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I want to show that a function that decays slower than $1/x$ is not integrable and I tried it the following way:

Assume the positive, strictly decreasing and continuous function $g(x)$ decays slower than $1/x$, i.e. $$xg(x)\to\infty\,\, (x\to\infty).$$ Then there exists a large $c>0$ such that $g(x)\geq 1/x\,\, \forall x\geq c$. Hence, $$\int\limits_0^\infty g(x) dx \geq \int\limits_c^\infty g(x) dx \geq \int\limits_c^\infty 1/x dx = \lim\limits_{x\to \infty} \ln x - \ln c = \infty.$$

However, I was given a heads up by a friend that $$xg(x)\to\infty\,\, (x\to\infty)$$ is not the negation of the limit being zero, which would correspond to a faster decay than $1/x$. Hence, what I need to show is:

If $\limsup x g(x) > 0$ then $g(x)$ is not integrable. Can somebody give me a hint here?

Tim
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  • I am confused. If you want to show that $g(x)>1/x$ for all sufficiently large $x$ implies the limit diverges, then you have done it. Do you want a stronger result? – almagest Sep 09 '14 at 07:27
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    The fact that $\lim \sup x g(x) > 0$ does not impply thta $g$ is non integrable. Just consider $g = \chi_{\mathbb{Z}}$ – PenasRaul Sep 09 '14 at 07:29
  • Okay, well, what I need in the end is the converse result: If $g$ is integrable, then $xg(x)\to 0$ for $x\to\infty$.

    For that I want to show that a strictly decreasing, continuous function which decays slower than 1/x is not integrable.

    Does the fact that $g(x)$ decays slower than $1/x$ imply that $g(x)>1/x$ for all sufficiently large $x$? Then, we're done here.

     – Tim Sep 09 '14 at 07:36
  • Clearly not as stated. Suppose $g(x)=\frac{1}{2x}$, or what about $g(x)=\frac{1}{x}-\frac{1}{x^2}$ – almagest Sep 09 '14 at 07:56
  • Then, how would you prove the result? – Tim Sep 09 '14 at 08:14

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We show: if $g(x)$ is a positive, decreasing function satisfying $\limsup_{x\to\infty} xg(x) > 0$, then $\int_0^\infty g(x)\,dx$ diverges. (Continuity isn't required, other than as a way to ensure the partial integrals $\int_0^X g(x)\,dx$ are defined.)

By the lim sup hypothesis, there exists a constant $\delta>0$ and an increasing sequence $\{x_n\}$ tending to infinity such that $g(x_n) > \frac\delta{x_n}$. In particular, $g(x) > \frac\delta{x_n}$ for all $x\le x_n$, since $g$ is decreasing. Therefore (setting $x_0=0$ for convenience) \begin{equation*} \int_0^{\infty} g(x)\,dx = \sum_{j=1}^\infty \int_{x_{j-1}}^{x_j} g(x)\,dx > \sum_{j=1}^\infty \int_{x_{j-1}}^{x_j} \frac\delta{x_j} \,dx = \delta \sum_{j=1}^\infty \bigg( 1 - \frac{x_{j-1}}{x_j} \bigg). \end{equation*} If infinitely many of the $\frac{x_{j-1}}{x_j}$ are less than $\frac12$ then the right-hand side clearly diverges. Otherwise, $\frac{x_{j-1}}{x_j} \ge \frac12$ for $j\ge J$. Using the inequality $1-t \ge \frac{-\log t}{2\log2}$ for $t\in[\frac12,1]$, we see that \begin{equation*} \int_0^{\infty} g(x)\,dx > \delta \sum_{j=J}^\infty \bigg( 1 - \frac{x_{j-1}}{x_j} \bigg) \ge \frac\delta{2\log2} \sum_{j=J}^\infty \log\frac{x_j}{x_{j-1}} = +\infty, \end{equation*} since the sum is a telescoping series: $\sum_{j=J}^K \log\frac{x_j}{x_{j-1}} = \log\frac{x_K}{x_{J-1}}$.

Greg Martin
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  • Wow, thanks. However, I still have some questions:
    1. Why do you choose the boundary of $\frac{1}{2}$? What is your particular argument for the first divergence assertion?
    2. Why are the $\frac{x_{j-1}}{x_j}$ increasing? I don't see this.
    3. Where does the inequality for $t\in[\frac{1}{2}$,1]$ come from?

    In general, where does the result come from, do you have a reference for that? I'm probably going to use it in my thesis and a reference would be great. Best, M

     – Tim Sep 09 '14 at 19:45
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    (1) $\frac12$ can be replaced by any constant strictly between $0$ and $1$, and the argument still holds, except that you'd need to change the $2\log2$ in the denominator of the function in (3) to some other constant. As for the divergence assertion: all summands are positive, so what happens if you throw out all the terms except the infinitely many for which the ratio exceeds $\frac12$? – Greg Martin Sep 09 '14 at 22:52
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    (2) I don't think I claimed, or used, the assertion that the $\frac{x_{j-1}}{x_j}$ are increasing. (3) I wanted $1-t\ge-\log t$ for $t\in[0\frac12]$; this isn't true, but clearly it will be true if I divide by a large enough constant; so I did. (4) I made it up myself! (although I seem to remember similar problems here on math.stackexchange, maybe one of them has a reference) – Greg Martin Sep 09 '14 at 22:53
  • (1) Okay, I see. The considered sum is $\geq$ the sum of the infinitely many summands which exceed $1/2$. The latter clearly diverges. (2) You say $$\frac{x_{j-1}}{x_j}\geq \frac{1}{2}, j\geq J.$$ Clearly, in the considered case, there are infinitely many ratios that exceed $1/2$ but how do you know that they all belong to an index greater than $J$? Or is it just $j\in J$ with $J\subset\mathbb{N}$ and then you reorder them? (4) Impressive! Thanks a lot! – Tim Sep 10 '14 at 09:08
  • If there are infinitely many somethings in a sequence, then there are infinitely many somethings in the sequence even if you throw away the first $J$ terms. – Greg Martin Sep 10 '14 at 19:33
  • I agree, but there still might be an index $j \geq J$ where the fraction is smaller than $1/2$ or not? – Tim Sep 10 '14 at 21:41
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    I'm not sure any more that I understand your question. The proof has two cases. The first case is if infinitely many of the $x_{j-1}/x_j$ are less than $\frac12$. The second case is if only finitely many of them are less than $\frac12$ - this is equivalent to all of them past some $J$ being greater than $\frac12$. Either case might hold for a particular sequence, and the proofs are different in the two cases, but either way the divergence is established. – Greg Martin Sep 11 '14 at 05:55
  • I was questioning the mentioned equivalence of finitely many less than a half and ratios past some J greater than a half. But the more I think about, it makes sense to me!

    Thank you for your effort and time!

     – Tim Sep 11 '14 at 07:47