The difference between 'weak limit point' and 'converge weakly'

Question

For the following theorem.

Let $S$ be a nonempty subset of $H$ and let $x:[0,+ \infty) \rightarrow H$. Assume that

$\quad$ (i) for every $z\in S$, $\lim_{t\rightarrow \infty} \left\|x(t)-z\right\|$ exists;

$\quad$ (ii) every weak sequential limit point of $x(t)$, as $t\rightarrow \infty$, belongs to $S$.

Then $x(t)$ converges weakly as $k\rightarrow \infty$ to a point in $S$.

My questions are:

What is the difference between 'weak limit point' and 'converge weakly'?
What is the difference between 'weak limit point' and 'weak sequential limit point'?

The first one is the main question, thanks!

score 0 · Accepted Answer · answered Nov 23 '20 at 11:55

0

A weak sequential limit point is a point $h$ such that $x(t_n) \to h$ weakly for some sequence $t_n \to \infty$. $x(t)$ converges weakly to $h$ if $x(t) \to h$ weakly as $t \to \infty$ which means $F(x(t)) \to F(h) $ as $t \to \infty$ for every continuous linear map $F$.

answered Nov 23 '20 at 11:55

Kavi Rama Murthy

359,332

Thank you for your answer! While I have another question, the weak sequential limit point of $x(t)$ may not be unique, right? – RhX1999 Nov 23 '20 at 12:36
Right, it may not be unique. @RenHaixin – Kavi Rama Murthy Nov 23 '20 at 12:37
I have completely understood. Thank you for your answer! – RhX1999 Nov 23 '20 at 12:39
After thinking about it for a while, I still have a question. Does $x(t_n) \to h$ weakly means $F(x(t_n)) \to F(h)$ as $t_n \to \infty$ for every continuous linear map $F$? – RhX1999 Nov 23 '20 at 12:54
Yes, that is correct. – Kavi Rama Murthy Nov 23 '20 at 13:46
Great, I really understand now, thank you for your patience to answer my questions! – RhX1999 Nov 23 '20 at 14:02

The difference between 'weak limit point' and 'converge weakly'

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