sorry it's wrong

f(X)=2X use the continuous property see that f(f(x))=4x find f(1)

oty wrote: I hope that is correct ; let $P(x,y)$ the assertion $f(yf(x+y)+f(x))=4x+2yf(x+y)$ ,$P(x,0)$ imply $f(f(x))=4x$ , this relation prove that f are bijective . put $f(0)=a$, and let $c\in{\mathbb R}$ such that $f(c)=0$ , $P(c,0)$ imply , $c=\frac{a}{4}$ , since $f(a)=f(c)$ (by $P(0,0)$) we get $a=\frac{a}{4}$ (by injectivity) and so $a=0$ , hence $f(0)=0$ . Now , $P(0,x)$ give $f(xf(x))=2xf(x)$(*) , and so , $f(y)=2y$ ${\forall{y\in{\mathbb R}}}$ ; which verify the FE $R = \left\{ {xf(x)\left| {x \in R} \right.} \right\}$? if $f(y) = 2y$, $R \ne \left\{ {xf(x)\left| {x \in R} \right.} \right\}$.

yes, you are Right , sorry , I am just a beginner in FE .

I hope that is correct ; let $P(x,y)$ the assertion $f(yf(x+y)+f(x))=4x+2yf(x+y)$ ,$P(x,0)$ imply $f(f(x))=4x$(**) , this relation prove that f are bijective . put $f(0)=a$, and let $c\in{\mathbb R}$ such that $f(c)=0$ , $P(c,0)$ imply , $c=\frac{a}{4}$ , since $f(a)=f(c)$ (by $P(0,0)$) we get $a=\frac{a}{4}$ (by injectivity) and so $a=0$ , hence $f(0)=0$ . Now , $P(0,x)$ give $f(xf(x))=2xf(x)$(*) , take $x=1$ , we get : $f(f(1))=2f(1)$ (***) , again take $x=1$ in (**) and comparing with (***) we find $f(1)=2$ . Now ,$P(x,1-x)$ give : $f(f(x)+2(1-x))=4=f(b)$ for an unique $b\in{\mathbb{R}}$ and by injectivity , we get : $f(x)=2x+b-2$ , $\forall{x\in{\mathbb R}}$ , replace in the FE , we get $b=2$ and so $f(x)=2x$, $\forall{x\in{\mathbb R}}$ .

and now ? , can someone check my second attempt please . Thank's .

Setting $y=0$ means $f(f(x))=4x$, whence $f$ is bijective. Also, $f(f(f(x)))=f(4x) \implies 4f(x)=f(4x) \implies f(0)=0$ Setting $x=0,y=1$ gives $f(f(1))=2f(1)$. As $f(f(1))=4$, $f(1)=2 \implies f(2)=4, f(4)=8$. Setting $x=1-y$, we have $f(2y+f(1-y))=4-4y+4y=4$ $f(f(2y+f(1-y))=f(4) \implies 8y+4f(1-y)=8 \implies f(1-y)=2-2y$ $\implies \boxed{f(x)=2x}$ for all real $x$.

Please someone check my solution. For $y=0$ we have $f(f(x))=4x$(1),so f is bijective. Now take $a$ such as $f(a)=0$.For $x=y=0$ we take $f(f(0))=0$.But $f(a)=0$.So $f(f(0))=f(a) \Leftrightarrow f(0)=a$,since f is bijective. Now for $x=a,y=0$ we have $f(0)=4a \Leftrightarrow f(0)=4f(0) \Leftrightarrow f(0)=0$ Setting $x=0y=f(y)$ we have $f(4y^2)=8y^2$.For $y \geq 0$ let $y=\frac{\sqrt{x}}{2}$ we get $f(x)=2x$ for all $x \geq 0$ This is the step that i'm not sure if it's right. But since f is bijective we have $f(x)=2x$ for all $x \in \mathbb{R}$ Please tell me somebody if my concept is right. PS.Sorry for my bad English

Unfortunately your proof is incomplete, since you only proved that $f(x) = 2x$ for $x\geq 0$. The fact that $f$ is bijective does not by itself prolong this to all $x\in \mathbb{R}$.

