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	<title>Comments on: Can Bourgain&#8217;s argument be usefully modified?</title>
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		<title>By: Gil Kalai</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2863</link>
		<dc:creator><![CDATA[Gil Kalai]]></dc:creator>
		<pubDate>Tue, 22 Feb 2011 21:28:46 +0000</pubDate>
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		<description><![CDATA[Sure, here it is: http://www.ma.huji.ac.il/~kalai/line5.pdf]]></description>
		<content:encoded><![CDATA[<p>Sure, here it is: <a href="http://www.ma.huji.ac.il/~kalai/line5.pdf" rel="nofollow">http://www.ma.huji.ac.il/~kalai/line5.pdf</a></p>
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		<title>By: gowers</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2862</link>
		<dc:creator><![CDATA[gowers]]></dc:creator>
		<pubDate>Tue, 22 Feb 2011 12:08:42 +0000</pubDate>
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		<description><![CDATA[Gil, are you prepared to make public the proof that the conjecture implies the bound you say? Or is it meant to be an easy exercise? (Just staring at it I feel a bit mystified, but perhaps if I made a serious attempt to prove the implication -- and not in my head -- I would realize that it was not too hard.) Or is it possible to give a sketch that makes the implication seem reasonable?]]></description>
		<content:encoded><![CDATA[<p>Gil, are you prepared to make public the proof that the conjecture implies the bound you say? Or is it meant to be an easy exercise? (Just staring at it I feel a bit mystified, but perhaps if I made a serious attempt to prove the implication &#8212; and not in my head &#8212; I would realize that it was not too hard.) Or is it possible to give a sketch that makes the implication seem reasonable?</p>
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		<title>By: Gil Kalai</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2858</link>
		<dc:creator><![CDATA[Gil Kalai]]></dc:creator>
		<pubDate>Sun, 20 Feb 2011 19:22:22 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2858</guid>
		<description><![CDATA[Maybe, given the amount of discussion, this is a good place to concentrate both polymath6ing and openly discussing both Roth and the cupset problem, both lower bounds and upper bounds. 

About 10-12 years ago Roy Mshulam and I thought quite extensively (without success) about trying to improve the Fourier argument even slightly. (We had more avant-garde ideas about the problem more recently that I discussed over my blog.) Anyway, we had a certain proposed conjecture (whose strength depends on various parameters) about general functions and I wonder how this conjecture survive the killer examples described by Tim and if the argument by Michael and Nets gives also it. 

So lets consider $latex F_3^n$ and for a function $latex f$ and $latex q \ge 1$ and $latex 1 \le k \le n$ we denote by $latex \&#124;f\&#124;_{q,k}$ the maximum over affine k-dimensional flats $latex L$ of $latex (3^{-k}\sum_{x\in L}&#124;f(x)&#124;^q)^{1/q}$. 

Now the conjecture is that for some constant $latex C$ and some negotiable $latex \gamma,\alpha,q$ every set $latex E \subset F_3^n$ of size $latex n^{\gamma}$ the following holds: 

(**) $latex \sum_{x\in E}~&#124;\hat f(x)&#124;\le c_1\&#124;f\&#124;_{q,n^\alpha}^2$

It turns out that if (**) holds with $latex \gamma \to \infty$, and $latex \alpha/q \to 1$ then the $latex 3^n/n$ upper bound for cupsets can be improved to $latex 3^n/n^\beta$ for every $latex \beta$. 

