Let $P(x,y)$ denote the assertion that $f(xf(y))+yf(x)=xf(y)+f(xy)$. First note that $P(0,y)\implies f(0)=0$ and $P(x,1)\implies f(xf(1))=xf(1)\implies f(1)=0$ (by the first condition). Now $P(1,x)\implies f(f(x))=2f(x)$ for all $x$ and so $f^{n}(x)=2^{n-1}f(x)$ for $n\ge2$. By the first condition, $f(x)\le0$ for all $x$. Suppose that $f(r)\ne0$ somewhere for some $r$ ($f(x)=0$ everywhere is a trivial solution). If $f(t)=0$, then $P(r,t)\implies f(tr)=tf(r)$. Since $f(r)<0$ and $f(tr)\le0$, $t$ must be nonnegative. (*) But \begin{align*} P\left(x,\frac{1}{x}\right)+P\left(\frac{1}{x},x\right) &\implies 0+0\ge f\left(xf\left(\frac{1}{x}\right)\right)+f\left(\frac{f(x)}{x}\right)=0 \\ &\implies f\left(xf\left(\frac{1}{x}\right)\right)=f\left(\frac{f(x)}{x}\right)=0 \end{align*}for all nonzero $x$. For $x>0$, $f(x)/x\le0$, whence $f(x)=0$ according to (*). Furthermore, $P(x,-1)\implies f(-x)=xf(-1)$ for all $x>0$. Finally, $f(r)<0\implies 2f(r)=f(f(r))=-f(r)f(-1)\implies f(-1)=-2$. It's easy to check that $f(x)=2x[x<0]$ works.

math154, you are wrong! you said that $ f(x) $ is not positive from the first condition. from the first condition we have $ M>0 $ so... it can't be said anything about $ f(x) $.

anonymouslonely wrote: you said that $ f(x) $ is not positive from the first condition. from the first condition we have $ M>0 $ so... it can't be said anything about $ f(x) $. Sorry if I wasn't clear. Assume for the sake of contradiction that $f(r)>0$ for some $r$. Then by the relation $f^n(r)=2^{n-1}f(r)$, we get that $f(x)$ is not bounded above, which contradicts the first condition.

sorry,you're right. I read your solution very fast... my mistake.

Step 1: $f(x) \le 0$ for all $x$. Denote $P(x,y)$ as the assertion that $f(xf(y))+yf(x)=xf(y)+f(xy)$. $P(0,y)$ gives $f(0)+yf(0)=f(0) \implies yf(0)=0 \implies f(0)=0$. $P(x,1)$ gives $f(xf(1))+f(x)=xf(1)+f(x) \implies f(xf(1))=xf(1)$. If $f(1) \not= 0$, $xf(1)$ is surjective over $\mathbb{R}$, contradiction. This gives us $f(1)=0$, and $P(1,y)$ gives us $f(f(y))=2f(y)$. If there exists an $x$ that $f(x) >0$, we will have $f^n(x)=2^{n-1}f(x)$, which is unbounded. Contradiction. Done. Step 2. $f(x)=0$ for all $x \ge 0$. Take $P(x,f(y))$. This gives $f(xf(f(y))+f(y)f(x)=xf(f(y))+f(xf(y))$, or $f(2xf(y))+f(x)f(y)=2xf(y)+f(xf(y))$. This gives us $-f(2xf(y))+2xf(y) = f(x)f(y)-f(xf(y)) \ge 0$, so $f(xf(y)) \le xf(y)$. This gives $yf(x) \ge f(xy)$. If $x$ is positive, we can let $y=\frac{1}{x}$ to get $\frac{f(x)}{x} \ge 0$, or $f(x) \ge 0$. This gives $f(x)=0$. Done. $f(x) \equiv 0$ is a trivial solution. Assume that there exists an $a<0$ such that $f(a)<0$. Step 3. $f(x)=2x$ for $x<0$. Take $f(a)=k$. $P(x,a)$ gives you $f(xk)+af(x)=xk+f(ax)$. For all $x<0$, as $2xk, ax>0$, we have $f(xk)=f(ax)=0$, so $f(x)=\frac{k}{a}x$. Therefore, $f(x)=cx$ for some fixed constant $c$. Now $f(f(x))=2f(x) \implies c^2=2c \implies c=2$, or $f(x)=2x$ for $x<0$. If $x,y \ge 0$, we have $f(xf(y))+yf(x)=xf(y)+f(xy)=0$. If $x \le 0, y \ge 0$, we have $f(xf(y))+yf(x)=xf(y)+f(xy)=2xy$. If $x \ge 0, y \le 0$, we have $f(xf(y))+yf(x)=xf(y)+f(xy)=4xy$. If $x,y \le 0$, we have $f(xf(y))+yf(x)=xf(y)+f(xy)=2xy$. Therefore, the solution set is $\boxed{f(x) \equiv 0 \text{ }\forall x \in \mathbb{R}}$ and $\boxed{f(x)=0\text{ if } x\ge 0, f(x)=2x\text{ if } x\le 0}$

APMO 2011/5 wrote: Determine all functions $f:\mathbb{R}\to\mathbb{R}$, where $\mathbb{R}$ is the set of all real numbers, satisfying the following two conditions: 1. There exists a real number $M$ such that for every real number $x,f(x)<M$ is satisfied. 2. For every pair of real numbers $x$ and $y$, \[ f(xf(y))+yf(x)=xf(y)+f(xy)\]is satisfied.