$y=0$ yields $f(f(x)) = 4x$ for all $x$. $f$ must therefore be bijective, and we must have $f(f(0)) = 0$. Thus, $f(0) = f(f(f(0))) = 4f(0) \implies f(0) = 0$. Setting $x=0$ now gives $f(yf(y)) = 2y f(y)$ for all $y$. Note that $f(0) = 2 \cdot 0$. Now take any $x \neq 0$, and pick $y$ so that $f(x+y) = \frac{f(x)}{x}$ (such a $y$ must exist since $f$ is surjective.) Then $yf(x+y) + f(x) = (x+y) f(x+y)$, so $4x + 2yf(x+y) = f(yf(x+y) + f(x)) = 2(x+y)f(x+y)$, whence $f(x+y) = 2$, so we must have $\boxed{f(x) = 2x}$ for all $x$.

Setting $y=0$ implies $f(f(x))=4x$, so $f$ is bijective. So there exists $b$, such that $f(b)=2$. Setting $x=b-y$ implies $f(yf(b)+f(b-y))=4(b-y)+2yf(b)=4b-4y+4y=4b$. Since function is injective, $yf(b)+f(b-y)$ is a constant, so $f(x)=ax+c$. Substitution shows that $f(x)=2x$.

Correct me if I'm wrong. We have $f(f(x))=4x$ so $f$ is bijective. Now, there exist $x, y \in \mathbb{R}$ such that $yf(x+y)=k$ for all $ k \in \mathbb{R}$ So let's $yf(x+y)=k$ and $f(x)=r$ to get $f(k+r) = f(r) + 2k \implies \frac{f(k+r)-f(r)}{k}=2$. This last equality is the slope of the line between points $(r+k,f(r+k))$ and $(r,f(r))$. But this points could be any points, and since the slope remains constant, it is that all points of $f$ are on the same line, thus $f$ is linear with slope $2$, i.e. $f(x)=2x+b \implies f(f(x))=4x+3b$, but $f(f(x))=4x \implies b=0$ and $f(x)=2x$

Let $P(x, y)$ mean plugging in the values of $x, y$ into the equation. $P(x, 0)$ gives $f(f(x)) = 4x$. Letting $x = f(0)$ tells us that $f(f(f(0))) = 4f(0)$. But $f(f(0)) = 0$ so $f(f(f(0))) = f(0)$ so we must have $f(0) = 0$. Now let $P(0, y)$ to get $$f(yf(y)) = 2yf(y) \qquad{(1)}$$ Taking $y = 1$ gives us $f(f(1)) = 2f(1)$. But we also know that $f(f(1)) = 4 \cdot 1 = 4$. Thus, $f(1) = 2$. Now, we let $P(x, 1-x)$ and we get $f(2(1-x) + f(x)) = 4$. Therefore \begin{align*} f(f(2 - 2x + f(x))) &= f(4) \\ \Rightarrow 8 - 8x + 4f(x) &= f(4) \end{align*} Which means $f(x) = 2x + c$ for some constant $c$. (Specifically $c = \frac{f(4) - 8}{4}$). Plugging this form into the original functional equation: \begin{align*} 2(y(2x+2y+c) + (2x+c)) + c &= 4x + 2y(2x+2y +c) \\ \Rightarrow c &= 0\end{align*} Thus our solution is $f(x) = 2x$.

My first FE in along time Let $P(x,y)$ be the given assertion. Put $y=0$ we get $f^2(x)=4f(x)$ so $f$ is bijective. Let $c$ be a number such that $f(c)$ =$2$ Now applying $f$ again on the equation gives $f(4x+2yf(x+y)) = 4yf(x+y)+4f(x)$ now subsequently put $x=c-y$ and then get $f(x)=2x-2c+2$. Put $x=2$ we get $f(x)=2x$ for all $x$. is the only solution and clearly it works.

