Suppose $p\in (1,\infty)$ and $\frac{1}{p}+\frac{1}{p'}=1$. Let $J: W^{-1,p'}\rightarrow W_0^{1,p}$ be the duality map i.e. $$\langle f,J(f)\rangle=\|f\|_{W^{-1,p'}}^2\text{ and } \|J(f)\|_{W_0^{1,p}}=\|f\|_{W^{-1,p'}}$$
Is true that $J(-f)=-J(f)$?
My attempt: By using the characterization above we have that $\langle f, J(-f)+J(f)\rangle=0$ and hence $J(-f)+J(f)$ is in the kernel of $f$... How to proceed?
I suggest to ignore the particulars of the spaces. Let $E$ be a real Banach space with dual $E^*$. Suppose that $J:E\to E^*$ is a duality map; that is, $$\langle Jx,x\rangle_{E^*,E} =\|x\|_E^2 = \|Jx\|_{E^*}^2 \ \text{ for all } \ x\in E$$ The the map $\widetilde J(x)=-J(-x)$ is also a duality map, because $$\langle -J(-x),x\rangle_{E^*,E} =\langle J(-x),-x\rangle_{E^*,E} = \|-x\|_E^2 = \|x\|_{E}^2 = \|Jx\|_{E^*}^2$$ When $E$ is smooth, such as the space in your example, the duality map is unique and it follows that $J(x)=-J(-x)$.