If $f(k) = 1$, then $f(x) = f(xf(k)) = kf(x)$, so $k =1$. Let $y = 1/f(x)$ and set $k = xf(y)$, then $f(k) = f(xf(y)) = yf(x) = 1$. Hence $f(1) = 1$ and $f(1/f(x)) = 1/x$. Also $f(f(y)) = f(1f(y)) = y$. Hence $f(1/x) = 1/f(x)$. Finally, let $z = f(y)$, so that $f(z) = y$. Then $f(xy) = f(xf(z)) = zf(x) = f(x)f(y)$. Now notice that $f(xf(x)) = xf(x)$. Let $k = xf(x)$. We show that $k = 1$. $f(k^2) = f(k)f(k) = k^2$ and by a simple induction $f(kn) = kn$, so we cannot have $k > 1$, or $f(x)$ would not tend to $0$ as $x$ tends to infinity. But $f(1/k) = 1/k$ and the same argument shows that we cannot have $1/k > 1$. Hence $k = 1$. So the only such function $f$ is $f(x) = 1/x$.

Hmm i had a slightly different solution: Consider $x=1$, which gives $$f(f(y))=yf(1)\implies f(f(f(y)))=f(yf(1))=f(y)f(1).$$When we plug in $y=1$ into the last two equalities we get $f(f(1))=f(1)^2$ which instantly yields $f(1)=1$ due to the codomain restriction. Then due to the original expression that we derived we have $f(f(y))=y$, which we can very easily prove injectivity and surjectivity, and thus show that $f$ is bijective. Now consider $x=y$, which yields $f(xf(x))=xf(x)$. However since we have shown bijectivity, we claim that $f$ is monotonically decreasing. (due to the second restriction) Then the only possible value for $xf(x)$ is identically $1$, so $xf(x)=1 \implies f(x)=\frac{1}{x}$. Checking gives that this is a valid solution.

can we use fixed points?i think that would provide the solution

e.lopes wrote: If $f(k) = 1$, then $f(x) = f(xf(k)) = kf(x)$, so $k =1$. Let $y = 1/f(x)$ and set $k = xf(y)$, then $f(k) = f(xf(y)) = yf(x) = 1$. Hence $f(1) = 1$ and $f(1/f(x)) = 1/x$. Also $f(f(y)) = f(1f(y)) = y$. Hence $f(1/x) = 1/f(x)$. Finally, let $z = f(y)$, so that $f(z) = y$. Then $f(xy) = f(xf(z)) = zf(x) = f(x)f(y)$. Now notice that $f(xf(x)) = xf(x)$. Let $k = xf(x)$. We show that $k = 1$. $f(k^2) = f(k)f(k) = k^2$ and by a simple induction $f(kn) = kn$, so we cannot have $k > 1$, or $f(x)$ would not tend to $0$ as $x$ tends to infinity. But $f(1/k) = 1/k$ and the same argument shows that we cannot have $1/k > 1$. Hence $k = 1$. So the only such function $f$ is $f(x) = 1/x$. Good idea but wrong logic. Do you consider the existence of $k$ such that $f(k)=1$?

- Taking $x=1$ gives $ff(y)=f(1)y$ hence $f$ is bijective. - The number $xf(x)$ is a fixed point for all $x>0$. In particular $ff(1)=f(1)$. Then $f(1)=1$. - If $p$ is a fixed point then so is $p^2$. Taking $x=1/p,y=p$ gives $f(1/p)=1/p$ hence $1/p$ is also fixed point. - If $p>1$ then $p,p^2,p^4,\cdots$ are fixed points with limit $\infty$, contradiction. Similarly we can't have $0<p<1$. - $p=1$ is the only fixed point and $f(x)=1/x$ for all $x$.

Let $P(x,y)$ be the given assertion. $\lim_{y\to\infty}P(x,y)\Rightarrow\lim_{x\to0^+}f(x)=\infty$ Let $f$ have a fixed point $k$. If $k>1$, then $P(k,k)\Rightarrow f\left(k^2\right)=k^2$, so $f$ has arbitrarily large fixed points (simple induction). But then $\lim_{x\to\infty}f(x)=\infty$, contradiction. If $k<1$, then $f$ has arbitrarily small fixed points, so we reach a similar contradiction. Thus, any fixed point of $f$ must be equal to one, but $P(x,x)\Rightarrow f(xf(x))=xf(x)\Rightarrow xf(x)=1\Rightarrow\boxed{f(x)=\frac1x}$, which satisfies both conditions.