Let $f(0)=c$, $P(x,y)$ be the assertion. $P(x,0)\implies f(f(x))=4x$, thus $f(x)$ is bijective and subbing $x=0$ gives $f(c)=0$. Subbing $x=c$ into $f(f(x))=4x$ gives $c=f(0)=4c$ hence $c=0$. $P(0,y)\implies f(yf(y))=2yf(y)$, sub $y=\frac{1}{2}$ and using injectivity gives $f\left(\frac{1}{2}\right)=1$. Sub $x=\frac{1}{2}$ into $f(f(x))=4x$ to get $f(1)=2$, doing this again we get $f(2)=4$. $P(x,1-x)\implies f(2-2x+f(x))=4x+2(2-2x)=4=f(2)$, hence by injectivity $f(x)=2x$ for all $x$ which clearly works.

$y=0$ then $f(f(x))=4x$. $x=0$ then $f(yf(y)+f(0))=2yf(y)$. Take $f(0)=c$ then $f(c)=ff(0)=0$. But $f(f(c))=f(0)$ while $f(f(c))=4c=4f(0)$, so $f(0)=0$. So $f(yf(y))=2yf(y)$. Say set $S$ be set of $z \in \mathbb{R}$ which $f(z)=2z$. We have $yf(y) \in S$ and $0 \in S$. Take $y=1$ then $f(1) \in S$, so $2f(1)=f(f(1))=4 \Rightarrow f(1)=2$. Observe if $x \in S$, then $yf(x+y)+f(x)=yf(x+y)+2x \in S$ for any $y \in \mathbb{R}$.Take $y=1-x$, then $1-x+2x \in S$ whenever $x \in S$. So $1 \in S$ means $2 \in S$ means $3 \in S$ means $4 \in S$. Now take $y=2-x$ in original equation then $f((2-x)f(2)+f(x))=4x+2(1-x)f(2) \Rightarrow f(4-2x+f(x))=8=f(4)$. Injectivity of $f$ means $f(x)=2x$ for all $x$.

yunxiu wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f\left( {yf(x + y) + f(x)} \right) = 4x + 2yf(x + y)\]for all $x,y\in\mathbb{R}$. Netherlands (Birgit van Dalen) My solution: By $P(0,0)$ we get $f(f(0))=0$ By $P(x,0)$ we get $f(f(x))=4x$(1);now put $x=f(0)$ we get $f(0)=0$ Now note that if $f(t)=0$ for some $t$ By (1) we get that $t=0$(2) Now by $P(x,f(x)-x)$ and using (1) we get: $f(4xy+f(x))=4x(1+2y)$ now put $y=-\frac{1}{2}$ we get $f(-2x+f(x))=0$ for all $x$ so by (2) we get $f(x)=2x$ for all $x$

Let \[f\left( {yf(x + y) + f(x)} \right) = 4x + 2yf(x + y)\]be denoted by $Q(x,y)$ $Q(x,0)\rightarrow f(f(x))=4x$ so function is bijective.Let $w$ be such number that $f(w)=0$ then $f(0)=4w$ and $f((0))=0=f(w)\rightarrow f(0)=w\rightarrow f(0)=4f(0)\rightarrow f(0)=0$ $Q(0,y)\rightarrow f(yf(y))=2yf(y)\rightarrow f(f(1))=2f(1)=4\rightarrow f(1)=2$ $Q(x,1-x)\rightarrow f((1-x)2+f(x))=4x+4(1-x)=4=f(f(1))=f(2)\rightarrow$ by injectivity $(1-x)2+f(x)=2\rightarrow f(x)=2x$

