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This is not homework.

I'd love your help proving that if $f$ is an unbounded monotonic increasing function, then $$\lim_{x \to \infty} \frac{1}{x}\int_{0}^{x} \ f(t)\, dt=\infty.$$ I want to use it in a couple of proofs, but I can't prove it by myself.

Thanks a lot.

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If you say "$f$ is an unbounded monotonic increasing function on $[0,\infty)$", then you rule out the special cases that make your statement false. In particular, this guarantees that $f$ is bounded below, and therefore unbounded above. –  TonyK Dec 7 '11 at 19:14

4 Answers 4

up vote 4 down vote accepted

I think the l'Hôpital's rule may be helpful here if your $f$ satisfies the requirements of the rule. And say the antiderviative of $f$ is $F(x)$, then

$$ \lim_{x\to\infty}\frac{1}{x}\int_{0}^{x}f(t)dt = \lim_{x\to\infty}\frac{F(x) - F(0)}{x} = \lim_{x\to\infty}f(x) .$$

Since your tag is calculus, not something like real analysis, I assume your function is a relatively normal function involving Riemann integral.

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You need to assume that the integral $\int_0^x f(t) dt$ is always finite ; it can be possible that it is $-\infty$, because $f$ might go to $-\infty$ when $x \to 0^+$ and this might cause problems. Assume that the integral from $0$ to some point $a$ is finite, so that we don't have any problems. Thus without loss of generality we can assume that $f$ is positive, because at some point $c$, $f(x)$ must be positive for all $x > c$ and we can separate the integral in two parts : the one before $c$ and the one after. Since the integral from $0$ to $c$ will be finite, we have that

Then $$ \frac 1x \int_c^x f(t) dt \ge \frac 1x \int_{x/2}^x f(t) \, dt \ge \frac {(x/2) f(x/2)}{x} = \frac {f(x/2)}2 \to \infty. $$

Hope that helps,

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Hint If $f$ is unbounded and increasing function, prove that $\lim_{x \to \infty} f(x) =\infty$.

Then L'Hôpital solves the problem.

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You named it first. –  dfeuer Dec 5 '11 at 7:41

$\newcommand{\d}{\;\mathrm{d}}$My try.

First of all, note that you have to assume that: $$\forall x>0,\quad \int_0^x f(t)\d t>-\infty\; ,$$ for otherwise the limit is $-\infty$ and the claim is not true.

Moreover, w.l.o.g. you can assume that $f\,$ has finite integral over each interval $[0,x]$ (because otherwise your limit is trivial).

Since $f\,$ is unbounded from above and increasing, when you choose $n\in \mathbb{N}$ you can always find $x_n\geq 0$ such that for all $x\geq x_n$, $$f(x)\geq n\qquad \text{and}\qquad \int_0^x f(t)\d t\geq 0\; ;$$ then for $x\geq x_n$ you get: $$\frac{1}{x}\int_{x_n}^x f(t)\d t \geq \frac{n(x-x_n)}{x}\; .$$ Thus: $$\frac{1}{x}\int_0^x f(t)\d t \geq \frac{1}{x}\int_{x_n}^x f(t)\d t\geq \frac{n(x-x_n)}{x}$$ and: $$\tag{1} \liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t \geq \liminf_{x\to \infty} \frac{n(x-x_n)}{x} =n\; ;$$ inequality (1) proves that $\liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\ \text{d} t$ exceeds $n$ for all $n\in \mathbb{N}$, hence you necessarily have: $$\liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t =+\infty$$ therefore: $$\lim_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t =+\infty$$ as you claimed.


Just a small remark. Under your assumptions on $f$, the function $F(x):=\int_0^x f(t)\ \text{d} t$ is convex. Thus relation: $$\lim_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t=\infty$$ expresses the fact that $F$ is also coercive.

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