Consider the function $g(x)=xf(x)$. This has the same domain and codomain as $f(x)$. We claim that the solution is $f(x)=\frac{1}{x}$; that is, $g(x)=1$ for all positive reals $x$. This is easily verifiable, because $\frac{1}{\frac{x}{y}}=y\left(\frac{1}{x}\right)$. We can simplify the given equation into $$g(x)g(y)=g\left(g(y)\left(\frac{x}{y}\right)\right).$$In particular, we have $$g(x)^2=g(g(x)).$$Consider some number $a$ in the range of $g(x)$. Then $a^2=g(a)$ so that $f(a)=a$. Additionally, this means that $a^2$ is in the range of $g(x)$, so that $f(a^2)=a^2$, $f(a^4)=a^4$, and $f(a^{2^k})=a^{2^k}$ for all nonnegative integers $k$. Suppose, for the sake of contradiction, that $a>1$. Then we know that $f(a^{2^k})=a^{2^k}$. If we let $k$ grow arbitrarily large, we get $f(a^{2^k})\to \infty$ as $a^{2^k}\to \infty$, which contradicts the problem statement. Then we have that $a \leq 1$; that is, $g(x) \leq 1 \rightarrow f(x) \leq \frac{1}{x}$. Again, suppose for the sake of contradiction that $a<1$. Then plugging $y=a$ into the original equation gives $$f(ax)=af(x).$$Letting $x=1$ gives $f(1)=1$, and letting $x=\frac{1}{a}>1$ gives $\frac{1}{a}=f\left(\frac{1}{a}\right)$. But since $f(\frac{1}{a}) \leq a$, we have $\frac{1}{a} \leq a$, which is impossible as $a < 1$ and $\frac{1}{a} > 1$. Then the only possible value for $a$ is $1$. This is equivalent to $g(x)=1$ for all positive reals $x$, or $f(x)=\boxed{\frac{1}{x}}$. $\blacksquare$

$y\to x \implies \exists f(z)=z \forall z=xf(x).$ Let $a$ be such that satisfies the condition on $z.$ $$\text{Then } f(a^2)= f(af(a))=af(a)=a^2.$$By induction it follows $f(a^n)=a^n.$ Assume $a>1$. Thus $n\to \infty \implies a^n\to +\infty.$ A contradiction. By similar argument, we can show that $f(a^{-n})=a^{-n}.$ And also get a contradiction with $a<1.$ So it must be that $a=1.$ Which follows that $$xf(x)=1\implies \boxed{f(x)=1/x},$$which clearly fits.

I thought I didn't do it, so I resolved it. Let $P(x,y)$ be the assertion. $P(x,x)\implies f(xf(x))=xf(x),$ which is a fixed point. Case 1: $\exists z>1: f(z)=z.$ $P(z,z)\implies f(z^2)=z^2,$ thus by induction, $f$ has infinite many fixed points. Then $\lim_{z\to \infty} z=\infty \implies \lim_{z\to \infty}f(z)=\infty,$ a contradiction. Case 2: $\exists z<1: f(z)=z.$ $P(z,1/z)\implies f(1/z)=1/z>1,$ a contradiction. Thus all fixed points $xf(x)=1\implies f(x)=1/x,$ which indeed fits.

Let $P(x,y)$ denote the given assertion. The answer is $\boxed{f(x)=\frac{1}{x}}$, which clearly works. Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$. $P(x,a): f(xf(a))=af(x)$. $P(x,b): f(xf(a))=bf(x)$. So $af(x)=bf(x)\implies a=b$. $\blacksquare$ $P(1,x): f(f(x))=xf(1)$. $P(x,1): f(xf(1))=f(x)$. So $f(f(f(x)))=f(xf(1))=f(x)$. Now since $f$ is injective, we get $f(f(x))=x$. $P(x,f(y)): f(xy)=f(x)f(y)$. Set $g(x)=\ln f(e^x)$. Note that $f(e^{x}e^y )=f(e^x)f(e^y)$, so $f(e^{x+y})=f(e^x)f(e^y)$. Take $\ln$ of both sides. We get $g(x+y)=g(x)+g(y)$. Since $f(x)\to0$ as $x\to \infty$, $g(x)\to -\infty$ as $x\to \infty$. This implies that $g$ is bounded on some interval, so $g$ is linear. Let $g(x)=cx$, we get $f(x)=x^c$. So $f(x\cdot y^c)=x^c\cdot y\implies x^c\cdot y^{c^2}=x^c\cdot y\implies y^{c^2}=y\implies c^2=1$. Clearly $f(x)=x$ doesn't satisfy $f(x)\to 0$ as $x\to \infty$, so $f(x)=\frac{1}{x}$.