We claim that the only solution is $\boxed{f(x) = 2x}$. Let $P(x, y)$ denote the assertion. Now, $P(x, 0)$ gives \begin{eqnarray*} f(f(x)) & = & 4x, \end{eqnarray*}which implies that $f$ is bijective. Then, $P(f(0), 0)$ gives \begin{eqnarray*} f(f(f(0))) & = & 4f(0) \\ f(0) & = & 4f(0) \\ f(0) & = & 0. \end{eqnarray*}Next, $P(0, 1)$ gives \begin{eqnarray*} f(f(1)) & = & 2f(1) \\ 4 & = & 2f(1) \\ f(1) & = & 2 \mbox{.} \end{eqnarray*}Finally, $P(x, 1 - x)$ gives \begin{eqnarray*} f(f(x) + 2 - 2x) & = & 4 \mbox{.} \end{eqnarray*}Since $f$ is bijective, we must have \begin{eqnarray*} f(x) + 2 - 2x & = & a, \end{eqnarray*}for some real number $a$ such that $f(a) = 4$. This implies that \begin{eqnarray*} f(x) & = & 2x + c, \end{eqnarray*}for some real number $c$. Plugging this into the original equation, we find that $c = 0$. Thus, the only solution is $f(x) = 2x$.

Let $P(x,y)$ denote the given assertion. $P(x,0): f(f(x))=4x$, which implies $f$ bijective. So by tripling the involution, we have $f(f(f(x)))=f(4x)=4f(x)$. So $f(0)=4f(0)\implies f(0)=0$. $P(0,x): f(xf(x))=2xf(x)$. So $f(f(1))=2f(1)=4\implies f(1)=2$ $P(x,1-x): f(2-2x+f(x))=4$. Since $f$ is injective and $f(2)=4$, $2-2x+f(x)=2\implies \boxed{f(x)=2x}$, which works. Edit: Didn't realize I'd already solved this.

yunxiu wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f\left( {yf(x + y) + f(x)} \right) = 4x + 2yf(x + y)\]for all $x,y\in\mathbb{R}$. Netherlands (Birgit van Dalen) Setting $y=0$ gives $$ f(f(x))=4x \Longrightarrow 4f(x)=f(f(f(x)))=f(4x) \Longrightarrow f(0)=0 $$Now, letting $g\equiv \frac{f}{2}$ yields $$ P(x,y): \thickspace g(yg(x+y)+g(x))=x+yg(x+y), $$where $g(0)=0$ and $g(g(x))=x$, so $g$ is bijective. Now, $P(0,1)$ gives $g(g(1))=g(1) \Longrightarrow g(1)=1$. And $P(x,1-x)$ finally gives $$ g(1-x+g(x))=x+1-x=1=g(1) \Longrightarrow g(x)=x $$Hence $f(x)=2x$ is the only solution, which indeed fits.

$P(x,0)\implies f(f(x)) = 4x$ so $f$ is bijective. Let $u = f(0)$. $P(u,1)\implies f(f(u+1)) = 4u + 2f(u+1)\implies 4u + 4 = 4u + 2f(u+1)\implies f(u+1) = 2$. $P(2,u-1)\implies f(2u - 2 + 4u + 4) = 8 + 4u - 4\implies f(6u + 2) = f(2)\implies u = 0$. $P(x,1-x)\implies f(2-2x+f(x)) = 4x + 4 - 4x\implies f(2 - 2x + f(x)) = f(2)\implies f(x)\equiv 2x$ which is a solution.

We let $P(x, y)$ denote the assertion. For notational purposes and simplicity, I am declaring $f^3(x) := f(f(f(x)))$ Show Bijection: $P(x,0) \implies f(f(x))=4x \forall x \in \mathbb{R}$. This is obviously a bijection if we are mapping across the reals, in a similar manner to an involution. Show $f(0)=0:$ Because of the above, if we pass $f(f(x))$ again into the parameters of the function, then $f^3(x) = f(4x)$ However, substituting $x \mapsto f(x)$ into the body of $P(x,0)$ would show that $f^3(x) = 4f(x)$, so we have $4f(x) = f(4x)$ Substituting $0$ into this would yield $4f(0) = f(0) \implies f(0)=0$ Show that the only solution is $f(x) = 2x:$ $P(0, x) \implies f(xf(x)) = 2xf(x)$ By $P(0, x)$, we hence have that $f(f(1)) = 4 = 2f(1) \iff f(1) = 2 \implies f(2) = 4$ $P(x, 1-x) \implies f(2-2x+f(x)) = 4(x + 1-x) = 4$ Hence, because $f(2) =4$, $2-2x + f(x) = 2 \implies \color{red}\boxed{f(x) = 2x}$, which we could immediately check works, QED $\blacksquare$

