Fix $1<p<N$ and let $(u_n)\subseteq W^{1,p}(\mathbb{R}^N)$ be a sequence satisfying the following:
- $u_n \to 0$ and $\nabla u_n \to 0$ pointwise almost everywhere,
- $u_n \to 0$ in $L_{loc}^p(\mathbb{R}^N)$.
- $\nabla u_n$ is bounded in $L^p(\mathbb{R}^N)$
I am trying to show that for any $\phi \in C_c^\infty(\mathbb{R}^N)$ one has: $$ \int_{\mathbb{R}^N} \lvert \nabla \left( \phi u_n \right)\rvert^p =\int_{\mathbb{R}^N} \lvert \phi \rvert^p\lvert \nabla \left(u_n \right)\rvert^p + o(1). $$ So far, I have only been able to obtain the following \begin{align*} \int_{\mathbb{R}^N} \left\vert \nabla(\phi u_n)\right\vert^p = \left(\left[\int_{\mathbb{R}^N} \lvert \phi \rvert^p\lvert \nabla \left(u_n \right)\rvert^p\right]^{1/p}+ o(1)\right)^p. \end{align*}
My intuition says that I should be able to finish the proof by using some generalization of the binomial theorem. If anyone can either point me in the right direction or provide a source that I can cite showing that this is true I would highly appreciate it!
The function $x \mapsto x^p$ is convex. Thus, $$(a+b)^p = (\frac{2\,a}2 + \frac{2\,b}2)^p \le \frac12 \, (2\,a)^p + \frac12 \, (2\,b)^p = 2^{p-1}\,(a^p + b^p)$$ for all $a,b \ge 0$.
Edit: As noted by Quoka, this does not yield the desired result, but only an inequality involving the constant $2^{p-1}$.
Now, I think that your identity is not enough. Let us denote $a_n := [\int |\phi|^p \, |\nabla u_n|^p ]^{1/p}$. Then, $a_n$ might go to infinity. Let us, for simplicity, consider the case $p = 2$. Then, $$ (a_n + o(1))^2 = a_n^2 + 2 \, a_n \, o(1) + o(1)^2.$$ However, since $a_n$ might be unbounded, the second addend might no longer be $o(1)$. To give a precise example: $$(n + 1/n) ^2 = n^2 + 2 + 1/n^2 \ne n^2 + o(1).$$