Product of two functions that are not in $ L^1(\mathbb R)$

Question

Is it true that if two functions are not $ L^1$ functions, then their product is not a $ L^1$ function? How can we prove it?

Answer 1

I don't think so. Let $\theta(x)=\cases{1 \quad x>0\\0\quad x\leq0}$

we have $\int dx\, \theta(\pm x)=\infty$, but $\theta(x)\theta(-x)=0$ which is clearly $L^1$

Answer 2

It's false. Consider $f(x)=0$ if $x < 1$ and $f(x)=1/x$ if $x\geq 1$. Then $f \not \in L^1(\mathbb{R})$ but $f^2 \in L^1(\mathbb{R})$.

Answer 3

No.

Take $f(x)=\frac{1}{x}1_{[1,+\infty)}$ and $g(x)=\frac{\sin{x}}{x}1_{[1,+\infty)}$

$f,g$ are not in $L^1$ but $fg$ is.