# The limit $\lim_{n \to \infty} \int_0^1 f(\{n x\}) g(\{n/x\}) \, \mathrm{d} x$

fractional-partintegrationlimitsreal-analysisriemann-integration

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?

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 $$\sum_{k=1}^{\infty} \frac{1}{k^2} = \zeta(2) = \frac{\pi^2}{6}$$. $$\square$$