Let $n\in\mathbb{N},\ T>0$ and $$\Sigma:=\{\sigma\in L^{2}(0,T;\mathbb{R}^{n}):\sum_{i=1}^{n}\sigma_{i}(t)=1,\ \sigma_{i}(t)\geq 0\ \hbox{almost everywhere}\}.$$

Do we then have $$\Sigma\overset{c}{\hookrightarrow} L^{2}(0,T;\mathbb{R}^{n}),$$ that is $\Sigma$ is compact in $L^{2}(0,T;\mathbb{R}^{n})$.

A popular paper concerning this topic is the paper "Compact sets in the space $L^{p}(0,T;B)$" written by Jacques Simon. Therein it is stated that a set is compact if and only if

  • the set $\left\{\int_{t_{1}}^{t_{2}}\sigma(t)dt,\ \sigma\in\Sigma\right\}$ is relatively compact in $\mathbb{R}^{n}\ \forall\ 0<t_{1}<t_{2}<T$,

  • $\sup\limits_{\sigma\in\Sigma}\|\sigma(\cdot+h)-\sigma(\cdot)\|_{L^{2}(0,T;\mathbb{R}^{n})}\rightarrow 0$ for $h\rightarrow 0$.

The first statement can easily be proved, but we fail in proving the second statement. Maybe someone has an idea or has seen such a problem...or maybe the second statement does not hold for $\Sigma$ at all.

We are thankful for every hint or comment and thank You very much in advance



1 Answer 1


The second statement is false for $n \geq 2$ (for $n = 1$ the compactness of the one point set is trivially true).

Without loss of generality take $T = 1$. Let $\sigma^{(k)}(x)$ be the following function: write $x$ is base two, and let $\sigma^{(k)}_1(x)$ be the $k$th bit of the base two expansion of $x$, and $\sigma^{(k)}_2(x) = 1 - \sigma^{(k)}_1(x)$. Take the rest of the $\sigma^{(k)}_i = 0$.

For $h = 2^{-k}$, we have that $\|\sigma^{(k)}(\cdot + h) - \sigma^{(k)}(\cdot)\| = \sqrt{2}$. Contradicting the second statement.

In the spirit of concentration compactness, by putting your domain to be $[0,T]$ you've prevented noncompactness of domain. By putting the integrals to lie in a compact set you ruled out some degree of noncompactness of codomain. (Typically we just require that the integrals are bounded in some sense.) The final criterion is to prevent noncompactness of Fourier domain (shifting up in frequency space). Your definition of $\Sigma$ cannot prevent this last defect.

  • $\begingroup$ Wow, impressive, a very nice counterexample, this completely answers my question. Thank you very much! $\endgroup$
    – Alex
    Commented Mar 13, 2013 at 12:13

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