$f_n\to f$ in $L^2$ and $fg\in L^2(\Omega)\implies f_n\,g\in L^2?$

Question

Let $f,g\in L^2(\Omega),\,$ $f_n\in L^p\,\,\forall 1\leq p<\infty$ such that $f_n\to f$ in $L^2$ and $fg\in L^2(\Omega)$. I was trying to understand if we can derive that $f_n\,g\in L^2?$ My first question was about $p=2$ but now I am wondering what happens in the general case. I would apreciate any idea. Thank you.

Answer 1

$f_n \in L^2$: No, in general this is not true. Consider for example $\Omega = (0,1)$ endowed with the Lebesgue measure and $$f_n(x) := \frac{1}{n} x^{-1/4}.$$ Then $f_n \to f:=0$ in $L^2$ and therefore $f \cdot g=0 \in L^2$ for any $g \in L^2$. On the other hand, if we choose $g(x) := x^{-1/4} \in L^2$, then $f_n \cdot g \notin L^2$ for all $n \in \mathbb{N}$.

---

$f_n \in L^p$ for all $p \in [1,\infty)$: This is also not true. Consider e.g. $\Omega = (1,\infty)$ endowed with the measure

$$\mu(dx) := e^{-x} \, dx.$$

Then $f_n(x) := \frac{1}{n} x^2 \in L^p(\mu)$ for any $p \geq 1$ and $f_n \to f:=0$ in $L^2$. Hence, $f \cdot g=0 \in L^2(\mu)$ for any $g \in L^2(\mu)$. On the other hand, if we set

$$g(x) := e^{x/2} \frac{1}{x},$$

then $g \in L^2(\mu)$, but

$$g(x) f_n(x) = \frac{1}{n} e^{x/2} x \notin L^2(\mu) \qquad \text{for all $n \in \mathbb{N}$}.$$