$L^1$ convergence

Question

Setting

• For $i \in \mathbb{N}$, consider two sequences $f_i,g_i \in L^1(\mathbb{R})$ such that $$ f_i \rightarrow_{L^1} f \in L^1(\mathbb{R}) $$ and also $$ g_i \rightarrow_{L^1} g \in L^1(\mathbb{R}).$$

• Assume that all the functions involved are (1) striclty positive and that for any $i \in \mathbb{N}$, we have (2) $$ 0 < \int_\mathbb{R} f_i^2 g_i^{-1} < \infty$$ and also (3) $$ 0 < \int_\mathbb{R} f^2 g^{-1} < \infty $$

Question

Do we have the following convergence $$\int_\mathbb{R} f_i^2 g_i^{-1} \rightarrow \int_\mathbb{R} f^2 g^{-1} $$

Answer 1

No, this is not true. The following argument is about spaces of finite measure, but it can easily be adapted to your question about $\mathbb R$.

Let $(X, m)$ be a space with $m(X) < \infty$. Take $g_i = g = 1$, and $f=0$. You are then asking the following:

if $f_i \in L^2 (X) \subseteq L^1 (X)$, and if $f_i \to 0$ in $L^1 (X)$, does it follow that $f_i \to 0$ in $L^2 (X)$?

No, it does not: take $X = [0,1] \subset \mathbb R$, and $f_i = i \; 1_{[0, \frac 1 {i^2}]}$ (with $1_A$ the indicator function of the subset $A$). Then $f_i \to 0$ in $L^1$ but $\| f_i \| _{L^2} = 1 \not\to 0$ in $L^2$.