My Solution :

What a nice problem

Let $P(x,y)$ denote the given assertion. Suppose that $f \neq 0$. First of all it is easy to see that $f(1)=f(0)=0$ by plugging in $P(x,1)$ and $P(x,0)$ respectively. $$P(1,x) \implies f(f(x))=2f(x) \implies f(2^kf(x))=2^{k+1}f(x)$$The last implication follows trivially after an induction, and is true for any $k \in \mathbb{N}$, now let $m\in \mathbb{Z}$ and $n\in \mathbb{N}$ such that $|n| > |m|+100$. $$P(2^m,2^nf(x)) \implies f(2^{m+n+1}f(x))+2^nf(2^m)f(x)=2^mf(2^nf(x))+f(2^{m+n}f(x)) \implies$$$$\implies 2^{m+n+2}f(x)+2^nf(2^m)f(x)=2^{m+n+2}f(x) \implies f(2^m)=0$$$$P(x,2^m) \implies f(2^mx)=2^mf(x)$$Thus we clearly have that $f(x) \le 0$ for any real $x$, because $f$ is upper-bounded. $$P\left(x,\frac{1}{x}\right)+P\left(\frac{1}{x},x\right) \implies f\left(xf\left(\frac{1}{x}\right)\right)+f\left(\frac{f(x)}{x}\right)=0\implies f\left(xf\left(\frac{1}{x}\right)\right)=f\left(\frac{f(x)}{x}\right)=0$$$$P\left(x,\frac{f(x)}{x}\right)\implies \frac{f(x)^2}{x}=f(f(x))=2f(x) \implies f(x) \in \{0,2x\}$$We clearly have that $f(x)=0$ for any non-negative real, now suppose there is a real $u<0$ such that $f(u)=0$, plug in $x$ a negative real and $y=u$, we have that $uf(x)=f(ux)=0$, which means that $f \equiv 0$, or $f(x)=2x$ for any $x<0$ and $f(x)=0$ for any $x\ge 0$

I initially thought this problem to be really easy for a P5, then I realised it is not...

We claim the solutions are $f\equiv 0$ and $f(x)=0$ for $x\ge 0$ and $f(x)=2x$ for $x<0$. It is easy to check that these indeed satisfy the FE. Let $P(x,y)$ be the given FE. Note that \[P(x,1)\implies f(xf(1))=xf(1).\]Thus, if $f(1)\ne 0$, we have $f\equiv\mathrm{id}$, which is impossible since $f$ is bounded above. So WLOG, suppose $f(1)=0$. Now, \[P(1,y)\implies \boxed{f(f(y))=2f(y)},\]so in particular, $y=1\implies f(0)=0$. Note that \[f(f(y))=2f(y)\implies f^n(y)=2^{n-1}f(y),\]so if $f(y)>0$, we eventually get a $u$ such that $f(u)\ge M$, which is a contradiction. Thus, $\boxed{f(x)\le 0}$ for all $x$. Now, \[P(x,1/x)+P(1/x,x)\implies f(xf(1/x))+f(f(x)/x)=0,\]so we must have them both be $0$. In particular, $\boxed{f(f(x)/x)=0}$. Thus, \[P(x,f(x)/x)\implies f(x)^2/x=f(f(x))=2f(x),\]so $\boxed{f(x)\in\{0,2x\}}$. More generally, if $f(a)=0$, then $P(x,a)\implies af(x)=f(ax)$. However, since $f(x)\le 0$ always, we have that $f(x)=0$ for all $x\ge 0$. Thus, $f(ax)=af(x)$ for all $a\ge 0$, so $f(-a)=af(-1)$ for all $a\ge 0$. If $f(-1)=0$, then we get $f\equiv 0$, otherwise if $f(-1)=2$, then we get $f(x)=0$ for $x\ge 0$ and $f(x)=2x$ for $x\le 0$.

