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.