Prove that there is no sequence $(X_n)$ such that $X_n\overset{P}{\rightarrow}0$ and $\mathbb{E}(X_n)\to 2$ and also $\sup_n\mathbb{E}(X_n^2)<\infty.$
Attempt. If we didn't have the last restriction, on $([0,1],\mathcal{B}_{[0,1]},\lambda)$, sequence $X_n:=2n{1}_{[0,1/n]}$ would give $X_n\overset{P}{\rightarrow}0$ and $\mathbb{E}(X_n)= 2$ (whereas $\sup\mathbb{E}(X_n^2)= +\infty$). Back to the nonexistence part, Markov's inequality: $$\mathbb{P}(|X_n|>\epsilon)\leqslant \frac{\mathbb{E}(X_n^2)}{\epsilon^2}\leqslant \frac{\sup_n\mathbb{E}(X_n^2)}{\epsilon^2}$$ seems compelling to use, but I have not reached a contradiction.
Thanks in advance.
Assume that $X_n \to 0$ in probability and $\sup_n \mathbb{E}(X_n^2) < \infty$. Then $\mathbb{E}(X_n) \to 0$.
Indeed, let $\varepsilon >0$ and write $$ \begin{align}|\mathbb{E}(X_n)| &\leq \mathbb{E}\left(|X_n| \mathbf{1}_{|X_n| > \varepsilon}\right) + \mathbb{E}\left(|X_n| \mathbf{1}_{|X_n|\leq \varepsilon}\right)\\ &\leq \sqrt{\mathbb E \left(X_n^2\right)\mathbb P\left(|X_n|>\varepsilon\right)}+\varepsilon \end{align} $$ where in the last inequality I use Cauchy-Schwarz. The assumptions imply that $$ \limsup_n \left|\mathbb E (X_n)\right| \leq \limsup_n \sqrt{\mathbb E \left(X_n^2\right)\mathbb P\left(|X_n|>\varepsilon\right)}+\varepsilon = \varepsilon. $$ Since this is true for every $\varepsilon >0$, letting $\varepsilon \to 0$ gives $\mathbb{E} (X_n) \to 0$.
The above argument is elementary. As suggested in Oliver Diaz's comment, you can also argue that the sequence $(X_n)_n$ is uniformly integrable (as it is bounded in $L^2$) and it converges to $0$ in probability, hence it converges to $0$ in $L^1$.