Let $k\geq 1$ be an integer, $f:[0,1]\to\mathbb R$ be a Riemannian integrable function and $g:[0,1]\to \mathbb R$ be a continuous function. Calculate
$$\lim_{n\to\infty}\int_0^1 f(\{nx^k\})g(x^n)\,dx,$$
where $\{x\}$ denotes the fractional part of the real number $x$.
This is W33 of József Wildt International Mathematical Competition, 2021. Yesterday, I planned to ask this question here because I could only handle the case where $k=1$ at that time. However, when I was ready to click the button "Review your question", an idea suddenly came to my mind: Why not consider $g\equiv 1$ first? After 1 hour or 2, that idea became a full answer to this problem. Now I write it here. It is a little bit long, because I proved two lemmas that is useful for us.
My question. Is my proof presented below correct or not? Does there exist a shorter and more elegant proof?
Lemma 1. If $f:[0,1]\to\mathbb R$ is measurable and bounded, then
$$\lim_{n\to\infty}\int_0^1 f(\{nx^k\})\,dx=\int_0^1f(x)\,dx.$$
Proof of Lemma 1.
Firstly,
$$\int_0^1 f(\{nx^k\})\,dx=\sum_{m=0}^{n-1}\int_{\left(\frac mn\right)^{1/k}}^{\left(\frac {m+1}n\right)^{1/k}}f(nx^k-m)\,dx=\int_0^1\sum_{m=0}^{n-1}\frac1{nk}f(t)\left(\frac{m+t}n\right)^{\frac1k-1}\,dt.$$
Let
$$f_n(t)=\sum_{m=0}^{n-1}\frac1{nk}f(t)\left(\frac{m+t}n\right)^{\frac1k-1},\qquad n\geq 1, t\in(0,1].$$
For fixed $t\in(0,1]$, we have
$$f_n(t)\to f(t)\int_0^1 \frac1k u^{\frac1k-1}\,du=f(t),\qquad \text{as }\ \ n\to\infty.$$
On the other hand,
\begin{align*}
|f_n(t)|&\leq |f(t)|\frac1{nk}\left(\frac tn\right)^{\frac1k-1}+|f(t)|\sum_{m=1}^{n-1}\frac1{nk}\left(\frac{m}n\right)^{\frac1k-1}\\
&\leq |f(t)|\frac1{k}t^{\frac1k-1}+|f(t)|\frac1k\int_{1/n}^1\left(u-\frac1n\right)^{\frac1k-1}\,du\\
&\leq \frac{\|f\|_{L^\infty}}{k}t^{\frac1k-1}+|f(t)|.
\end{align*}
Therefore, DCT implies that $\int_0^1 f_n(t)\,dt\to\int_0^1 f(t)\,dt$, concluding the proof of Lemma 1.
Lemma 2. If $f\in L^\infty(0,1)$, and $0<\eta\ll1$, then there exists a positive cosntant relying only on $k$, $C(k)>0$ such that
$$\int_{1-\eta}^1|f(\{nx^k\})|\,dx\leq \left(\eta+\frac{C(k)}{n}\right)\|f\|_{L^1}.$$
Proof of Lemma 2.
Assume that $\left(\frac mn\right)^{1/k}\leq 1-\eta<\left(\frac {m+1}n\right)^{1/k}$ for some $m\geq \frac n2+1$, then
\begin{align*}
\int_{1-\eta}^1|f(\{nx^k\})|\,dx&\leq \int_{\left(\frac mn\right)^{1/k}}^1|f(\{nx^k\})|\,dx\\
&=\int_0^1|f(t)|\sum_{p=m}^{n-1}\frac1{nk}\left(\frac{p+t}n\right)^{\frac1k-1}\,dt\\
&\leq \|f\|_{L^1}\sum_{p=m}^{n-1}\frac1{nk}\left(\frac{p}n\right)^{\frac1k-1}\\
&\leq \|f\|_{L^1}\frac1k\int_{\frac mn}^1\left(u-\frac1n\right)^{\frac1k-1}\,du\\
&\leq \|f\|_{L^1}\left[\left(1-\frac1n\right)^{\frac1k}-\left(\frac{m-1}n\right)^{\frac1k}\right].
