Weak and weak star convergence

Conversation started Nov 20, 2011 at 19:46.

Matt

Nov 20, 2011 19:46

Does f_n converges strongly to f (in l^1) really mean lim_{n \rightarrow \infty} \sum_{k=0}^\infty | f_n(k) - f(k)| = 0?

Gigili

Hello @tb, you around?

Matt

And similarly, does f_n converges weakly to f really mean \lim_{n \rightarrow \infty} \sup _{\| f_n - f\| \leq 1} |\phi(f_n) - \phi(f)| for all \phi \in (l^1)^\ast?

Gigili

I need to know what does this sentence mean: In der Genievorstellung der Romantik waren große Dichter über allem stehende Persönlichkeiten.

t.b.

@Gigili: Loose translation: Using the conception of a genius during romanticism the poets were above everyone else.

Jonas Teuwen

@Matt Phi(f_n) -> phi(f) for all phi in your dual.

Matt

Nov 20, 2011 19:56

@JonasTeuwen: Yes and now I'm trying to figure out what the arrow exactly stands for. Is what I wrote correct?

Jonas Teuwen

@Matt <f_n, phi> -> <f, phi> if that makes things clearer.

t.b.

@Matt no, it means |\phi(f_n) - \phi(f)| -> 0 (note that this is a sequence of real numbers).

Matt

@JonasTeuwen: No that makes things less clear. My phi is already in the dual. If I'm not mistaken you should have a g in the product?

Jonas Teuwen

I'm sorry then :-). Listen to t.b..

t.b.

And this should hold for all \phi from the dual of course.

Matt

Nov 20, 2011 19:59

@JonasTeuwen: actually, I don't know. You tell me: what's your phi there?

Jonas Teuwen

An element from your dual.

Gigili

@tb Thank you, but I don't understand it =\

t.b.

@Gigili: The sentence asserts that poets were considered the greatest geniuses of all.

Matt

@tb: Confusing. The norm in the dual is sup \{|\phi(f)| \mid \| f \| \leq 1 \} but convergence in the dual is just convergence like in R instead of being sup \{|\phi(f) - \phi(f_n)| \mid \| f - f_n\| \leq 1 \}.

I should call it a day, probably.

t.b.

@Matt But we're not talking about convergence with respect to any norm, we're talking about convergence of real (or complex) numbers. If you evaluate a linear functional at a point in the Banach space, you get such a number. Now weak convergence f_n -> f means that for all continuous linear functionals \phi we have |\phi(f_n) - \phi(f)| -> 0

Matt

Nov 20, 2011 20:08

@tb: Thank you.

Gigili

@tb Got it, thank you.

Matt

At least no one can see you blush here when you realise your mistake : ) I like.

t.b.

Given the color of your gravatar it is hard to imagine to see it blushing :)

Matt

: ) I was talking about my real face. I hate when I blush but I can't control it.

Conversation ended Nov 20, 2011 at 20:14.