$P(x,0) : f(f(x)) = 4x \implies f$ is bijective. $f(f(x)) = 4x \implies f(4x) = 4f(x) \implies f(0) = 0$. $P(0,1) : f(f(1)) = 2f(1) = 4 \implies f(1) = 2 \implies f(2) = 4 \implies f(4) = 8$. $P(1-y,y) : f(yf(1) + f(1-y)) = 4 - 4y + 2yf(1) \implies f(2y + f(1-y)) = 4 \implies 8y + 4f(1-y) = f(4) = 8 \implies f(1-y) = 2(1-y) \implies f(x) = x$. by checking in main equation we see it works.

Let $P(x,y)=f(yf(x+y)+f(x))=4x+2yf(x+y)$ $P(0,0)$ yields: $f(f(0))=0$ $P(f(0),0)$ gives us: $f(f(0))=4f(0) \Longleftrightarrow f(0)=0$ $P(x,0)$ gives $f(f(x))=4x$ Claim:The function is bijective:

Claim:$f(x)=2x$

$f(yf(x+y)+f(x))=4x+2yf(x+y)$ $P(0,0)$ $ff(0)=0$ $P(f(0),0)$ $f(0)=0$ $P(x,-x)$ $ff(x)=4x$ $\implies$ $f$ bijective $P(0,1)$ $ff(1)=2f(1)=4$ $\implies$ $f(1)=2$ $P(2,-1)$ $f(f(2)-2)=4$ $P(x,1-x)$ $f(2-2x+f(x))=4$ $2-2x+f(x)=f(2)-2$ $f(x)=2x-4+f(2)$ replace for all $f(x)$ then you will get $f(2)=4$ and $f(x)=2x$

Clearly, $f(x)=2x$ satisfies the problem condition. It is also easy to check this is the only linear solution. Plugging in $y=0$ gives $f(f(x))=4x$. Let $c=f(0)$. We have $f(c)=0$. Plug in $(x, y)=(c, 1)$ to obtain $f(f(c+1))=4c+2f(c+1)$. Thus, $f(c+1)=2$. This also implies that $f(2) = 4c+4$. Finally, let $y=c+1-x$. We obtain $f(f(x)+2(c+1-x))=4x+4(c+1-x)=4(c+1)$. Taking $f$ on both sides, we see $4(f(x)+2(c+1-x))$ is a constant. Thus, $f(x) must be linear.

Solution from Twitch Solves ISL: The only solution is $f(x) \equiv 2x$ which obviously works. Let $P(x,y)$ be the given condition. Then: Note $P(x,0) \implies f(f(x)) = 4x$; in particular $f$ is bijective. This also implies $f(4x) = f(f(f(x))) = 4f(x)$. Taking $x=0$ gives $f(0) = 4f(0) \implies f(0) = 0$. Now $P(0,2) \implies f(2f(2)) = 4f(2) = f(8) \implies f(2) = 4$. Then $P(0,1) \implies f(f(1)) = 2f(1) \implies 4 = 2f(1) \implies f(1)=2$. Finally, $P(x,1-x)$ gives \[ f\left( 2(1-x) + f(x) \right) = 4x + 4(1-x) = 4. \]Since $f$ is a bijection and $f(2) = 4$, this means $2(1-x)+f(x) = 2$, so $f(x) = 2x$ as desired.

