Not compact embedding

• Nov 20th 2011, 01:44 AM
AlexanderW
Not compact embedding
Hi,

how to prove, that for $n=1$ and $\Omega:=(0,1)$ the embedding $C^0(\overline{\Omega}) \to L^2(\Omega)$ is not compact?

I've got a hint, to use the sequence $f_k(x)=\sin(\pi k x)$ with $k \in \mathbb N$. So, I should prove, that the space is not sequentially compact.

Bye,
Alexander
• Nov 20th 2011, 04:03 AM
girdav
Re: Not compact embedding
The sequence $\{f_k\}$ is weakly convergent in $C^0(\overline{\Omega})$ but it's not strongly convergent in $L^2(\Omega)$.
• Nov 20th 2011, 02:13 PM
Jose27
Re: Not compact embedding
An argument without weak topologies would be: $\| f_k\| _{C^0} = 1$, but $\| f_k\|_{L^2} = 2\pi$ (I think, haven't checked this is the correct value, the fact is that this norm is constant) and $f_k$ is orthogonal in $L^2$.
• Nov 21st 2011, 02:32 AM
AlexanderW
Re: Not compact embedding
Hi,
thank you for your hints. By definition, a sequence $\{x_k\}$ in a banach space $X$ is weakly convergent to an element $x\in X$ , if $f(x_k) \to f(x)$ for every $f \in X^*$, where $X^*$ denotes the dual space of $X$.

The dual space of $C^0([0,1])$ is the space of all linear functions $C^0([0,1]) \to \mathbb R$. So I have to prove that there exists a $f$, with $g(f_k)= g(\sin (\pi k))\rightarrow g(f)$ for all $g:C^0([0,1]) \to \mathbb R$ ?
( $f_k(x)=\sin(\pi k x)$)
How to find this $f$?

Now to the other statement:
Do I have to prove that there is no $\widetilde{f}$ with $\left \| \sin(\pi k) - \widetilde{f}\right \|_2 \to 0$ ?

Bye,
Alexander
• Nov 21st 2011, 12:59 PM
girdav
Re: Not compact embedding
You can show that $\{f_k\}$ converges weakly to $0$. Then use the characterization of the dual space of $C^0(\overline{\Omega})$.