In fact, if you fix $latex r$ between 2 and 3  you can have $latex \beta &gt; min\{(3-r)/r(1-\alpha/q), (6-2r+\gamma(r-2)/r)\}.$ So you can adjust the above parameters in the conjecture to imply stronger and weaker forms of the cap conjecture. We could say something for very very specific E&#039;s but not for anything near the general case. We can adjust (**) with $latex \alpha = q(1-c_2\log n/n)$ and $latex \gamma=c_3n/\log n$ to have (**) implies $latex (3-t)^n$ upperbounds so maybe some killer examples or some other examples already kill this. Also we can adjust the parameters to make $latex \beta -1$ tiny and one wonders if then (**) follows from the new results by Michael and Nets.]]></description>
		<content:encoded><![CDATA[<p>Maybe, given the amount of discussion, this is a good place to concentrate both polymath6ing and openly discussing both Roth and the cupset problem, both lower bounds and upper bounds. </p>
<p>About 10-12 years ago Roy Mshulam and I thought quite extensively (without success) about trying to improve the Fourier argument even slightly. (We had more avant-garde ideas about the problem more recently that I discussed over my blog.) Anyway, we had a certain proposed conjecture (whose strength depends on various parameters) about general functions and I wonder how this conjecture survive the killer examples described by Tim and if the argument by Michael and Nets gives also it. </p>
<p>So lets consider <img src='http://s0.wp.com/latex.php?latex=F_3%5En&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='F_3^n' title='F_3^n' class='latex' /> and for a function <img src='http://s0.wp.com/latex.php?latex=f&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f' title='f' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=q+%5Cge+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q &#92;ge 1' title='q &#92;ge 1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=1+%5Cle+k+%5Cle+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1 &#92;le k &#92;le n' title='1 &#92;le k &#92;le n' class='latex' /> we denote by <img src='http://s0.wp.com/latex.php?latex=%5C%7Cf%5C%7C_%7Bq%2Ck%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;|f&#92;|_{q,k}' title='&#92;|f&#92;|_{q,k}' class='latex' /> the maximum over affine k-dimensional flats <img src='http://s0.wp.com/latex.php?latex=L&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L' title='L' class='latex' /> of <img src='http://s0.wp.com/latex.php?latex=%283%5E%7B-k%7D%5Csum_%7Bx%5Cin+L%7D%7Cf%28x%29%7C%5Eq%29%5E%7B1%2Fq%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(3^{-k}&#92;sum_{x&#92;in L}|f(x)|^q)^{1/q}' title='(3^{-k}&#92;sum_{x&#92;in L}|f(x)|^q)^{1/q}' class='latex' />. </p>
<p>Now the conjecture is that for some constant <img src='http://s0.wp.com/latex.php?latex=C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='C' title='C' class='latex' /> and some negotiable <img src='http://s0.wp.com/latex.php?latex=%5Cgamma%2C%5Calpha%2Cq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;gamma,&#92;alpha,q' title='&#92;gamma,&#92;alpha,q' class='latex' /> every set <img src='http://s0.wp.com/latex.php?latex=E+%5Csubset+F_3%5En&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='E &#92;subset F_3^n' title='E &#92;subset F_3^n' class='latex' /> of size <img src='http://s0.wp.com/latex.php?latex=n%5E%7B%5Cgamma%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n^{&#92;gamma}' title='n^{&#92;gamma}' class='latex' /> the following holds: </p>