let $P(x,y)$ the assertion $f(yf(x + y) + f(x)) = 4x + 2yf(x + y)$ $f(0) = a$ $P(0,0) => f(a) = 0$ $P(x,0) => f(f(x)) = 4x => bijective (1)$ in $(1): x->a => a=4a => a=0 => f(0) = 0$ $P(0,1) => f(1) = 2$ in $(1): x-> 1 => f(f(1))=f(2)=4$ $P(x-y,y) => f((x-y)f(x)+f(x))=4(x-y)+2yf(x)$ $x->1 => f((1-y)2+2)=4(1-y)+4y = 4$ $=> 2(1-y)+2=2$ $=>f(x)=2x$

Let $P(x,y)$ be the assertion $f(yf(x+y)+f(x))=4x+2yf(x+y)$. $P(x,0)$ gives that $\boxed{f(f(x))=4x \ (1)}$. From $(1)$ we get that $f$ is injective $(*)$ ($f(a)=f(b) \implies f(f(a))=f(f(b)) \implies 4a=4b \implies a=b$). Substituting $(1)$ in $P(x,y)$ we obtain $f(yf(x+y)+f(x))=4x+2yf(x+y)\underset{(1)}{=}f(f(x+\dfrac{yf(x+y)}{2})) \underset{(*)}{\implies} \boxed{yf(x+y)+f(x)=f(x+\dfrac{yf(x+y)}{2}) \ (2)}$ by injectivity. In $(2)$, make $y=f(\frac{1}{2})-x$ so $(f(\frac{1}{2})-x)f(f(\frac{1}{2}))+f(x)=f(x+\dfrac{yf(f(\frac{1}{2}))}{2}) \underset{(1)}{\implies} 2(f(\frac{1}{2})-x)+f(x)=f(x+\dfrac{2y}{2})=f(x+y)=f(f(\frac{1}{2}))=2 \implies f(x)=2x+\underbrace{2-2f(\frac{1}{2})}_{\text{constant}}$ because $f(f(\frac{1}{2}))=4\cdot\frac{1}{2}=2$. So $f(x) \equiv 2x+c$ for some constant $c$, but $f(f(x))=4x$ by $(1)$ so $f(f(x))=2f(x)+c=2(2x+c)+c=4x+3c=4x \implies c=0$ and testing, $f(x) \equiv 2x$ in fact works and it's the only solution.

$P(x, 0)$ gives $f(f(x)) = 4x$, which gives bijectivity. Substitute $f(x)$ here, we have $f(4x) = 4f(x) \implies f(0) = 0$. $P(0, y)$ gives $f(yf(y)) = 2yf(y) \implies f(f(1)) = 2f(1) \implies f(1) = 2$. Substitute $y = \frac{1}{2}$ here, we have $\frac{1}{2} \cdot f(\frac{1}{2}) = \frac{1}{2} \implies f(0.5) = 1$. So, $f(4\cdot0.5) = f(2) = 4\cdot f(0.5) = 4$. $P(x, -x + 1)$ gives: $$f(f(x) - 2x + 2) = 4 = f(2) \implies f(x) - 2x + 2 = 2 \implies f(x) = 2x$$and, we're done...

#MakeFECreativeAgain. We claim the only solution is $f(x) = 2x \quad \forall x$ which works. We now prove it's the only solution. Let $P(x, y)$ denote the given assertion. $P(x, 0) \implies f^2(x) = 4x$, so we get both that $f$ is bijective and $f(4x) = 4f(x)$. Further from here we have $4f(0) = f(0) \implies f(0) = 0$. Now $P(0, 1) \implies 4 = f(f(1)) = 2f(1) \implies f(1) = 2,\implies f(2) = f(f(1)) = 4$. Finally, $P(1-y, y) \implies f(2y+f(1-y)) = 4 = f(2) \implies f(1-y) = 2(1-y) \implies \boxed{f(x) = 2x \forall x}$. $\square$