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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,

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  • $\begingroup$ If you are in a Hilbert Space, write $\|x_n-x\|^2 = \|x_n\|^2 - 2\langle x, x_n\rangle + \|x\|^2$ $\endgroup$ Nov 3 '21 at 14:28
  • $\begingroup$ Yes, but I do not have $x_n → x$. I have a weak convergence. $\endgroup$
    – Mihai.Mehe
    Nov 3 '21 at 14:33
  • $\begingroup$ Yes- you can use that fact. Notice the term $-2\langle x , x_n \rangle$ in the above expansion o fhte norm. $\endgroup$ Nov 3 '21 at 14:58
  • $\begingroup$ Thanks @rubikscube09 , but I cannot see clearly how to get the limsup to play a role here. If I $lim... \geq liminf ...$ of this expression $0\leq lim ||x_n-x||^2\leq 2||x||^2-2liminf<x,x_n> = 0$. $\endgroup$
    – Mihai.Mehe
    Nov 3 '21 at 15:08
  • $\begingroup$ Everything seems right, the fact that the $\liminf \leq \limsup$ follows from the definition, and then you are using the squeeze theorem. Then you finish the proof as in math.stackexchange.com/questions/1326803/… $\endgroup$
    – LL 3.14
    Nov 3 '21 at 15:43
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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 $+\infty$)

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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.

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  • $\begingroup$ Clarification: By definition $x_{n}\rightarrow x$ weakly iff $\langle x_{n}-x,y\rangle\rightarrow0$ for all $y\in H$. Given that $x_{n}\rightarrow x$ weakly, we put $y=x$ in above, then $\langle x_{n}-x,x\rangle\rightarrow0$. Hence, $\langle x_{n},x\rangle\rightarrow\langle x,x\rangle$. $\endgroup$ Nov 3 '21 at 20:10

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