# Green's formula and traces in weighted Sobolev spaces

Let $$B_1$$ denote the unit ball in $$\mathbb{R}^d$$, let $$$$\rho(x) = 1-|x| \quad \text{ for } x \in B_1,$$$$ and let $$\sigma >0$$ be given. As per the comments, notice that $$\rho(x)$$ is in fact the distance of $$x \in B_1$$ to the boundary $$\partial B_1$$. For $$k \geq 0$$, consider the following norm $$$$\| u\|_k^2 = \sum_{j=0\\ |\alpha|=j}^k \int_{B_1} \rho^{\sigma+j} |D^\alpha u|^2 dx,$$$$ and we finally let $$\mathcal{H}^k$$ denote the completion of $$C^\infty(\overline{B_1})$$ with respect to the above norm, for $$k \geq 0$$. In particular, it can be shown that $$$$\mathcal{H}^0 = L^2(B_1, \rho^\sigma dx) := L^2_\sigma,$$$$ and also $$$$\mathcal{H}^1 = L^2_\sigma \cap \dot{H}^1_{\sigma+1},$$$$ where $$\dot{H}^1_{\sigma+1}$$ is the homogeneous Sobolev space consisting of all $$f \in L^1_{loc}(B_1)$$ for which the seminorm $$$$\int_{B_1} \rho^{\sigma+1} |\nabla u|^2 dx$$$$ is finite, and similarly $$\mathcal{H}^2 = L^2_\sigma \cap \dot{H}^1_{\sigma+1} \cap \dot{H}^2_{\sigma+2}$$. An analogous characterization holds for the higher order $$\mathcal{H}^k$$ spaces.

Now, using standard integration by parts, it can be seen that

$$$$\int_{B_1} \nabla u \cdot \nabla v \ \rho^{\sigma+1}dx = -\int_{B_1} \nabla \cdot (\rho^{\sigma+1} \nabla u) v dx \quad \forall u, v \in C^\infty(\overline{B_1}),$$$$

since the boundary term $$$$\int_{B_1} \rho^{\sigma+1} \nabla u \cdot \nu \ v \ dS$$$$ vanishes, due to the form of $$\rho^{\sigma+1}$$ given above.

My question is the following:

May one extend the above integration by parts formula to the case where, say, $$u \in \mathcal{H}^2$$ and $$v \in \mathcal{H}^1$$? Moreover, would this mean that $$\rho^{\sigma+1} \nabla u \cdot \nu = 0$$ on $$\partial B_1$$ in some "trace sense", a condition which clearly holds for any $$u \in C^\infty(\overline{B_1})$$?

Many thanks.

• note that the weight is essentially the same as $\delta(x)$ (the Euclidean distance from $x$ to the boundary. ) People look at various spaces with these weights...so googling should find something – Math604 Mar 2 at 5:00
• If my (probably inappropriately superficial) estimates were not wrong, I would suppose you will need something like $|\nabla \rho|^2 \lesssim \rho$ to prove the identity by a density argument.. but you probably already tried that anway. – Hannes Mar 5 at 14:15
• @Hannes This is a very good remark. So it might not be possible to argue by density for $u \in \mathcal{H}^2$, but I think considering the class of functions $u \in \mathcal{H}^1$ with $\nabla \cdot (\rho^{\sigma+1} \nabla u) \in L^2(B_1, \rho^{-\sigma}dx)$ might suffice. – bgsk Mar 5 at 16:23
• Yes, that also would correspond to the way one would define the divergence-gradient operator in $L^2$ with the weight. It will however probably remain difficult to actually characterize that space, I guess. But maybe the abstract domain of definition as you suggested is enough for what you actually want to do. – Hannes Mar 6 at 12:07
• @Hannes I think that for such $u$, the formula does indeed hold. But I don't see what this would mean for the term $\rho^{\sigma+1} \nabla u \cdot \nu$ on $\partial B_1$. – bgsk Mar 6 at 18:18