Does $\left|\text E\left[X_n-X\right]\right|\to0$ imply $\left|\text E\left[(X_n-X)Y\right]\right|\to 0$ when $Y$ is bounded? Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(X_n)_{n\in\mathbb N}\subseteq\mathcal L^1(\operatorname P)$ and $X\in\mathcal L^1(\operatorname P)$ with $$\left|\operatorname E\left[X_n-X\right]\right|\xrightarrow{n\to\infty}0\tag1$$ and $Y:\Omega\to\mathbb R$ be bounded and $\mathcal A$-measurable. Are we able to conclude that $$\left|\operatorname E\left[(X_n-X)Y\right]\right|\xrightarrow{n\to\infty}0?\tag2$$
We have $$\left|\operatorname E\left[(X_n-X)Y\right]\right|\le\|Y\|_\infty\operatorname E\left[|X_n-X|\right]\tag3,$$ but this is clearly not enough.
 A: You can use (3) provided it is given that $E|X_n-X| \to 0$. But with the given hypothesis the conclusion fails.
Let $EX=0$ and $ X_n=-X$ for all $n$. Then $E(X_n-X)=0$ for all $n$.  If $Y=X$ then $E(X_n-X)Y=-EX^{2}$. So any bounded (non-zero) random variable with $EX=0$ will serve as  a counter-example.  Specifically, you can take $X$ with uniform distribution on $(-1,1)$.
A: Take $Y=\text{sgn}(X_{n}-X)$ . Then $Y$ is clearly bounded . Where $\text{sgn}$ denotes the signum function defined as below
$\text{sgn}(X)=\begin{cases}1\,, X\geq 0\\  -1\,, X<0 \end{cases}$
Then $(X_{n}-X)Y= (X_{n}-X)\text{sgn}(X_{n}-X)=|X_{n}-X|$
Now take $X_{n}$ to be given by the pmf $P(X_{n}=-1)=\frac{1}{2}$ and $P(X_{n}=1)=\frac{1}{2}$ .
And let $X=0$ .
Then $E[X_{n}-X]=E[X_{n}]=0\,,\forall n\in\Bbb{N}$
And also $E[(X_{n}-X)Y]=E[|X_{n}-X|]=E|X_{n}|=1\,,\forall n\in\Bbb{N}$ .
Note :- If instead you had $L^{1}(P)$ convergence of $X_{n}$ to $X$ , that is $E[\big|X_{n}-X\big|]\to 0$ , then it follows by Holder's Inequality as you have done.
