# Show that $\int_{\mathbb{R}^3}\nabla u\cdot\nabla a\, dx=-\int_{\mathbb{R}^3}u\Delta a\, dx$.

Let $$u\in H^1(\mathbb{R}^3)$$ and $$a\in C^\infty(\mathbb{R}^3)$$ such that $$|\nabla a|,\Delta a \in L^\infty(\mathbb{R}^3)$$, where the norm in the Sobolev space $$H^1(\mathbb{R}^3)$$ is given by $$\|u\|_{H^1}=\|u\|_{L^2}+\|\nabla u\|_{L^2}$$. I would like to show that $$\int_{\mathbb{R}^3}\nabla u\cdot\nabla a\, dx=-\int_{\mathbb{R}^3}u\Delta a\, dx.$$ My thoughts are to use integration by parts and show that the integral over the boundary is zero, which intuitively makes sense because $$|\nabla a |$$ is bounded and $$u$$ decays. In the reference given in this answer the authors states, in particular, that if $$B_r\subset \mathbb{R}^3$$ is an open ball with radius $$r$$ and $$f,g\in C^1(\mathbb{R}^3)$$, then $$\int_{B_r} D_j f(x)g(x)dx=\int_{\partial {B_r}}(fg)(y)\nu_j(y)d_{2}y-\int_{B_r} f(x)D_jg(x)dx,$$ where $$d_{2}y$$ is the euclidean $$2$$-dimensional density on $$\partial B_r$$ and $$\nu (y)$$ the outer normal at $$y\in\partial B_r$$, which in this case i believe is $$y/r$$. Since $$u$$ is not necessarily $$C^1$$, i thought that i would need to approximate $$u$$ by smooth functions and take the limit $$r\rightarrow \infty$$, does this work? Also, i currently don't know how to evaluate the integral $$\int_{\partial {B_r}}(uD_ja)(y)\nu_j(y)d_{2}y.$$

Hints or references will be appreciated.

As $$S(\mathbb{R}^3)\subset L^1(\mathbb{R}^3)$$, taking $$u_n\rightarrow u$$ a.e., $$u_n\in S(\mathbb{R}^3)$$ Schwartz functions, for all $$n\in\mathbb{N}$$, we have $$\left|\int_{\partial {B_r}}(u_nD_ja)(y)\frac{y_j}{r}dy\right|\leq \|{\nabla a}\|_{\infty}\int_{\partial {B_r}}|u_n|dy\stackrel{r\rightarrow\infty}{\longrightarrow} 0.$$ it follows that $$\int_{\mathbb{R}^3} D_j u_n(x)D_ja(x)dx=-\int_{\mathbb{R}^3} u_n(x)D^2_ja(x)dx.$$

but now i don't know what ensures that i can take the limit $$n\rightarrow\infty$$ inside the integrals. The dominated convergence theorem seems to require $$u\in L^1(\mathbb{R}^3)$$, i cannot think of any $$L^1$$ function to bound the sequence.

On the other hand, i could take $$a_n \rightarrow a$$ with $$a_n$$ smooth and compactly supported. As $$u\in H^1(\mathbb{R}^3)$$, integration by parts holds by the definition of weak derivate, but i think i cannot take $$n\rightarrow \infty$$.

• Why is $u$ not $C^1$? Nov 27, 2020 at 20:42
• @WhoKnowsWho, all i know is that $u \in H^1$, so that $u$ and $|\nabla u|$ are defined a.e. and are square integrable.
– Alan
Nov 27, 2020 at 20:52

The idea is indeed to approximate $$u$$ by smooth functions, and to let $$r\to\infty$$. To deal with the integral over $$\partial B_r$$, you do not need to evaluate the integral - it suffices to show it tends to $$0$$ as $$r\to\infty$$.
The basic reason is that you can choose an approximating sequence $$(u_n)$$ to $$u$$ with each $$u_n\in C^\infty\cap H^1$$ such that $$u_n(x)\to 0$$ as $$|x|\to\infty$$. In particular, the space of Schwartz functions $$\mathcal{S}$$ is a dense subspace of $$C^\infty\cap H^1$$ (with respect to the $$H^1$$ norm) and every Schwartz function satisfies this property. You could also work with the space of smooth compactly supported functions if you want - these are dense in $$C^\infty\cap H^1$$.
If you choose an approximating sequence $$(u_n)$$ of functions in $$C^\infty\cap H^1$$ in this way, then you can show that for fixed $$n$$, $$\int_{B_r} u_n(x)D_jg(x)dx\to 0$$ as $$r\to\infty$$, since $$u_n$$ restricted to $$\partial B_r$$ tends to $$0$$; the claim then basically follows from dominated convergence. The proof then boils down to choosing your approximating sequence $$(u_n)$$ and the order in which to take the double limits $$r\to\infty,n\to\infty$$ carefully.