Any idea?

$f^n(x)=f(f^{n-1}(x))$ ? Or $f^n(x)=(f(x))^n$?

ratatuy wrote: $f^n(x)=f(f^{n-1}(x))$ ? Or $f^n(x)=(f(x))^n$? It's not specified in the original paper but I am 100% sure that it is $f(x)^n$ (this naturally makes both sides homogeneous of degree $1010 \cdot 2021$).

ratatuy wrote: $f^n(x)=f(f^{n-1}(x))$ ? Or $f^n(x)=(f(x))^n$? We can consider the two options it will give two problems to solve

Suppose $g,h:[0,1] \rightarrow {\bf R}$ are continuous and comparable, i.e. $g(a)\le g(b)\iff h(a)\le h(b)$. Then $$\int_0^1 fg dx-\int_0^1 f dx\cdot \int_0^1 g dx={1\over 2}\int_{[0,1]^2} (g(x)-g(y))(h(x)-h(y)) dA \ge 0$$Moreover the inequality is strict if there exist $a,b\in (0,1)$ with $g(a)<g(b),h(a)<h(b)$ since then $((g(x)-g(y))(h(x)-h(y))>0$ on a nbhd of $(a,b)$ in $(0,1)^2$. Now take $k\in [1,1010], g:=f^k, h:=f^{2021-k}$, then $g,h$ are comparable, hence $\int_0^1 f^k dx\cdot \int_0^1 f^{2021-k}dx\le \int_0^1 f^{2021}dx$ with strict inequality if $f$ is not constant. Multiplying all these inequalities we find $LHS \le RHS$ in the original problem. If $f$ is not constant, then $f\not \equiv 0$, each factor on the LHS is positive and for each $k\in [1,1010]$ the corresponding inequality is strict. Hence $LHS<RHS$. It follows that $f$ must be constant if we have equality, and indeed $f=a\ge 0$ is a solution.

alexheinis wrote: Suppose $g,h:[0,1] \rightarrow {\bf R}$ are continuous and comparable, i.e. $g(a)\le g(b)\iff h(a)\le h(b)$. Then $$\int_0^1 fg dx-\int_0^1 f dx\cdot \int_0^1 g dx={1\over 2}\int_{[0,1]^2} (g(x)-g(y))(h(x)-h(y)) dA \ge 0$$Moreover the inequality is strict if there exist $a,b\in (0,1)$ with $g(a)<g(b),h(a)<h(b)$ since then $((g(x)-g(y))(h(x)-h(y))>0$ on a nbhd of $(a,b)$ in $(0,1)^2$. Now take $k\in [1,1010], g:=f^k, h:=f^{2021-k}$, then $g,h$ are comparable, hence $\int_0^1 f^k dx\cdot \int_0^1 f^{2021-k}dx\le \int_0^1 f^{2021}dx$ with strict inequality if $f$ is not constant. Multiplying all these inequalities we find $LHS \le RHS$ in the original problem. If $f$ is not constant, then $f\not \equiv 0$, each factor on the LHS is positive and for each $k\in [1,1010]$ the corresponding inequality is strict. Hence $LHS<RHS$. It follows that $f$ must be constant if we have equality, and indeed $f=a\ge 0$ is a solution. How did you get the equation on the third line?

@luka: I can read your question in 2 ways: how did I come up with it or how does one prove it? The proof is easy: expanding the RHS gives ${1\over 2} (\int_0^1 g(x)h(x)dx-\int_0^1 g(x)dx\cdot \int_0^1 h(y)dy-\int_0^1 g(y)dy\cdot \int_0^1 h(x)dx+\int_0^1 g(y)h(y)dy)$. Changing a few symbols gives the LHS. I didn't see this at first: first I proved a discrete analogue with the Rearrangement Theorem, details below. Then I took the limit in Riemann sums. Proving strict inequality with this method is possible but cumbersome, and then I looked for a direct proof with integrals, which is the result in the the third line. And welcome to AOPS. Suppose $a_1,\cdots,a_n $ are real numbers and $b_1,\cdots,b_n$ also with comparable order, hence $a_k\le a_l\iff b_k\le b_l$. The rearrangement property with the vectors $x:=(a_1\cdots a_n \cdots a_1\cdots a_n),y:=(b_1\cdots b_n\cdots b_1\cdots b_n), z:=(b_1\cdots b_1 \cdots b_n\cdots b_n)$ gives $(x,y)\ge (x,z)$ hence $n\sum_1^n a_k b_k \ge \sum_1^n a_k \cdot \sum_1^n b_k$. Now divide by $n^2$, let $a_k=g(k/n),b_k=h(k/n)$ and take the limit to obtain $\int_0^1 gh dx\ge \int_0^1 g dx\cdot \int_0^1 h dx$. One can try to prove the sum-inequality by taking the difference: this gives $n\sum_1^n a_k b_k -\sum_1^n a_k \cdot \sum_1^n b_k=\sum_{k<l} (a_k-a_l)(b_k-b_l)$. Translating this line of proof to integrals over $[0,1]^2$ gives the formula that we we used.

alexheinis wrote: @luka: I can read your question in 2 ways: how did I come up with it or how does one prove it? The proof is easy: expanding the RHS gives ${1\over 2} (\int_0^1 g(x)h(x)dx-\int_0^1 g(x)dx\cdot \int_0^1 h(y)dy-\int_0^1 g(y)dy\cdot \int_0^1 h(x)dx+\int_0^1 g(y)h(y)dy)$. Changing a few symbols gives the LHS. I didn't see this at first: first I proved a discrete analogue with the Rearrangement Theorem, details below. Then I took the limit in Riemann sums. Proving strict inequality with this method is possible but cumbersome, and then I looked for a direct proof with integrals, which is the result in the the third line. And welcome to AOPS. Suppose $a_1,\cdots,a_n $ are real numbers and $b_1,\cdots,b_n$ also with comparable order, hence $a_k\le a_l\iff b_k\le b_l$. The rearrangement property with the vectors $x:=(a_1\cdots a_n \cdots a_1\cdots a_n),y:=(b_1\cdots b_n\cdots b_1\cdots b_n), z:=(b_1\cdots b_1 \cdots b_n\cdots b_n)$ gives $(x,y)\ge (x,z)$ hence $n\sum_1^n a_k b_k \ge \sum_1^n a_k \cdot \sum_1^n b_k$. Now divide by $n^2$, let $a_k=g(k/n),b_k=h(k/n)$ and take the limit to obtain $\int_0^1 gh dx\ge \int_0^1 g dx\cdot \int_0^1 h dx$. One can try to prove the sum-inequality by taking the difference: this gives $n\sum_1^n a_k b_k -\sum_1^n a_k \cdot \sum_1^n b_k=\sum_{k<l} (a_k-a_l)(b_k-b_l)$. Translating this line of proof to integrals over $[0,1]^2$ gives the formula that we we used. Thanks, everything is clear for me.