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?
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Yes, this is correct. Existence of some weak derivatives of order $n$ does not imply that all weak derivatives of lower order exist. However there is a theorem (under the assumption that ALL weak derivatives of order $n$ exist) stating something similar. – Crostul Feb 10 '15 at 08:40
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2Hi, you are right. One reason for this is that the definition of a weak derivative of order $n$ only depends on the original function but not on the derivative of order $n-1$. Although this alone wouldn't imply that it is possible that a certain weak derivative exists without the according lower order derivatives. But you find your statement to be true by simply giving a example where the higher derivative exists but the lower one doesn't. See for example http://www.iadm.uni-stuttgart.de/LstAnaMPhy/Weidl/fa-ws04/Suslina_Sobolevraeume.pdf, Chapter 7, Example 1. (Sry for not elaborating it here) – Murp Feb 10 '15 at 14:32
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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.
warsaw31
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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 $0\le |\beta|\le |\alpha|$.
Tomás
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