$g_n$ converges to $0$ in $L1$ implies $fg_n$ converges to $0$ in $L1$

Question

Let $f:X\rightarrow \mathbb{C}$ be an integrable function and $g_n:X\rightarrow \mathbb{C}$ be a sequence of integrable functions so that $\|g_n\|_1\rightarrow 0$ and $|g_n(x)|\leq 1$ for every $n, x$. Show that $\|fg_n\|_1\rightarrow 0$.

I think the $|g_n(x)|\leq 1$ part is only so we can say $g_n \geq g_n^2$, but that's just intuition. There is no obvious inequality (Holder came to my mind...)

Any suggestions?

Answer 1

Let $(g_{n_k})_{k=1}^\infty$ be a subsequence of $(g_n)_{n=1}^\infty$ Since $\|g_n\|_1\to 0$, $(g_{n_k})$ has a further subsequence $(g_{n_{k_j}})_{j=1}^\infty$ such that $g_{n_{k_j}} \to 0$ pointwise a.e. on $X$. Use dominated convergence to show that $\|fg_{n_{k_j}}\|_1 \to 0$. Since $g_{n_k}$ was an arbitrary subsequence of $g_n$, the result follows.

Answer 2

The previous answer is very elegant. A more pedestrian approach is to use: $$ \int |fg_n| \leq \int |f| 1_{|f|>R} + R \int |g_n| $$ And take the limits in the right order.