Cool problem, used only standard techniques but was certainly fun to solve, and, weirdly enough, had a naturally occurring semi-pathological solution. Easier than this year's (2019) P5, but still ~ IMO P2/P5 level, I'd say... Well, let's get the basics over with: Letting $P(x,y)$ denote the assertion given in condition 2, $P(0,y): yf(0)=0 \implies f(0)=0$. $P(y,x): xf(y)-yf(x)=f(xy)-f(yf(x))=xf(y)-yf(x)=f(xf(y))-f(xy)$ (by $P(x,y)$), hence: $f(xf(y))+f(yf(x))=2f(xy)$. $P(x,1): f(xf(1))=xf(1) \implies f(1)=0$ (as $f$ is bounded from top). $P(1,x): f(f(x))=2f(x)$. Say, for some $y$, $f(y)=t \neq 0 \implies f(t)=2t$. $P(1,t): f(2t)=2f(t)=4f(t)$. Extending for $2^nt$, $P(1,2^{n-1}t): f(2^nt)=2^{n+1}t$, thus $f(2^nt)=2^{n+1}t$. $P(2^n,t): f(2^{n+1}t)+tf(2^n)=2^nf(t)+f(2^nt)=2^{n+2}t \implies tf(2^n)=0 \implies f(2^n)=0$. $P(x,2^n): f(x \cdot 2^n)=2^nf(x)$. (This directly implies $f(x) \leq 0$ for all $x$ because of the upper bound). Putting $(x, \frac 1x)$ in $f(xf(y))+f(yf(x))=2f(xy)$, we get $f(xf(\frac 1x))=-f(\frac{f(x)}x)$, and as both must be $\leq 0$: $\implies f(xf(\frac 1x))=f(\frac{f(x)}x)=0$. $P(x,\frac 1x): \frac{f(x)}x = xf( \frac 1x)$. Say, for $y$, $f(y) \neq 0$. Thus, $P(\frac{1}{f(y)},y): yf(\frac{1}{f(y)})=\frac{f(y)}{f(\frac 1y)} +f(\frac{y}{f(y)})$. But we've seen both terms on the RHS before- we know $\frac{f(x)}x=xf(\frac 1x) \implies \frac{f(y)}{f(\frac 1y)}=y^2$. Moreover, as $f(\frac{f(x)}x)=0$ and $f(\frac 1x)=\frac{f(x)}{x^2} \implies f(\frac 1{\frac{f(y)}y})=\frac{f(\frac{f(y)}y)}{(\frac{f(y)}y)^2} \implies f(\frac{y}{f(y)})=0$. Thus we get $yf(\frac{1}{f(y)})=y^2 \implies f(\frac 1{f(y)})=y$. Clearly, for large enough $y$, we'll need $f(\frac{1}{f(y)})$ to be not defined $\implies f(y)=0$ for all $y>M$, and as $f(2^nx)=2^nf(x) \implies f(y)=0$ for all $y \geq 0$. We're almost done here, but right at the end a semi-pathological type solution pops up- but it actually comes out quite naturally, so I'm not going to complain- Letting $m,n>0$, $P(-m,-n): nf(-m)=mf(-n)$ (using $f(x)=0$ if $x \geq 0$ and $f(x) \leq 0$). Here we see that if for any $m$, $f(-m)=0 \implies f \equiv 0$. Assuming $f$ is non-zero for all negative reals, $\frac{f(-n)}{-n}=k$, for some constant $k$, fixing $m$. Thus $f(-n)=k \cdot (-n)$. But remember $f(f(y))=2f(y)$? This implies $k=2$, and sure enough, $f(x)=2x$ for $x<0$ indeed turns out to work. Hence our solutions are: $f \equiv 0$ and $f(x)=0$ if $x \geq 0$,$f(x)=2x$ if $x<0$.

We divide the solution into three parts. Let $P(x,y)$ denote the assertion $f(xf(y))+yf(x)=xf(y)+f(xy)$. Part A: Show that $f(x)\leq 0$ for all $x$. $P(0,1)\implies f(0)=0$. Moreover, $P(x,1)\implies f(xf(1))=xf(1)$. But identity function is not bounded so $f(1)=0$. Hence $P(1,y)$ implies $f(f(y))=2f(y)$. The previous equation means $f(x) = \tfrac{f(f(x))}{2} < \tfrac{M}{2}$ for all $x$. Hence iterating yields $f(x) < \tfrac{M}{2^n}$ for any $n\in\mathbb{Z}^+$. This means $f(x)\leq 0$ for all $x$ as desired. Part B: Show that $f(x)=0$ for all $x>0$. We divide this part into three steps. If $f(x)=0$ for all negative $x$, then plugging in $y<0$, $x>0$ yields $yf(x) = f(xy)=0$ or $f(x)=0$ for all $x>0$ as desired. Otherwise, let $y_0<0$ such that $f(y_0)\ne 0$. Then $$P(x,y_0)\implies f(xf(y_0)) \leq xf(y_0) - y_0f(x) \leq xf(y_0).$$Thus $f(x)\leq x$ for all $x$. Finally, plug in $P(x,\tfrac{1}{x})$ and use the above bound $$f(1)-\frac{f(x)}{x} = f(xf(\tfrac 1x)) - xf(\tfrac 1x) \leq 0$$but we had $f(1)=0$ so $f(x)\geq 0$ for all $x\geq 0$, hence we are done. Part C: Finishing Once we have part B, the rest is easy. For $t>0$, $P(-1,t)$ gives $f(-t) = tf(-1)$. Hence $f$ must be in form $$f(x) = \begin{cases}0 & x\geq 0 \\ cx & x\leq 0\end{cases}$$for some constant $c$. From $f(f(x))=2f(x)$, we get $c\in\{0,2\}$. Both solutions are valid hence we are done.

