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.

  • $\begingroup$ This is really nice.Thanks!! $\endgroup$ – 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 $x<u$, $P(x,\frac{u-x}{f(x)})$ $\implies$ $f(x)f(\frac{u-x}{f(x)})=2^2$ 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.


  • 1
    $\begingroup$ This is also very nice.Thanks! $\endgroup$ – Yes it's me Aug 4 '20 at 15:01
  • 1
    $\begingroup$ Great solution! $\endgroup$ – Sunaina Pati Aug 4 '20 at 16:39
  • $\begingroup$ @Shubhangi Thanks! $\endgroup$ – Shubhrajit Bhattacharya Aug 4 '20 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.