<p>(**) <img src='http://s0.wp.com/latex.php?latex=%5Csum_%7Bx%5Cin+E%7D%7E%7C%5Chat+f%28x%29%7C%5Cle+c_1%5C%7Cf%5C%7C_%7Bq%2Cn%5E%5Calpha%7D%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sum_{x&#92;in E}~|&#92;hat f(x)|&#92;le c_1&#92;|f&#92;|_{q,n^&#92;alpha}^2' title='&#92;sum_{x&#92;in E}~|&#92;hat f(x)|&#92;le c_1&#92;|f&#92;|_{q,n^&#92;alpha}^2' class='latex' /></p>
<p>It turns out that if (**) holds with <img src='http://s0.wp.com/latex.php?latex=%5Cgamma+%5Cto+%5Cinfty&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;gamma &#92;to &#92;infty' title='&#92;gamma &#92;to &#92;infty' class='latex' />, and <img src='http://s0.wp.com/latex.php?latex=%5Calpha%2Fq+%5Cto+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha/q &#92;to 1' title='&#92;alpha/q &#92;to 1' class='latex' /> then the <img src='http://s0.wp.com/latex.php?latex=3%5En%2Fn&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='3^n/n' title='3^n/n' class='latex' /> upper bound for cupsets can be improved to <img src='http://s0.wp.com/latex.php?latex=3%5En%2Fn%5E%5Cbeta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='3^n/n^&#92;beta' title='3^n/n^&#92;beta' class='latex' /> for every <img src='http://s0.wp.com/latex.php?latex=%5Cbeta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;beta' title='&#92;beta' class='latex' />. </p>
<p>In fact, if you fix <img src='http://s0.wp.com/latex.php?latex=r&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r' title='r' class='latex' /> between 2 and 3  you can have <img src='http://s0.wp.com/latex.php?latex=%5Cbeta+%3E+min%5C%7B%283-r%29%2Fr%281-%5Calpha%2Fq%29%2C+%286-2r%2B%5Cgamma%28r-2%29%2Fr%29%5C%7D.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;beta &gt; min&#92;{(3-r)/r(1-&#92;alpha/q), (6-2r+&#92;gamma(r-2)/r)&#92;}.' title='&#92;beta &gt; min&#92;{(3-r)/r(1-&#92;alpha/q), (6-2r+&#92;gamma(r-2)/r)&#92;}.' class='latex' /> So you can adjust the above parameters in the conjecture to imply stronger and weaker forms of the cap conjecture. We could say something for very very specific E&#8217;s but not for anything near the general case. We can adjust (**) with <img src='http://s0.wp.com/latex.php?latex=%5Calpha+%3D+q%281-c_2%5Clog+n%2Fn%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha = q(1-c_2&#92;log n/n)' title='&#92;alpha = q(1-c_2&#92;log n/n)' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cgamma%3Dc_3n%2F%5Clog+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;gamma=c_3n/&#92;log n' title='&#92;gamma=c_3n/&#92;log n' class='latex' /> to have (**) implies <img src='http://s0.wp.com/latex.php?latex=%283-t%29%5En&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(3-t)^n' title='(3-t)^n' class='latex' /> upperbounds so maybe some killer examples or some other examples already kill this. Also we can adjust the parameters to make <img src='http://s0.wp.com/latex.php?latex=%5Cbeta+-1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;beta -1' title='&#92;beta -1' class='latex' /> tiny and one wonders if then (**) follows from the new results by Michael and Nets.</p>
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		<title>By: Michael Bateman</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2840</link>
		<dc:creator><![CDATA[Michael Bateman]]></dc:creator>
		<pubDate>Tue, 15 Feb 2011 01:13:25 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2840</guid>
		<description><![CDATA[(My apologies in advance for the poor formatting; I realized my faux pas with not enough time to fix it before tomorrow.)