If $f$ is constant, then only $f\equiv 0$ works. Assume $f$ is nonconstant. Let $P(x,y)$ denote the FE. We begin with some easy observations. $P(x,0)$ gives $f(xf(0))=xf(0)+f(0)$. If $f(0)\not = 0$, this implies $f$ can grow arbitrarily large, contradiction. So $f(0)=0$. $P(x,1)$ gives $f(xf(1)) + f(x) = xf(1)+f(x)$, so $f(xf(1))=xf(1)$. Again, by the same logic, this implies $f(1)=0$. Now, $P(1,y)$ gives $f(f(y))=f(y)+f(y)=2f(y)$. In particular, $f^k(y)=2^{k-1}f(y)$ for any $k\ge 1$. If there exists $y_0$ with $f(y_0)>0$, then $f^k(y_0)$ grows arbitrarily large, contradiction. Therefore, $\boxed{f(x)\le 0 \ \forall x}$. Claim 1: $f\left(\tfrac{f(x)}{x}\right)=0$ for all $x\not = 0$. Proof: We take advantage of the lack of symmetry in $P(x,y)$. We have \begin{align*} P(x,y)&: f(xf(y)) + yf(x)=xf(y)+f(xy) \\ P(y,x)&: f(yf(x))+xf(y)=yf(x)+f(xy). \end{align*}Adding these two gives $f(xf(y))+f(yf(x))=2f(xy)$. Plug in $y=1/x$ into this; we get $f(xf(\tfrac{1}{x}))=-f(\tfrac{1}{x}f(x))$. But $f$ is always nonpositive! Therefore, $f(f(x)/x)=0$, as desired. $\blacksquare$ Claim 2: $f(a)=0 \implies a\ge 0$ for any $a$. Proof: Suppose $f(a)=0$ for some $a$. Then $P(x,a)$ gives $af(x)=f(ax)$. If $a<0$, then $f(ax)$ and $f(x)$ have different sign, so $f(x)=0$, contradiction since we are assuming $f$ nonconstant. Therefore, $a\ge 0$. $\blacksquare$ Claim 3: if $x$ is positive, then $f(x)=0$. Proof: Since $f(f(x)/x)=0$, then Claim 2 gives $f(x)/x\ge 0$. If $x$ is positive, then $f(x)/x \ge 0$ implies $f(x)\ge 0$, which implies $f(x)=0$ since $f$ is nonpositive. $\blacksquare$ Claim 4: If $x$ is negative, then $f(x)=2x$. Proof: We need to find $f$ on the negatives. We have \[ P\left(x,\frac{f(x)}{x}\right): \ \frac{f(x)}{x} \cdot f(x) = f(f(x)) = 2f(x) \implies f(x)^2 = 2xf(x),\]so $f(x)=0$ or $f(x)=2x$ for each $x$. But Claim 2 tells us that $a<0\implies f(a)\not = 0$, so since $x$ is negative, we conclude $f(x)=2x$. $\blacksquare$ Finally, the general form of the nonconstant solution is \[ f(x) = \begin{cases} 0 &x \ge 0 \\ 2x &x<0\end{cases}\]and we can check that this works. In conclusion, the solutions are the above and $f\equiv 0$.

Let $P(x,y)$ denote the assertion. We can check that $f\equiv 0$ is a solution, so assume $f\not\equiv 0.$ $\textbf{Claim: }$ $f(f(x))=2f(x)$ for all $x\in\mathbb{R}.$ $\textbf{Proof: }$ Assume $f(1)\ne 0.$ Then, $P(1,x)$ gives $f(f(y))+yf(1)=2f(y),$ which implies that $f$ is injective. Therefore, since $P(x,x)$ yields $f(xf(x))=f(x^2),$ we have $f(x)=x$ for all $x.$ But this is impossible, as $f$ is bounded, so $f(1)=0.$ In particular, $P(1,x)$ gives $f(f(x))=2f(x)$ for all $x\in\mathbb{R}.$ $\blacksquare$ $\textbf{Claim: }$ $f(2x)=2f(x)$ for all $x\in\mathbb{R}.$ $\textbf{Proof: }$ Let $c\ne 0$ be a real number in the codomain of $f.$ By our first claim, we know $f(2c)=4c$ and $f(4c)=8c.$ $P(2,c)$ yields $$8c+cf(2)=4c+4c\implies f(2)=0.$$Now, $P(x,2)$ gives $2f(x)=f(2x),$ as desired. $\blacksquare$ $\textbf{Claim: }$ $f(x)\le 0$ for all $x\in\mathbb{R}.$ $\textbf{Claim: }$ Suppose $f(x)>0$ for some $x\in\mathbb{R}.$ By our first claim, $f^{i}(x)=2^{i-1}f(x)$ for all $i\ge 2,$ so $f$ can be arbitrarily large, contradiction. $\blacksquare$ $\textbf{Claim: }$ $f(x)\in\{0,2x\}$ for all $x\in\mathbb{R}$ $\textbf{Proof: }$ Consider some $x\in\mathbb{R},$ and let $k=\frac{f(x)}{x}.$ From $P(1/x,x)$ and $P(x,1/x),$ we have $$f(k)+xf(1/x)=k,$$$$f(xf(1/x))+k=xf(1/x).$$Substituting the second equation into the first yields $$f(xf(1/x))+f(k)=0.$$Therefore, since $f(xf(1/x)),f(k)\le 0,$ we have $f(xf(1/x))=f(k)=0.$ Now, $P(x,k)$ gives $k^{2}x=kf(x)=f(kx).$ But we know $f(kx)=f(f(x))=2f(x)=2kx,$ so $k^{2}x=2kx.$ This implies $k\in\{0,2\},$ as desired. $\blacksquare$ $\textbf{Claim: }$ If $f(x)=0$ and $f(y)=2y,$ then $f(xy)=2xy$ $\textbf{Proof: }$ From $P(x,y),$ we obtain $$f(2xy)=2xy+f(xy).$$Since $f(2xy)=2f(xy),$ we have $f(xy)=2xy,$ as desired. $\blacksquare$ Now let $A=\{x|f(x)=0\}$ and $B=\{x|f(x)=2x\}.$ From our work above, we know (1) $A\cup B=\mathbb{R}$ (2) If $b\in B,$ then $b\le 0$ (3) If $a\in A$ and $b\in B,$ then $ab\in B$ These conditions are enough to imply that $A$ is the set of nonnegative real numbers and $B$ is the set of nonpositive real numbers. This produces the solution \[ f(x) = \begin{cases} $0$ & \text{if $x> 0$} \\ $2x$ & \text{if $x\le 0$} \end{cases}. \]

