The limit $\lim_{n \to \infty} \int_0^1 f(\{n x\}) g(\{n/x\}) \, \mathrm{d} x$ This is problem 1.82 in Ovidiu Furdui's book 'Limits, Series, and Fractional Part Integrals'. Given $f,g \in \mathrm{C}([0,1])$, let
$$ a_n = \int \limits_0^1 f(\{n x\}) g\left(\left\{\frac{n}{x}\right\}\right) \, \mathrm{d} x \, , \, n \in \mathbb{N} \, , $$
where $\{\cdot\} = \cdot - \lfloor \cdot \rfloor$ is the fractional part function. We want to find the limit $\lim_{n \to \infty} a_n$ (if it exists).
The known results (problems 1.70 and 1.71 in the same book)
$$ \lim_{n \to \infty} \int \limits_0^1 f(x) g\left(\left\{\frac{n}{x}\right\}\right) \, \mathrm{d} x = \lim_{n \to \infty} \int \limits_0^1 f(\{n x\}) g(x) \, \mathrm{d} x = \int \limits_0^1 f(x) \, \mathrm{d} x \int \limits_0^1 g(x) \, \mathrm{d} x $$
suggest that the desired limit might also just be the product of the two integrals (which is true if either function is constant).
We can write
$$ a_n = \sum \limits_{k=0}^{n-1} \int \limits_{k/n}^{(k+1)/n} f(n x - k) g \left(\left\{\frac{n}{x}\right\}\right) \, \mathrm{d} x \overset{n x = k + u}{=} \int \limits_0^1 f(u) \frac{1}{n} \sum \limits_{k=0}^{n-1} g\left(\left\{\frac{n^2}{k + u}\right\}\right) \, \mathrm{d} u \, , \, n \in \mathbb{N} \, . $$
This shows that, by the dominated convergence theorem, the conjecture is true if
$$ \lim_{n \to \infty} \frac{1}{n} \sum \limits_{k=0}^{n-1} g\left(\left\{\frac{n^2}{k + u}\right\}\right) = \int \limits_0^1 g(x) \, \mathrm{d} x $$
holds for almost every $u \in [0,1]$. The left-hand side looks almost like a Riemann sum and the equation is reminiscent of the Riemann integral criterion for equidistributed sequences. However, the situation is slightly different and I do not know how to proceed from here, which is why I want to ask you:

Can we complete this step of the proof? Is there another way to show that $\lim_{n \to \infty} a_n = \int_0^1 f(x) \, \mathrm{d} x \int_0^1 g(x) \, \mathrm{d} x$? Or is this conjecture in fact false in some cases?

 A: The following lemmas will be useful:

Lemma 1. For any bounded interval $I$ and for any $C^1$ function $f : I \to \mathbb{R}$ that has only finitely many critical points,
$$ \lim_{\lambda \to \infty} \int_{I} e^{i\lambda f(x)} \, \mathrm{d}x = 0 $$


Lemma 2. Let $f : \mathbb{R} \to \mathbb{R}$ be $1$-periodic and locally integrable. Then for any $n \in \mathbb{N}$,
$$ \int_{0}^{1} f(nx) \, \mathrm{d}x = \int_{0}^{1} f(x) \, \mathrm{d}x
\qquad\text{and}\qquad
\int_{0}^{1} |f(n/x)| \, \mathrm{d}x \leq \frac{\pi^2}{6} \int_{0}^{1} |f(x)| \, \mathrm{d}x. $$