Here are some half-baked thoughts about possibly increasing the size of epsilon.  

My knowledge of covering lemmas and related technology is a bit thin.  We already have a structural theorem for sets S such that S is not smoothing.  Could we possibly use this to prove a structural theorem for sets S such that, say, 2S is not smoothing?  how about 4S?  etc.  This seems like an interesting question because of the following sketch.  


As mentioned above, when epsilon grows the spectrum may become additively smoothing, which is a problem (at least for the existing technology).  What I notice, however, is that even when epsilon is large, we may still conclude (something like?) one of the iterated difference sets is not additively smoothing.  By using the random selection argument as in [BK], we can conclude that the number of 2m-tuples is less than 

N^{4m + mO(epsilon)} . 

Since the number of 4-tuples is known to be 

N^{7-O(epsilon)} 

in this case, it is possible for the number of 8-tuples to be as large as 

N^{15+ O(epsilon)} 

without reaching a contradiction.  As epsilon grows, this ruins the non-smoothing of the spectrum.  

Write E_{2m} to denote the number of 2m-tuples in the spectrum S.  Suppose we have a smoothing-like condition at every &quot;step&quot;.  More precisely, suppose E_{2m} is larger than the trivial Holder bound over E_m plus an additional factor of E_m ^{beta} for some small number beta.  If this is true for m=3,4,5,..., M, where M depends on beta and epsilon, then we contradict the estimate E_{2M} &lt; N^{4M + MO(epsilon)} mentioned above that we obtained from the random selection argument.  Hence we must not have this additional beta smoothing for every m.  If we write S for the spectrum, this seems to say something an awful lot like &quot;2mS is additively non-smoothing&quot;.]]></description>
		<content:encoded><![CDATA[<p>(My apologies in advance for the poor formatting; I realized my faux pas with not enough time to fix it before tomorrow.)</p>
<p>Here are some half-baked thoughts about possibly increasing the size of epsilon.  </p>
<p>My knowledge of covering lemmas and related technology is a bit thin.  We already have a structural theorem for sets S such that S is not smoothing.  Could we possibly use this to prove a structural theorem for sets S such that, say, 2S is not smoothing?  how about 4S?  etc.  This seems like an interesting question because of the following sketch.  </p>
<p>As mentioned above, when epsilon grows the spectrum may become additively smoothing, which is a problem (at least for the existing technology).  What I notice, however, is that even when epsilon is large, we may still conclude (something like?) one of the iterated difference sets is not additively smoothing.  By using the random selection argument as in [BK], we can conclude that the number of 2m-tuples is less than </p>
<p>N^{4m + mO(epsilon)} . </p>
<p>Since the number of 4-tuples is known to be </p>
<p>N^{7-O(epsilon)} </p>
<p>in this case, it is possible for the number of 8-tuples to be as large as </p>
<p>N^{15+ O(epsilon)} </p>
<p>without reaching a contradiction.  As epsilon grows, this ruins the non-smoothing of the spectrum.  </p>
<p>Write E_{2m} to denote the number of 2m-tuples in the spectrum S.  Suppose we have a smoothing-like condition at every &#8220;step&#8221;.  More precisely, suppose E_{2m} is larger than the trivial Holder bound over E_m plus an additional factor of E_m ^{beta} for some small number beta.  If this is true for m=3,4,5,&#8230;, M, where M depends on beta and epsilon, then we contradict the estimate E_{2M} &lt; N^{4M + MO(epsilon)} mentioned above that we obtained from the random selection argument.  Hence we must not have this additional beta smoothing for every m.  If we write S for the spectrum, this seems to say something an awful lot like &quot;2mS is additively non-smoothing&quot;.</p>
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		<title>By: Christian</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2838</link>
		<dc:creator><![CDATA[Christian]]></dc:creator>
		<pubDate>Mon, 14 Feb 2011 21:00:43 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2838</guid>
		<description><![CDATA[What I meant with the last paragraph was just this: It is not even clear why one should assign strict zeros instead of just very small numbers to the points outside the Bohr set under consideration, which conceivably might then turn out to be &quot;technically easier&quot;, though it should&#039;t be essential either.]]></description>
		<content:encoded><![CDATA[<p>What I meant with the last paragraph was just this: It is not even clear why one should assign strict zeros instead of just very small numbers to the points outside the Bohr set under consideration, which conceivably might then turn out to be &#8220;technically easier&#8221;, though it should&#8217;t be essential either.</p>
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		<title>By: Christian</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2837</link>
		<dc:creator><![CDATA[Christian]]></dc:creator>
		<pubDate>Mon, 14 Feb 2011 20:35:14 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2837</guid>
		<description><![CDATA[Dear Tim,

here are just some vague thoughts I had after reading your post and .pdf-file, which I&#039;d like to write down here - basically in order to see whether I understood those things properly. Yet before getting started I should perhaps say that although this comment is unfortunately somewhat long, it doesn&#039;t give any new ideas, it&#039;s just sort of a reformulation of some things you&#039;ve already said.

