$X_n \to X$ in $L_2$, show that $\lim_{n \to \infty}E[X_n^2]=E[X^2]$ $X,X_1,...$ are random variables, $X_n \to X$ in $L_2$. Show that $\lim_{n \to \infty}E[X_n^2]=E[X^2]$.
My attempt:
$X_n \to X$ in $L_2 \implies \lim_{n \to \infty} E[(X_n-X)^2]=0 \implies \lim_{n \to \infty} E[X_n^2]-2\lim_{n \to \infty} E[X_nX]=-E[X^2]$
Intuitively I see that $-2\lim_{n \to \infty}E[X_nX]=-2E[X^2]$ (and then the theorem is proved) but can not prove it. 
Any ideas?  
 A: Here is an answer that doesn't (explicitly) use normed vector spaces.
Nevertheless, I'd like to mention that, for a given probability space $(\Omega, \mathcal F, P),$ we have the Hilbert space $L^2(\Omega) := \{Y:\Omega\rightarrow\mathbb R\ |\ Y\ \mathcal F$-measurable and $E(Y^2) < \infty \}$ with norm $\|Y\| = \bigl(E(Y^2)\bigr)^\frac12$ and scalar product $\langle Y,Z\rangle = E(YZ).$ Along the way I will make some pedagogical remarks about this object. They are completely irrelevant to the understanding of the answer.
So we are concerned with real-valued random variables on $(\Omega, \mathcal F, P).$ We assume that all random variables are finite almost everywhere, because otherwise arithmetic is complicated. We are given that $\lim_{n\rightarrow\infty}E((X_n - X)^2) = 0$ and want to show that $\lim_{n\rightarrow\infty}E(X_n^2) = E(X^2).$ We require some basic facts.
Fact 1: Suppose that $E(Y^2) < \infty$ and $E(Z^2) < \infty$. Then we have that $E(YZ)$ exists and
$$
|E(YZ)| \leq E|YZ| < \infty.
$$
Proof: We have
$$
0 \leq (Y+Z)^2 = Y^2 + 2YZ + Z^2\qquad and\ thus\qquad -2YZ \leq Y^2 + Z^2
$$
and also
$$
0 \leq (Y-Z)^2 = Y^2 - 2YZ + Z^2\qquad and\ thus\qquad 2YZ \leq Y^2 + Z^2.
$$
This together shows that
$$
|YZ| \leq \frac12(Y^2 + Z^2).
$$
Taking expectations, we get
$$
E|YZ| \leq \frac12(E(Y^2) + E(Z^2)) < \infty.
$$
Now existence of $E(YZ)$ and $|E(YZ)| \leq E|YZ|$ are clear from general facts about Lebesgue integration. End of proof.
Fact 1 shows that the inner product on $L^2(\Omega)$ is well-defined.
Fact 2: Suppose that $E(Y^2) < \infty$ and $E(Z^2) < \infty$. Then we have
$$
E((Y+Z)^2) < \infty.
$$
Proof: According to fact 1, $E(YZ)$ exists and is finite. So we have
$$
E((Y+Z)^2) = E(Y^2 + 2YZ + Z^2) = E(Y^2) + 2 E(YZ) + E(Z^2) < \infty.
$$
End of proof.
Fact 2 basically shows that $L^2(\Omega)$ is a real vector space.
Fact 3: Suppose that $E(Y^2) < \infty$ and $E(Z^2) < \infty$. Then we have
$$
(E(YZ))^2 \leq E(Y^2) E(Z^2).
$$
Proof: According to fact 1, $E(YZ)$ exists and is finite. So we have
$$
\forall\lambda\in\mathbb R: 0 \leq E((Y+\lambda Z)^2) = E(Y^2 + \lambda 2YZ + \lambda^2Z^2) = E(Y^2) + \lambda 2E(YZ) + \lambda^2E(Z^2).
$$
This implies that the quadratic equation
$$
0  = E(Y^2) + \lambda 2E(YZ) + \lambda^2E(Z^2)
$$
in $\lambda$ has no or one real solution, but not two real solutions. Now a glance at the well-known quadratic formula tells us that this in turn implies
$$
0 \geq 4(E(YZ))^2 - 4 E(Y^2)E(Z^2),
$$
i.e. the value under the square root in the quadratic formula must not be positive. But this last inequality is exactly what we want. End of proof.
