2
$\begingroup$

I am trying to solve the following exercise (2.3.16) from Tao-Vu book.

Let $A,B$ be additive sets with common ambient group $Z$. Show that $\sigma[A\cup B]\leq e^{d(A,B)}+2e^{4d(A,B)}$. In the converse direction, establish the estimate $$d(A,B)\leq \log \sigma[A\cup B]+\dfrac{1}{2}\log \dfrac{|A\cup B|}{|A|}+\dfrac{1}{2}\log \dfrac{|A\cup B|}{|B|}.$$

My approach: i) We note that the first inequality can be written as follows: $$\dfrac{|A\cup B+A\cup B|}{|A\cup B|}\leq \dfrac{|A-B|}{|A|^{1/2}|B|^{1/2}}+2\dfrac{|A-B|^4}{|A|^2|B|^2}.$$ It is easy to show that $|A\cup B+A\cup B|\leq |A+A|+|B+B|+|A+B|.$ By Ruzsa triangle inequality we obtain: $d(A,-A)\leq d(A,-B)+d(-B,-A) \Leftrightarrow |A+A|\leq \dfrac{|A+B||A-B|}{|B|}$. In the same way one can show that $|B+B|\leq \dfrac{|A+B||A-B|}{|A|}.$ Since $d(A,-B)\leq 3d(A,B)$ (see Corollary 2.12) then $|A+B|\leq \dfrac{|A-B|^3}{|A||B|}.$ Hence we obtain $$|A+A|\leq \dfrac{|A-B|^4}{|A||B|^2} \ \ \text{and} \ \ |B+B|\leq \dfrac{|A-B|^4}{|A|^2|B|}.$$

Therefore, $$\dfrac{|A\cup B+A\cup B|}{|A\cup B|}\leq 2\dfrac{|A-B|^4}{|A|^2|B|^2}+\dfrac{|A+B|}{|A\cup B|},$$ since $|A|,|B|\leq |A\cup B|.$ However, I cannot prove that $\dfrac{|A+B|}{|A\cup B|}\leq \dfrac{|A-B|}{|A|^{1/2}|B|^{1/2}}$ and I firmly believe that this bound is not true at all because if we take $A=B$ then it is equivalent to $\sigma[A]\leq \delta[A],$ where $\sigma[A]=\dfrac{|2A|}{|A|}$ and $\delta[A]=\dfrac{|A-A|}{|A|}$ are doubling and difference constants.

ii) The second inequality is equivalent to $|A-B|\leq |A\cup B+A\cup B|$ and I have no idea how to prove that. I was trying to prove using Ruzsa triangle inequality but I failed.

It would be great to see how to solve this exercise.

$\endgroup$

1 Answer 1

0
$\begingroup$

The inequality in the "converse direction" has a typo since, for example, if $A=B$ it says $\delta[A]\leq \sigma[A]$, which need not be true. The inequality should instead involve $c\log\sigma[A\cup B]$, where $c$ is some absolute constant. I leave the identification of this constant as an exercise (use Exercise 2.3.4).

As for the first inequality, you are of course correct that one cannot have $|A+B|/|A\cup B|\leq e^{d(A,B)}$ in general. I don't see offhand how to improve your argument to effectively absorb more data into the $2e^{4d(A,B)}$ term; nor do I have a counterexample to show that it cannot be done. On the other hand, the format of the proposed inequality is very suggestive of your strategy, in which one shows that $|A+A|/|A\cup B|$ and $|B+B|/|A\cup B|$ are both bounded by $e^{4d(A,B)}$, while $|A+B|/|A\cup B|$ is bounded by some other term. So I wonder if there isn't also a typo here, and instead of $e^{d(A,B)}$, it should be something that actually does bound $|A+B|/|A\cup B|$ (for example, $e^{3d(A,B)}$). This would fit with the spirit of the exercise, which is just to show that if $A$ and $B$ are close in Rusza distance, then $A\cup B$ has small doubling. So to summarize, what I am saying is that the following inequality is true (essentially by what you've already written), and suffices to establish the purpose of the exercise: $$ \sigma[A\cup B]\leq e^{3d(A,B)}+2e^{4d(A,B)}. $$

$\endgroup$

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.