The answer is $f(x)=x-|x|$ or $f(x)\equiv 0$. They both work. Let $P(x,y)$ denote the assertion in the question $P(0,y)$ gives $f(0)=0$. $P(x,1)$ gives $f(xf(1))=xf(1)$. If $f(1)\ne 0$, we can select $x$ st $xf(1)>M$ so $f(1)=0$ Now $P(1,x)$ gives $f(f(x))=2x$. Using this, we can get $f(x)\le 0$ by noting $f^k(x)=2^{k-1}f(x)$ This motivates us to consider $P(x/2, f(y))$ which gives us $f(xf(y))+f(y)f(\frac x2) = xf(y)+f(\frac x2 f(y))$ So we get $f(xy)-yf(x)=f(\frac x2 f(y)) - f(y)f(x/2)$ Taking $(x,f(y))$ into the above equation gives $f(xf(y))-f(y)f(x)=f(xf(y))-2f(y)f(x/2)$ so $f(x)=2f(x/2)$ We can algebraically solve $f(xf(y))=f(y)(2x-f(x))$ but that is not necessary. While bashing, we can see $f(xy)-yf(x)=f(x/2 f(y))-f(y)f(x/2)$ implies $f(xy)\le yf(x)$. If $y$ is positive, then $f((xy)(1/y))\le (1/y)f(xy)$, implying $f(xy)=yf(x)$ for all $y\in \mathbb{R}^+$. From here we can simply bash to get the solutions.

We claim the only functions are $f\equiv 0$ and $f(x)=0$ for $x\leq 0$ otherwise $f(x)=2x,$ we can verify these work. We present the proof in steps. Denote the assertion by $P(x,y).$ Step 1: We prove that $f(1)=0.$ $P(0,x)$ gives $f(0)=0.$ Then $P(x,1)$ forces $f(1)=0$ since otherwise $f$ is the identity, contradicting the conditions. Q.E.D. Step 2: We prove that $f(x)\leq 0$ for all x. Induction on $P(1,x)$ yields $f^k(x)=2^{k-1}f(x)$ where $k\leq 2$ is an integer. Notice that we have a contradiction with (1) if not $f(x)\leq 0.$ Q.E.D. Step 3: We show that $f(x)\in \{0,2x\}$ for all $x.$ By combining $P(x,x^{-1})$ with $P(x^{-1},x)$ we get $f(xf(x^{-1}))=f(f(x)/x)=0.$ Using this $P(x,f(x)/x)$ gives $f(x)=0$ or $f(x)=2x.$ Q.E.D. Step 4: We show that the aforementioned solutions are the only solutions. Clearly $f\equiv 0$ is a solution. Assume that not. Then evidently we have $f(x)=0$ for all $x\geq 0.$ If $f(z)<0$ for some $z<0,$ then note that for any $x<0$ we have $f(xf(z))=f(2xf(z))=0.$ Then $P(x,f(z))$ implies $f(x)=2x.$ Q.E.D.

