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I was studying Evans book (Partial Differential Equations) and in page 279 he use the fact that if a sequence $u_{n}\in L^{\infty}(\mathbb{R}^{n})$ is such that $$\|u_{n}\|_{\infty}\leq C$$ $C$ constant, then there exist $u\in L^{\infty}(\mathbb{R}^{n})$ such that a subsequence of $u_{n}$ converges weakly in $L^{2}_{Loc}(\mathbb{R}^{n})$ to $u$.

Now my question is: If $\Omega$ is a bounded domain, then $L^{p}(\Omega)\hookrightarrow L^{q}(\Omega)$, where $1\leq q<p$ and "$\hookrightarrow$" stands for compact immersion?

The answer of the question or any reference is appreciate.

Thanks

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up vote 6 down vote accepted

No.

Take $\Omega = (-\pi, \pi)$, $p=2$, $q=1$, and consider the sequence $f_n(x) = e^{inx}$, which is bounded in $L^2$. This sequence is orthogonal in $L^2$ and so it converges to 0 weakly in $L^2$, i.e. $\int f_n g \to 0$ for every $g \in L^2$. (use Bessel's inequality).

Suppose a subsequence $f_{n_k}$ converges in $L^1$ to some $f$. Let $g = \operatorname{sgn} f \in L^\infty \subset L^2$. Then as argued, $\int f_{n} g \to 0$ so $\int f_{n_k} g \to 0$ as well. But on the other hand $\int f_{n_k} g \to \int fg = \int |f|$, so we must have $f=0$, i.e. $f_{n_k} \to 0$ in $L^1$. This is absurd because $\|f_{n}\|_{L^1} = 2\pi$ for all $n$.

We have produced a bounded sequence in $L^2$ with no $L^1$-convergent subsequence, so the embedding of $L^2$ into $L^1$ is not compact.

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Interesting. This sill valid if we consider only real functions? –  Tomás Oct 5 '12 at 16:11
    
@Tomás: Sure; take $f_n(x)=\cos(nx)$ on the same space. –  Nate Eldredge Oct 5 '12 at 21:27
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