# Sobolev spaces are smooth? Their dual is strictly convex?

Do you know any reference which says something about the:

• Smoothness of the Sobolev space $$W^{1,p}(\Omega)$$ i.e. if the duality mapping $$J\colon W^{1,p}(\Omega)\to W^{1,p}(\Omega)^*$$ is a singleton.

• Is $$W^{1,p}(\Omega)^*$$ strictly convex?

Here $$\Omega\subset\mathbb{R}^N$$ is a bounded domain and $$\infty>p>1$$ is any exponent.

• For nice $\Omega$, Pelczynski and Wojciechowski proved that the Sobolev space is isomorphic to $L^p([0,1])$, see doi.org/10.1016/S0304-0208(01)80041-4 Commented Jul 22 at 7:55

• For a reflexive Banach space $$X$$, $$X$$ is strictly convex [smooth] if and only if $$X^*$$ is smooth [strictly convex]. This is Corollary II.1.4. So it is enough to establish that $$W^{1,p}$$ is smooth.
• Theorem I.3.5 says that $$X$$ is smooth if and only if the norm is G-differentiable on $$X \setminus \{0\}$$.
• In Chapter II.4 the classical result of Clarkson is established, that the $$L^p$$ spaces are uniformly convex, in particular strictly convex, so they are also smooth (go back and forth between $$L^p$$ and its dual). In particular, their norm is G-differentiable except for in $$0$$.
• The norm on $$W^{1,p}$$ is a composition of $$L^p$$ norms and continuous linear operators $$W^{1,p} \to L^p$$ and thus inherits $$G$$-differentiability from the $$L^p$$ ones, so $$W^{1,p}$$ is smooth and its dual is strictly convex.