IMC 2017 · Day 2 · P6

Case 1: $L$ is finite. Take an arbitrary $\varepsilon > 0$. We construct a number $K \ge 0$ such that $\Bigl| \int_0^1 f(nx)\,dx - L \Bigr| < \varepsilon$.

Since $\lim\limits_{x \to +\infty} f(x) = L$, there exists a $K_1 \ge 0$ such that $\bigl| f(x) - L \bigr| < \frac{\varepsilon}{2}$ for every $x \ge K_1$. Hence, for $n \ge K_1$ we have $$\begin{align*} \Bigl| \int_0^1 f(nx)\,dx - L \Bigr| &= \Bigl| \frac1n \int_0^n f(x)\,dx - L \Bigr| = \frac1n \Bigl| \int_0^n \bigl( f - L \bigr) \Bigr| \le \\ &\le \frac1n \int_0^n |f - L| = \frac1n \left( \int_0^{K_1} |f - L| + \int_{K_1}^n |f - L| \right) < \frac1n \left( \int_0^{K_1} |f - L| + \int_{K_1}^n \frac{\varepsilon}{2} \right) = \\ &= \frac1n \int_0^{K_1} |f - L| + \frac{n - K_1}{n} \cdot \frac{\varepsilon}{2} < \frac1n \int_0^{K_1} |f - L| + \frac{\varepsilon}{2}. \end{align*}$$ If $n \ge K_2 = \frac{2}{\varepsilon} \int_0^{K_1} |f - L|$ then the first term is at most $\frac{\varepsilon}{2}$. Then for $x \ge K := \max(K_1, K_2)$ we have $$\Bigl| \int_0^1 f(nx)\,dx - L \Bigr| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

Case 2: $L = +\infty$. Take an arbitrary real $M$; we need a $K \ge 0$ such that $\int_0^1 f(nx)\,dx > M$ for every $x \ge K$.

Since $\lim\limits_{x \to +\infty} f(x) = \infty$, there exists a $K_1 \ge 0$ such that $f(x) > M + 1$ for every $x \ge K_1$. Hence, for $n \ge 2 K_1$ we have $$\begin{align*} \int_0^1 f(nx)\,dx = \frac1n \int_0^n f(x)\,dx = \frac1n \int_0^n f &= \frac1n \left( \int_0^{K_1} f + \int_{K_1}^n f \right) = \\ &= \frac1n \left( \int_0^{K_1} f + \int_{K_1}^n (M + 1) \right)$$

= \frac1n \left( \int_0^{K_1} f - K_1 (M + 1) \right) + M + 1. \end{align*} If $n \ge K_2 := \Bigl| \int_0^{K_1} f - K_1 (M + 1) \Bigr|$ then the first term is at least $-1$. For $x \ge K := \max(K_1, K_2)$ we have $\int_0^1 f(nx)\,dx > M$.

Case 3: $L = -\infty$. We can repeat the steps in Case 2 for the function $-f$.

Let $F(x) = \int_0^x f$. For $t > 0$ we have $$\int_0^1 f(tx)\,dx = \frac{F(t)}{t}.$$ Since $\lim\limits_{t \to \infty} t = \infty$ in the denominator and $\lim\limits_{t \to \infty} F'(t) = \lim\limits_{t \to \infty} f(t) = L$, L'Hospital's rule proves $\lim\limits_{t \to \infty} \frac{F(t)}{t} = \lim\limits_{t \to \infty} \frac{F'(t)}{1} = \lim\limits_{t \to \infty} \frac{f(t)}{1} = L$. Then it follows that $\lim\limits_{n \to \infty} \frac{F(n)}{n} = L$.