khan.academy wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for any real $x, y$ holds equality $$f(xf(y)) + f(xy) = 2f(x)y$$Proposed by Arseniy Nikolaev Is the answer in terms of f(1)?(just to confirm my solution)

Math-wiz wrote: khan.academy wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for any real $x, y$ holds equality $$f(xf(y)) + f(xy) = 2f(x)y$$Proposed by Arseniy Nikolaev Is the answer in terms of f(1)?(just to confirm my solution) No... $f(1)$ is either $0$ or $1$... to prove that you could just prove that the function is injective by substituting $x=1$ and then prove that, if $f(1) \neq 0$, then substitute $y$ as $1$ to get $f(xf(1))=f(x)=> f(1)=1$.

For convenience, let $ P\left(x,y\right) $ be the assertion of the functional equation. First of all, $$ P\left(1,y\right) \rightarrow f\left(f\left(y\right)\right)+f\left(y\right) = 2f\left(1\right)y $$So $ f\left(1\right) = 0 $ or $ f $ is injective. If $ f\left(1\right) = 0 $, then $ P\left(x,1\right) \rightarrow f\left(0\right) = f\left(x\right) $, which means $ f $ is constant $ 0 $. Otherwise, $ f $ is injective. In this case, $$ P\left(x,1\right) \rightarrow f\left(xf\left(1\right)\right) = f\left(x\right) \rightarrow f\left(1\right) = 1 $$by injectivity and thus, $ f\left(f\left(y\right)\right)+f\left(y\right) = 2y $. Now, $$ P\left(x,x\right) \rightarrow f\left(xf\left(x\right)\right)+f\left(x^2\right) = 2xf\left(x\right) = f\left(f\left(xf\left(x\right)\right)\right)+f\left(xf\left(x\right)\right) $$Therefore, $ f\left(xf\left(x\right)\right) = x^2 $ and $ f\left(x^2\right)+x^2 = 2xf\left(x\right) $. It seems that we have enough results to bash it $$ x^2f\left(x\right)^2 = f\left(x^3f\left(x\right)\right) = 2xf\left(x^3\right)-f\left(x^4\right) = 2xf\left(x^3\right)-2x^2f\left(x^2\right)-x^4 \rightarrow xf\left(x\right)^2+4x^2f\left(x\right) = 2f\left(x^3\right)+3x^3 $$Finally, count $ 2f\left(x^6\right) $ in two different ways. First method: Cubic then square $$ 2f\left(x^6\right) = x^2f\left(x^2\right)^2+4x^4f\left(x^2\right)-3x^6 = 4x^4f\left(x\right)^2-4x^5f\left(x\right)+x^6+8x^5f\left(x\right)-4x^6-3x^6 = 4x^4f\left(x\right)^2+4x^5f\left(x\right)-6x^6 $$ Second method: Square then cubic $$ f\left(x^6\right) = 2x^3f\left(x^3\right)-x^6 = x^3\left(xf\left(x\right)^2+4x^2f\left(x\right)-3x^3\right)-x^6 = x^4f\left(x\right)^2+4x^5f\left(x\right)-4x^6 $$ We conclude that $ f\left(x\right)^2-2xf\left(x\right)+x^2 = 0 $. This implies $ f\left(x\right) = x $, as desired.