It seems that one may look at any argument proving a version of Roth&#039;s theorem that involves density increments as a method (in a not too algorithmic sense of this word) to actually find a 3AP in a set satisfying the requested density bound by making successively better guesses as to where such an 3AP could be. Thus in the original argument, we start with a set $latex A$ living in $latex \mathbb{Z}_N$ of size $latex CN/loglogN$, compute certain Fourier coefficients, and then guess that (one of) the 3AP(s) of $latex A$ can actually be found in a certain subprogression, then we do the very same computations there, make a further guess, and so forth until our guesses become precise enough to actually locate 3AP inside $latex A$, at which point we are done. The verious improvements over Roth&#039;s orinal result can then be regarded as providing descriptions of much better strategies for making these guesses.

So far our guesses can also be thought of as assigning zeros and ones to the elements of $latex A$ in each step, where those elements among which we still look for a 3AP get assigned ones, while the other ones receive zeros. If we formulate the strategy behind the proof like this, the idea of assigning reals from the interval $latex [0, 1]$ instead of just zeros and ones doesn&#039;t appear unnatural: So in each stage we could construct a function $latex f$ from $latex A$ to $latex [0, 1]$, and an equation like $latex f(a)=0$ intuitively means &quot;For the time being I won&#039;t consider 3AP(s) through $latex a$ any further&quot;, while $latex f(a)=1/10$ could mean &quot;At the moment it&#039;s not particularly likely, but still possible that I will end up producing a 3AP through $latex a$&quot;, an equation like &quot;f(a)=2f(a&#039;)&quot; roughly means &quot;Currently it seems twice as likely that our eventual 3AP will pass through $latex a$ than that it passes through $latex a&#039;$&quot;, and so on. 

Now for Roth&#039;s original argument it is definitely essential for technical reasons to just work with $latex \{0, 1\}$-valued functions (as one &quot;changes $latex N$&quot; throughout the proof), even though this appears to lead to some more or less arbitrary but at the same time rather immaterial choices concerning the boundaries of the subprogressions to which one passes when iterating. But already for Bourgains argument (my understanding of which on a conceptual level is surely by far more limited than yours), it doesn&#039;t seem to be clear why one has to do so, and if I happen to understand what you suggest to some extent correctly, then you propose to work with a guessing function $latex f$ whose values at the center of the Bohr set under consideration is much greater then the value of $latex f$ near the boundary. (Instead of taking $latex f$ to be the characteristic function of the Bohr set involved.)

