IMC 1994 · Day 1 · P5

Set $\|g\|_1 = \int\limits_0^b |g(x)|\,dx$ and $$\omega(f,t) = \sup \{ |f(x) - f(y)| : x, y \in [0,b],\ |x - y| \le t \}.$$ In view of the uniform continuity of $f$ we have $\omega(f,t) \to 0$ as $t \to 0$. Using the periodicity of $g$ we get $$\begin{align*} \int_0^b f(x) g(nx)\,dx &= \sum_{k=1}^{n} \int_{b(k-1)/n}^{bk/n} f(x) g(nx)\,dx \\ &= \sum_{k=1}^{n} f(bk/n) \int_{b(k-1)/n}^{bk/n} g(nx)\,dx + \sum_{k=1}^{n} \int_{b(k-1)/n}^{bk/n} \{ f(x) - f(bk/n) \} g(nx)\,dx \\ &= \frac{1}{n} \sum_{k=1}^{n} f(bk/n) \int_0^b g(x)\,dx + O\bigl( \omega(f, b/n) \|g\|_1 \bigr) \\ &= \frac{1}{b} \sum_{k=1}^{n} \int_{b(k-1)/n}^{bk/n} f(x)\,dx \int_0^b g(x)\,dx \\ &\quad + \frac{1}{b} \sum_{k=1}^{n} \left( \frac{b}{n} f(bk/n) - \int_{b(k-1)/n}^{bk/n} f(x)\,dx \right) \int_0^b g(x)\,dx + O\bigl( \omega(f, b/n) \|g\|_1 \bigr) \\ &= \frac{1}{b} \int_0^b f(x)\,dx \int_0^b g(x)\,dx + O\bigl( \omega(f, b/n) \|g\|_1 \bigr). \end{align*}$$ This proves a). For b) we set $b = \pi$, $f(x) = \sin x$, $g(x) = (1 + 3\cos^2 x)^{-1}$. From a) and $$\int_0^{\pi} \sin x\,dx = 2, \qquad \int_0^{\pi} (1 + 3\cos^2 x)^{-1}\,dx = \frac{\pi}{2}$$ we get $$\lim_{n \to \infty} \int_0^{\pi} \frac{\sin x}{1 + 3\cos^2 nx}\,dx = 1.$$