5
$\begingroup$

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 $\epsilon >0$?

$\endgroup$
6
  • 1
    $\begingroup$ 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? $\endgroup$ Feb 6, 2023 at 18:36
  • $\begingroup$ 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. $\endgroup$
    – Adi
    Feb 6, 2023 at 19:32
  • $\begingroup$ So if there's a counterexample, i would be very much interested to know what features it might have. $\endgroup$
    – Adi
    Feb 6, 2023 at 19:33
  • $\begingroup$ There you go! ;-) $\endgroup$ Feb 6, 2023 at 20:22
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 9, 2023 at 17:07

3 Answers 3

7
$\begingroup$

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 $p > 1$.
$\endgroup$
3
  • $\begingroup$ Thank you very much for this explanation, this is very helpful. $\endgroup$
    – Adi
    Feb 8, 2023 at 15:49
  • $\begingroup$ Thank you, I will ready it carefully. $\endgroup$ Feb 8, 2023 at 17:52
  • $\begingroup$ 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 $L^1$. $\endgroup$ Feb 8, 2023 at 22:17
5
$\begingroup$

No. Consider a function $f\in L^1(\mathbb R)$, $f\ge 0$, with $f(x) = 2^{n^2}$ on $2^{-n}<x<2^{-n}+2^{-n^2-n}$ 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.

$\endgroup$
10
  • $\begingroup$ How is your $f$ related to the $u$ in the OP? $\endgroup$ Feb 6, 2023 at 20:06
  • $\begingroup$ 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? $\endgroup$ Feb 6, 2023 at 20:22
  • $\begingroup$ @IosifPinelis: $f=|\nabla u|^2$, except that we also need to modify to go from one to two dimesions. $\endgroup$ Feb 6, 2023 at 22:09
  • 1
    $\begingroup$ 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 $n$. $\endgroup$ Feb 7, 2023 at 9:34
  • 1
    $\begingroup$ @GiorgioMetafune: I think we can fix it though by just spreading out the interval over many separated smaller intervals, still of total length $2^{-{n^2}-n}$. $\endgroup$ Feb 7, 2023 at 19:25
4
$\begingroup$

$\newcommand\R{\mathbb R}\newcommand\ep\epsilon\newcommand{\de}{\delta} $For $(s,t)\in\R^2$, let \begin{equation} u(s,t):=\sum c_k g\Big(\frac{R-r_k}{h_k}\Big), \end{equation} where $g(z):=\max(0,1-|z|)$ for real $z$, $R:=\sqrt{s^2+t^2}$, \begin{equation} c_k:=h_k^{1-\de/2},\quad h_k:=k^{-3/(1-\de)}, \quad r_k:=k^2/\ln^2k, \end{equation} 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 $0<r_k-h_k<r_k+h_k<r_{k+1}-h_{k+1}-2$.

Then \begin{equation} |\nabla u(s,t)|^2=\sum \frac{c_k^2}{h_k^2}\, 1(|R-r_k|<h_k) =\sum h_k^{-\de}\, 1(|R-r_k|<h_k) \end{equation} 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 $r\in(0,1]$ \begin{equation} \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}; \end{equation} 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, \begin{equation} \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 \end{equation} and \begin{equation} \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, \end{equation} 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 \begin{equation} \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, \end{equation} so that $\nabla u\notin L^{2+\ep}$. (This answers the original version of the question, before the replacement of $L^{2+\ep}$ by $L^{2+\ep}_{loc}$.)

$\endgroup$
5
  • $\begingroup$ Can you modify the argument to prove a local counterexample? $\endgroup$ Feb 7, 2023 at 9:36
  • $\begingroup$ @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 $L^{2+\epsilon}_{loc}$. $\endgroup$ Feb 7, 2023 at 13:54
  • $\begingroup$ I have some doubts on Christian construction, see the comments to his answer. Unnless I misunderstood somerhing the bound looks not uniform. $\endgroup$ Feb 7, 2023 at 14:18
  • $\begingroup$ @GiorgioMetafune : I see. I will read the answer more carefully now. $\endgroup$ Feb 7, 2023 at 14:35
  • $\begingroup$ 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. $\endgroup$
    – Adi
    Feb 8, 2023 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.