At this point, it seems that maybe one can even dispose of working with Bohr sets on a strictly technical level and instead merely &quot;think of them in the back of our head&quot;, transforming at each step of the iteration the definition of the Bohr set of which we &quot;really think&quot; into an appropriate definition of a certain guessing function (which probably then has to somehow &quot;codify&quot; certain pieces of &quot;regularity information&quot; on that Bohr set, etc.)]]></description>
		<content:encoded><![CDATA[<p>Dear Tim,</p>
<p>here are just some vague thoughts I had after reading your post and .pdf-file, which I&#8217;d like to write down here &#8211; basically in order to see whether I understood those things properly. Yet before getting started I should perhaps say that although this comment is unfortunately somewhat long, it doesn&#8217;t give any new ideas, it&#8217;s just sort of a reformulation of some things you&#8217;ve already said.</p>
<p>It seems that one may look at any argument proving a version of Roth&#8217;s theorem that involves density increments as a method (in a not too algorithmic sense of this word) to actually find a 3AP in a set satisfying the requested density bound by making successively better guesses as to where such an 3AP could be. Thus in the original argument, we start with a set <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='A' title='A' class='latex' /> living in <img src='http://s0.wp.com/latex.php?latex=%5Cmathbb%7BZ%7D_N&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbb{Z}_N' title='&#92;mathbb{Z}_N' class='latex' /> of size <img src='http://s0.wp.com/latex.php?latex=CN%2FloglogN&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='CN/loglogN' title='CN/loglogN' class='latex' />, compute certain Fourier coefficients, and then guess that (one of) the 3AP(s) of <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='A' title='A' class='latex' /> can actually be found in a certain subprogression, then we do the very same computations there, make a further guess, and so forth until our guesses become precise enough to actually locate 3AP inside <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='A' title='A' class='latex' />, at which point we are done. The verious improvements over Roth&#8217;s orinal result can then be regarded as providing descriptions of much better strategies for making these guesses.</p>
<p>So far our guesses can also be thought of as assigning zeros and ones to the elements of <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='A' title='A' class='latex' /> in each step, where those elements among which we still look for a 3AP get assigned ones, while the other ones receive zeros. If we formulate the strategy behind the proof like this, the idea of assigning reals from the interval <img src='http://s0.wp.com/latex.php?latex=%5B0%2C+1%5D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='[0, 1]' title='[0, 1]' class='latex' /> instead of just zeros and ones doesn&#8217;t appear unnatural: So in each stage we could construct a function <img src='http://s0.wp.com/latex.php?latex=f&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f' title='f' class='latex' /> from <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='A' title='A' class='latex' /> to <img src='http://s0.wp.com/latex.php?latex=%5B0%2C+1%5D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='[0, 1]' title='[0, 1]' class='latex' />, and an equation like <img src='http://s0.wp.com/latex.php?latex=f%28a%29%3D0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f(a)=0' title='f(a)=0' class='latex' /> intuitively means &#8220;For the time being I won&#8217;t consider 3AP(s) through <img src='http://s0.wp.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> any further&#8221;, while <img src='http://s0.wp.com/latex.php?latex=f%28a%29%3D1%2F10&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f(a)=1/10' title='f(a)=1/10' class='latex' /> could mean &#8220;At the moment it&#8217;s not particularly likely, but still possible that I will end up producing a 3AP through <img src='http://s0.wp.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' />&#8220;, an equation like &#8220;f(a)=2f(a&#8217;)&#8221; roughly means &#8220;Currently it seems twice as likely that our eventual 3AP will pass through <img src='http://s0.wp.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> than that it passes through <img src='http://s0.wp.com/latex.php?latex=a%27&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a&#039;' title='a&#039;' class='latex' />&#8220;, and so on. </p>
<p>Now for Roth&#8217;s original argument it is definitely essential for technical reasons to just work with <img src='http://s0.wp.com/latex.php?latex=%5C%7B0%2C+1%5C%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;{0, 1&#92;}' title='&#92;{0, 1&#92;}' class='latex' />-valued functions (as one &#8220;changes <img src='http://s0.wp.com/latex.php?latex=N&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='N' title='N' class='latex' />&#8221; throughout the proof), even though this appears to lead to some more or less arbitrary but at the same time rather immaterial choices concerning the boundaries of the subprogressions to which one passes when iterating. But already for Bourgains argument (my understanding of which on a conceptual level is surely by far more limited than yours), it doesn&#8217;t seem to be clear why one has to do so, and if I happen to understand what you suggest to some extent correctly, then you propose to work with a guessing function <img src='http://s0.wp.com/latex.php?latex=f&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f' title='f' class='latex' /> whose values at the center of the Bohr set under consideration is much greater then the value of <img src='http://s0.wp.com/latex.php?latex=f&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f' title='f' class='latex' /> near the boundary. (Instead of taking <img src='http://s0.wp.com/latex.php?latex=f&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f' title='f' class='latex' /> to be the characteristic function of the Bohr set involved.)</p>
<p>At this point, it seems that maybe one can even dispose of working with Bohr sets on a strictly technical level and instead merely &#8220;think of them in the back of our head&#8221;, transforming at each step of the iteration the definition of the Bohr set of which we &#8220;really think&#8221; into an appropriate definition of a certain guessing function (which probably then has to somehow &#8220;codify&#8221; certain pieces of &#8220;regularity information&#8221; on that Bohr set, etc.)</p>
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		<title>By: Michael Bateman</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2835</link>
		<dc:creator><![CDATA[Michael Bateman]]></dc:creator>
		<pubDate>Mon, 14 Feb 2011 19:58:38 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2835</guid>
		<description><![CDATA[Gil--

