Clearly $\forall x\in \mathbb{R}\,:\, f(x)\ge 0$. Suppose on the contrary $f(x_0)=a >1$. Take some $x<x_0$. Define $(x_n)_{n=1}^{\infty}$ as follows: $x_{i+1}=(x+x_i)/2\,,\, i=0,1,\ldots$. We have $f(x)> f(x_i)\ge a^{2^i}\,, i=1,2,\ldots$, a contradiction.

Another Solution?

for 1 >= f(x) >= 0 , f(x)^2 < f(x). But if for f(x) > 1 the above condition was true, the function should have been vigorously decreasing function. Now the decrease should have been in the positive side of number line as if it was negative, f(x)^2 would be positive which will be greater than f(y) which would be negative violating the condition. When a decreasing function even in the positive side it must once again reach the negative side again violating the condition as the way discussed above. Thus f(x) belongs to [0,1] Please tell me anyone if this solution is correct !!! Or else please correct me !!

$f(x)^2 \le f(y)$ for constants $x>y$. Then taking $n$ constants in between $x,y$ and applying the condition gives $f(x)^{2^{n+1}} \le f(y)$. Suppose $f(x) > 1$. Then $f(x)^{2^{n+1}}\to\infty$ as $n\to \infty$, but $f(y)$ is constant, contradiction. Thus $f(x) \le 1$ for all $x\in\mathbb{R}$ Also, $f(y)\ge f(x)^2 \ge 0$ for all $y\in\mathbb{R}$. Thus $f(x)\in [0,1]$ for all $x\in\mathbb{R}$.

Clearly we get $f(x) \geq 0$ for all $x \in \mathbb{R}$. It now suffices to show $f(x) \le 1$ $\forall x \in \mathbb{R}$. Now, choose $n+2$ reals such that $y < r_1< \cdots <r_n < x$. Note that $b^2>a^2$ whenever $b>a>0$ and so $$f(x)^{2^n} \le f(r_n)^{2^{n-1}} \le f(r_{n-1})^{2^{n-2}} \le \cdots \le f(y)$$Hence $f(x)^{2^n} \le f(y)$ for any $n \in \mathbb{N}$. Clearly, the LHS is strictly increasing whenever $f(x) >1$, and so for a fixed $y$, the inequality cannot hold. Hence $f(x) \le 1$ for all $x \in \mathbb{R}$, as desired.

$f(y)\geq f(x)^2\geq 0.$ $x>a_1>a_2>\dots>a_k>y\implies f(y)\geq f(x)^{2^k}\implies f(x)\leq 1.$

then take arbitrary large k, now if its greater than 1, then its absurd so must be 0 <= fx) <= 1 and the degrees of freedom of x,y finishes the problem 2^k ZETA_in_olympiad wrote: $f(y)\geq f(x)^2\geq 0.$ $x>a_1>a_2>\dots>a_k>y\implies f(y)\geq f(x)^{2^k}\implies f(x)\leq 1.$