3
$\begingroup$

Here's a little question I saw in a book recently, which I can see but can't set out my own formal proof and it's annoying me.

Say $\lim_{x\to\infty} g(x) = a$

and $g$ is continuous,

so now prove that

$\lim_{x\to\infty} \frac{1}{x} \int_0^x g(y) \mathrm{d}y = a$ .

I can see that if we define $\int_0^x g(y)\mathrm{d}y = G(x) - G(0)$

then $\frac{1}{x} \int_0^x g(y) \mathrm{d}y = \frac{G(x) - G(0)}{x}$

which clearly looks a lot like the limit of a differential, but I'm not sure how to handle the limit?!

$\endgroup$
  • $\begingroup$ Is $g$ continuous? $\endgroup$ – Davide Giraudo Oct 10 '12 at 9:19
  • $\begingroup$ Yes, we can assume $g$ is continuous. $\endgroup$ – Donald Peters Oct 10 '12 at 9:29
  • $\begingroup$ Can you include it in the OP? Thanks! $\endgroup$ – Davide Giraudo Oct 10 '12 at 9:32
2
$\begingroup$

Perhaps you assume $g$ is continuous, at least some kind of integrability has to be assumed.

Hints:

  1. Since $\lim_{x\to\infty} g(x) =a$ exists, we can for each $\varepsilon>0$ find $x_0$ such that $$|g(x)-a|<\varepsilon$$ for all $x\geq x_0$.

  2. Split the integral $$\int_0^x=\int_0^{x_0}+\int_{x_0}^x$$

  3. What can we say about $$\frac{1}{x}\int_0^{x_0} ?$$

  4. What can we say about $$\frac{1}{x}\int_{x_0}^x ?$$

$\endgroup$
  • $\begingroup$ So for 3, we know that the integral will be greater than $a\epsilon$ and for 4 the opposite? $\endgroup$ – Donald Peters Oct 10 '12 at 9:37
  • $\begingroup$ 3: No, we cannot say that. However, that integral does not depend on $x$, what can we say when $x\to\infty$? $\endgroup$ – AD. Oct 10 '12 at 9:44
  • $\begingroup$ 4: Try to estimate the integral using 1, before letting $x\to\infty$ $\endgroup$ – AD. Oct 10 '12 at 9:46
  • $\begingroup$ So maybe I have it: for each $\epsilon > 0$ I can find $x_0$ s.t.: $a-\epsilon \le g(x) \le a$ for all $x \ge x_0$. So.... $(a-\epsilon)x = \int_0^x (a-\epsilon)\mathrm{d}y \le \int_0^x g(y) \mathrm{d}y \le \int_0^x a \mathrm{d}y$ so: $\frac{(a-\epsilon)x}{x} \le \frac{1}{x}\int_0^x g(y)\mathrm{d}y \le a$ And so the limit is $a$. $\endgroup$ – Donald Peters Oct 10 '12 at 9:46
  • $\begingroup$ Right, (typo $g(x)\leq a+\varepsilon$). $\endgroup$ – AD. Oct 10 '12 at 9:48
2
$\begingroup$

We don't need $g(x)$ to be continous, we only need $g(x)$ to be integrable.

Take any $\varepsilon > 0$. Since $\lim_{x\to +\infty} g(x) = a$, for any $x>M_\varepsilon$ we have $|g(x)-a|<\varepsilon$. Take

$$I_\varepsilon = \int_{0}^{M_\varepsilon}\left|g(x)-a\right|\,dx$$

and

$$n_\varepsilon = \left\lceil\frac{I_\varepsilon}{\epsilon}\right\rceil.$$

For any $x>0$ we have:

$$-a+\frac{1}{x}\int_{0}^{x}g(x)\,dx=\frac{1}{x}\int_{0}^{x}(g(x)-a)\,dx,$$

so, for any $x> \max(n_\varepsilon, M_\varepsilon)$ we have:

$$\left|-a+\frac{1}{x}\int_{0}^{x}g(x)dx\right|\leq \frac{I_\varepsilon}{x}+\frac{1}{x}\int_{M_\varepsilon}^{x}|g(x)-a|\,dx \leq 2\varepsilon.$$

Since $\varepsilon$ can be taken arbitrarily small, the last inequality gives:

$$\lim_{x\to +\infty}\frac{1}{x}\int_{0}^{x}g(x)\,dx = a,$$

QED.

$\endgroup$
  • $\begingroup$ @BillDubuque: honestly? Because I am a bit disgruntled by the amount of trivial questions on MSE these days. It's becoming Yahoo Answers v2. To tell a beginner "you should probably study on your own, before desperately ask for help" is not really polite, I agree, but that does not mean it is the wrong thing to do. You have no idea? Well, make one, then ask. $\endgroup$ – Jack D'Aurizio Nov 1 '18 at 1:05
  • $\begingroup$ That's true, but when the noise/signal ratio raises over the 80% also a general-level site ceases to be useful. $\endgroup$ – Jack D'Aurizio Nov 1 '18 at 1:23
  • $\begingroup$ Judging "useful" will of course be very subjective. It seems to me that many students are learning much from the site (at least in the tags I'm active in). $\endgroup$ – Bill Dubuque Nov 1 '18 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.