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,
Alex