Determine all functions $f:\mathbb{R}\to\mathbb{R}$, where $\mathbb{R}$ is the set of all real numbers, satisfying the following two conditions: 1) There exists a real number $M$ such that for every real number $x,f(x)<M$ is satisfied. 2) For every pair of real numbers $x$ and $y$, \[ f(xf(y))+yf(x)=xf(y)+f(xy)\]is satisfied. Solved with Ritwin, CT17 a couple of months ago. The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x)=0\text{ if } x\ge 0 \text{ and } f(x) = 2x \text{ if }x<0}$. These solutions can be verified to work. We now prove they are the only ones. It's easy to see that the only constant solution is $f\equiv 0$, so we henceforth assume that $f$ is not constant. $P(0,1): 2f(0) = f(0)\implies f(0)=0$. $P(x,1): f(xf(1)) + f(x) = xf(1) + f(x)\implies f(xf(1)) = xf(1)$. If $f(1)\ne 0$, then $xf(1)$ can take any real value, so $f(x)=x\forall x\in \mathbb{R}$, which contradicts condition 1. So $f(1)=0$. $P(1,x): f(f(x)) = 2f(x)$. In other words, $f(k) = 2k$ if $k$ is in the range of $f$. $P(2,f(x)): f(4f(x)) + f(x)f(2) = 8f(x)$. Since $2f(x)$ is in the range of $f$, we have $f(2f(x)) = 4f(x)$, so $4f(x)$ is in the range of $f$. Therefore, $f(4f(x)) = 8f(x)$. So \[f(x) f(2) = 0\implies f(2)^2 = 0\implies f(2)=0\] Claim: $f(x)\le 0$ for all reals $x$. Proof: $P(2,x): 4f(x) = 2f(x) + f(2x)\implies f(2x) = 2f(x)$. Inductively, we find $f(2^n x) = 2^n f(x)$. If $f(x)>0$ for some $x$, then setting $n$ sufficiently large gives contradiction to the first condition. $\square$ $P\left(x,\frac{1}{x}\right): f\left(xf\left(\frac{1}{x}\right)\right)+ \frac{f(x)}{x} = xf\left(\frac{1}{x}\right) $ $P\left(\frac{1}{x},x\right): f\left(\frac{f(x)}{x}\right) + xf\left(\frac{1}{x}\right) = \frac{f(x)}{x} $. So $f\left(x f\left(\frac{1}{x}\right )\right) + f\left(\frac{f(x)}{x}\right) + xf\left(\frac{1}{x}\right) = xf\left(\frac{1}{x}\right),$ so \[f\left(xf\left(\frac{1}{x}\right)\right) + f\left(\frac{f(x)}{x}\right)=0\] Since both $f\left(xf\left(\frac{1}{x}\right)\right)$, and $f\left(\frac{f(x)}{x}\right)$ are $\le 0$, we have \[f\left(xf\left(\frac{1}{x}\right)\right) =f\left(\frac{f(x)}{x}\right)=0\] This also gives us $xf \left(\frac{1}{x}\right) = \frac{f(x)}{x}$, or $f\left(\frac{1}{x}\right) = \frac{f(x)}{x^2}$. $P\left(x,\frac{f(x)}{x}\right): \frac{f(x)^2}{x} = f(f(x)) = 2f(x)$. So if $f(x)\ne 0$, then $f(x) = 2x$. In fact, this gives that $f(x)=x$ iff $x=0$. $P\left(-x,\frac{1}{x^2}\right): f\left(-f\left(\frac{1}{x^2}\right)\right) + \frac{f(-x)}{x^2} = -f\left(\frac{1}{x^2}\right) + f\left(\frac{1}{-x}\right)$ However, \[\frac{f(-x)}{x^2} = \frac{f(-x)}{(-x)^2} = f\left(\frac{1}{-x}\right)\] So $f\left(-f\left(\frac{1}{x^2}\right)\right) = -f\left(\frac{1}{x^2}\right)$, which implies $f\left(\frac{1}{x^2}\right) = 0$. Since $\frac{1}{x^2}$ can take any real value greater than $0$, we have $f(x) = 0\forall x>0$. Suppose $f(k) = 0$ for some $k<0$. If $x<0$, $P(x,k): kf(x) = f(xk)$. Since $xk>0$, $f(xk) = 0$, which implies $kf(x)=0\implies f(x) =0$. Then $f\equiv 0$ would hold, however this contradicts our assumption that $f$ isn't constant. So $f(x) \ne 0\forall x<0$, which implies $f(x) = 2x\forall x<0$, as desired.

