Let $f \in W^{1,1} (\mathbb{R}^N)$.
In the one-dimensional case, we have the following simple proof:
$\int_{\mathbb{R}} \partial_{x} f(x) \text{d}x = [f(x)]^{\infty}_{-\infty} = 0 $
This follows from the fact that, for $f$ to be integrable over all $\mathbb{R}$, it must vanish at infinity, in the sense that, for almost all $x$, $^{\ \ \text{lim}}_{n \rightarrow \infty} f(x \pm n) = 0$.
I would then like to generalise this to the $N-$dimensional case in the following way:
Let $\bar{x}$ denote the vector $x$ without the $i^{\text{th}}$ component. That is, $\bar{x} = [x_1, x_2, ..., x_{i-1}, x_{i+1}, ..., x_N]$.
$\int_{\mathbb{R}} \partial_{x_i} f(x) \text{d}x = \int_{\mathbb{R}^{N-1}} \int_{\mathbb{R}} \partial_{x_i} f(x) \text{d}x_i \text{d}\bar{x} = \int_{\mathbb{R}^{N-1}} [f(x)]^{x_i = \infty}_{x_i = -\infty} \text{d}\bar{x} = \int_{\mathbb{R}^{N-1}} 0 \ \text{d}\bar{x} = 0 $
Could someone please confirm that this proof actually works? And if not, I would appreciate a suggestion for a real proof.
My next question regards the following integral:
Let $\varphi: \mathbb{R} \rightarrow \mathbb{R}$, $\varphi \in C(\mathbb{R}) $, and $u:\mathbb{R}^N \rightarrow \mathbb{R} $, $u \in W^{2,p}(\mathbb{R}^N)$, for all $p \in [1,\infty]$.
Consider $\int_{\mathbb{R}^N} u(x)^2 (\partial_{x_i} u(x)) \varphi(u(x)) \text{d}x$.
We know this integral is finite, as $u \in L^\infty$ implies $u$ bounded, and thus $\varphi(u)$ is also bounded. Then we can take:
$\int_{\mathbb{R}^N} u(x)^2 (\partial_{x_i} u(x)) \varphi(u(x)) \text{d}x \leq ||\varphi(u(x))||_{L^\infty}\int_{\mathbb{R}^N} u(x)^2 (\partial_{x_i} u(x)) \text{d}x$ which is finite as $u \in W^{2,p}(\mathbb{R}^N)$ for all $p$.
We wish to show that $\int_{\mathbb{R}^N} u(x)^2 (\partial_{x_i} u(x)) \varphi(u(x)) \text{d}x = 0 $.
We define $\Phi(x) = \int^{x}_{0}s^2 \varphi(s) \text{d}s$. Then of course $\partial_x \Phi(x) = x^2 \varphi(x)$.
Then we get $ \int_{\mathbb{R}^N} u(x)^2 (\partial_{x_i} u(x)) \varphi(u(x)) \text{d}x = \int_{\mathbb{R}^N} (\partial_{x_i} u(x)) \partial_u \Phi(u(x)) \text{d}x = \int_{\mathbb{R}^N} \partial_{x_i} \Phi(u(x)) \text{d}x = 0$.
The final step equating it to $0$ comes from the equality discussed at the top of the post.
My question here is whether the function $\Phi(u(x))$ is well defined, and whether the derivative $\partial_u \Phi(u(x))$ makes sense. It seems fine, but I'm just lacking confidence as it's a type of concatenation I've never used before. Thank you.