# Solve the functional equation $f(x)f(y)=f(y)f(xf(y))+\frac{1}{xy}$

Find all mappings $$f$$ from $$\mathbb{R}_+$$ to $$\mathbb{R}_+$$ such that for $$\forall x,y \in \mathbb{R}_+$$:

$$f(x)f(y)=f(y)f(xf(y))+\frac{1}{xy}$$

Since the domain is $$\mathbb{R}_+$$, we can't use the trick $$x=-y$$ or something like this.

Also I've tried $$x = 0, 1, 2$$ and found nothing useful.

What I already got is that there isn't $$x$$ such that $$f(x) = 1$$, which could be easily deduced by contradiction.

EDIT

Thanks to @Sil, we can actually use this site https://approach0.xyz to search for maths formula.

• Using approach0.xyz we can find some off-site solutions, e.g. here or here
– Sil
Commented Nov 9, 2022 at 20:20
• @Sil OH this site is extrmely useful. It's very difficult to search for maths formula. I will edit it in for the convenience of future comers. Commented Nov 10, 2022 at 8:38
• Sure, you might want to see also How to search on this site? for more details
– Sil
Commented Nov 10, 2022 at 9:42

Consider the statement $$P(x,y):f(x)f(y)=f(y)f(xf(y))+\frac1{xy}$$

We first look at

$$P(1,x):f(1)f(x)=f(x)f(f(x))+\frac{1}{x}$$ $$\Rightarrow f(f(x))=f(1)-\frac1{xf(x)}\ (\star)$$

To use this, we need to find another $$f(f(x))$$. We then look at $$P(f(x),y):f(f(x))f(y)=f(y)f(f(x)f(y))+\frac1{yf(x)}$$ $$\Rightarrow f(f(x))=f(f(x)f(y))+\frac1{yf(x)f(y)}$$

Equating them, we get $$f(1)-\frac1{xf(x)}=f(f(x)f(y))+\frac{1}{yf(x)f(y)}$$

The right-hand side is almost symmetric wrt to $$x$$ and $$y$$. Then we look at what happens if we swap $$x$$ and $$y$$

$$f(1)-\frac1{yf(y)}=f(f(x)f(y))+\frac{1}{xf(x)f(y)}$$

Subtracting them we have $$\frac1{yf(y)}-\frac1{xf(x)}=\frac{1}{f(x)f(y)}\left(\frac1y-\frac1x\right)\Rightarrow xf(x)-x=yf(y)-y$$

Since the left-hand side is just a function of $$x$$ and the right-hand side is just a function of $$y$$, they both must evaluate to a constant, say $$\lambda$$. Thus $$xf(x)-x=\lambda\Rightarrow f(x)=\frac{\lambda}{x}+1$$

Plugging this into $$(\star)$$, we get

$$\frac{\lambda x}{\lambda+x}+1=\lambda+1-\frac{1}{\lambda+x}$$

A bit of cross-multiplication, and a bit of expansion with some cancellations gives us $$\lambda^2=1\Rightarrow \lambda=\pm1$$. Now if $$\lambda=-1$$, then $$f(x)=1-\frac1x$$, but that is negative for $$0 and thus it is not admissible. Hence $$\lambda=1$$ and the only solution is $$f(x)=1+\frac1x$$

• Thanks so much! I once found that if we let $F(x) = xf(x) + \frac{1}{1-f(1)}$ then we can get: $f(1)F(x) = F(f(1)x)$, which seems very close to answer. But finally this method needs $F(x)$ to be contiguous to further solve the equation... Commented Nov 9, 2022 at 8:32

Based on this solution. Write the equation as $$P(x,y):f(x) - f(xf(y)) = \frac{1}{xyf(y)}.$$ Then adding $$P(xf(y),z)$$ to $$P(x,y)$$ we get $$f(x)-f(xf(y)f(z))=\frac{1}{xyf(y)}+\frac{1}{xf(y)zf(z)}.$$ Since LHS is symmetric in $$y,z$$, RHS must be too, hence $$\frac{1}{xyf(y)}+\frac{1}{xf(y)zf(z)}=\frac{1}{xzf(z)}+\frac{1}{xf(z)yf(y)}.$$ Multiplying by $$xyzf(y)f(z)$$ and subtracting $$y+z$$ we find $$zf(z)-z=yf(y)-y.$$ So $$xf(x)-x$$ must be a constant and $$f(x)=\frac{c}{x}+1$$. Substituting back we find $$c^2=1$$ and hence $$c=1$$ as we need only solutions in positive reals. So $$f(x)=\frac{1}{x}+1$$ is the only solution.

• Thanks a lot. It seems that the key is to get $xf(x)-x=yf(y)-y$. I have been in the wrong way. Commented Nov 10, 2022 at 7:20