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Define the function $\|\cdot\|$ by $\|x\|=\min_{n\in \mathbb{Z}}|x-n|$. This is of course periodic with period $1$. Now let $f(x)=\|x\|^{-\alpha}$ and $g(x)=\|x\|^{-\beta}$, and assume $\alpha+\beta>1$.

A book that I'm reading claims that $(f*g)(x)\sim \|x\|^{1-\alpha-\beta}$ where $\sim$ is defined as follows around a point $a$:

$$h(x)\sim k(x)\quad \text{ iff }\quad h=O(k)\ \&\ k=O(h).$$

To show this the author says that "as this [convolution] is an even function, it suffices to consider $0\le x\le 1/2$" and writes

$$(f*g)(x)=\int_0^1 f(u)g(x-u)=\int_{I_1}+\int_{I_2}+\int_{I_3}$$

where $I_1=[-x/2, x/2], I_2=[x_2, 3x/2], I_3=[3x/2, 1-x/2]$. Then for $u\in I_1$ we have $|x-u|\sim x$ so $g(x-u)\sim x^{-\beta}$ for such $u$. Hence

$$\int_{I_1}\sim x^{-\beta}\int_0^{x/2}u^{-\alpha}\, du\sim x^{-\beta}\cdot x^{1-\alpha}.$$

If $u\in I_2$, then $u\sim x$, so $f(u)\sim x^{-\alpha}$, and so $$\int_{I_2}\sim x^{-\alpha}\int_0^{x/2}u^{-\beta}\, du\sim x^{-\alpha}\cdot x^{1-\beta}.$$

For $u\in I_3$ we have both $\|u\|\sim x$ and $x-u\sim x$, so

$$\int_{I_3}\sim \int_{x/2}^{1-x/2}u^{-\alpha-\beta} \, du\le \int_{x/2}^{\infty}u^{-\alpha-\beta}\, du\sim x^{1-\alpha-\beta}.$$

  1. I would appreciate it if someone could help with clarifying at least one of the three integrals. (I think I can prove the final conclusion with a different route, although I've not tried writing it down in full detail. However, I'd really like to understand the method of the author too.)

  2. Here is another paragraph from the same source:

We note that $\int_0^1f(x)^p\, dx<\infty$ precisely when $p<1/\alpha$, $\int_0^1 g(x)^q\, dx<\infty$ when $q<1/\beta$, and $\int_0^1(f*g)(x)^r\, dx<\infty$ when $r<1/(\alpha+\beta-1)$. Since $\alpha+\beta-1<\alpha$, we see that $r>p$. Similarly, $r>q$.

Of course, $\alpha+\beta-1<\alpha$ implies $1/(\alpha+\beta-1)>1/\alpha$, but how does this imply $r>p$ ? Am I missing something silly, or the author has made a mistake?

Thanks!

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  • $\begingroup$ About the second question. Is the aim the following? Given the inequalities $\alpha+\beta>1$, $p<\frac{1}{\alpha}$, $r<\frac{1}{\alpha+\beta-1}$ and $\alpha+\beta-1<\alpha$, prove the inequality $p<r$. Because without further information, that cannot be done, e.g. let $\alpha=1$ and $\beta=\frac{1}{2}$, then $\alpha+\beta>1$ and $\alpha+\beta-1<\alpha$ are satisfied, while $p<1$ and $r<2$ does not necessarily lead to $p<r$... $\endgroup$ – Eric S. Dec 4 '15 at 13:12
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  1. The integrals are only breakdown of the whole, and have no errors.

  2. The following section can be easily seen to be incorrect for $\alpha=0.5$, $\beta=0.6$, $r=0.1$ and $p=1$.

We note that $\int_0^1f(x)^p\, dx<\infty$ precisely when $p<1/\alpha$, $\int_0^1 g(x)^q\, dx<\infty$ when $q<1/\beta$, and $\int_0^1(f*g)(x)^r\, dx<\infty$ when $r<1/(\alpha+\beta-1)$. Since $\alpha+\beta-1<\alpha$, we see that $r>p$. Similarly, $r>q$.

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