SHORTLISTED PROBLEMS FOR THE 68th NMO

Let $f:[0,1]\to [0,1]$ be integrable. For each $n\in \mathbb{N}$, consider $$I_n=n{\int }_0^1(f(x){)}^n(1-f(x))dx.$$

Let $\epsilon >0$. Define $A=\{x\in [0,1]:f(x)>1-\epsilon \}$ and $B=[0,1]\setminus A=\{x\in [0,1]:f(x)\le 1-\epsilon \}$.

Then $$I_n=n{\int }_A(f(x){)}^n(1-f(x))dx+n{\int }_B(f(x){)}^n(1-f(x))dx.$$

For $x\in B$, $f(x)\le 1-\epsilon$, so $(f(x){)}^n\le (1-\epsilon {)}^n$ and $1-f(x)\le 1$: $$n{\int }_B(f(x){)}^n(1-f(x))dx\le n(1-\epsilon {)}^n\cdot m(B),$$ where $m(B)$ is the Lebesgue measure of $B$ (at most $1$). Since $0<1-\epsilon <1$, $n(1-\epsilon {)}^n\to 0$ as $n\to \infty$.

For $x\in A$, $0<1-f(x)<\epsilon$, and $(f(x){)}^n\le 1$: $$n{\int }_A(f(x){)}^n(1-f(x))dx\le n\epsilon \cdot m(A).$$ But also, for $x\in A$, $(f(x){)}^n\le (1{)}^n=1$, so $$n{\int }_A(f(x){)}^n(1-f(x))dx\le n\epsilon \cdot m(A).$$ But $m(A)\le 1$, so $n\epsilon \cdot m(A)\le n\epsilon$.

However, for fixed $\epsilon$, as $n\to \infty$, the integral over $A$ can be made small as follows. For $x\in A$, $f(x)>1-\epsilon$, so $(f(x){)}^n>(1-\epsilon {)}^n$, but $1-f(x)<\epsilon$.

Alternatively, note that for all $x\in [0,1]$, $0\le f(x)\le 1$, so $(f(x){)}^n\le 1$, and $1-f(x)\le 1$.

But for $f(x)<1$, $(f(x){)}^n\to 0$ as $n\to \infty$.

Let $E=\{x\in [0,1]:f(x)=1\}$. Then for $x\in E$, $(f(x){)}^n=1$, but $1-f(x)=0$, so the integrand is $0$.

For $x\notin E$, $f(x)<1$, so $(f(x){)}^n\to 0$ as $n\to \infty$.

Therefore, for any $\delta >0$, there exists $N$ such that for all $n>N$, $(f(x){)}^n<\delta$ for all $x\notin E$.

Thus, $$I_n=n{\int }_{[0,1]\setminus E}(f(x){)}^n(1-f(x))dx.$$ But $0\le (f(x){)}^n(1-f(x))\le \delta$ for $x\notin E$, and the measure of $[0,1]$ is $1$.

Therefore, $$I_n\le n\delta .$$ But as $n\to \infty$, $\delta$ can be made arbitrarily small, and $n\delta \to 0$ as $n\to \infty$ for fixed $\delta$ only if $\delta$ decays faster than $1/n$.

But more precisely, for any $\epsilon >0$, choose $0<\eta <1$ such that $f(x)\le 1-\eta$ except on a set of measure less than $\epsilon$.

Then $$I_n\le n(1-\eta {)}^n+n{\int }_A(f(x){)}^n(1-f(x))dx,$$ where $A=\{x:f(x)>1-\eta \}$ and $m(A)<\epsilon$.

On $A$, $1-f(x)<\eta$, so $$n{\int }_A(f(x){)}^n(1-f(x))dx\le n\eta \cdot m(A)\le n\eta \epsilon .$$ Thus, $$I_n\le n(1-\eta {)}^n+n\eta \epsilon .$$ As $n\to \infty$, $n(1-\eta {)}^n\to 0$, so for large $n$, $I_n<2n\eta \epsilon$.

Since $\epsilon$ is arbitrary, $I_n\to 0$ as $n\to \infty$.

Therefore, $$\underset{n\to \infty }{\lim }n{\int }_0^1(f(x){)}^n(1-f(x))dx=0.$$