Let's see how these lemmas apply to the problem. We introduce two bilinear forms on $L^2([0,1])$:
\begin{align*}
B_n(f, g) &= \int_{0}^{1} f(\{nx\})g(\{n/x\}) \, \mathrm{d}x, \\
B_{\infty}(f, g) &= \left( \int_{0}^{1} f(x) \, \mathrm{d}x \right)\left( \int_{0}^{1} g(x) \, \mathrm{d}x \right).
\end{align*}
Then the claim reduces to proving $B_n(f, g) \to B_{\infty}(f, g)$ for each $f, g \in C([0, 1])$. To this end, for each $\varepsilon > 0$, find trigonometric polynomials $p(x) = \sum_{k} a_k e^{2\pi i k x}$ and $q(x) = \sum_k b_k e^{2\pi i k x}$ such that
$$ \|f - p\|_{L^2[0,1]} < \varepsilon \qquad\text{and}\qquad \|g - q\|_{L^2[0,1]} < \varepsilon, $$
where $\| f \|_{L^2[0,1]} := \left( \int_{0}^{1} |f(x)|^2 \, \mathrm{d}x \right)^{1/2} $ is the $L^2$-norm on $[0, 1]$. Then
$$ B_n(p, q)
= \int_{0}^{1} p(nx)q(n/x) \, \mathrm{d}x
= a_0 b_0 + \sum_{(k, l) \neq (0, 0)} a_k b_l \int_{0}^{1} e^{2\pi i n (kx + l/x)} \, \mathrm{d}x. $$
By Lemma 1, we find that all the terms, except for $a_0b_0$, vanish as $n\to\infty$. So it follows that
$$ \lim_{n\to\infty} B_n(p, q)
= a_0 b_0
= B_{\infty}(p, q). \tag{1} $$
To show that the same conclusion holds for $f$ and $g$ in place of $p$ and $q$, respectively, we show that $(B_n)_{n\in\mathbb{N}\cup\{\infty\}}$ is a family of uniformly bounded bilinear forms. Indeed, by the Cauchy–Schwarz inequality and Lemma 2, for $f, g \in L^2([0, 1])$,
\begin{align*}
|B_n(f, g)|
&\leq \left( \int_{0}^{1} |f(\{nx\})|^2 \, \mathrm{d}x \right)^{1/2}\left( \int_{0}^{1} |g(\{n/x\})|^2 \, \mathrm{d}x \right)^{1/2} \\
&\leq \left( \int_{0}^{1} |f(x)|^2 \, \mathrm{d}x \right)^{1/2}\left( \frac{\pi^2}{6} \int_{0}^{1} |g(x)|^2 \, \mathrm{d}x \right)^{1/2}
= \frac{\pi}{\sqrt{6}} \|f\|_{L^2[0,1]} \|g\|_{L^2[0,1]}.
\end{align*}
Moreover, by the Jensen's inequality,
\begin{align*}
|B_{\infty}(f, g)|
\leq \left( \int_{0}^{1} |f(x)|^2 \, \mathrm{d}x \right)^{1/2}\left( \int_{0}^{1} |g(x)|^2 \, \mathrm{d}x \right)^{1/2}
= \|f\|_{L^2[0,1]} \|g\|_{L^2[0,1]}.
\end{align*}
Therefore we have $|B_n(f, g)| \leq C \|f\|_{L^2[0,1]} \|g\|_{L^2[0,1]}$ for all $n \in \mathbb{N}\cup\{\infty\}$ with $C = \frac{\pi}{\sqrt{6}}$. Finally, we find that
\begin{align*}
&|B_n(f, g) - B_{\infty}(f, g)| \\
&\quad\leq |B_n(f - p, g)| + |B_n(p, g - q)| + |B_{\infty}(f - p, g)| + |B_{\infty}(p, g - q)| \\
&\qquad + |B_n(p, q) - B_{\infty}(p, q)| \\
&\quad\leq 2C \bigl( \|f - p\|_{L^2[0,1]}\|g\|_{L^2[0,1]} + \|p\|_{L^2[0,1]}\|g - q\|_{L^2[0,1]} \bigr)
+ |B_n(p, q) - B_{\infty}(p, q)| \\
&\quad\leq 2C \varepsilon \bigl( \|g\|_{L^2[0,1]} + \|f\|_{L^2[0,1]} + \varepsilon \bigr)
+ |B_n(p, q) - B_{\infty}(p, q)|.
\end{align*}
So by letting $\limsup$ as $n\to\infty$ and using $\text{(1)}$,
$$ \limsup_{n\to\infty} |B_n(f, g) - B_{\infty}(f, g)| \leq 2C \varepsilon \bigl( \|g\|_{L^2[0,1]} + \|f\|_{L^2[0,1]} + \varepsilon \bigr). $$
However, since the left-hand side does not depend on $\varepsilon$, letting $\varepsilon \to 0^+$ proves the desired claim.

Proof of Lemma 1. By the assumption, we may write $I = I_1\cup\cdots\cup I_m$ for some non-overlapping intervals $I_1, \ldots, I_m$ such that $f' \neq 0$ in the interior $\mathring{I}_k$ of each $I_k$. Then by writing
$$ \int_{I} e^{i\lambda f(x)} \, \mathrm{d}x = \sum_{k=1}^{m} \int_{I_k} e^{i\lambda f(x)} \, \mathrm{d}x, $$
it suffices to prove that each term in the right-hand side converges to $0$ as $\lambda \to \infty$. However, by substituting $y = f|_{I_k}(x)$, we get
$$ \int_{I_k} e^{i\lambda f(x)} \, \mathrm{d}x
= \int_{f(I_k)} e^{i\lambda y} \left|\bigl(f|_{I_k}^{-1}\bigr)'(y)\right| \, \mathrm{d}y. $$
Since $ \int_{f(I_k)} \left|\bigl(f|_{I_k}^{-1}\bigr)'(y)\right| \, \mathrm{d}y = \int_{I_k} \mathrm{d}x = |I_k| < \infty$, Riemann–Lebesgue lemma tells that the above integral converges to $0$ as $\lambda \to \infty$ as desired. $\square$
Proof of Lemma 2. For the first identity,
$$ \int_{0}^{1} f(nx) \, \mathrm{d}x
= \frac{1}{n} \int_{0}^{n} f(x) \, \mathrm{d}x
= \frac{1}{n} \cdot n\int_{0}^{1} f(x) \, \mathrm{d}x
= \int_{0}^{1} f(x) \, \mathrm{d}x, $$
where the second step follows from the $1$-periodicity of $f$. Then the second inequality follows from
\begin{align*}
\int_{0}^{1} |f(n/x)| \, \mathrm{d}x
= \int_{1}^{\infty} \frac{|f(nx)|}{x^2} \, \mathrm{d}x
\leq \sum_{k=1}^{\infty} \frac{1}{k^2} \int_{k}^{k+1} |f(nx)| \, \mathrm{d}x
= \left( \sum_{k=1}^{\infty} \frac{1}{k^2} \right) \int_{0}^{1} |f(x)| \, \mathrm{d}x
\end{align*}
and $ \sum_{k=1}^{\infty} \frac{1}{k^2} = \zeta(2) = \frac{\pi^2}{6} $. $\square$
