Conversation started Nov 30, 2011 at 9:34.
Matt
Matt
Nov 30, 2011 09:34
So if I have x_n -> x weakly implies Tx_n -> Tx weakly, where T : X -> Y is a linear map and X,Y are normed spaces, how can I show that T is continuous?
Asaf Karagila
Asaf Karagila
You can force it to be.
Matt
Matt
: P
Jonas Teuwen
Jonas Teuwen
What does x_n -> x weakly mean?
(to you)
Matt
Matt
For all phi in X^*: phi(x_n) -> phi(x)
Asaf Karagila
Asaf Karagila
It barely gets there... it takes forever!
Matt
Matt
Nov 30, 2011 09:37
I guess I need to do something with phi(Tx_n) -> phi(Tx)...
To show that it's continuous, do I try to show that it's bounded or use the epsilon delta definition?
Jonas Teuwen
Jonas Teuwen
Did you already try any of them? :-).
I would suggest that you just try and see if it works, that will help you more. If you get stuck you can ask for help (unless you're already stuck of course).
Matt
Matt
I thought I was stuck already.
It's homework so I have basically zero energy to try stuff. Must be psychological.
Jonas Teuwen
Jonas Teuwen
Do you know PUB?
Asaf Karagila
Asaf Karagila
The place you go to drink?
Jonas Teuwen
Jonas Teuwen
So then you know that weakly convergent sequences are bounded.
Principle of uniform boundedness.
Matt
Matt
Nov 30, 2011 09:48
@JonasTeuwen I only know Pubs...
Uniform boundedness principle? I tried that already
I don't think it fits here
@JonasTeuwen: Thanks, I'll try again!
Jonas Teuwen
Jonas Teuwen
Assume T is unbounded, then there exists a sequence x_n converging weakly to 0 such that ||T(x_n)|| -> infty.
Asaf Karagila
Asaf Karagila
@Matt: If pubs won't do the work, try pubes.
Daniil
Daniil
kinky
Jonas Teuwen
Jonas Teuwen
(And what do you know about convergent real sequences?)
Matt
Matt
@JonasTeuwen I think your previous comment answers the question. Or am I missing something?
Jonas Teuwen
Jonas Teuwen
Nov 30, 2011 09:54
Well, those are bounded hence Tx_n cannot converge weakly.
Matt
Matt
I think I'm confused.
 
Conversation ended Nov 30, 2011 at 9:56.