Let $P(x,y)$ be the assertion in the question , firstly by $P(x,1)$ we can get $f(xf(1))=xf(1)$ , which implies that $f(1)=0$ by the first condition. Thus by $P(1,x)$ we have $f(f(x))=2f(x)$ for all numbers $x \in \mathbb{R}$ and also $f(2f(x))=4f(x)$ too. Note that for all numbers $x$ we have $f(x) \le 0$ ( because otherwise if for a real number $x_0$ we have $f(x_0) > 0$ , then there exist a natural number $n$ such that $f^{n+1}(x_0)=2^{n}x_0 \ge M$ which is a contradiction.) one can see that : $$f(f(x)y))-2yf(x)=f(f(y)x)-2xf(y)=f(xy)-xf(y)-yf(x) \implies f(f(2)x)-2f(2)x=f(2f(x))-4f(x)=0 , \forall x \in \mathbb{R} \implies f(2)=0$$$$P(x,2) \implies f(2x)=2f(x), \forall x \in \mathbb{R}$$$$P(x,f(y)) \implies f(2xf(y))+f(x)f(y)=2f(xf(y))+f(x)f(y)=2xf(y)+f(xf(y)) \implies f(xf(y))=2xf(y)-f(x)f(y)=xf(y)-yf(x)+f(xy)$$$$\implies f(xy)+f(x)f(y)=xf(y)+yf(x) \implies (x-f(x))(y-f(y))=(xy-f(xy)) , \forall x , y \in \mathbb{R}$$Thus if we define a new function $g:\mathbb{R} \to \mathbb{R}$ as $g(x)=x-f(x)$ , then $g(x)$ is multiplicative and for all real numbers $x$ , we have $g(x) \ge x$. So by rewriting the main equation of problem , we have $g(x-g(x))=g(x)-x$. Now while for all numbers $x \ge 1$ we have $g(x) \ge 1$ , as the result for all numbers $x \in \mathbb{R^+}$ , $g(x)=x$ holds and since $g(xy)=g(x)g(y)$ , for all real numbers $x \le 0$ we have $g(x)=-xg(-1)$. Now note that $g(x) \ge x$ and $g(x-g(x))=(g(x)-x)g(-1)=g(x)-x$ which means $g(x)=x$ for all $x \in \mathbb{R}$ or $g(-1)=1$ and $g(x)=|x|$. Thus , functions $f(x)=0$ and $f(x)=x-|x|$ are answers of this problem which clearly work. So we're done.

Let $P(x,y)$ denote the proposition $f(xf(y))+yf(x)=xf(y)+f(xy)$. From $P(0,x)$ we get $f(0)=0$. If $f(1)\ne 0$, then from $P\left(\frac{x}{f(1)},1\right)$ we get $f(x)=x,$ which is a contradiction. Thus $f(1)=0$. From $P(1,x)$ we get $f^2(x)=2f(x)$, and we can easily prove by induction that $f^k(x)=2^{k-1}f(x)$, for all $k\geq2$. This implies that $f(x)\leq0$, $\forall x\in\mathbb{R}$. Now taking $P(x,-1)$ we get $$f(xf(-1))-f(x)=xf(-1)+f(-x)\implies f(xf(-1))-xf(-1)=f(x)+f(-x)\leq0$$$$\implies f(xf(-1))\leq xf(-1).$$ If $f(-1)=0$, from $P(x,-1)$ we get $f(x)+f(-x)=0$, $\forall x\in\mathbb{R}$, from which we get that $f\equiv 0$. If $f(-1)\ne 0$, we have $f(x)\leq x$, $\forall x\in\mathbb{R}$. Taking $P(x,y)$ $$xf(y)+f(xy)=f(xf(y))+yf(x)\leq xf(y)+yf(x)$$$$\implies f(xy)\leq yf(x), \forall x,y\in\mathbb{R}.$$By taking $x>0$ and $y=\frac{1}{x}$ in the above inequality, we get $f(x)\geq 0$ for $x>0$, and thus $f(x)=0$ for $x$ positive. Now take $x,y<0$ in $P(x,y)$ $$f(\underbrace{xf(y)}_{\geq0})+yf(x)=xf(y)+f(\underbrace{xy}_{>0})$$$$\implies yf(x)=xf(y)\implies f(x)=Cx,$$for some constant $C$. One can easily check that we must have $C=2$, and the only solutions are $f\equiv 0$ or $f(x)=0$, for $x\geq0$ and $f(x)=2x$, for $x<0$. $\blacksquare$

Solved with cj13609517288 and lrjr24 Denote the assertion with $P(x, y)$. We claim that the solution set is $f(x) = 0$ and \[ f(x) = \begin{cases} 0 & x \ge 0 \\ 2x & x \le 0 \end{cases} \] Claim: $f(0) = f(1) = 0$ and $f(f(x)) = 2f(x)$ Proof. Subsitute $P(x, 0)$ to get that \[ f(xf(0)) = (x+1)f(0) \]which implies by the bound that $f(0) = 0$. By $P(x, 1)$ it follows that \[ f(xf(1)) = xf(1) \]which implies that $f(1) = 0$. Thus, by $P(1, x)$ \[ f(f(x)) = 2f(x) \]$\blacksquare$ Claim: $f(x) \le 0$. Proof. Suppose that $f(c) > 0$. Then \[ f^{n+1}(c) = 2^n f(c) \]contradicts the bound. $\blacksquare$ We now show that $f(x) = 0$ for $x \ge 0$. Suppose that $c > 0$ such $f(c) \ne 0$. Claim: $f(x) \ne 0$ for $x \le 0$. Proof. Substitute $P(x, c)$ for $x < 0$ to get \[ f(xf(c)) + cf(x) = xf(c) + f(cx) \]Since the LHS is nonpositive and $xf(c) > 0$ it follows that $f(cx) < 0$ for all $cx$. $\blacksquare$ Claim: There exists a $d < 0$ such $f(c) = f(d)$ Proof. By $P(x, x)$ it follows that \[ f(xf(x)) = f(x^2). \]Then set $x = \sqrt{c}$. $\blacksquare$ Now, by comparing $P(x, c)$ and $P(x, d)$ we get that $f(cx) - cf(x) = f(dx) - df(x)$ or $f(dx) - f(cx) = (d - c)f(x)$ If we now set $x = \frac{1}{c}$ it follows that \[ f\left(\frac{d}{c}\right) = (d - c)f\left(\frac{1}{c}\right) \]However, then the LHS is strictly less than zero while the RHS is nonnegative, contradiction, so $f$ is zero on the positives. Claim: $f$ is either $0$ or $2x$ on the negatives. Proof. Now, substitute $P(-1, y)$ for $y > 0$ to get \[ yf(-1) = f(-y) \]so $f$ is linear on the negatives. If $f(-1) = 0$ then $f$ is uniformly $0$. Else, since $f(f(x)) = 2f(x)$ from earlier $f(x) = -2x$ on the negatives. $\blacksquare$

