(-2f(x))=g(0)=0\]by injectivity, we can see that $f(x)=0 \ \forall x\in \mathbb{R}$ in this case. Otherwise, if $f(0)\ne 0$, then we claim that $f$ is bijective, indeed, take $P(0,-\frac{x-g(0)}{f(0)})$, we have \[f(-2f(-\frac{x-g(0)}{f(0)}))=x \ \forall x\in \mathbb{R}.\]Now, there exist $t\in \mathbb{R}$ such that $f(t)=0$, we have \[P(x,t)\implies g(x)=(1+t)f(x)\]then \[P(x+2f(1+t),1+t)\implies f(x)=(x+2f(t+1))f(1+t)\]and $f$ is linear. Let $f(x)=mx+c, m,c\in \mathbb{R}$, then by substituting $f,g$ back to our equation and equating coefficients, we get that $m=-\frac{t}{2}, c=\frac{t^2}{2}$. Thus, $f,g$ can be written in the form of \[f\equiv -\frac{t}{2}x+\frac{t^2}{2}, g\equiv (1+t)\left(-\frac{t}{2}x+\frac{t^2}{2}\right)\]and it's easy to see that these are indeed the solutions. $\square$

Let $P(x,y):=f(x-2f(y))=xf(y)-yf(x)+g(x)$ Claim: $f$ is injective Proof: Let $f(a)=f(b)$ $P(x,a)$ yields $af(x)=xf(a)+g(x)-f(x-2f(a))$ $P(x,b)$ yields $bf(x)=xf(b)+g(x)+f(x-2f(b))$ $\therefore af(x)=bf(x)$ thus $f$ is injective if and only if $f\not\equiv0$ $\square$ If $f\equiv0$ we obtain $g\equiv0$ So from now on assume that $f\not\equiv0$ and thus is injective. If $f(0)=0$ $P(x,x)$ yields $f(x-2f(x))=g(x)$ $P(x,0)$ yields $f(x)=g(x)=f(x-2f(x))\Longrightarrow f(x)=f(x-2f(x))\overset{\text{injectivity}}{\Longrightarrow}2f(x)=0\Longleftrightarrow f(x)=0$ which clearly contradicts our assumption. Thus $f(0)\neq0$ $P\left(0,\frac{f(0)}{g(0)}\right)$ yields $f\left(-2f\left(\frac{f(0}{g(0)}\right)\right)=0$ thus $\exists c\text{ such that }f(c)=0$ $P(x,c)$ yields $g(x)=f(x)(x+1)$ furthermore using this in $P(x,y)$ we obtain the following $f(x-2f(y))=xf(y)+f(x)(c+1-y):=Q(x,y)$ $Q(0,x)$ yields $f(-2f(x))=f(0)(c+1-y)$ thus $f$ is also surjective. $Q(1,c+1)$ yields $f(c+1)=-\frac{c}{2}$ $Q(x,c+1)$ yields $f(x+c)=-\frac{cx}{2}\text{ now letting }x\to x-c\text{ we obtain }f(x)=-\frac{c(x-c)}{2}, \forall x\in\mathbb{R}$ Plugging this result into $P(x,x)$ yields $g(x)=-\frac{(c^2+c)(x-c)}{2}, \forall x\in\mathbb{R}$ So to sum up $\boxed{f(x)=-\frac{c(x-c)}{2}, g(x)=-\frac{(c^2+c)(x-c)}{2}\text{ and }f\equiv0, g\equiv0, \forall x\in\mathbb{R}\text{ and for }c\neq0}$ $\blacksquare$.