# Functions satisfying $f(x)f(y)=2f(x+yf(x))$ over the positive reals

From the IMO shortlist:

We denote by $$\mathbb{R}^+$$ the set of all positive real numbers.

Find all functions $$f: \mathbb R^+\rightarrow\mathbb R^+$$ which have the property: $$f(x)f(y)=2f(x+yf(x))$$ for all positive real numbers $$x$$ and $$y$$.

$$\textbf{My progress: }$$At first assume $$f$$ is not injective.

Then there exist $$z,y$$ such that $$f(z)=f(x)$$ assume WLOG $$z >x$$.Now,We can choose some appropriate $$y$$ such that $$x+yf(x)=z$$.

Then it follows that $$f(y)=2$$ which means $$2$$ has an inverse. Now, substituting inverse of $$2$$ in place of $$y$$ from which we could get a bunch of values for which the function assumes the same value. I then thought of somehow proving this would show the function is constant which I failed to do.

The case with $$f$$ injective is pretty easy.interchaging $$x,y$$ we get, $$f(x+yf(x))=f(y+xf(y))$$ Using injectivity we could easily deduce from here that $$f$$ is linear and get a contradiction.

But I cannot do the first case. I do believe that $$f(x)=2$$ is the only function that works.

Any kind of hint or solution is appreciated.

Like you wrote, we can show that $$f$$ is not injective and there is some $$a$$ such that $$f(a) = 2$$. Suppose towards a contradiction that there exists a $$b$$ with $$f( b) < 2$$.

$$P(a, x)$$ gives $$f(x) = f(2x + a)$$ so we can find a $$c > a$$, with $$f(c ) = f(b)$$ Let $$y \in \mathbb{R}^+$$ such that $$c + f(c)y = 2y + a$$

Then $$P(c, y)$$ gives $$f(c) = 2$$ which is a contradiction.

Similarly, suppose that there exists a $$b$$ with $$f(b ) > 2$$. Then we can find a $$c < a$$ with $$f(c) = f(b)$$ and a $$y$$ with $$c + f(c)y = 2y + a$$ and $$P(c,y)$$ gives the same contradiction.

• This is really nice.Thanks!! – Yes it's me Aug 4 '20 at 14:56

Let $$P(x,y)$$ be the statement $$\color{red}{f(x)f(y)=2f(x+yf(x))}$$

We will solve this functional equation dividing it into $$4$$ steps.

$$\boxed{\text{(1) f(x)\ge 2 \forall x}}$$

If $$f(x)<1$$ for some $$x$$, then $$P(x,\frac x{1-f(x)})$$ $$\implies$$ $$f(x)=2>1$$, contradiction. So $$f(x)\ge 1$$ $$\forall x$$

If $$f(x)\ge 2^t$$ $$\forall x$$ and for some $$t\ge 0$$; then $$P(x,x)$$ $$\implies$$ $$f(x)^2\ge 2^{1+t}$$ and so $$f(x)\ge 2^{\frac{1+t}2}$$ $$\forall x$$

Setting then $$a_0=0$$ and $$a_{n+1}=\frac{1+a_n}2$$, we get $$f(x)\ge 2^{a_n}$$ $$\forall x$$, $$\forall n$$ and so $$f(x)\ge 2$$ $$\forall x$$.

$$\boxed{\text{(2) If f(u)=2 for some u>0, then f(x)=2 \forall x}}$$

If $$f(u)=2$$ for some $$u>0$$, then, for $$x, $$P(x,\frac{u-x}{f(x)})$$ $$\implies$$ $$f(x)f(\frac{u-x}{f(x)})=2^2$$ and so $$f(x)=2$$ $$\forall x\le u$$

But $$P(u,u)$$ $$\implies$$ $$f(u(1+2))=2$$ and so $$f(u(1+2)^n)=2$$ and so, using previous line, $$f(x)=2$$ $$\forall x$$.

$$\boxed{\text{(3) f is not injective}}$$

If $$f(x)$$ is injective, comparaing of $$P(x,1)$$ with $$P(1,x)$$ we get $$f(x+f(x))=f(1+xf(1))$$ and so $$x+f(x)=1+xf(1)$$ and so $$f(x)=1+x(f(1)-1)$$ Plugging this back in the original equation, we get $$2=1$$, contradiction!

$$\boxed{\text{(4) f(x)=2 \forall x}}$$

Since non injective, let $$a>b$$ such that $$f(a)=f(b)$$. Then let $$u=\frac{b-a}{f(a)}$$. Hence $$P(a,u)$$ $$\implies$$ $$f(u)=2$$

Therefore we have all solutions.

$$\tag*{\square}$$

• This is also very nice.Thanks! – Yes it's me Aug 4 '20 at 15:01
• Great solution! – Sunaina Pati Aug 4 '20 at 16:39
• @Shubhangi Thanks! – Shubhrajit Bhattacharya Aug 4 '20 at 17:16