Find all $f:\mathbb{R}_+\to\mathbb{R}_+$ such that $(x+y) \, f\big(f(x) \, y\big)=x^2 f\big(f(x)+f(y)\big)$ for all $x,y>0$.

Question

Find all functions $f\colon\Bbb R_+\to\Bbb R_+$ that satisfy the functional equation $$(x+y)\, f\bigl(f(x)\, y\bigr)=x^2 \,f\bigl(f(x)+f(y)\bigr)$$ for all $x,y\in\mathbb{R}_+$. Here, $\mathbb{R}_+$ is the set of positive real numbers.

Attempt: There is no continuous solution $f$ such that $\lim_{x\to 0^+}f(x)$ exists. If such a solution exists, then let $L=\lim_{x\to 0^+}f(x)$. If $L>0$, then by taking $x\to 0^+$, we get $$y\,f(Ly)=0.$$ Therefore $f(Ly)=0$ for all $y>0$. This is a contradiction because the codomain of $f$ is $\mathbb{R}_+$. Therefore, $L=0$.

Now, taking $y\to 0^+$, we get $$0=x^2\,f\big(f(x)\big)$$ for all $x>0$. This means $f\big(f(x)\big)=0$. This is again a contradiction.

How do we solve the functional equation if the continuity and limit assumptions are dropped?

Answer 1

There is no such function $f$.

Assume $f\colon \Bbb R_+\to\Bbb R_+$ satisfies the functional equation $$(x+y)f\bigl(f(x)y\bigr)=x^2f\bigl(f(x)+f(y)\bigr) $$ for all $ x,y>0$.

Lemma. $f$ is injective.

Proof. Assume $f(x_1)=f(x_2)=a$ for some $x_1\ne x_2$. Then with $A=f(a+f(y))$, $B=-f(ay)$, $C=-yf(ay))$, we know that $x_1,x_2$ are distinct roots of $ AX^2+BX+C$. As $A\ne 0$, this implies $$x_1x_2=\frac CA=-y\cdot\frac{f(ay)}{f(a+f(y))}=y\frac BA=-y(x_1+x_2) $$ for all $y$, which is absurd. $\square$

Now whenever $x+y=x^2$, i.e., for $x>1$ and $y=x^2-x$, we can cancel $x^2$ on both sides and find $$f\bigl(f(x)(x^2-x)\bigr)=f\bigl(f(x)+f(x^2-x)\bigr) $$ and by the lemma, $$f(x)(x^2-x)=f(x)+f(x^2-x)$$ and ultimately $$\tag1f(x^2-x)=(x^2-x-1)f(x)$$ for all $x>1$. However, for any $x\in\left(1,\frac{\sqrt5+1}2\right)$, this leads to a contradiction. For example with $x=\frac32$, equation $(1)$ tells us that $$0<f\left(\tfrac34\right)=-\tfrac14 f\left(\tfrac32\right)<0,$$ contradiction.

Answer 2

Let me know if I made a mistake in the following

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First of all, $f$ must be unbounded near $0$: $$\lim_{y\to0^+}\frac{f(f(x)y)}{f(f(x)+f(y))}=x;$$clearly $f$ can't converge to $0$, and it can't converge to $\ell\in\Bbb R^+$ either because, taking $x\to0^+$, $\ell/f(2\ell)=0$ is an absurdity. Finally, it can't oscillate:$$\lim_{x\to0^+}\frac{f(f(x)y)}{f(f(x)+f(y))}=0 $$for all $y$ is an absurdity. Therefore $\lim_{x\to0^+}f(x)=+\infty$.

As a consequence, if we can show that $f(y)$ does not tend to $0$ as $y\to1^+$ (which would make $1$ a discontinuity since $f(1)>0$), there must exist $x,y$ such that $f(x)y=f(x)+f(y)$, i.e. $$f(x)=\frac{f(y)}{y-1}. $$This is because the RHS goes to $+\infty$ as $y\to1^+$, and given a large enough $RHS$ we can always make $x$ small enough to have the equality. Indeed, we have$$\lim_{y\to1^+} \frac{f(f(y)+f(1))}{f(f(y))}=2$$which ensures $f(y)\not\longrightarrow_{t\to1^+}0.$ Now, if $x$ and $y$ satisfty $$f(x)=\frac{f(y)}{y-1} $$then the functional equation yields $x+y=x^2.$Combining them, $$f(x^2-x)=(x^2-x-1)f(x).$$We may as well write $y=x^2-x=1+h$, and thus $x=\frac{1+\sqrt{5+4h}}{2}=:\varphi_h$, where $\varphi=\varphi_0$ is the golden ratio. This way we get $$f(1+h)=hf(\varphi_h). $$We have seen $f(1+h)$ does not go to $0$ as $h\to0^+$, therefore, going also back to the functional equation, we must have for all $c$:$$+\infty=\lim_{h\to0^+}f(\varphi_h)=\frac{c^2}{c+\varphi/f(c)}f(f(c)+f(\varphi/f(c)))$$,which is impossible. So we can conclude no $f:\Bbb R^+\to\Bbb R^+$ satisfies the functional equation.