Using the fundamental theorem of calculus, we can show that the Sobolev space $W^{2,1}(\mathbb R^2)$ embeds into $L^\infty(\mathbb R^2)$.

If we attempt to prove this by applying Sobolev embedding twice, we run into an issue: we can only get $W^{2,1}(\mathbb R^2) \hookrightarrow W^{1,2}(\mathbb R^2) \hookrightarrow BMO(\mathbb R^2)$.

My question: Is there a function space $X$ that is "slightly better" than $W^{1,2}(\mathbb R^2)$ such that we have the embeddings $W^{2,1}(\mathbb R^2) \hookrightarrow X \hookrightarrow L^\infty(\mathbb R^2)$?

My first thought was to use Besov spaces $B^{s,p}_{q}(\mathbb R^2)$, since we do have $B^{2,1}_{1}(\mathbb R^2) \hookrightarrow B^{1,2}_{1}(\mathbb R^2) \hookrightarrow L^\infty(\mathbb R^2)$. However, $B^{2,1}_{1}(\mathbb R^2) \neq W^{2,1}(\mathbb R^2)$ (since $2$ is an integer), so this does not work.

  • 1
    $\begingroup$ Any interpolation space $X$ between $W^{2,1}(\mathbb{R}^2)$ and $L^\infty(\mathbb{R}^2)$ would surely do the job, but identifying these with a (more or less) known precise function space might be nontrivial due to the limit cases of integrability, I suppose. (The Besov space idea goes in that direction of course.) $\endgroup$
    – Hannes
    Apr 12, 2022 at 9:38
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    $\begingroup$ I just stumbled over this answer by Giovanni Leoni which says that $W^{2,1}(\mathbb{R}^2) \subset H^{2,1}(\mathbb{R}^2)$, so things seem rather spicy. $\endgroup$
    – Hannes
    Apr 14, 2022 at 8:59
  • $\begingroup$ I'm not familiar with Bessel potential spaces. What can we do with the embedding $W^{2,1}(\mathbb R^2) \subset H^{2,1}(\mathbb R^2)$? $\endgroup$
    – Alan C
    Apr 14, 2022 at 9:58
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    $\begingroup$ Sorry for the late reply; unfortunately we cannot do very much. I only wanted to indicate that the fact that the spaces mentioned do not coincide will in only complicate things further. :-) $\endgroup$
    – Hannes
    Apr 25, 2022 at 8:42
  • $\begingroup$ Perhaps not what you want, but one can take $X$ to be the space of bounded continuous functions on ${\mathbb R}^2$. Or the functions that are absolutely continuous in say the $y$ variable, uniformly in the $x$ variable. $\endgroup$
    – Terry Tao
    May 2, 2022 at 19:16

1 Answer 1


I learned from Lenka Slavíková that we have the embeddings $$W^{2,1}(\mathbb R^2) \hookrightarrow V^1 L^{2,1}(\mathbb R^2) \hookrightarrow L^\infty(\mathbb R^2),$$ where $V^1 L^{2,1}(\mathbb R^2)$ is the space of functions $u$ such that both $u$ and its derivative $Du$ belong to the Lorentz space $L^{2,1}(\mathbb R^2)$.

See, for example, equation (2.10) of Higher-order Sobolev embeddings and isoperimetric inequalities, by Andrea Cianchi, Luboš Pick, Lenka Slavíková. https://arxiv.org/abs/1311.0153


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