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 $\epsilon >0$?



Higher integrability for Sobolev functions
Since the [OP asked for a discussion of features](https://mathoverflow.net/questions/440271/higher-integrability-for-sobolev-functions#comment1135653_440271), I provide one by way of an explanation of Christian Remling's counterexample:


Holder's inequality states that
$$ | \int fg | \leq \| f\|\_p \|g\|\_q $$
if $p^{-1} + q^{-1} = 1$, and the dual characterization of $L^p$ states that
$$ f \in L^p \iff \exists M \forall g\in L^q, \int fg \leq M \|g\|\_q $$


Your question can be interpreted as "does the dual characterization of $L^p$ work if, instead of all $g$ in $L^q$, we only look at those $g$s which are characteristic functions of balls (here we use that for $g$ the characteristic function of $B(x,r)$ in $\mathbb{R}^n$, its $L^q$ norm is $\approx r^{n/q}$).


While one may expect that the set of characteristic functions of balls form a nice "basis" for $L^q$, the problem however is that we don't see some convexity properties: knowing that the bounds hold for $g\_1$ and $g\_2$ only gives


$$ \int f (g\_1 + g\_2) \leq M (\|g\_1\|\_q + \|g\_2 \|\_q) $$


the right hand side of which is not $\leq \|g\_1 + g\_2\|\_q$. In Christian Remling's answer his $f$ is built out of functions
$$ \chi\_{B(0, 2^{-k} + 2^{-k^2 - k})} - \chi\_{B(0,2^{-k})} $$
We have for this function, in $\mathbb{R}^n$
$$ \|\chi\_{B(0, 2^{-k} + 2^{-k^2 - k})} - \chi\_{B(0,2^{-k})}\|\_q \approx 2^{-nk/q - k^2/q} \ll 2^{-nk/q} \approx \|\chi\_{B(0, 2^{-k} + 2^{-k^2 - k})}\|\_q + \|\chi\_{B(0,2^{-k})}\|\_q $$
These functions are, in some sense, "orthogonal" to the characteristic functions of balls, and so their largeness are not detectable by integration against balls.




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**Edit**: since there are some questions about the actual construction of the counterexample, here's an outline of how one can construct one.


1. We will work in $\mathbb{R}$ for simplicity. You can do the same on $\mathbb{R}^2$ using balls instead of intervals.
2. We will study $f = |\nabla u|^2$ to simplify notation. Throughout $f$ will be non-negative. The requirements are now
$$ \int\_{I} f \leq r^{1-\delta} \tag{req}$$
for all $r\in (0,1]$, and $I$ an interval of width $r$.
3. Given an interval $I$ with with $|I| \leq 1$. Let $J$ be the interval with the same center as $I$, but width $|J| = \frac1{3^{1/(1-\delta)}} |I|^{1/(1-\delta)} \leq\frac13 |I|$. Observe that as $I$ shrinks in length, the interval $J$ shrinks faster. Denote by
$$ f\_I(x) = \begin{cases} \frac{1}{|J|^\delta} & x\in J \\ 0 & x\not\in J \end{cases} $$
One can check that $f\_I$ satisfies the requirement (req), as given any interval $I'$ we have $$\int\_{I'} f\_I \leq |J|^{-\delta} \min(|I'|,|J|) \leq |I'|^{1-\delta}$$
4. Now take $\mathscr{I}$ a collection of disjoint intervals, then $f = \sum\_{I\in \mathscr{I}} f\_I$ is a function that satisfies (req): Given an arbitrary interval $I'$, if it only intersects one of the $J$s from our construction, then by the previous part we have that the inequality holds. When $I'$ spans multiple $J$s, let $\mathscr{I}'$ be the set of the corresponding $I$s, we have
$$ \int\_{I'} f \leq \sum\_{I\in \mathscr{I}'} |J|^{1-\delta} = \sum\_{I\in \mathscr{I}'} \frac13 |I| $$
But the length of $I'$ must be at least $\frac13 \sum\_{I\in \mathscr{I}'} |I|$ as the $I$ in $\mathscr{I}'$ are disjoint intervals, and the $J$s are in their centers.
5. The $L^1$ norm of $f\_I$ is exactly $\frac13 |I|$. For $p > 1$, the $L^p$ norm of $f\_I$ is
$$ \| f\_I\|\_{p}^p = \left(\frac13|I|\right)^{\frac{1-\delta p }{1-\delta}} $$
Note that the exponent is $< 1$.
6. Finally, take $w\_n$ a sequence of numbers, each $\leq \frac13$, such that $\sum w\_n$ converges but $\sum (w\_n)^{\alpha}$ diverges for every $\alpha < 1$ (so something like $w\_n = ( n \ln(n)^2)^{-1}$). Choose a disjoint family of intervals $I\_n$ such that $|I\_n| = w\_n$, and setting $f = \sum f\_{I\_n}$. This gives an example of an $L^1$ function satisfying condition (req) that is not in any $L^p$ for $p > 1$.


