Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
up vote 6 down vote accepted


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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.