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Question

Math 6210 • Real Analysis

Fall 1997 • Question 1

Let \(f(x)=x^{-1/2}\) if \(0<x<1\), \(f(x)=0\) if \(x\le 0\) or \(x\ge 1\), and \(\{r_n\}_{n=1}^{\infty }\) be an enumeration of the rationals. Show that if \[ g(x)=\sum _{n=1}^{\infty }\frac {f(x-r_n)}{2^n} \] then \(g\in L^1(\mathbb R)\) and hence \(g(x)<\infty \) \(m\)-a.e. Show however, that even though \(g^2(x)<\infty \) \(m\)-a.e., that \(g\notin L^2(\mathbb R)\).

Raw LaTeX

Let \(f(x)=x^{-1/2}\) if \(0<x<1\), \(f(x)=0\) if \(x\le 0\) or
\(x\ge 1\), and \(\{r_n\}_{n=1}^{\infty}\) be an enumeration of the rationals. Show
that if
\[
g(x)=\sum_{n=1}^{\infty}\frac{f(x-r_n)}{2^n}
\]
then \(g\in L^1(\mathbb R)\) and hence \(g(x)<\infty\) \(m\)-a.e. Show however, that
even though \(g^2(x)<\infty\) \(m\)-a.e., that \(g\notin L^2(\mathbb R)\).