We claim $f(x)=\frac{1}{x}$ is the only solution, which indeed satisfies both the functional equation and the limit condition. Claim: $f(1)=1$. Proof: $P(1,1):f(f(1))=f(1)$ $P(\frac{1}{f(1)},1):f(1)=f(\frac{1}{f(1)})$ $P(\frac{1}{f(1)},\frac{1}{f(1)}):f(1)=(\frac{1}{f(1)})f(\frac{1}{f(1)})=1\Rightarrow f(1)=1$. Claim: $x=1$ is the only fixed point. Proof: Assume $k\neq 1$ is fixed. Then from $P(k,k)$ we know $f(k^2)=k^2$ and from induction $f(k^n)=k^n$ for all positive integers $n$. But then $f$ has infinitely large (if $k>1$) or infinitely small (if $k<1$) fixed points, which contradicts the limit condition. Thus only $k=1$ is possible. We finish by noting that from $P(x,x)$ we get $f(xf(x))=xf(x)$, which means that $xf(x)$ is a fixed point for all $x$, but then $xf(x)=1$ and $f(x)=\frac{1}{x}$, as desired.

Let $P(x,y):=f(xf(y))=yf(x)$ Claim: $f$ is injective Proof: Let $f(a)=f(b)$ $P(x,a)$ yields $f(xf(a))=af(x)$ $P(x,b)$ yields $f(xf(b))=bf(x)$ Thus $af(x)=bf(x)\Longrightarrow a=b$, therefore $f$ is injective $\square$. Claim: $f$ is surjective Proof: $P(x,\frac{x}{f(x)})$ yields $f\bigg(xf\left(\frac{x}{f(x)}\right)\bigg)=x$, therefore $f$ is surjective $\square$. Since $f$ is injective and surjective, it is bijective. $P(x,1)$ yields $f(xf(1))=f(x)\overset{\text{from injectivity}}{\Longrightarrow} xf(1)=x\Longrightarrow f(1)=1$ $P(1,f(x))$ yields $f(f(x))=x$ $P(x,f(y))$ yields $f(xy)=f(x)f(y)$ thus the function is multiplicative and is a Cauchy function. Furthermore $\exists z\text{ such that } f(z)=z$ $P(z,z)$ yields $f(z^2)=z^2$ Furthermore by induction we obtain that $f(z^n)=z^n$ If $z>1$, we have that $\lim_{n\to\infty}f(z^n)=\lim_{n\to\infty}z^n=0$ which is clearly a contradiction, since $\lim_{n\to\infty}z^n$ diverges and explodes to infinity. Thus $z\le1\Longleftrightarrow xf(x)\le1\Longrightarrow f(x)\le\frac{1}{x}$ Therefore $f(x)\le1\text{ for }x\ge1$, thus $f$ is monotonically decreasing. Furthermore, since $f$ is a Cauchy function and it's monotonically decreasing, $f(x)=x^c$ for some constant $c$. By plugging into our original function we obtain $f(x\cdot y^c)=y\cdot x^c\Longrightarrow x^c\cdot y^{c^2}=y\cdot x^c\Longrightarrow y^{c^2}=y$ $\therefore c^2=1\Longrightarrow c=\pm1$ $c=1$ implies that $f(x)=x,\forall x\in \mathbb{R}^+$, however this is clearly a contradiction from our second condition since $\lim_{x\to\infty}x$ blows up to infinity, and does not converge to a finite value. Thus $c=-1$, $c=-1$ implies $f(x)=\frac{1}{x}$ which clearly satisfies both conditions. So, to sum up $\boxed{f(x)=\frac{1}{x}, \forall x \in \mathbb{R}^+}$ $\blacksquare$.