When Integrals Approximate One

In the following problems, let $f$ be a function that is continuous on $\mathbb R$ .

Problem 1. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ if the limit exists.

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Solution. We first claim that $\displaystyle \int_0^x f(t)\, \mathrm dt \to \infty$ . Fix $M > 0$ . By hypothesis, for any $\epsilon > 0$ , there exists $K > 0$ such that $x > K$ implies

$\displaystyle \left|\frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1\right| < \epsilon \quad \Rightarrow \quad \int_0^x f(t)\, \mathrm dt > (1 - \epsilon)x.$

Setting $\epsilon = 1/2$ , choose $N := \max\{K, 2M\}$ so that $x > N$ implies

$\displaystyle \int_0^x f(t)\, \mathrm dt > (1-1/2) \cdot 2M = M.$

Therefore $\displaystyle \int_0^x f(t)\, \mathrm dt \to \infty$ . By L’Hôpital’s rule and the fundamental theorem of calculus,

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\frac{\mathrm d}{\mathrm dx} \int_0^x f(t)\, \mathrm dt}{\frac{\mathrm d}{\mathrm dx} (x)} = \lim_{x \to \infty} \frac{\int_0^x f(t)\, \mathrm dt}{x} = 1.$

But what if we don’t know that the limit exists? The added feature of $f$ being increasing helps.

Problem 2. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ if $f$ is increasing.

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Solution. Fix $\epsilon > 0$ . By the given condition, for any $k > 0$ to be chosen, there exists $K > 0$ such that for $x > K$ ,

$\displaystyle -k \epsilon < \frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1 < k \epsilon.$

We claim that $f(x) \leq 1$ for any $x$ . Suppose $f(N) > 1$ for some $N$ . Define $\epsilon_0 := (f(N)-1)/2$ so that for $x > N$ , $f(x) \geq f(N) > 1 + \epsilon_0$ . Then

$\displaystyle \begin{aligned} \frac{1}{x} \int_0^x f(t)\, \mathrm dt &= \frac{1}{x} \int_0^N f(t)\, \mathrm dt + \frac{1}{x} \int_N^x f(t)\, \mathrm dt \\ &\geq \frac{1}{x} \int_0^N f(t)\, \mathrm dt + (1+\epsilon_0) (1-N/x) \\ &\geq \frac{N}{x} \cdot (f(0) - (1+\epsilon_0)) + (1+\epsilon_0) . \end{aligned}$

Taking $x \to \infty$ yields

$\displaystyle 1 = \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt \geq 0 \cdot (f(0) - (1+\epsilon_0)) + (1+\epsilon_0) > 1,$

a contradiction. Therefore, $f(x) \leq 1$ for any $x$ . On the other hand, apply the mean value theorem and the fundamental theorem of calculus to find $c \in (0, x)$ such that

$\displaystyle f(c) = \frac{1}{x} \int_0^x f(t)\, \mathrm dt > 1 - k\epsilon.$

Setting $k = 1$ and using the fact that $f$ is increasing,

$1 - \epsilon < f(c) \leq f(x) \leq 1 < 1 +\epsilon\quad \Rightarrow \quad |f(x) - 1| < \epsilon.$

It turns out that we can we relax our hypothesis. However, this will require more rudimentary ideas.

Problem 3. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ .

(Click for Solution)

Solution. Fix $\epsilon > 0$ . By the given condition, for any $k > 0$ to be chosen, there exists $K > 0$ such that for $x > K$ ,

$\displaystyle -k \epsilon < \frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1 < k \epsilon.$

By algebra, since $x > K > 0$ ,