# Higher integrability for Sobolev functions

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 $$\epsilon >0$$?

• This does not see at all the fact that the integrand is a gradient, does it? So you could ask if a function $v\in L^2$ whose square average cannot blow up too fact is in fact better than $L^2$. In which case the answer is probably no, just fiddling around with the usual logarithms and borderline integrability should give a counterexample, I guess? Feb 6 at 18:36
• You are right in the sense that it does not see the gradient and that the gradient replaced with any other generic function, the question remains the same.
Feb 6 at 19:32
• So if there's a counterexample, i would be very much interested to know what features it might have.
Feb 6 at 19:33
• There you go! ;-) Feb 6 at 20:22
• If you are changing the goals, I would suggest asking this as a new question, and not just editing the current one. When asking the new question you can link back to this one. Feb 9 at 17:07

Since the OP asked for a discussion of features, 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.

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 $$p > 1$$.
• Thank you very much for this explanation, this is very helpful.
Feb 8 at 15:49
• Thank you, I will ready it carefully. Feb 8 at 17:52
• I checked, it works. It is very nice that the inequality improves in case of multiple intersections. By the way, I discovered (asking!) a trivial 2 dimensional example. Take a function which is $L^1$ (locally) and no more and think at is as a funcion of 2 variables. It satisfies the estimate with $Cr$ (in 2 d) but it is no more than $L^1$. Feb 8 at 22:17

No. Consider a function $$f\in L^1(\mathbb R)$$, $$f\ge 0$$, with $$f(x) = 2^{n^2}$$ on $$2^{-n} and essentially $$f=0$$ otherwise. Then $$\int_{-r}^r f(x)\, dx \simeq \sum_{n\gtrsim (-\log r)} 2^{-n}\lesssim r^{\alpha}$$, but $$f\notin L^p$$ for $$p>1$$.

We can also make $$f$$ continuous at all $$x\not=0$$, so $$(1/2r)\int_{x-r}^{x+r} f(t)\, dt$$ stays bounded at these points (it converges to $$f(x)$$).

A radial function of this type gives a counterexample in any dimension.

• How is your $f$ related to the $u$ in the OP? Feb 6 at 20:06
• I must be dense, but where did $r^\alpha$ come from? For $\int_{-r}^r f$, you sum over dyadic intervals with $2^k < r$, on each interval the integral contribute $2^k$, so the integral should be of size $r^1$. Is there supposed to be an $\alpha$ in the definition of $f$ somewhere? Feb 6 at 20:22
• @IosifPinelis: $f=|\nabla u|^2$, except that we also need to modify to go from one to two dimesions. Feb 6 at 22:09
• I have doubts on this example. If you take $a=2^{-n}, b=a+2^{-n-n^2}$ the integral of $f$ on the interval is $2^{-n}$ and cannot be bounded by a positive power of $b-a$, independently of $n$. Feb 7 at 9:34
• @GiorgioMetafune: I think we can fix it though by just spreading out the interval over many separated smaller intervals, still of total length $2^{-{n^2}-n}$. Feb 7 at 19:25

$$\newcommand\R{\mathbb R}\newcommand\ep\epsilon\newcommand{\de}{\delta}$$For $$(s,t)\in\R^2$$, let $$$$u(s,t):=\sum c_k g\Big(\frac{R-r_k}{h_k}\Big),$$$$ where $$g(z):=\max(0,1-|z|)$$ for real $$z$$, $$R:=\sqrt{s^2+t^2}$$, $$$$c_k:=h_k^{1-\de/2},\quad h_k:=k^{-3/(1-\de)}, \quad r_k:=k^2/\ln^2k,$$$$ and $$\sum:=\sum_{k\ge k_0}$$, where in turn $$k_0$$ is an integer large enough so that $$k\ge2$$ and for all $$k\ge k_0$$ we have $$0.

Then $$$$|\nabla u(s,t)|^2=\sum \frac{c_k^2}{h_k^2}\, 1(|R-r_k| almost everywhere (a.e.). So, for each $$x=(s,t)\in\R^2$$ there is some integer $$k\ge k_0$$ such that for all $$r\in(0,1]$$ $$$$\frac1{|B_r|}\int_{B_r(x)} |\nabla u|^2 \ll \frac1{r^2} h_k^{-\de}\,\min(h_k,r)r\le \frac1{r^\de};$$$$ here and in what follows, $$A\ll B$$ means $$A\le CB$$ for some universal real constant $$C>0$$. So, the condition displayed in the OP holds (up to a universal positive real constant factor, which can obviously be removed by rescaling $$u$$). Also, $$$$\int_{\R^2} |\nabla u|^2\ll \sum \frac{c_k^2}{h_k^2}\,r_k h_k =\sum h_k^{-\de}\,r_k h_k=\sum\frac1{k\ln^2k}<\infty$$$$ and $$$$\int_{\R^2} |u|^2\ll \sum c_k^2\,r_k h_k\le\sum \frac{c_k^2}{h_k^2}\,r_k h_k <\infty,$$$$ so that $$u\in W^{1,2}(\R^2)$$.

Thus, all the conditions on $$u$$ hold. However, for any real $$\ep>0$$ there some real $$\eta>0$$ such that $$$$\int_{\R^2} |\nabla u|^{2+\ep}\asymp \sum \frac{c_k^{2+\ep}}{h_k^{2+\ep}}\,r_k h_k =\sum h_k^{-(1+\ep/2)\de}\,r_k h_k =\sum r_k h_k^{(1-\de)(1-\eta)}= \sum\frac1{k^{1-3\eta}\ln^2k}=\infty,$$$$ so that $$\nabla u\notin L^{2+\ep}$$. (This answers the original version of the question, before the replacement of $$L^{2+\ep}$$ by $$L^{2+\ep}_{loc}$$.)

• Can you modify the argument to prove a local counterexample? Feb 7 at 9:36
• @GiorgioMetafune : Such a modification would be something like Christian Remling's example. Both examples are based on the same "lacunary" idea. Christian Remling worked near $0$ and I worked near $\infty$. Somehow, it took me too much time to implement the idea, fiddling with the $c_k$'s, $r_k$'s, and $h_k$'s, and in the meantime the OP changed $L^{2+\epsilon}$ to $L^{2+\epsilon}_{loc}$. Feb 7 at 13:54
• I have some doubts on Christian construction, see the comments to his answer. Unnless I misunderstood somerhing the bound looks not uniform. Feb 7 at 14:18
• @GiorgioMetafune : I see. I will read the answer more carefully now. Feb 7 at 14:35
• This is very helpful answer, the local or global aspect of my question is not particularly important for what I am trying to understand. In this sense, the global example is very helpful.