Let $X:=W^{1,1}(\mathbb{R}^{2})\cap W^{1,\infty}(\mathbb{R}^{2})$ be given. Is it true that $$ X\overset{c}\hookrightarrow L^{1}(\mathbb{R}^{2}),$$ meaning that $X$ is compactly embedded in $L^{1}(\mathbb{R}^{2})$.

The typically Frechet-Kolmogorov Compactness Theorem cannot be applied here directly since we would need the measure to be bounded (here $\mathbb{R}^{2}$).

There is a result by Adams stating that we might have for $p\in [1,\infty)$ $$ W^{1,p}_{0}(\mathbb{R^{2}})\overset{c}{\hookrightarrow} L^{p}(\mathbb{R}^{2}) $$ and I was hoping that the additional Lipschitz-continuity in $X$ could make the compact embedding hold even for $X$!

Does anyone have an idea where I could find such a result, or -- even better -- if I have missed something obvious and the result is quite easy to prove or to disprove?

Thank you very much and best,


  • 2
    $\begingroup$ Why don't you simply apply Adam's result with $p = 1$ (using $W^{1,1}(\mathbb R^2) = W_0^{1,1}(\mathbb R^2)$)? $\endgroup$ – gerw Feb 11 '17 at 12:31

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