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?!

  • 0
    Can you include it in t$h$e OP? T$h$anks!2012-10-10

2 Answers 2

2

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 $\frac{1}{x}\int_{x_0}^x ?$

  • 0
    Happy to help, just note that $a-\varepsilon\leq g(x)\leq a+\varepsilon$ and not $a-\varepsilon\leq g(x)\leq a....$2012-10-10
2

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.

  • 0
    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).2018-11-01