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This question is about the content of this paper by J. Bourgain, N. Katz, T. Tao.


In the final step (page 18) of the proof of Szemerédi-Trotter type theorem, we have already known

$$|A''+A''|\lesssim N^{\frac{1}{2}+C\epsilon}$$

Similarly, by the Balog-Szemerédi-Gowers Theorem, there exists $A^{\star\star}\subseteq A'\setminus\{0\}$ with $|A^{\star\star}|\gtrsim N^{\frac{1}{2}-C\epsilon}$ such that

$$|A^{\star\star}\cdot A^{\star\star}|\lesssim N^{\frac{1}{2}+C\epsilon}$$

My question is why we have $A^{\star\star}=A''$. Do I understand it wrong?

Actually, I have tried to let the above $A^{\star\star}$ be a subset of $A’'\setminus\{0\}$ (by using the Balog-Szemerédi-Gowers Theorem) but it seems that we don’t have (or I cannot prove)

$$\left|\left\{(t,x_0,x_1)\in B\times A\times A’’:(1-t)x_0+tx_1\in A,t\neq0,1\right\}\right|\gtrsim N^{\frac{3}{2}-C\epsilon}$$

(The similar inequality (16) for $A’'$)

​​Could anyone help me understand the proof? Thanks for any help!

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  • $\begingroup$ I think some of your notations is off (i.e. in the pdf you linked, (18) deals with A’). $\endgroup$
    – Asaf
    Commented Jul 10, 2023 at 1:29
  • $\begingroup$ @Asaf Right. But I think it might be a typo in paper? Since BSG theorem gives us an upper bound of sum-subsets, I mean we should replace $A'$ with $A''$ in the previous two inequalities if I am not mistaken. $\endgroup$ Commented Jul 10, 2023 at 1:49

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I think you're correct that (18) should read:

$$|A'' + A''| \lesssim N^{1/2+C\epsilon}.$$

The display before (16) tells you that for each $x_1 \in A' $, (and hence $x_1 \in A'' \subseteq A'$) one has:

$$|\{(t,x_0) \in B \times A : (1-t)x_0 + t x_1 \in A; t \neq 0, 1 \}| \gtrsim N^{1-C \epsilon}$$

Since $|A''|\gtrsim N^{1/2}$, after adjusting implicit constants, summing over elements in $A'$, gives us the estimate

$$|\{(t,x_0,x_1) \in B \times A \times A'' : (1-t)x_0 + t x_1 \in A; t \neq 0, 1 \}| \gtrsim N^{3/2-C \epsilon}$$

On the other hand, using (14) (e.g. $|A| \lesssim N^{1/2}$), the pigeonhole principle gives us (after adjusting implicit constants again) that for some $x_0$ (in fact the typical $x_0 \in A$) one has:

$$|\{(t,x_1) \in B \times A'' : (1-t)x_0 + t x_1 \in A; t \neq 0, 1 \}| \gtrsim N^{1-C \epsilon}.$$

This is the claim in the display following (18). One can then return to the argument as written.

Also let me point out that these arguments have been simplified and improved since the breakthrough paper of Bourgain-Katz-Tao. For instance, one can replace the repeated applications to BSG with a geometric argument. See, for instance, the paper of Stevens and de Zeeuw: https://arxiv.org/abs/1609.06284.

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  • $\begingroup$ They derive (18) from $A’$ and BSG, so I think there might be a typo in (18). I mean maybe we should replace $A'$ with $A''$ if I am not mistaken. $\endgroup$ Commented Jul 10, 2023 at 1:54
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    $\begingroup$ I think I understand your question now. I've updated the answer. $\endgroup$
    – Mark Lewko
    Commented Jul 10, 2023 at 3:09
  • $\begingroup$ Oh! I totally forgot the restrictions on $A'$ and $A''$. It seems that this question turns out to be a silly question Orz. Anyway, thank you very much for your detailed and patient response. $\endgroup$ Commented Jul 10, 2023 at 4:18
  • $\begingroup$ I just met the same problem and found an answer here. Thanks for the post! By the way, you said that "summing over $A^\prime$", shouldn't it be summing over $A^{\prime\prime}$? It seems that the inequality is $|\{(t,x_0,x_1)\in B\times A\times A^{\prime\prime}:(1-t)x_0+tx_1\in A,\textrm{ and } t\neq0,1\}|=\sum_{x\in A^{\prime\prime}}|\{(t,x_0)\in B\times A:(1-t)x_0+tx_1\in A,\textrm{ and } t\neq 0,1\}|\gtrsim|A^{\prime\prime}|N^{1-C\epsilon}\gtrsim N^{3/2-C\epsilon}$. $\endgroup$ Commented Nov 12 at 11:41

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