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,
Best Answer
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 $+\infty$)