Showing a series converges absolutely almost everywhere

602 Views Asked by At

Let $f:\mathbb{R}^m\rightarrow\bar{\mathbb{R}}$ be a Lebesgue integrable function with $\int |f|>0$. Show that the infinite series $\sum_n\frac{f(n\vec{x})}{n^p}$ converges absolutely almost everywhere if and only if $p>1-m$.

I'm not sure how to even approach this. I assume that this infinite series must equal some kind of Lebesgue integral involving $f$, but I don't see the connection. I tried to get some kind of contradiction from assuming that the set on which $\sum_n|\frac{f(n\vec{x})}{n^p}|$ diverges has positive measure, but I didn't get anywhere. I also don't know what $p>1-m$ has to do with anything, though that may be clearer once it's converted into an integral somehow.

2

There are 2 best solutions below

2
On BEST ANSWER

'only if' is false. Let $m=1, f=I_{(1,2)}$. Then the series converges absolutely for every $x$ for any $p \in \mathbb R$, even $p<0$, so we may have $p<1-m$.

For the if part you can proceed as follows:

$\sum_n \int |\frac {f(nx)} {n^{p}}|\, dx =\int |f(y)|\, dy \sum \frac 1 {n^{p+m}}$ by the substitution $y=nx$. If $\int |f(y)|\, dy \neq 0$ and $p+m>1$ it follows that $\sum_n |\frac {f(nx)} {n^{p}}| <\infty$ almost everywhere.

Remark: if it is assumed that $p>0$ then the condition $p>1-m$ holds always, so the statement is false. It is essential, therefore, to allow negative values of $p$.

2
On

Observe that $$ \int_{\mathbb{R}^m} \sum_{n=1}^\infty \frac{|f(nx)|}{n^p} dx = \sum\limits_{n=1}^\infty \frac{1}{n^p} \int\limits_{\mathbb{R}^m} |f(nx)| dx = \sum\limits_{n} \frac{1}{n^p} \frac{1}{n^m} \int\limits_{\mathbb{R}^m} |f(x)|dx, $$ where we changed the variables in the second integral by setting $nx = y$. The last sum converges in view of $p+m > 1$, hence the original are series converge almost everywhere.