FEcreater wrote: For convenience, let $ P\left(x,y\right) $ be the assertion of the functional equation. First of all, $$ P\left(1,y\right) \rightarrow f\left(f\left(y\right)\right)+f\left(y\right) = 2f\left(1\right)y $$So $ f\left(1\right) = 0 $ or $ f $ is injective. If $ f\left(1\right) = 0 $, then $ P\left(x,1\right) \rightarrow f\left(0\right) = f\left(x\right) $, which means $ f $ is constant $ 0 $. Otherwise, $ f $ is injective. In this case, $$ P\left(x,1\right) \rightarrow f\left(xf\left(1\right)\right) = f\left(x\right) \rightarrow f\left(1\right) = 1 $$by injectivity and thus, $ f\left(f\left(y\right)\right)+f\left(y\right) = 2y $. Now, $$ P\left(x,x\right) \rightarrow f\left(xf\left(x\right)\right)+f\left(x^2\right) = 2xf\left(x\right) = f\left(f\left(xf\left(x\right)\right)\right)+f\left(xf\left(x\right)\right) $$Therefore, $ f\left(xf\left(x\right)\right) = x^2 $ and $ f\left(x^2\right)+x^2 = 2xf\left(x\right) $. I'll present another way to finish after this. First, $P(0,0)$ easily implies that $f(0)=0$. Consider $P(x,yf(y))$ and $P(xy,y)$, we can see that \begin{align*} 2yf(x)f(y) &=f(xf(yf(y)))+f(xyf(y))\\ &=f(xyf(y))+f(xy^2)\\ &=2yf(xy)\\ \end{align*}Thus, combining with $f(0)=0$, we get that $f(xy)=f(x)f(y)$ for all pairs of real numbers $x,y$. In particular, $f(x)\ge 0$ for all $x\ge 0$ and $f(-1)=-1$, which further implies that $f(x)$ is odd. Finally, from $f(f(y))+f(y)=2y$, we can also get $f^{n+1}(y)+f^n(y)=2f^{n-1}(y)$ for all $n\in\mathbb{N}$. Thus, it's straightforward to prove that $$f^n(y)=\frac{(2y+f(y))+(-2)^n(y-f(y))}{3}$$by induction. Hence, for any $y\ge 0$, the fraction above must also be non-negative for any $n\in\mathbb{N}$, which easily implies that $f(y)=y$ for all non-negative $y$. Since $f(x)$ is odd, we must have $f(x)\equiv x$ as desired.

We have f(xf(y)) +f(xy)=2f(x)y Let x=0 [NOTE THIS STEP] The above expression reduces to: f(0)=f(0)y Either f(0)=0 or y=1 For f(0)=0, We can get a hint that f(x)=x.. But for time being, Let's try y=1 or any n=1... But it would be better to take x=1 [NOTE THIS STEP] Substituting x=1 and y =1 in parental equation, we get: f(f(1)) + f(1) = 2f(1) Therefore: f(f(1)) = f(1) Hence, we can conclude: f(x)=x

@above is incorrect, it's not a proof. Let $P(x,y)$ be the assertion $f(xf(y))+f(xy)=2yf(x)$. $P(0,0)\Rightarrow f(0)=0$ $P(1,x)\Rightarrow f(f(x))+f(x)=2f(1)x$ If $f(1)=0$ then $P(x,1)\Rightarrow\boxed{f(x)=0}$, which fits. Assume now that $f(1)\ne0$, so $f$ is injective. $P(1,1)\Rightarrow f(f(1))=f(1)\Rightarrow f(1)=1$ by injectivity So we now have $f(f(x))=2x-f(x)$. Note that $f(k)=0$ immediately gives $k=0$. $P(x,x)\Rightarrow f(xf(x))+f(x^2)=2xf(x)=f(f(xf(x)))+f(xf(x))$, so we get $f(xf(x))=x^2$ and $f(x^2)+x^2=2xf(x)$. The latter gives that $f$ is odd. Note that $f(f(x)f(f(x)))=f(x)^2$. $P(x,f(x)f(f(x)))-P(xf(x),f(x))\Rightarrow2f(x)^2f(f(x))=2x^2f(x)\Rightarrow f(x)f(f(x))=x^2\forall x\ne0$, but it also holds for $x=0$. Then we note that $x^2=f(x)f(f(x))=f(x)(2x-f(x))=2xf(x)-f(x)^2$, or $(f(x)-x)^2=0$. Then $\boxed{f(x)=x}$, which works.

Wrong

@bove, your $P(1,x)$ is incorrect (so is the last paragraph).

