Examples of product of two $L^1_\text{loc}$ functions that is not $L^1_{\text{loc}}$

Question

Let $f\in L^1_\text{loc}$ and $g\in L^1_\text{loc}$, does $fg \in L^1_\text{loc}$?

My textbook says it isn't in a general case. However if $g\in \mathcal{E} = \mathscr{C}^\infty$, then $fg\in L^1_\text{loc}$.

I can't seem to find a example of two locally integrable functions where the product is not locally integrable. I would need to find some function $g$ which is isn't $\mathscr{C}^\infty$ which would create a singularity? Could someone help?

Answer 1

Take $$f(x) = g(x) = x^{-\alpha} \chi_{(0, 1)(x)}$$ for some $\alpha \in (1/2, 1)$ on $(0, 1)$.

Then $f$ and $g$ are not only locally integrable but also integrable, yet their product is not.

---

If this example is bothersome because of the singularity only being at one point, then a minor adjustment can be made so that there are many more: For example,

$$\sum_{i = 1}^{\infty} 2^{-i} |x - q_i|^{-\alpha}$$

where $\{q_i\}$ is some enumeration of the rationals.

Answer 2

What about $f(x)=g(x)=\frac{1}{\sqrt{x}}$ if $x\neq 0$ and $f(0)=g(0)=0$ ?