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 $n\rightarrow \infty$.

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

1 Answer 1


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.


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