\end{align*}
Let $\phi(s)=s^{1/k}$ for $s\in(0,1]$, then
\begin{align*}
\phi\left(\frac mn+\frac1n\right)-\phi\left(\frac mn-\frac1n\right)&\leq \frac2n\max_{1/2\leq s\leq 1}|\phi'(s)|=\frac{2^{2-\frac1k}}{nk}\\
&=\frac1{nk}+\frac{C(k)}{n}=\frac1n\min_{1/2\leq s\leq 1}|\phi'(s)|+\frac{C(k)}{n}\\
&\leq \phi(1)-\phi\left(1-\frac1n\right)+\frac{C(k)}{n}.
\end{align*}
Hence
\begin{align*}
\int_{1-\eta}^1|f(\{nx^k\})|\,dx&\leq \left(\phi\left(1-\frac1n\right)-\phi\left(\frac mn-\frac1n\right)\right)\|f\|_{L^1}\\
&\leq \left(\phi(1)-\phi\left(\frac mn+\frac1n\right)+\frac{C(k)}{n}\right)\|f\|_{L^1}\\
&\leq \left(\eta+\frac{C(k)}{n}\right)\|f\|_{L^1}.
\end{align*}
This completes the proof of Lemma 2.
Proof of the main problem.
We prove
$$\lim_{n\to\infty}\int_0^1 f(\{nx^k\})g(x^n)\,dx=g(0)\int_0^1f(x)\,dx.\tag{1}$$
For each $\epsilon>0$ and $0<\eta\ll1$, since $g$ is continuous at $x=0$, we can find $\delta>0$ such that $|g(x)-g(0)|\leq \epsilon$ for all $x\in[0,\delta]$. For $n$ large enough, we have $(1-\eta)^n<\epsilon$ and thus
$$|g(x^n)-g(0)|\leq \epsilon,\qquad \forall x\in[0,1-\eta].$$
Therefore, for all large $n$, we have
\begin{align*}
&\ \ \ \left|\int_0^1 f(\{nx^k\})g(x^n)\,dx-g(0)\int_0^1f(x)\,dx\right|\\
&\leq\left|g(0)\int_0^1f(\{nx^k\})\,dx-g(0)\int_0^1f(x)\,dx\right|+\int_0^1|f(\{nx^k\})||g(x^n)-g(0)|\,dx\\
&\leq|g(0)|\left|\int_0^1f(\{nx^k\})\,dx-\int_0^1f(x)\,dx\right|+\epsilon \int_0^{1-\eta}|f(\{nx^k\})|\,dx+2\|g\|_{L^\infty}\int_{1-\eta}^1|f(\{nx^k\})|\,dx\\
&\leq|g(0)|\left|\int_0^1f(\{nx^k\})\,dx-\int_0^1f(x)\,dx\right|+\epsilon \int_0^{1}|f(\{nx^k\})|\,dx+2\|g\|_{L^\infty}\int_{1-\eta}^1|f(\{nx^k\})|\,dx.
\end{align*}
By Lemma 1 and Lemma 2, taking $\limsup_{n\to\infty}$ on each side gives that
$$\limsup_{n\to\infty}\left|\int_0^1 f(\{nx^k\})g(x^n)\,dx-g(0)\int_0^1f(x)\,dx\right|\leq \epsilon\|f\|_{L^1}+2\eta\|g\|_{L^\infty}\|f\|_{L^1}.$$
Letting $\epsilon,\eta\to0$, we get $(1)$.
I would like to express my gratitude to anyone who are reading this long post. Comments or remarks on a easier proof, or improved results (e.g. convergence rates, the same conclusion under weaker conditions), or anything else related to this problem, are very welcome.
Best Answer
Your second lemma is not really necessary. Note that a Riemannian integrable function is always bounded, you can bound the remaining as follows: For all $\epsilon>0$, \begin{align}&\limsup_n \left| \int_{0}^1 f\left( \left\{nx^k \right\} \right)\left( g(x^n) -g(0)\right)\mathrm{d} x \right|\\ \le &2\epsilon \|f\|_{\infty} \|g\|_{\infty}+\limsup_n \int_{0}^{1-\epsilon} \| f\|_{\infty} \left|g(x^n) -g(0) \right|\mathrm{d} x= 2\epsilon \|f\|_{\infty} \|g\|_{\infty}.\end{align} Thus, $$ \lim_n \left| \int_{0}^1 f\left( \left\{nx^k \right\} \right)\left( g(x^n) -g(0)\right)\mathrm{d} x \right|=0.$$ For the first lemma, I think your solution is already okay.