In unitary space $\|x_n\| \to \|x\|$ and $\langle x_n, x\rangle \to \|x\|^2$ implies $x_n \to x$. 
Prove that in unitary space if $\|x_n\| \to \|x\|$ and $\langle x_n, x\rangle \to \|x\|^2$ implies $x_n \to x$.


Now, so far I have just written
$\|x_n\| \to \|x\| \implies \lim_\limits{n\to \infty} |\|x_n\|-\|x\|| = 0$
and
$\langle x_n, x\rangle \to \|x\|^2 \implies \langle x_n, x\rangle \to \langle x, x\rangle \implies \lim_\limits{n\to \infty}  | \langle x_n, x\rangle - \langle x, x\rangle | = \lim_\limits{n\to \infty} | \langle x_n - x, x\rangle |$
I am however, to show that $\|x_n - x \| \to 0$.
How can I do that?
 A: It can be done easier. Observe that
\begin{equation*}
\begin{split}
\|x_n-x\|^2 & =  \langle x_n-x, x_n-x\rangle = \langle x_n, x_n\rangle - \langle x, x_n\rangle - \langle x_n, x\rangle + \langle x, x\rangle \\
& = \|x_n\|^2 - \langle x, x_n\rangle - \langle x_n, x\rangle + \|x\|^2
\end{split}
\end{equation*}
From the hypothesis,
$$\|x_n\| \to \|x\|, \quad \langle x_n, x\rangle \to \|x\|^2$$
so we conclude that $$\|x_n-x\|^2  \to 0 \Rightarrow x_n \to x. $$
A: $$\|x_n-x_0\|^2 = | \langle x_n-x_0, x_n-x_0\rangle| = | \langle x_n, x_n-x_0\rangle| - \langle x_0, x_n-x_0\rangle| \leq | \langle x_n, x_n-x_0\rangle | + | \langle x_0, x_n-x_0\rangle|$$
Now because $\langle x_n, x_0\rangle \to \|x_0\|^2 \implies \lim_\limits{n\to \infty} | \langle x_n - x_0, x_0\rangle | = 0$ then the second term:
$$
| \langle x_0, x_n-x_0\rangle| = \overline{| \langle x_n-x_0, x_0\rangle|} \to 0
$$
The first term:
$$
| \langle x_n, x_n-x_0\rangle | = \overline{| \langle x_n-x_0, x_n\rangle |} = \overline{| \langle x_n, x_n\rangle -  \langle x_0, x_n\rangle |} = \overline{|\|x_n\|^2 - \overline{\langle x_n, x_0\rangle}|} = \overline{|\|x_n\|^2 - \|x_0\|^2} \to 0
$$
because of $\|x_n\| \to \|x_0\| \implies \|x_n\|^2 \to \|x_0\|^2 =\implies \lim_\limits{n\to \infty} |\|x_n\|^2-\|x_0\|^2| = 0$
Finally
$$
\|x_n-x_0\|^2 \to 0
$$
