Does the embedding $W^{2,1}(\mathbb R^2) \to L^\infty(\mathbb R^2)$ factor through some space that is "slightly better" than $W^{1,2}(\mathbb R^2)$?

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.

• 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.) Apr 12, 2022 at 9:38
• 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. Apr 14, 2022 at 8:59
• I'm not familiar with Bessel potential spaces. What can we do with the embedding $W^{2,1}(\mathbb R^2) \subset H^{2,1}(\mathbb R^2)$? Apr 14, 2022 at 9:58
• 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. :-) Apr 25, 2022 at 8:42
• 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. May 2, 2022 at 19:16

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 $$L^{2,1}(\mathbb R^2)$$.