Let $f(x)$ be a monotonic increasing function on $[0,\infty)$ and for every $x>0$ it is integrable in $[0,x]$, so that $\lim_{x\to \infty}\frac{1}{x}\int_{0}^{x}f(t)dt=a$. I need to prove that $\lim_{x\to \infty}f(x)=a$.

I tried to use the limit definition: $|\frac{1}{x}\int_{0}^{x}f(t)dt -a|<\epsilon$ and to use the fact that $\int_{0}^{x}f(t)dt\geq x\inf f(x)$, but I can't see how does it help me eventually.

I would love your help with this.

Thanks a lot!

  • 3
    $\begingroup$ Hint: $a = \frac{1}{x}\int_0^x a dt$ $\endgroup$ – Thomas Andrews Dec 2 '11 at 21:16

Here's a proof sketch:

  1. If $f$ is unbounded from above, then show that $\frac1x \int_{0}^{\infty} f(x) ~dx \to +\infty$, as $x \to \infty$ (which, of course, contradicts the hypothesis). This shows that $f$ is bounded above.

  2. Define $b := \sup \{ f(x) \ :\ x \geqslant 0 \}$. (This indeed exists and is real, thanks to (1.).) Now, using the hypothesis that $f$ is monotone increasing, show that $f(x) \to b$ as $x \to \infty$.

  3. If $f(x) \to b$ as $x \to \infty$, then show that $\frac1x \int_{0}^{\infty} f(x) ~dx \to b$.

  4. From (3.) and the given hypothesis, since the limit of a function is always unique, conclude that $b = a$. Therefore, from (2.), conclude that $f(x) \to a$ as $x \to \infty$.

Note. It might be possible to simplify the proof by not discussing the unbounded case separately, but I deal with it separately because it helps me focus on the more interesting parts later. If it helps intuition, you could ignore (1.) initially and then come back to it after the rest of the proof.

  • $\begingroup$ How could it be that the function is monotone decreasing and bounded from below? $\endgroup$ – Jozef Dec 2 '11 at 19:40
  • $\begingroup$ @Jozef Oh well, I thought I missed that the function is monotone increasing in my first read, and I edited my answer to make my $f$ monotone increasing as well. // Reg. your question, e.g., $1/x$ is monotone decreasing and bounded below. Similarly, $1 - e^{-x}$ is monotone increasing and bounded above. $\endgroup$ – Srivatsan Dec 2 '11 at 19:43
  • $\begingroup$ Yeah,Thanks, but I can't see what guarantee that in our question $\endgroup$ – Jozef Dec 2 '11 at 19:45
  • $\begingroup$ @Jozef I don't understand you. Which item (1-4) do you have a question about? $\endgroup$ – Srivatsan Dec 2 '11 at 19:45
  • $\begingroup$ I don't understand 1. How come showing this proves that f is bounded from above? $\endgroup$ – Jozef Dec 2 '11 at 19:55

For increasing $f$:

For $b$ fixed, we have $\lim\limits_{x\rightarrow\infty}{1\over x}\int_0^b f =0$. From this and your hypothesis, we have $\lim\limits_{x\rightarrow\infty}{1\over x}\int_b^x f =a$.

Now: $$ \tag{1}{f(b)\over x}(x-b)\le {1\over x}\int_b^x f\le {f(x)\over x}(x-b). $$ Taking the limit as $x$ tends to infinity on both sides of the first inequality in (1) gives $$ f(b)\le a . $$ This is true for all $b$; thus, since $f$ is monotone, $\lim\limits_{x\rightarrow\infty} f(x)$ exists and $$ \tag{2}\lim_{x\rightarrow\infty} f(x)\le a. $$

Taking the limit as $x$ tends to infinity of both sides of the second inequality in (1) now produces $$ a\le \lim_{x\rightarrow\infty}{f(x)}; $$ whence the result follows.

  • $\begingroup$ Why does the limit in (2) is smaller or eqaule to $a$? $\endgroup$ – Jozef Dec 3 '11 at 20:36
  • $\begingroup$ Right above, it's shown $f(b)\le a$ for any $b$. So $\lim\limits_{x\rightarrow\infty }f(x)\le a$ $\endgroup$ – David Mitra Dec 3 '11 at 20:47

A simple interpretation:

Since if $F'(x) = f(x)$, then $$\int\limits_0^x f(t) dt = F(x)-F(0)$$ write the problem as

$$\lim\limits_{x \to \infty} \frac{F(x)-F(0)}{x}=a$$

Since $f$ is integrable in $[0,x)$, $F(0)$ must be finite, so that

$$\lim\limits_{x \to \infty} \frac{F(x)}{x}=a$$

this means that $F(x)$ is approaches $a \cdot x$ asymptotically for $x$ sufficiently large, so that

$$f(x) = F'(x) \to a \text{ for } x \to \infty$$

You might also want to go $\Leftarrow$ and prove that the limit in the problem exists only if $f$ is bounded as $x \to \infty$.


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.