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The following question is just a toy model:

Let $f:[0,1] \rightarrow \mathbb{R}$ be Lebesgue integrable, and suppose that for any $0\le a<b \le1$, $$\int_a^b |f(x)|dx \le \sqrt{b-a}$$ then prove or disprove that $$ \sup \left\{\frac{\int_E |f|dx}{|E|^{1/2}}: E \subset [0,1]\right\}<+\infty$$


If the claim above is false, then is it possible to prove that for any fixed $0<t<1/2$, $$ \sup \left\{\frac{\int_E |f|dx}{|E|^{t}}: E \subset [0,1]\right\}<+\infty$$


Motivation: This is a long story. Throughout the following, we assume that $f$ is a measurable function from a bounded regular open set $\Omega \subset \mathbb{R}^n$ to $\mathbb{R}$.

We know that if $f$ is $L^p(\Omega) \space(p>1)$, then by Holder's inequality, $$ \sup \left\{\frac{\int_E |f|dx}{|E|^{1-1/p}}: E \subset \Omega\right\}<+\infty \quad \quad\quad\quad(*)$$Then naturally I wanted to ask the inverse question: $\space$ Does $(*)$ imply $f \in L^p(\Omega)$?

One of my smart friends figured out that $(*)$ is equivalent to $f$ is weak $L^p$. See the answer here: a characterization of $L^p$ space.

Then naturally I want to know in $(*)$, instead of taking supremum over all measurable sets, what would happen if taking supremum over all cubes or balls? This leads to the toy model I asked at the beginning: the toy model is for $p=2, n=1$, $f$ is integrable, and cube is thus just an interval.

Now let me formulate my question neatly as follows: Set

$M_p:=\left\{f:\sup \left\{\frac{\int_E |f|dx}{|E|^{1-1/p}}: E \subset \Omega\right\}<+\infty\right\}$

$\tilde{M_p}:=\left\{f:\sup \left\{\frac{\int_B |f|dx}{|B|^{1-1/p}}: \text{$B$ is a ball $\subset \Omega$}\right\}<+\infty\right\}$

$L_p^w$ := the weak-$L^p$ space.

$\tilde{L_p}:=\{f \in L^q(\Omega): \forall 1\le q<p\}$

Then by my friend's result and interpolation theorem, $$L_p^w=M_p \subset \tilde{L_p}$$Also trivially, $M_p \subset \tilde{M_p}$. So the ultimate goal is that I want to know the relationship between $M_p, \tilde{M_p}$, and $\tilde{L_p}$. In particular, the toy model I asked at the beginning focuses on whether $M_p = \tilde{M_p}$.

An equivalent statement of whether $M_p=\tilde{M_p}$ is the following:

Let $0<s<1$. If $\mu$ is a finite measure on $\Omega$ and absolutely continuous with respect to Lebesgue measure in $\mathbb{R}^n$, and $\lim\sup _{r \rightarrow 0} \frac{\mu({B_r(x)})}{r^{ns}} \le 1, \forall x\in \Omega$ , is it true that $sup \{\frac{\mu(E)}{|E|^s}: E \subset \Omega\}<+\infty$ ?

I also want to understand the following question:

$\space$ If $M_p = \tilde{M_p}$ and $f \in M_p$, is it true that $\sup \left\{\frac{\int_E |f|dx}{|E|^{1-1/p}}: E \subset \Omega\right\}=\sup \left\{\frac{\int_B |f|dx}{|B|^{1-1/p}}: \text{$B$ is a ball $\subset \Omega$}\right\}?$ Or what can we say about the ratio?

By the way, the definition $\tilde{M_p}$ here is the same as $M^p$ defined in Gilbarg and Trudinger on Page 164, which is the so called Morrey Space. I looked up some references but didn't find any claims whether or not $M^p \subset L^q \quad \forall 1 \le q < p$.

Maybe I'm thinking too much. I should focus on solving one problem and then go step by step.


My effort:

In terms of the possible approaches, I think the approximate continuity of any measurable function and a nice covering argument would be helpful. Also, if $f$ is integrable, then one can observe that if $$T_pf(x):=\lim\sup_{r \rightarrow 0} \frac{1}{|B_r(x)|^{1−1/p}}\int_{B_r(x)}|f(y)|dy$$ is bounded in $\Omega$, then $$\mathcal{M}_pf(x):=\sup_{r > 0} \frac{1}{|B_r(x)|^{1−1/p}}\int_{B_r(x)}|f(y)|dy$$ is also bounded, and vice versa. Also, $$T_pf(x)=0,\mathcal{H}^{s}-a.e, \forall s\ge 1-1/p$$ So the size of the blow-up points should be very small, and thus a nice covering argument may be applied, at least we don't need to worry about cover the singular sets by balls or other arbituary sets. Maybe at least $\tilde{M_p} \subset M_q, \forall 1\le q<p$ can be provable. I have a lot of other observations, but it is cumbersome to type them down. Overall, I think these problems should be related to geometric measure theory and are not trivial.

