IMO 2002 Problem 5 Solution doubt

Question

Question -

(IMO-2002) Find all functions from the set $\mathbb{R}$ of real numbers to itself such that $$ (f(x)+f(z))(f(y)+f(t))=f(x y-z t)+f(x t+y z) $$

So solution goes like these

The only solutions are $f(x)=0$ for all $x ; f(x)=1 / 2$ for all $x ;$ and $f(x)=x^{2}$ for all $x$ .

Put $x=y=z=0$ to get $f(0)=f(0)(f(0)+f(t))$ for all $t$ . In particular, $f(0)=2 f(0)^{2}$ so that $f(0)=0$ or $f(0)=1 / 2$

If $f(0)=1 / 2,$ then $f(t)=1 / 2$ for all $t$

Suppose $f(0)=0 .$ Putting $z=t=0,$ we get

$f(x y)=$ $f(x) f(y)$ for all $x, y .$ Hence $f$ is multiplicative and $f(1)^{2}=$ $f(1) .$

Thus $f(1)=0$ or $1 .$ If $f(1)=0,$ it is easy to check that $f(x)=0$ for all $x .$

Assuming $f(1)=1,$ taking $x=0, y=$ $t=1,$ we get $f(z)=f(-z)$ for all $z .$ Using $f\left(x^{2}\right)=f(x)^{2}$ for all $x$ and $f(z)=f(-z)$ for all $z,$ infer that $f(x) \geq 0$ for all $x$

Taking $t=x, z=y,$ we get

$f\left(x^{2}+y^{2}\right)=(f(x)+f(y))^{2}$

Thus $f\left(x^{2}+y^{2}\right) \geq f(x)^{2}=f\left(x^{2}\right) .$

This shows that $f(u) \geq$ $f(v)$ if $u \geq v \geq 0 .$

Hence $f$ is an increasing function on the positive reals.

Taking $y=z=t=1,$ we see that $$ f(x-1)+f(x+1)=2(f(x)+1) $$ This implies that $f(n)=n^{2}$ for all non-negative integers $n .$ since $f$ is even, $f(n)=n^{2}$ for all integers $n .$

Using multiplicativity of $f,$ we see that $f(r)=r^{2}$ for all rationals $r .$

since $f$ is increasing on positive reals and $f(r)=r^{2}$ for all rationals, it is easy to show that $f(x)=x^{2}$ for all $x \geq 0$ ...since $f$ is even, we infer that $f(x)=x^{2}$ for all real $x$

now i did not understand why these above lines are true (bold one)

thankyou

Answer 1

Suppose you had an irrational $x$ with $f(x)>x^2$.

Consider a rational $q$ with $$x<q<\sqrt {f(x)}$$

Then $$f(q)=q^2<f(x)$$ contrary to the assumption that $f(x)$ is increasing.

The case $f(x)<x^2$ is similar.

Answer 2

The idea is that the rationals are dense in $\mathbb R$, so if the function is continuous, the argument is true. Since here the function is increasing (and not yet continuous), this implies continuity by the fact that you can't jump to any other value and come back to $r^2$.

E.G., you have in $x=0$ the function is equal to $0$. If you look at any real number close to $0$, and you decide (by absurd) to give it a different value than the square function, for instance $5$, you could find a rational bigger than that real number, and such that the value of the function on that rational is strictly smaller, thus the function is not increasing anymore: contradiction.

Do you need more details?