Non compact embedding

Could someone please explain to me, why the embedding $\iota \colon C^0( \overline{\Omega}) \to L^2(\Omega)$ is not compact?

$\Omega=(0,1)$ and $\overline{\Omega}$ denotes the closure of $\Omega$.

I already got the hint to use the sequence $f_k(x):=\sin(k\pi x)$ with $k \in \mathbb N$. Then I can prove, that $(f_k)$ converges weakly to $0$. But how to continue the proof?

• If this injection $i$ was compact, then $i(\{f_k,k\})$ would have a compact closure in $L^2(0,1)$. In particular, we would be able to extract a converging subsequence (for the norm). – Davide Giraudo Nov 30 '11 at 11:40
• To see how to implement Davide's suggestion: I worked out some similar examples in section 2.4 of these lecture notes, it should give you a general thread to follow. (You may especially want to look at Lemma 29.) – Willie Wong Nov 30 '11 at 11:52

I offer this so that your previous work is not wasted:

In order to be compact, your embedding would have to map a bounded sequence to a sequence with a norm convergent subsequence. Since you've shown $\{f_n\}$ converges weakly to $0$, any norm convergent subsequence of $\{f_n\}$ would have to converge in norm to 0 (since norm convergence to $y$ implies weak convergence to $y$). But $\Vert f_n\Vert_{L_2}^2={1\over 2}-{\sin 2\pi n\over 4\pi n}$ for each $n$.

More directly, though, just use Davide's and Willie's suggestions.