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Let $u \in W^{1,2}(\mathbb{R}^2)$ be a given function satisfying $$\frac{1}{|B_r|}\int_{B_r(x)} |\nabla u|^2 dy \leq \frac{1}{r^{\delta}}$$ for all $r \leq 1$, $x \in \mathbb{R}^2$ and some fixed $\delta \in (0,1)$.

Hueristically, this looks like that the gradient cannot go to infinity too quickly. In this sense, does it rigorously imply that $\nabla u \in L^{2+\epsilon}_{loc}$ for some $\epsilon >0$?

Updated question: Given the very helpful counterexamples and the ideas, I have the following question: Suppose for each $r>0$, there holds $$\int_{B_r(x)} |\nabla u|^2 dy \leq \frac{r}{c}\int_{\partial B_r(x)} |\nabla u|^2 d\sigma$$ for some $c \in (0,2)$ which is a uniform constant, then does this imply some form of higher integrability?

I am still yet to completely understand the hueristics by Willie-Wong, but it appears to me that the new condition seems to be a bit stronger.

Clearly, when $c=2$, it is easy to see that $u \in W^{1,\infty}_{loc}(\mathbb{R}^2)$, so does $c<2$ still retain some form of higher integrability?

Let $u \in W^{1,2}(\mathbb{R}^2)$ be a given function satisfying $$\frac{1}{|B_r|}\int_{B_r(x)} |\nabla u|^2 dy \leq \frac{1}{r^{\delta}}$$ for all $r \leq 1$, $x \in \mathbb{R}^2$ and some fixed $\delta \in (0,1)$.

Hueristically, this looks like that the gradient cannot go to infinity too quickly. In this sense, does it rigorously imply that $\nabla u \in L^{2+\epsilon}_{loc}$ for some $\epsilon >0$?

Let $u \in W^{1,2}(\mathbb{R}^2)$ be a given function satisfying $$\frac{1}{|B_r|}\int_{B_r(x)} |\nabla u|^2 dy \leq \frac{1}{r^{\delta}}$$ for all $r \leq 1$, $x \in \mathbb{R}^2$ and some fixed $\delta \in (0,1)$.

Hueristically, this looks like that the gradient cannot go to infinity too quickly. In this sense, does it rigorously imply that $\nabla u \in L^{2+\epsilon}_{loc}$ for some $\epsilon >0$?

Let $u \in W^{1,2}(\mathbb{R}^2)$ be a given function satisfying $$\frac{1}{|B_r|}\int_{B_r(x)} |\nabla u|^2 dy \leq \frac{1}{r^{\delta}}$$ for all $r \leq 1$, $x \in \mathbb{R}^2$ and some fixed $\delta \in (0,1)$.

Hueristically, this looks like that the gradient cannot go to infinity too quickly. In this sense, does it rigorously imply that $\nabla u \in L^{2+\epsilon}_{loc}$ for some $\epsilon >0$?

Updated question: Given the very helpful counterexamples and the ideas, I have the following question: Suppose for each $r>0$, there holds $$\int_{B_r(x)} |\nabla u|^2 dy \leq \frac{r}{c}\int_{\partial B_r(x)} |\nabla u|^2 d\sigma$$ for some $c \in (0,2)$ which is a uniform constant, then does this imply some form of higher integrability?

I am still yet to completely understand the hueristics by Willie-Wong, but it appears to me that the new condition seems to be a bit stronger.

Clearly, when $c=2$, it is easy to see that $u \in W^{1,\infty}_{loc}(\mathbb{R}^2)$, so does $c<2$ still retain some form of higher integrability?