Prove that $\sum_{k=1}^{\infty} a_k g(x-r_k) \not\in L^2(\mathbb{R})$

Question

Problem. Let $$g(x)=\begin{cases} \frac{1}{\sqrt{x}} &x\in(0,1)\\0 &\text{Otherwise} \end{cases}$$ And let $\{r_k, k\in \mathbb{N}\}$ be enumeration of rational numbers in $\mathbb{R}$ and $\{a_k\}\subset (0,\infty)$ be a sequence such that $$\sum_{k=1}^{\infty}a_k<\infty$$. Accepting the fact that $f(x)=\sum_{k=1}^{\infty} a_k g(x-r_k)\in L^1(\mathbb{R})$, prove that $$ \sum_{k=1}^{\infty} a_k g(x-r_k)\not\in L^2([a,b])\hspace{5mm} \text{for any $a,b$ with } -\infty<a<b<\infty$$

How should I approach this problem?

Answer 1

Hint: Choose $r_k\in (a,b)$. Show that $g(x-r_k)\notin L^2[a,b]$.