$L^1$ convergence of PDFs vs $L^2$ convergence of CDFs Let $f_n$ denote a sequence of PDFs, and $F_n$ denote the corresponding sequence of CDFs. Given $L^1$ convergence of the PDFs to some PDF $f$,
$$\int_\mathbb{R} |f_n(x) -f(x)| dx \rightarrow 0$$
does this imply $L^2$ convergence of the corresponding CDFs to the corresponding CDF $F$,
$$\int_\mathbb{R} \left(F_n(x) - F(x)\right)^2 dx \rightarrow 0$$
 A: First note that $0\leqslant F(x),F_n(x)\leqslant 1$ for all $n,x$ so $\sup_{n,x}|F_n(x)-F(x)|=2$. It follows that $$\sup_{n,x}(F_n(x)-F(x))^2\leqslant 2|F_n(x)-F(x)|.$$ 
From $L^1$ convergence of $f_n$, continuity of $F_n, F$, and nonnegativity of $f_n,f$, we have for each $t\in\mathbb R$
\begin{align}
|F_n(t)-F(t)| &= \left| \int_{-\infty}^t f_n(x)\ \mathsf dx -\int_{-\infty}^t f(x)\ \mathsf dx\right|\\
&\leqslant \int_{-\infty}^t |f_n(x)-f(x)|\ \mathsf dx\\
&\leqslant \int_{\mathbb R} |f_n(x)-f(x)|\ \mathsf dx\stackrel{n\to\infty}\longrightarrow 0,
\end{align}
and hence $F_n$ converges in distribution to $F$. In fact, as $F$ and $F_n$ are bounded, continuous, and monotone, it follows from Pólya's extension to Dini's theorem (cf. *Problems and Theorems in Analysis, I, p. 270) that $F_n$ converges uniformly to $F$ on any compact subset of $\mathbb R$. Since $F$ is a CDF, we may extend it to a continuous function $\overline F$ on the extended real line $[-\infty,\infty]$ by $\overline F(-\infty)=0$, $\overline F(\infty)=1$ (and similarly for $F_n$, and therefore $F_n$ converges uniformly to $F$ on $[-\infty,\infty]$. Since $F$ and $F_n$ are uniformly continuous, given $\varepsilon>0$ we may choose $N$ so that $n\geqslant N$ implies $$\sup_{x\in\mathbb R}|F_n(x)-F(x)|<\varepsilon. $$ Then $$\int_{\mathbb R} (F_n(x)-F(x))^2\ \mathsf dx\leqslant 2 \int_{\mathbb R}|F_n(x)-F(x)|\ \mathsf dx\stackrel{n\to\infty}\longrightarrow 0$$
so that $F_n$ converges to $F$ in $L^2$.