Also, my smart friend suggests me try to apply the Littlewood-Paley Theory. He thinks of them as standard problems in harmonic analysis.


Any ideas, comments and partial result would be fully appreciated. I've no idea even about the toy model proposed.

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  • $\begingroup$ Well, I've thought quite a bit about this and got nowhere. My guess is some careful partition of the set needs to be done. Good problem! $\endgroup$ – abnry Apr 6 '15 at 2:13
  • $\begingroup$ @Nayrb, thanks for thinking this question. I agree with you that one needs a good partition of set, like the cube decomposition. Also, I added some further background in my post, in case people want to know why I asked this question. If you are interested in these questions, just keep an eye on them. In the future I'll definitely write an answer about what I get for all these questions. $\endgroup$ – student Apr 6 '15 at 4:04
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    $\begingroup$ @Jose27, my point is, if $T_pf(x):= \lim\sup_{r \rightarrow 0} \frac{1}{|B_r(x)|^{1-1/p}} \int _{B_r(x)}|f(y)|dy$ is bounded in $\Omega$, then $\mathcal{M_p}f(x) := \sup_{r>0} \frac{1}{|B_r(x)|^{1-1/p}} \int _{B_r(x)}|f(y)|dy$ is bounded, and vice versa. Also, I mean if $f$ is Lebesgue integrable, then $T_pf(x)=0, a.e.$ $\endgroup$ – student Apr 7 '15 at 2:03
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    $\begingroup$ @MagicMan, I've added what I thought should be added and deleted what I believed were not appropriate. Because previously there were some less rigorous and trivial argument. Also there were a lot of grammar mistakes. Even now I think there are a lot of problems, because I'm not an English speaker. Could you show me how to make the question better? Thanks! $\endgroup$ – student Apr 7 '15 at 23:03
  • $\begingroup$ This is a really fascinating problem. I am sure the conjecture is false, but I am struggling to find the counterexample. $\endgroup$ – Stephen Montgomery-Smith Apr 8 '15 at 23:00
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We want to mimic $f(x) = x^{-1/2}$ on $[0,1]$, but then cut $[0,1]$ into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller.

So let $x_n \to 0$ be a decreasing sequence with $x_0 = 1$. Write $x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose that $1 > \epsilon_n \to 0$ is a decreasing sequence. Then define $f$ on $[x_{n+1},x_n)$ to be $$ f(x) = \cases{\frac1{\epsilon_n} x_{n}^{-1/2} & if $x_{n+1} \le x \le (1 + \epsilon_n^2) x_{n+1}$ \cr 0 & if $ (1 + \epsilon_n^2) x_{n+1} \le x < (1 + \epsilon_n) x_{n+1}$\cr} $$ Check that $\int_a^b f(x) \, dx \le \sqrt{b-a}$.

Now set $E_N = \bigcup_{n=N}^\infty [x_{n+1},(1 + \epsilon_n^2) x_{n+1})$. Then $\int_{E_N} f(x) \, dx = \int_0^{x_N} f(x) \, dx \approx \sqrt{x_N}$, whereas $|E_N| \le \epsilon_N x_N$.

Suppose $x_n = 1/n^\alpha$ for $0<\alpha \le 1$. Then $\epsilon_n \approx \alpha/n$. Hence $$ \int_{E_N} f(x) \, dx \gtrsim (\alpha^{-1}|E_N|)^{\alpha/(2(\alpha+1))} .$$

Still lots of details to be checked. But I think this will provide a counterexample.

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  • $\begingroup$ I thought about a similar example before, and I agree that the ratio cannot be bounded universally regardless of the function chosen. I got stuck later on because this Morrey space does not have a compactness theorem, so the limit may not be in the space. I will go to sleep and tomorrow I will check your answer carefully. I am a little suspicious because at first glance there is no reason to find a limit due to lack of compactness. Thanks so much for your answer. If it is correct, then it will change my opinion on the role compactness of space plays. $\endgroup$ – student Apr 9 '15 at 6:43
  • $\begingroup$ The proof that $\int_a^b f(x) \, dx \le \sqrt{b-a}$ has quite a few details which I omitted, because I am rushed for time right now. Also thinking about it, I might have to change it to $\int_a^b f(x) \, dx \le 3\sqrt{b-a}$. $\endgroup$ – Stephen Montgomery-Smith Apr 9 '15 at 12:19
  • $\begingroup$ I checked all the details and it's correct. This is really a fascinating answer. Thank you so much! Honestly speaking, I was very close to the right answer, but messed up with some fussy details. I'll set a bounty because this answer deserves more credits. Later this weekend if I have time, I'll write down what I tried and why I failed. It is a valuable lesson for me at least. $\endgroup$ – student Apr 9 '15 at 15:30
  • $\begingroup$ by the way, would you like to share more motivations and details? Definitely I know all of them, because I've tried a lot. But in case others would like to know, because this is such a good answer that I don't want it to have some superficial flaws. $\endgroup$ – student Apr 9 '15 at 15:34
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    $\begingroup$ I am really busy for the next few days, so I won't be adding much detail for a while. I do want to provide more details for why $f$ is in the Morrey space. I also got caught on the fussy details several times when I tried to make this work. I think sleeping on it was what helped the most. I find my mind does a lot of unconscious thinking while I sleep or daydream. $\endgroup$ – Stephen Montgomery-Smith Apr 9 '15 at 19:42
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Through out the following, I use $P(E)$ to denote the perimeter of the set $E$. (See the definition of "sets of finite perimeter" in Giusti, minimal surfaces and functions of bounded variation. One may think of it as surface area.)

