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

integrationlebesgue-integrallebesgue-measuremeasure-theorysequences-and-series

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?

Best Answer

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 $x\in\mathbb{R}$.

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