The answer is $f(x) = 0$ and $f(x) = x,$ $ \forall x \in \mathbb{R}$.The injective part and $f(1) = 1$ is the same as #8. Note that $P(1,x) : f(f(x)) + f(x) = 2x.$ Letting $x \rightarrow y f(x),$ we have $$f(f(yf(x)) + f(yf(x)) = 2 yf(x) = f(xf(y)) + f(xy).$$ Now plug $y = x,$ we get $f(f(x f(x)) = f(x^2)$ and thus $f(xf(x)) = x^2$. Note that $x\rightarrow -x$ gives $f$ is odd. Now, $x \rightarrow f(x)$ and then $y = f(x)$ in $P(x,y)$ gives $f(f(x) f(f(x))) + f(f(x)^2) = 2f(x) ^2.$ Hence, $$f(x)^2 + f(f(x)^2) = 2f(x)^2 \implies f(f(x)^2) = f(x)^2.$$ Again, $x \rightarrow xf(x)$, we get $f(x^4) = x^4.$ Hence $f(x) = x$ for all positive $x.$ But as $f$ is odd we see $f(x) = x$ everywhere which clearly works.

@jasperE3 I tried to recreate how you chose $P(x, f(x)f(f(x)))$ and $P(xf(x), f(x))$ and it just seems like a miracle that you get 4 terms to cancel. Do you remember how you found these substitutions?

Let $P(x,y)$ denote the assertion: $f(xf(y)) + f(xy) = 2f(x)y$ $P(0,0)$ yields $f(0)=0$ Claim: The function is injective, if and only if $f(1)\neq0$ Proof: Let $f(a)=f(b)$ $P(1,a)$ yields $f(f(a))+f(a)=2f(1)a$ $P(a,b)$ yields $f(f(b))+f(b)=2f(1)b$ Thus $2f(1)a=2f(1)b$, furthermore $a=b$ if and only if $(1)\neq0$, which implies that the function is injective when $f(1)\neq0$ $\square$ Case 1: $f(1)=0$ $P(x,1)$ yields $f(xf(1))+f(x)=2f(x)\Longleftrightarrow f(x)=f(0)\implies f(x)=0$ Case 2: $f(1)\neq0$ $P(1,1)$ yields $f(f(1))+f(1)=2f(1)\Longleftrightarrow f(f(1))=f(1)\overset{\text{from inj.}}{\Longrightarrow}f(1)=1$ Furthermore, $P(x,x)\Longrightarrow f(xf(x))+f(x^2)=2f(x)x$ $P(1,xf(x))$ yields $f(f(xf(x)))+f(xf(x))=2f(x)x$ thus $f(f(xf(x))=f(x^2)\Longrightarrow f(xf(x))=x^2$ Notice that $P(x,yf(y))\Longrightarrow f(xf(yf(y))+f(xyf(y))=2yf(x)f(y)\Longleftrightarrow f(xyf(y))+f(xy^2)=2yf(x)f(y)$ $P(xy,y)\Longrightarrow f(xyf(y))+f(xy^2)=2yf(xy)$, thus $f(x)f(y)=f(xy) \forall x,y \in \mathbb{R}_{\neq0}$, when we combine with the fact that $f(0)=0$, we get that the function is multiplicative for all $x,y$ in reals. Thus $f(xf(x))=x^2$ becomes $f(x)f(f(x))=x^2$, furthermore notice that if we multiply both sides of the equation is transforms into: $f(x)^2f(f(x))=x^2f(x)\Longleftrightarrow f(x^2f(x))=x^2f(x)$, which implies that $f(x)=x,\forall x \in \mathbb{R}_{\neq0}$, however when we combine with the fact that $f(0)=0$, we get that $f(x)=x ,\forall x \in \mathbb{R}$ So to sum up, $\boxed{f(x)=x\text{, and }f(x)=0, \forall x\in \mathbb{R}}$ $\blacksquare$