The answer is $f \equiv 0$ and $$f=\begin{cases}0&x\geq 0\\2x&x<0\end{cases},$$which can be checked to work. We now show that they are the only ones. Let $P(x,y)$ denote the given assertion. From $P(0,y)$ we have $f(0)+yf(0)=f(0) \implies f(0)=0$. Then from $P(x,1)$ we have $f(xf(1))+f(x)=xf(1)+f(x) \implies f(1)=0$, otherwise by making $x$ large and of the opposite sign as $f(1)$ we contradict the first condition. From $P(1,x)$, we obtain $f(f(x))=f(x)+f(x)=2f(x)$, so by induction $f^k(x)=2^{k-1}f(x)$ for all $k \geq 2$. Therefore $f(x) \leq 0$ for all $x$, otherwise by plugging in $k$ large enough we contradict the first condition. If some $a$ has $f(a)=0$, then by $P(x,a)$ we obtain $af(x)=f(ax)$. If $a<0$, then we must have $f(x)=0$ always, else $f(x)=af(x)>0$ which is a contradiction. Therefore suppose that for all $x<0$, we have $f(x)<0$. I claim that for all $x>0$, we have $f(x)=0$. Indeed, consider $P(x,\tfrac{1}{x})$ and $P(\tfrac{1}{x},x)$. Since $f(1)=0$, adding the two equations gives $$f\left(xf\left(\frac{1}{x}\right)\right)+f\left(\frac{f(x)}{x}\right)=0 \implies f\left(\frac{f(x)}{x}\right)=0,$$since $f$ is always nonpositive. But if $f(x)\neq 0$, then $\tfrac{f(x)}{x}$ is strictly negative, so $f(\tfrac{f(x)}{x})<0$: contradiction. Finally, using a previous result, since $f(x)=0$ for all $x>0$, we have $xf(-1)=f(-x)$ for all $x>0$, so $f$ is linear on $\mathbb{R}^-$. Let $f(-1)=c$; from $P(x,-y)$ where $x,y>0$, we find that $$f(xf(-y))=xf(-y)+f(-xy) \implies -cxf(-y)=cxy+cxy \implies -c^2xy=2cxy \implies c=-2,$$which is what we wanted to prove. $\blacksquare$

Claim 1) If $f(x) = x$ for some $x$, $x$ must be $0$ pf) Assume there exists a nonnegative $y$ such that $f(y) = y$ $P(x,y): xy = yf(x)$, which implies $f(x)=x$. This causes a contradiction by condition (i) Now we will use substitution to calculate some basic values $P(1,1): f(f(1)) = f(1) \implies f(1) = 0$ $P(0,y): yf(0) = f(0) \implies f(0) = 0$ $P(1,y): f(f(y)) = 2f(y) \implies f(2f(y)) = f(f(f(y))) = 2f(f(y)) = 4f(y)$ $P(2,y): 4f(y) + yf(2) = 2f(y) + f(2y)$ Plug in $f(y)$ into $y$ in the above equation: $8f(y) + f(y)f(2) = 4f(y) + 4f(y) \implies f \equiv 0$ or $f(2) = 0$ $f \equiv 0$ is obviously a solution, so we now only have to deal with the case where $f(2) = 0$ and $f \not\equiv 0$ Claim 2) $f(\mathbb{R}) \subseteq (-\infty,0]$ pf) If there exists some $x$ such that $f(x)>0$, multiplying $x$ by $2$ sends $f(x)$ to positive infinity, causing a contradiction by condition (i) Claim 3) $f(x) \in {0,2x}$ pf) $P(x,\frac{1}{x})$ vs $P(\frac{1}{x},x): f(xf(\frac{1}{x})) = -f(\frac{1}{x}f(x))$ By Claim 2 this means that $f(\frac{f(x)}{x})$ must be $0$ for all $x$ $P(x,\frac{f(x)}{x}): \frac{1}{x}f(x)^2 = 2f(x) \implies f(x) = 0, 2x$ The rest is just standard FE technique of dealing with split functions