# Two sequences in L2 have product converging in L1

I'm studying for an analysis written exam and came across this problem:

Suppose that $\{f_k\}$ and $\{g_k\}$ are two sequences of functions in $L^2([0, 1])$. Suppose that $$\|f_k\|_2 \leq 1\ \text{for all}\ k$$ where the $L^p$ norm $\|\cdot\|_p,\ 1 \leq p \leq \infty$, is defined with respect to the standard Lebesgue measure $\mu$ on $[0, 1]$. Suppose further that there exist $f, g \in L^2([0, 1])$ such that $f_k(x) \to f(x)$ for a.e. $x \in [0, 1]$ and that $g_k \to g$ in $L^2([0, 1])$.

Prove that $f_kg_k \to fg$ in $L^1([0, 1])$.

I understand what the problem is asking and am familiar with all these types of convergence, but I can't figure out where to start. I would greatly appreciate any hints, ideas, or suggestions of what theorems would be useful!

• First, why is it the case that $fg\in L^1[0,1]$? (Hint: Cauchy—Schwarz).
• Then, note that $$\lvert f_kg_k - fg\rvert = \lvert f_kg_k - f_kg + f_kg - fg\rvert \leq \lvert f_kg_k - f_kg \rvert + \lvert f_kg - fg\rvert \tag{1}$$ and then that, by Cauchy—Schwarz, \begin{align} \int_{[0,1]}\lvert f_kg_k - f_kg \rvert &= \int_{[0,1]}\lvert f_k\rvert \lvert g_k - g \rvert \leq \sqrt{ \int_{[0,1]} f_k^2 } \cdot \sqrt{ \int_{[0,1]} (g_k - g)^2} \\ &\leq 1 \cdot \lVert g_k - g\rVert_2\tag{2} \end{align} (can you justify all inequalities?)
Similarly, \begin{align} \int_{[0,1]} \lvert f_kg - fg\rvert&=\int_{[0,1]} \lvert f_k - f\rvert\lvert g\rvert \leq \sqrt{ \int_{[0,1]} g^2 } \cdot \sqrt{ \int_{[0,1]} (f_k - f)^2}\\ &\leq \lVert g\rVert_2 \cdot \lVert f_k - f\rVert_2 \tag{3} \end{align}
Can you now (i) conclude from (1), (2), and (3)? (ii) justify the few parts not entirely spelled out? and importantly, (iii) see where the assumption on the $f_k$'s being uniformly bounded was used, and what may fail in the proof without it?