Do Products of L^2 Functions Converge in the Integral?

[NSAC]

Hi i have a question about $L^2$ spaces and convergence.
Here it goes:
Let $K\subset \mathbb{R}^2$ be bounded.
Let $g,h\in L^2(K)$, and a sequence $f_n\in L^2(K)$ such that $f_n$ converges strongly to $f\in L^2$.
Is it true that $\lim_{n\rightarrow \infty} \int_{K} f_n g h = \int_{K} f g h$? If it is how?
Thank you.

 

[zhangzujin]

$g,h \in L^2$, then $gh\in L^1$ by Holder inequality.

and so I do not know the integral $\int_K fgh$ is well-defined?

 

[NSAC]

and so I do not know the integral \int_K fgh is well-defined?

I didn't get what you mean. Are you asking it as a question?

 

[HallsofIvy]

$g,h \in L^2$, then $gh\in L^1$ by Holder inequality.

and so I do not know the integral $\int_K fgh$ is well-defined?

Yes. $L^2$ consists of functions whose square is integrable. The product of any two such functions is integrable but the product of three of them may not be.
($L^2$ is not closed under multiplication.)