Re: Not compact embedding

The sequence $\displaystyle \{f_k\}$ is weakly convergent in $\displaystyle C^0(\overline{\Omega})$ but it's not strongly convergent in $\displaystyle L^2(\Omega)$.

Re: Not compact embedding

An argument without weak topologies would be: $\displaystyle \| f_k\| _{C^0} = 1$, but $\displaystyle \| f_k\|_{L^2} = 2\pi$ (I think, haven't checked this is the correct value, the fact is that this norm is constant) and $\displaystyle f_k$ is orthogonal in $\displaystyle L^2$.

Re: Not compact embedding

Hi,

thank you for your hints. By definition, a sequence $\displaystyle \{x_k\}$ in a banach space $\displaystyle X$ is weakly convergent to an element $\displaystyle x\in X$ , if $\displaystyle f(x_k) \to f(x) $ for every $\displaystyle f \in X^* $, where $\displaystyle X^*$ denotes the dual space of $\displaystyle X$.

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

($\displaystyle f_k(x)=\sin(\pi k x)$)

How to find this $\displaystyle f$?

Now to the other statement:

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

Bye,

Alexander

Re: Not compact embedding

You can show that $\displaystyle \{f_k\}$ converges weakly to $\displaystyle 0$. Then use the characterization of the dual space of $\displaystyle C^0(\overline{\Omega})$.