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.)