The following example shows $h \in \tilde{M}_p$, but $h \notin M_q,\forall$ $\frac{1}{1-t}\le q<p$

Let $Q_k=[x_{k+1},x_k) \times [x_{k+1},x_k)$, and define $h$ on $Q_k$ to be $h_k=k^{1+\alpha}$, where $\alpha$ is to be specified later, and zero otherwise. Let $E_K=\bigcup_{k=K}^{\infty}Q_k$.

Since $x_k=k^{-\alpha}$, $x_k-x_{k+1}\approx k^{-\alpha-1}$, then we have the following:

$|Q_k| \approx k^{-2\alpha-2}, P(Q_k) \approx k^{-1-\alpha}, \int_{Q_k} h\approx k^{-1-\alpha}, |E_K| \approx K^{-1-2\alpha}$ and$\int_{E_K}h \approx K^{-\alpha}$

For any $t \in (0,1/2)$, choose $0<\alpha < \frac{1}{1-2t}$, then $$\frac{\int_{E_K}h}{|E_K|^t} \approx \frac{K^{-\alpha}}{(K^{-1-2\alpha})^t}=K^{1-\alpha(1-2t)}\rightarrow \infty$$ Therefore, since $M_p(\Omega) \subset M_q(\Omega), \forall 1 \le q \le p$ and let $p_0 = p_0(t)=\frac{1}{1-t}$, $h \notin M_p(\Omega)$ for any $p \ge p_0$.

However, for any $E \subset \Omega$ and $E$ has finite perimeter, since $\mathcal{H}^1(Q_k \cap Q_{k-1})=\mathcal{H}^1(\{(x_k,y_k)\})=0$,we have $$P(E) \ge P(E \bigcap (\cup_{k=1}^{\infty} Q_k) = \Sigma_{k=1}^{\infty} P(E\cap Q_k)$$ Also, since $h$ is supported in $\cup_{k=1}^{\infty} Q_k$, we have $$\int_E h=\Sigma_{k=1}^{\infty}\int_{E\cup Q_k}h$$ Therefore, $$\frac{\int_E h}{P(E)} \le \frac{\int_E h}{\Sigma_{k=1}^{\infty} P(E\cap Q_k)} \le \sup\{\frac{\int_F h}{P(F)}: F \subset Q_k, k=1,2, \dots\}$$For any $F \subset Q_k$, we have by the isoperimetric inequality that $$\frac{\int_F h}{P(F)} = h_k |F|/P(F) \le C h_k P(F) \le C h_k P(Q_k) \approx k^{1+\alpha}k^{-1-\alpha}=1 \quad (1)$$where C in the estimate above is the isoperimetric constant in $\mathbb{R}^2$, thus $$\sup\{\frac{\int_F h}{P(F)}: F \subset Q_k, k=1,2, \dots\} < \infty \quad \quad $$ and thus $$\sup\{\frac{\int_E h}{P(E)}: E \subset \Omega\} < \infty$$

By (1), we conclude that for any cube $Q \subset \Omega$, $\frac{\int_Q h}{P(Q)} \le C $. Since $Q$ is a cube, $P(Q) \approx |Q|^{1/2}$, hence $\frac{\int_Q h}{|Q|^{1/2}} \le C $. Therefore, we showed that $$ \sup\{\frac{\int_Q h}{|Q|^{1/2}}: Q \subset \Omega\} \le C,$$ thus $h \in \tilde{M}_2$


Remark: I did plan to summarize what I had done, and share some experiences, but I'm lack of time. So I just put another example here, whose proof is very sneaky. Maybe it's better to know this idea. Thanks to @Stephen Montgomery-Smith's inspiration, I found a whole bunch of examples by myself, and I only choose this one to post because it is quite interesting.

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