Let $f_1:\mathbb{R}^2 \to (0, \infty)$ and $f_2:\mathbb{R}^2 \to (0, \infty)$. Then consider the $L^1$ distance
$$ \Vert f_1 -f_2 \Vert_1=\int_{\mathbb{R}^2}|f_1(x,y)-f_2(x,y)|dxdy. $$
Now, fix any $y\in \mathbb{R}$, and consider the $L^1$ distance between the sections $f_1(\cdot,y)$ and $f_2(\cdot,y)$, i.e.
$$ \Vert f_1 -f_2 \Vert_{1,y}=\int_{\mathbb{R}}|f_1(x,y)-f_2(x,y)|dx. $$
It should hold that
$$ \Vert f_1 -f_2 \Vert_{1}=\int_{\mathbb{R}}\Vert f_1 -f_2 \Vert_{1,y}dy. $$ Consider now sequences of functions $f_{n,1}$ and $f_{n,2}$; under regularity conditions that allow to exchange limit and integral, we should have that $$\sup_{y \in \mathbb{R}}\Vert f_{n,1} -f_{n,2} \Vert_{1,y}\to 0 \quad (n\to \infty)$$
entails that also $\Vert f_{n,1} -f_{n,2} \Vert_{1}\to 0$. My question is the following: could you think of any condition on the sequences $(f_{n,1})$ and $(f_{n,2})$ such that a sort of converse implication holds true: i.e. $\Vert f_{n,1} -f_{n,2} \Vert_{1}\to 0$ entails that $$ \Vert f_{n,1} -f_{n,2} \Vert_{1,y}\to 0, \quad \forall y \in \mathbb{R}, \quad (n \to \infty) $$ or even that $\sup_{y \in \mathbb{R}}\Vert f_{n,1} -f_{n,2} \Vert_{1,y}\to 0 \quad (n\to \infty)$ (for the latter case, we could maybe restrict to the case of functions supported on the unit simplex or on the unit square). I'm particularly interested in the case in which $(f_{n,1})$ and $(f_{n,2})$ are sequences of probability densities.