Convergence of Product of Square Integrable Random Variables Let $Y\in\mathcal{L}^2$ and $\{X_{n}\}_{n=1}^{N}\subset\mathcal{L}^{2}$ be square integrable random variables. Moreover, let


*

*$(Y\cdot X_{n})\in\mathcal{L}^{2}$ for all $n$, and

*$X_{n}\rightarrow 0$ in $\mathcal{L}^{2}$.



Does this imply $(Y\cdot X_{n})\rightarrow 0$ in $\mathcal{L}^{2}$ ?

All I could come up with so far is $(Y\cdot X_{n})\rightarrow 0$ in $\mathcal{L}^{1}$ using Holders inequality:
$$\|\; Y\cdot X_{n} \;\|_{1} \quad\leq\quad \|\; Y \;\|_{2}\; \|\; X_{n} \;\|_{2} \quad.$$
 A: Let $0<\alpha<\frac12$.
Let $Y(x)=x^{-\alpha}$ on $(0,1]$ and whatever makes it $\mathcal L^2$ outside.
Choose some $\delta>0$ and let
$$a_n=n^{-\delta-\frac2{1-2\alpha}}$$
Let $\epsilon=4\alpha/(1-2\alpha)>0$ and let $b_n=a_n+n^{-(2+\epsilon)}$.
Define $X_n(x)=n\,\chi_{[a_n,b_n]}(x)$.
Then $\lVert X_n\rVert_2^2 = n^2\,(b_n-a_n)=n^{-\epsilon}$.
In particular, $X_n \in \mathcal{L}^2$ and $X_n\to 0$ in $\mathcal L^2$.
On the other hand, $Y\cdot X_n(x)= n\, x^{-\alpha}$ on $[a_n,b_n]$, and $0$ elsewhere.
Hence
\begin{align}
\lVert Y\cdot X_n\rVert_2^2
&=n^2\,\int_{a_n}^{b_n}\,\frac1{t^{2\alpha}}\,dt\\
&=\frac{n^2}{1-2\alpha}\,\left[t^{1-2\alpha}\right]^{b_n}_{a_n}\\
&=\frac{n^2}{1-2\alpha}\left(b_n^{1-2\alpha}-a_n^{1-2\alpha}\right)\\
\end{align}
We manipulate the expression above:
$$n^2\,b_n^{1-2\alpha}=\left(n^{\frac2{1-2\alpha}}b_n\right)^{1-2\alpha}=\left(n^{-\delta}+n^{\frac2{1-2\alpha}-(2+\epsilon)}\right)^{1-2\alpha}$$
Now, observe that $\frac2{1-2\alpha}-(2+\epsilon)=0$, and hence
$$n^2\,b_n^{1-2\alpha}=\left(n^{-\delta}+1\right)^{1-2\alpha},$$
which goes to $1$ as $n\to\infty$.
Moreover, similar manipulations show that $n^2\,a_n^{1-2\alpha}=\left(n^{-\delta}\right)^{1-2\alpha}$, which goes to $0$ as $n\to\infty$.
It follows that
$$\lVert Y\cdot X_n\rVert_2^2 \to \frac1{1-2\alpha}$$
as $n\to\infty$, so $Y\cdot X_n$ does not converge to $0$ in $\mathcal L^2$.
