Is it possible to define a Sobolev space with the function in $L^q$ and its gradient in $L^r$, $r>q$?

Question

Is there any definition of mixed Sobolev space, like $u \in L^p$ such that $\nabla u \in L^r$ with $r>q$? If so, is this a Banach space? as a norm $\max \{ ||u||_p, ||\nabla u ||_r \}$?

Answer 1

It is definitely possible. You can pick any two $p,r\in[1,\infty]$ and define $A^{p,r}$ as the space of (classes of a.e. equal) functions in $L^p$ such that they admit a weak derivative in $L^r$, with norm $\max\{\|u\|_p,\|\nabla u\|_r\}$ is finite. The norm always makes the space Banach for the same reason why standard Sobolev spaces are Banach spaces (the same classical proof works just fine).

If you work in $\mathbb R^n$, it is remarkable the “critical” case $p=r^*$, where $r^*$ is defined as $1/r^*:=1/r-1/n$ (if $r>n$, the exponent $p^*$ is not defined). In this case the Sobolev embedding ensures that you can bound the $L^p$ norm of $u$ with the $L^r$ norm of $\nabla u$, so that the first term in the maximum is redundant up to equivalent norms. In all other cases of $p,r$, the two norms are both needed to define the space (there is no norm stronger than the other). And by varying $p,r$, you always obtain different spaces.

In bounded domains $\Omega\subset \mathbb R^n$, the situation is a little different: the value of $p$ is irrelevant for $p\leq r^*$ (or for $r>n$), and the Banach space created above coincides with the usual Banach space $W^{1,r}$ up to equivalent norms, independently of $p$. This is essentially because a function with gradient in $L^r$ is locally in any $L^p$ with $p\leq p^*$ and you can prove what I said in the previous sentence thanks to the open mapping theorem. For $p>p^*$ (assuming $r<n$), the space strictly changes by varying $p$.