Fact 3 is tha famous Cauchy-Schwarz inequality in Hilbert spaces.
Now we have all the facts that we need. So let's consider the situation at hand, namely
$$
\lim_{n\rightarrow\infty}E((X_n-X)^2) = 0.
$$
Since we are interested in questions about convergence, we also can assume
$$
\forall n\in\mathbb N:E((X_n-X)^2)<\infty
$$
(discarding finitely many $n$ if necessary).
First we assume
$$
E(X^2) < \infty.
$$
Then we get from fact 2 that
$$
E(X_n^2) = E(((X_n-X)+X)^2) < \infty.
$$
With this, fact 1 shows us that $E(X_nX)$ exists and is finite. Moreover, we get from fact 3 that
$$
0 \leq (E(X_nX) - E(X^2))^2 = (E((X_n-X)X))^2 \leq E((X_n-X)^2)E(X^2).
$$
Here, the r.h.s tends to $0$ as $n$ tends to infinity. Thus, we can conclude
$$
\lim_{n\rightarrow\infty}E(X_nX) = E(X^2).
$$
This together with the fact that
$$
0 = \lim_{n\rightarrow\infty}E((X_n-X)^2) = \lim_{n\rightarrow\infty}E(X_n^2 - 2X_nX+X^2) = \lim_{n\rightarrow\infty}(E(X_n^2) - 2E(X_nX) + E(X^2))
$$
implies
$$
0 = \lim_{n\rightarrow\infty}(E(X_n^2) - E(X^2)),
$$
as desired.
Now let's consider the case that
$$
E(X^2) = \infty.
$$
Then we have to show that $E(X_n^2)$ tends to infinity as $n$ tends to infinity. Assume that's not the case. Then, there are a subsequence $n_k$ and a bound $M < \infty$ such that
$$
\forall k\in\mathbb N: E(X_{n_k}^2) < M.
$$
Since $\lim_{n\rightarrow\infty}E((X_n-X)^2) = 0,$ we can thus find an index $m$ such that
$$
E(X_m^2) < M\qquad and \qquad E((X-X_m)^2) < 1.
$$
Now we conclude from fact 2 that
$$
E(X^2) = E(((X-X_m) + X_m)^2) < \infty,
$$
a contradiction. Thus we really must have
$$
\lim_{n\rightarrow\infty}E(X_n^2) = \infty,
$$
as desired.
A: We can use the fact that, on any normed vector space, the norm is a continuous function with respect to the norm topology. Namely, if $V$ is a normed vector space with norm $\|\cdot\|,$ then we have for any $x,y\in V$ the "inverse" triangle inequalitiy
$$
\bigl| \|x\| - \|y\| \bigr| \leq \|x-y\|.
$$
If you haven't seen this before, it follows "easily" from the ordinary triangle inequality.
Now, if we have $z_n \rightarrow z$ in $V,$ then $\|z_n - z\| \rightarrow 0,$ and thus $\|z_n\| \rightarrow \|z\|$ by the inverse triangle inequality. So, indeed, $\|\cdot\|$ is continuous on $V$ with respect to the norm topology.
Of course, all this applies to your situation. Here, we're considering $L^2$ with norm
$$
\|X\| = \bigl(E(|X|^2)\bigr)^\frac12.
$$
So, if $X_n \rightarrow X$ in $L^2,$ then $\|X_n\| \rightarrow \|X\|$ and thus also
$$
E(|X_n|^2) = \bigl(\|X_n\|\bigr)^2 \rightarrow \bigl(\|X\|\bigr)^2 = E(|X|^2),
$$
as desired.
