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!


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

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

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?

  • $\begingroup$ 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
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    $\begingroup$ 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

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