Problems from Ukrainian Authors

Denote the statement by $P$, and let $P(x,y)$ denote the substitutions into it. Let $a$ be the real number such that $f(a)=1$. Let $S$ denote the set of $x>0$ such that $f(ax)=x$. Then $1\in S$. We prove several lemmas:

Lemma 1. If $x\in S$, then $2x^2\in S$. Proof. Use $P(ax,ax)$: $f(a2x^2)=2x^2\Rightarrow 2x^2\in S$.

Lemma 2. If $x\in S$, then $\sqrt{\frac{x}{2}}\in S$. Proof. Denote by $x_0$ the real number that satisfies $f(x_0)=\sqrt{\frac{x}{2}}$. Use $P(x_0,x_0)$: $f(x_0\sqrt{2}x)=x=f(ax)$, so $x_0\sqrt{2}x=ax$. We get $x_0=a\sqrt{\frac{x}{2}}$ and $f(a\sqrt{\frac{x}{2}})=\sqrt{\frac{x}{2}}$, so $\sqrt{\frac{x}{2}}\in S$.

Lemma 3. If $x,y\in S$, then $\frac{1}{2}(x+y)\in S$. Proof. From Lemma 2 we have $\sqrt{\frac{x}{2}},\sqrt{\frac{y}{2}}\in S$. Use $P(a\sqrt{\frac{x}{2}},a\sqrt{\frac{y}{2}})\Rightarrow f(\frac{1}{2}a(x+y))=\frac{1}{2}(x+y)$.

Lemma 4. The function $x\mapsto xf(x)$ is a surjection. Proof. Use $P(x,x)$ and get $f(2xf(x))=2f^2(x)$. Consider arbitrary $x>0$. Take $y$ such that $f(y)=\sqrt{\frac{1}{2}f(2x)}$. Substitute it into the last equality to get $f(2yf(y))=f(2x)\Rightarrow yf(y)=x$, as desired.

Lemma 5. If $xf(x)>yf(y)$, then $f(x)>f(y)$. Proof. Suppose the contrary, let $xf(x)>yf(y)$, but $f(x)\le f(y)$. Since $x\ne y$, we have $f(x)<f(y)$. Choose real $t$ with $xf(x)>t>yf(y)$. Then there is a real $u$ such that $uf(u)+yf(y)=t$. Then $f(uf(u)+yf(y))=f(t)=f^2(u)+f^2(y)>f^2(y)>f^2(x)$. Then there is a real $v$ such that $f(t)=f^2(v)+f^2(x)=f(xf(x)+vf(v))$. From here we get $t=xf(x)+vf(v)>xf(x)$, which is a contradiction, and the proof of the lemma is complete.

Lemma 6. The function $x\mapsto f(x)$ is increasing on $(0,+\infty )$. Proof. It suffices to prove that for any positive $a,b,c$ with $b>c$ we have $f(a+b)>f(a+c)$. Consider $x,y,z$ such that $a=xf(x),b=yf(y)$, and $c=zf(z)$. Since $b>c$, we have $f(y)>f(z)$. Then $$\begin{array}{rl}f(a+b)& =f(xf(x)+yf(y))=f^2(x)+f^2(y)>f^2(x)+f^2(z)\\ & =f(xf(x)+zf(z))=f(a+c),\end{array}$$ as desired.

From Lemma 1, $2^1\in S$, $2^3\in S$, $2^7\in S$, etc., so $S$ contains arbitrary large reals. Let us prove that $S$ contains arbitrary small (close to zero) reals. Consider arbitrary $\epsilon >0$. It suffices to prove that $\exists x\in S:0<x<\epsilon$. From Lemma 2, we have that $2^{\frac{n-1}{2}}\in S$ if $2^n\in S$. Since $2^0\in S$, we have $2^{\frac{1}{2}-1}\in S$, $2^{\frac{1}{4}-1}\in S$, $2^{\frac{1}{8}-1}\in S$, etc., so there is $x\in S:\frac{1}{2}<x<\frac{1}{2}+{\epsilon }^2$. Denote by $x_0$ the real such that $f(x_0)=\sqrt{x-\frac{1}{2}}$. Use $P(x_0,\frac{1}{\sqrt{2}}a)$. Since $\frac{1}{\sqrt{2}}\in S$, we have $f(\frac{a}{2}+x_0\sqrt{x-\frac{1}{2}})=x=f(ax)$. This means $\frac{a}{2}+x_0\sqrt{x-\frac{1}{2}}=x$.

Therefore $\sqrt{x-\frac{1}{2}}\in S$ and $\sqrt{x-\frac{1}{2}}<\epsilon$. We have proven that $S$ contains arbitrarily large and arbitrarily small numbers.

Suppose that $x\notin S$. Let us prove that there is $y\in S$, lying between $x$ and $f(ax)$. Suppose otherwise. We've proven that there are $y_1,y_2\in S$ such that $y_1<x,f(ax)<y_2$. Consider the following process: given $z_1,z_2\in S$ such that $z_1<x,f(ax)<z_2$, consider $z_3$, the midpoint of $[z_1,z_2]$. By Lemma 3, it lies in $S$ and by our assumption is not between $x$ and $f(ax)$. This means that the closed interval between $x$ and $f(ax)$ lies entirely either in $[z_1,z_3]$ or $[z_3,z_2]$. Then we can choose one of these intervals and repeat the operation. Since after each step the length of the interval halves, it will eventually be smaller than the length of the interval between $x$ and $f(ax)$, leading to a contradiction.

Thus, there is $y\in S$ between $x$ and $f(ax)$. Since $f$ is increasing, $0>(x-y)(f(ax)-y)=(x-y)(f(ax)-f(ay))>0$. This contradiction proves that $S=(0,+\infty )$, so $f(x)=\frac{x}{a}$. It's easy to check that all functions of the form $f(x)=cx$ with $c>0$ satisfy the statement.