Hi,

how to prove, that for and the embedding is not compact?

I've got a hint, to use the sequence with . So, I should prove, that the space is not sequentially compact.

Please can you help me?

Bye,

Alexander

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- Nov 20th 2011, 01:44 AMAlexanderWNot compact embedding
Hi,

how to prove, that for and the embedding is not compact?

I've got a hint, to use the sequence with . So, I should prove, that the space is not sequentially compact.

Please can you help me?

Bye,

Alexander - Nov 20th 2011, 04:03 AMgirdavRe: Not compact embedding
The sequence is weakly convergent in but it's not strongly convergent in .

- Nov 20th 2011, 02:13 PMJose27Re: Not compact embedding
An argument without weak topologies would be: , but (I think, haven't checked this is the correct value, the fact is that this norm is constant) and is orthogonal in .

- Nov 21st 2011, 02:32 AMAlexanderWRe: Not compact embedding
Hi,

thank you for your hints. By definition, a sequence in a banach space is weakly convergent to an element , if for every , where denotes the dual space of .

The dual space of is the space of all linear functions . So I have to prove that there exists a , with for all ?

( )

How to find this ?

Now to the other statement:

Do I have to prove that there is no with ?

Bye,

Alexander - Nov 21st 2011, 12:59 PMgirdavRe: Not compact embedding
You can show that converges weakly to . Then use the characterization of the dual space of .