Square integrable functions with singularities

Question

Is an $L^2(\mathbb{R})$ function allowed to have a singularity? For example, think of the following $$ f(x)=\frac{e^{-x^2}}{x^{1/3}}. $$ In principle, $f$ is square-integrable if we split the interval of integration into the two intervals $(-\infty,0)$ and $(0,\infty)$. Is $f$ in $L^2(\mathbb{R})$?

I understand a function can have a singularity at one of the endpoint of the interval as long as it is square integrable. For example $$g(x)=\frac{1}{x^{1/3}}$$ is in $L^2((0,1))$.

So really, my question is not only restricted to $L^2(\mathbb{R})$ but rather applies to any $L^2(I)$, where $I$ is an open interval of the reals. I am wondering if a square-integrable singularity is in the interior of the interval, then is the function in $L^2(\mathbb{R})$?

Answer 1

Yes, you can have singularities anywhere, even infinitely many of them, as long as $|f|^2$ is Lebesgue integrable on $\mathbb R$ (or $I$ as the case may be). Just look at the definition of $L^2$.

Answer 2

In any interval $(a,b)$ take a point $c$ and consider $f(x)=\frac 1 {|x-c|^{t}}$ where $0<t<1/2$. Then $f$ is square integrable with a singilarity.