Convergence of integral means of the gradient of a Sobolev function

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Let $B_R(x_0)\subset\mathbb{R}^n$ with $R<1$ for $n\geq3$ and suppose $u\in H^1\big(B_R(x_0);\mathbb{R}^N\big)\cap L^{\infty}\big(B_R(x_0)\big)$. Define, \begin{equation} \phi(R)\equiv R^{2-n}\int_{B_R(x_0)}1+|Du|^2\ \mathrm{d}x\quad 0<R<1. \end{equation} I am trying to show that $\lim_{R\rightarrow 0}\phi(R)=0$.

The best I could do is show that $\lim_{R\rightarrow 0}\phi(R)=0$ for a.e. $x_0\in\mathbb{R}^n$. This is as follows

We have \begin{equation} \phi(R)=\alpha(n)R^2\frac{1}{\alpha(n)R^n}\int_{B_R(x_0)}1+|Du|^2\ \mathrm{d}x, \end{equation}where $\alpha(n)$ is volume of the unit ball in $\mathbb{R}^n$. By Lebesgue differentiation theorem we have \begin{equation}\tag{1} \lim_{R\rightarrow 0}\frac{1}{\alpha(n)R^n}\int_{B_R(x_0)}1+|Du|^2\ \mathrm{d}x=1+|Du(x_0)|^2 \end{equation}for almost all $x_0$ in $\mathbb{R}^n$. Clearly, \begin{equation}\tag{2} \lim_{R\rightarrow 0}\alpha(n)R^2=0 \end{equation} Thus putting (1) and (2) together we have \begin{equation} \lim_{R\rightarrow 0}\phi(R)=0 \end{equation}for almost all $x_0$ in $\mathbb{R}^n$.

This expression (or one that is essentially similar to this) comes up in Giaquinta and Giusti's 1978 paper on "Nonlinear Ellpitic Systems with Quadratic Growth" on page 341, where I think they are using the above result to show that \begin{equation*} \chi(x_0, R)<\tau^n \end{equation*}where $\chi$ is a function that involves $\phi$ and $\tau$ is specified in the interval $(0, 1)$.

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No, this is not true for general (bounded) $H^1$ functions.

Short version: all we know about $|Du|^2$ is that it's integrable. So, the integral of $|Du|^2$ over $B_R$ is $o(1)$ as $R\to 0$, but nothing more concrete can be said. Then $\phi(R)= o(R^{2-n})$ which is not strong enough to support the claim.

Concrete example: $u(x)=\sin |x|^{-p}$ where $0<p<1/2$. We have $|Du|\sim |x|^{-p-1}|\cos |x|^{-p}|$, which is square integrable. The integral of $|Du|^2$ over $B_R$ is roughly $R^{3-2(p+1)} = R^{1-2p}$. This does tend to zero, but $\phi(R)$ has the factor $R^{-1}$ in front of the integral, hence $\phi(R)\sim R^{-2p}\to\infty$.

My guess is that the authors use some PDE satisfied by $u$.