Thread: Not compact embedding

1. Not compact embedding

Hi,

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

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

Please can you help me?

Bye,
Alexander

2. 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 $\displaystyle L^2(\Omega)$.

3. 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 $\displaystyle L^2$.

4. 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

5. 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 $\displaystyle C^0(\overline{\Omega})$.