4
$\begingroup$

So what I gathered from the givens about $f$, since $(\log\circ f)^\prime(x)=\frac{f^\prime(x)}{f(x)}$ it would mean that far enough, $f^\prime(x)<0$. I don't know how to go about this from here.


Another question in the same kind of area that I'm currently struggling with fruitlessly, assuming that $f$ is still positive and differentiable in $[0,\infty)$ - now the givens are different:

Assume that $\int_0^\infty f$ exists, and that $f^\prime$ is bounded. Show that $\lim \limits_{x\to\infty} f(x)=0$.

About this question I'm thinking about Lagrange's theorem, and using Cauchy's criterion for the convergence of improper integrals to show (somehow) that $f(x)$ can be made arbitrarily small far enough. That didn't get me very far, and frustration quickly ensued.

I appreciate any thoughts and hints, I feel like I'm missing something rather obvious..

Thank you!

$\endgroup$
5
$\begingroup$

Since $\displaystyle{\lim_{x\to\infty} (\log\circ f)^\prime(x) < 0}$, there is some $N\in[0,\infty)$ and some $c>0$ such that $(\log\circ f)^\prime(x)<-c$ for all $x\geq N$.

Let $x>N$, then by the mean value theorem there exists a $y\in[N,x]$ such that $$(\log\circ f)(x) = (\log\circ f)(N)+(x-N)(\log\circ f)^\prime(y)< (\log\circ f)(N)+(x-N)(-c).$$ Hence $f(x)<f(N)e^{-cx+cN}$ for all $x\geq N$.

Now $${\int_N^\infty} f(x)\,\mathrm{d}x < \int_N^\infty f(N)e^{-cx+cN}\,\mathrm{d}x = \left[f(N)\left(-\frac{1}{c}\right)e^{-cx+cN}\right]_N^\infty = \frac{f(N)}{c}.$$

Hence $\displaystyle{\int}_0^\infty f(x)\,\mathrm{d}x < \int_0^N f(x)\,\mathrm{d}x + \frac{f(N)}{c}<\infty$.

$\endgroup$
  • 1
    $\begingroup$ Shit+enter more. $\endgroup$ – Git Gud Apr 15 '13 at 21:32
  • 1
    $\begingroup$ @GitGud I'm sorry, what do you mean? $\endgroup$ – Abel Apr 15 '13 at 21:33
  • 1
    $\begingroup$ Oh, you meant to write Shift+enter I guess? I was a bit confused... $\endgroup$ – Abel Apr 15 '13 at 21:34
  • 3
    $\begingroup$ ahahah, yes. Didn't even notice it. I'm laughing so hard. $\endgroup$ – Git Gud Apr 15 '13 at 21:34
  • 1
    $\begingroup$ Do you know about the \displaystyle command? $\endgroup$ – Git Gud Apr 15 '13 at 21:39
1
$\begingroup$

The idea here is that we have a differential inequality of the form $f'(x) \leq Cf(x)$. What this says is that $f$ is growing more slowly than an exponential function, and so is integrable.

Some details:

Choose $N$ and $\epsilon_0 < 0$ so $\frac{f'(x)}{f(x)} \leq \epsilon_0$ for all $x \geq N$. Now consider the function $g(x) = f(x)e^{-\epsilon_0 x}$. Differentiate $g$ to find $g'(x) = \epsilon_0 f(x) e^{-ax} + f'(x) e^{-ax} \leq 0$ for $x$ large. So $g(x)$ decreases for $x$ large, and so we have a constant $C$ so that $g(x) \leq C$ for $x$ large. Hence $f(x) \leq Ce^{\epsilon_0 x}$. Since $\epsilon_0 < 0$, we have $f$ integrable.

======================================

For the second question, here's an idea that leads to a solution that feels inelegant, but works. If the function does not converge to zero, there is some $\epsilon > 0$ and a sequence $x_n$ numbers going to infinity (sufficiently spaced out... think about why this matters) so that $f(x_n) \geq \epsilon$. What does this condition and the derivative bound say about the area under the graph?

$\endgroup$
  • $\begingroup$ Thanks for the info! It's a cool way to look at it. *editting.. $\endgroup$ – Adar Hefer Apr 20 '13 at 19:12
  • $\begingroup$ As for the second question, I think I figured it out. The argument seems a little wasteful but here goes: Taking $x_n$ s.t $f(x_n)>\epsilon$, and another sequence $b_n$ s.t $f(b_n)=\epsilon/2$ ($b_n$ is chosen to be the first x that reaches $\epsilon/2$ which has to exist, otherwise the integral won't converge). Then I make another sequence $a_n=\epsilon$ that lies in $(x_n,b_n)$ (intermediate value). I propose that $|a_n-b_n|\to 0$ as $n\to\infty$. (otherwise I can show the integral won't converge). $\endgroup$ – Adar Hefer Apr 20 '13 at 19:20
  • $\begingroup$ And now I can show $f^\prime$ is unbounded using Lagrange's theorem, since I have two points that have different values on $f$, but that become arbitrarily close to one another. $\endgroup$ – Adar Hefer Apr 20 '13 at 19:22
  • $\begingroup$ This looks fine, given you show that $|a_n - b_n| \rightarrow 0$ follows from integrability properly. :) $\endgroup$ – Zach L. Apr 20 '13 at 19:24
  • 1
    $\begingroup$ This looks good. If you're writing this up to hand in for a class, there's something you should be careful about: make sure that you select the sequences recursively. You want to choose $x_n$, then $b_n$, then $a_n$, then $x_{n+1}$, etc. Otherwise, you can't rule out things like $x_n < x_{n+1} < a_n < b_n$. $\endgroup$ – Zach L. Apr 20 '13 at 19:56

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.