Does the existence of weak derivatives require the lower order derivatives also to exist?

Question

I just want to confirm that for weak derivatives you don't require the lower order derivatives to exist in order for the higher order derivatives to exist?

Answer 1

Remember that if $T$ is a distribution, $\alpha$ a multi-index then, $\partial^\alpha T$ is well defined as a distribution.

If $T$ is a $L^1_{loc}$ function, we say that it has a weak derivative of order $|\alpha|$, just by saying that the distribution $\partial^\alpha T$, can be identified with a function in $L^1_{loc}$. Note however the following theorem

Theorem. Let $L^1_{|\alpha|}(\Omega)$ be the space of distributions on $\Omega$ with derivatives of order $|\alpha|$ in the space $L^1(\Omega)$ then $$L^1_{|\alpha|}(\Omega)\subset L^1_{loc}(\Omega). $$

The proof can be found, for example, in Maz'ya's book, section 1.1.2.. As a corollary, we have that $L_{|\beta|}^1(\Omega)\subset L_{loc}^1(\Omega)$ for all multi-index $\beta$ with $0\le |\beta|\le |\alpha|$.

Answer 2

That is correct. Here are some examples where a higher order weak derivative exists but a lower order weak derivative does not exist:

(p.15 remark 2.19(ii)) http://bolzano.iam.uni-bonn.de/~beck/seltop/topics_pde.pdf

http://math.7starsea.com/post/308

Crostul mentioned that if ALL weak derivatives of order $n$ exist, then lower order weak derivatives exist. Crostul, do you (or anyone else), have a reference for this? If this is true, then for functions of a single variable, higher order weak differentiability would imply lower order weak differentiability.