Convergence a.e. of $\sum_{n=1}^{\infty} \frac{a_n}{\sqrt{\left|x-r_n\right|}}$, where $\sum_{n=1}^{\infty}\left|a_n\right|<\infty$

Question

Let $\left\{a_n\right\}_{n=1}^{\infty}$ and $\left\{r_n\right\}_{n=1}^{\infty}$ two sequences of real numbers, and suppose that: (i) $\sum_{n=1}^{\infty}\left|a_n\right|<\infty$; (ii) $r_i \neq r_j$ for all $i \neq j$. Prove that the series $$ \sum_{n=1}^{\infty} \frac{a_n}{\sqrt{\left|x-r_n\right|}} $$ converges absolutely Lebesgue almost everywhere on $\mathbb{R}$.

I tried to compute, for each $n\geq1$, $$\int_{\mathbb{R}} \frac{a_n}{\sqrt{\lvert x-r_n \rvert}}dx, $$ but this integral is divergent, so I'm out of ideas. Any hints?

Answer 1

It suffices to consider intervals of finite length, say $1$. Let $c\in\mathbb{R}$.

• If $r\in[c,1+c]$, then \begin{align} \int^{c+1}_c\frac{1}{\sqrt{|x-r|}}\,dx&=\int^r_c\frac{dx}{\sqrt{r-x}}+\int^{c+1}_r\frac{dx}{\sqrt{x-r}}\\ &=2\sqrt{r-c}+2\sqrt{1+c-r}<4 \end{align}

• If $r< c$, then \begin{align} \int^{c+1}_c\frac{1}{\sqrt{|x-r|}}\,dx&=\int^{1+c}_c\frac{dx}{\sqrt{x-r}}\\ &=2\big(\sqrt{1+c-r}-\sqrt{c-r}\big)\\ &=2\frac{1}{\sqrt{1+c-r}+\sqrt{c-r}}\leq2 \end{align}

• If $r> c+1$, then \begin{align} \int^{c+1}_c\frac{1}{\sqrt{|x-r|}}\,dx&=\int^{1+c}_c\frac{dx}{\sqrt{r-x}}\\ &=2\big(\sqrt{r-c} - \sqrt{r-1-c}\big)\\ &=2\frac{1}{\sqrt{r-c-1}+\sqrt{r-c}}\leq2 \end{align}

Therefore $$\int^{c+1}_c\sum_n\frac{|a_n|}{\sqrt{|x-r_n|}}\,dx\leq 4\sum_n|a_n|<\infty $$

Thus, for almost all $x\in [c,c+1]$, $\sum_n\frac{a_n}{\sqrt{|x-r_n|}}$ converges absolutely.

From this, the result extends to almost all $x\in\mathbb{R}$.

Answer 2

It suffices to prove a.e. convergence on $[-R,R]$ for any $R > 0$. Break the sum into two parts, depending on whether or not $|r_n| > R + 1$.

For ${\displaystyle \sum_{|r_n| < R + 1} {a_n \over \sqrt{|x - r_n|}}}$, you can use the integration argument you mentioned, over the interval $[-R,R]$.

For ${\displaystyle \sum_{|r_n| > R + 1} {a_n \over \sqrt{|x - r_n|}}}$, you can use that ${\displaystyle \bigg| {a_n \over \sqrt{|x - r_n|}}}\bigg| < |a_n|$ along with the fact that $\sum_n |a_n| < \infty$.

Answer 3

You don't need that integral. Simply notice that $\{r_n\mid n\in\Bbb N\}$ is countable hence a null set, and on its complement, the series is obviously absolutely convergent.