Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
For the classical Radon-Nikodym property, the answer is negative. In the book Vector measures by Diestel and Uhl , at the bottom of page 98 there is Theorem 1 which gives a characterisation of spaces with RNP. If the RNP were to hold, then we would have that the dual of $L^p({\bf R}^d; C^1({\rm S}^{d-1}))$ would be precisely $L^{p'}\left({\bf R}^d; (C^1({\rm S}^{d-1}))'\right)$, which is not the case.
@FanZheng: could you please elaborate on your statement that "he trace does live in any Sobolev space below L2". In what sense? I am genuinely interested in this, because some time ago I was looking at some mixed boundary elliptic problems where regularity goes strictly under $3/2$, but the thing that saved the day was that away from the boundary, the solution was in $H^1$.
Apologies for the late reply, but I believe that the author was a bit careless about the notation. I think he actually meant to say that "up to the multiplication with an arbitrary smooth function" where he would have spaces ${\rm L}^1(\Omega)$ and ${\rm W}^{-1,p}(\Omega)$. It is the only explanation I have managed to come up with and if I go through the proof with it, it works.
If it were normed, I could just use the standard Schauder's theorem. ${\rm L}^1_{loc}$ is complete metrizable locally convex space (the topology is generated by a countable family of seminorms). So, to say that a set is bounded it is to say that it is bounded with respect to all continuous seminorms.
