Weak convergence and $\limsup_{n→∞} \|x_n\| → \|x\|$ implies strong convergence. Let ${x_n}$ be a sequence in a Hilbert space $H$. Prove that $\{x_n\}$ converges strongly to
$x$, that is, $\|x_n − x\| → 0$ as $n → ∞$ if and only if

*

*$\{x_n\}$ converges weakly to an element $x ∈ H$; and

*$\limsup_{n→∞} \|x_n\| → \|x\|$.

Proof idea:
First implication (strong->weak) is obviously following from definition. I am not convinced about the second implication.
We know that weak convergence implies $\|x\|\leq \liminf \|x_n\|$.
But $\|x\|\leq \liminf\|x_n\|\leq \limsup \|x_n\|$ (is this always the case???)
and since $\limsup_{n→∞} \|x_n\| → \|x\|$ then $\limsup_{n→∞} \|x_n\| = \liminf_{n→∞} \|x_n\|= \lim_{n→∞} \|x_n\| = \|x\|$ Therefore $\|x_n\| → \|x\|$.
Thanks and Regards,
 A: Yes, from
$$x_n \overset{w}{\to} x$$
it follows
$$\lim \inf_n \|x_n\| \ge \|x\|$$
( if you find the implication
confusing, think of $x_n = e_n$ for $n$ odd, and $0$ for $n$ even).
Anyways, why is this true?  We have
$$\|x\|^2 = (x,x) = |(x,x)| = \lim_n |(x,x_n)|= \lim \inf_n |(x,x_n)| \le \\ \le  \lim \inf_n (\|x\|\cdot \|x_n\| )= \|x\| \cdot \lim \inf \|x_n\|$$
and divide by $\|x\|$ ( if $\|x\|=0$, then it was clear to start with).
This may look like rabbit out of the hat, the idea is that the norm of an element is the supremum of a function
$$\|x\| = \sup_{\|y\|=1} |(x,y)| = \sup_{\|y\|=1} f_x(y)$$
and since
$$f_{x_n}(\cdot) \to f_x(\cdot)$$
pointwise, we have
$$\lim \inf ( \sup f_{x_n} ) \ge \sup f_x$$
(again, if we don't know which way the inequality goes, just think of some bumps functions on $\mathbb{R}$, with the bumps moving away to $+\infty$)
A: We go to prove the "if" part because the "only if" part is trivial.
Claim 1: $\liminf_{n}||x_{n}||\geq||x||$.
Since $x_{n}\rightarrow x$ weakly, we have that $\langle x_{n},x\rangle\rightarrow\langle x,x\rangle$.
On the other hand, $\left|\langle x_{n},x\rangle\right|\leq||x_{n}||\cdot||x||$.
Therefore,
\begin{eqnarray*}
 &  & ||x||^{2}\\
 & = & \lim_{n}\left|\langle x_{n},x\rangle\right|\\
 & \leq & \liminf_{n}\left(||x_{n}||\cdot||x||\right)\\
 & = & ||x||\cdot\liminf_{n}||x_{n}||.
\end{eqnarray*}
If $||x||\neq0$, we divide both sides by $||x||$ and obtain $||x||\leq\liminf_{n}||x_{n}||$.
If $||x||=0$, it is trivial that $\liminf_{n}||x_{n}||\geq||x||$.

It is given that $\limsup_{n}||x_{n}||=||x||$. (Actually, this assumption
can be weaken as $\limsup_{n}||x_{n}||\leq||x||$.), we have that
$\limsup_{n}||x_{n}||\leq||x||\leq\liminf_{n}||x_{n}||$. Hence, $\liminf_{n}||x_{n}||=\limsup_{n}||x_{n}||=x$.
This shows that $\lim_{n}||x_{n}||=||x||$. Finally,
\begin{eqnarray*}
 &  & ||x_{n}-x||^{2}\\
 & = & \langle x_{n}-x,x_{n}-x\rangle\\
 & = & ||x_{n}||^{2}+||x||^{2}-\langle x_{n},x\rangle-\langle x,x_{n}\rangle\\
 & \rightarrow & ||x||^{2}+||x||^{2}-\langle x,x\rangle-\langle x,x\rangle\\
 & = & 0
\end{eqnarray*}
because $x_{n}\rightarrow x$ weakly. It follows that $x_{n}\rightarrow x$
respect to the norm-topology.