Someone asked similar questions via email earlier.  Here was my comment then:

------------

About the size of epsilon in the BK paper:  Section 6 in particular requires epsilon to be rather microscopic.  This is the structural theorem for sets with some additive energy but no additive smoothing.  As epsilon grows, the sharpness of the nonsmoothing hypothesis goes away; and over the course of the proof of this theorem, the nonsmoothing parameter gets magnified a number of times, leaving a vacuous theorem unless the nonsmoothing parameter (and consequently epsilon) is very small.
The reason the nonsmoothing hypothesis can disappear as epsilon grows is that the size of the spectrum can increase substantially (-- specifically it could be as large as (density)^{-3}).  Then the random selection argument (--at least in its current form -- which chooses sets of size about N) becomes a less effective way of estimating the number of m-tuples in the spectrum. 

-------------

I have no idea whether you have looked at our paper, so I&#039;ll just ramble a bit more in case it&#039;s helpful.  [BK] includes a theorem for sets that are not &quot;additively smoothing&quot;.  A set S is said to be additively smoothing if its difference set has more structure than itself.  One way to detect this is by comparing the L^p norms of the fourier transform of the set. (In fact this is how we define it precisely.)  (For p even) We know that the L^p norm is the number of p-tuples in the set S.  By Holder&#039;s inequality, an estimate on the number of (say) 4-tuples gives us a free estimate on the number of (say) 8-tuples.  If there are only this many 8-tuples, we say the set is NOT smoothing.  If there are many more, we say the set IS smoothing.  For example, consider a set of size M randomly distributed in a subspace of size M^{1+c} for some 0&lt;c&lt;1.  This is the additively smoothing example because M-M is all of the subspace, which has perfect additive structure.  An non-smoothing example is something like a bunch of (smaller) subspaces with no relation to each other.  Here the set of (popular) differences is just the set we started with, which has (trivially) no additional additive structure.  

Epsilon being small allows us to conclude that the spectrum is nonsmoothing.  The paper includes a structural theorem for nonsmoothing sets.  Large epsilon means smoothing is a possibility.  If there were an equally strong structural theorem for sets with additive smoothing, there might be some hope; but such a theorem would not be a trivial extension of our result.

Michael]]></description>
		<content:encoded><![CDATA[<p>Gil&#8211;</p>
<p>Someone asked similar questions via email earlier.  Here was my comment then:</p>
<p>&#8212;&#8212;&#8212;&#8212;</p>
<p>About the size of epsilon in the BK paper:  Section 6 in particular requires epsilon to be rather microscopic.  This is the structural theorem for sets with some additive energy but no additive smoothing.  As epsilon grows, the sharpness of the nonsmoothing hypothesis goes away; and over the course of the proof of this theorem, the nonsmoothing parameter gets magnified a number of times, leaving a vacuous theorem unless the nonsmoothing parameter (and consequently epsilon) is very small.<br />
The reason the nonsmoothing hypothesis can disappear as epsilon grows is that the size of the spectrum can increase substantially (&#8211; specifically it could be as large as (density)^{-3}).  Then the random selection argument (&#8211;at least in its current form &#8212; which chooses sets of size about N) becomes a less effective way of estimating the number of m-tuples in the spectrum. </p>
<p>&#8212;&#8212;&#8212;&#8212;-</p>
<p>I have no idea whether you have looked at our paper, so I&#8217;ll just ramble a bit more in case it&#8217;s helpful.  [BK] includes a theorem for sets that are not &#8220;additively smoothing&#8221;.  A set S is said to be additively smoothing if its difference set has more structure than itself.  One way to detect this is by comparing the L^p norms of the fourier transform of the set. (In fact this is how we define it precisely.)  (For p even) We know that the L^p norm is the number of p-tuples in the set S.  By Holder&#8217;s inequality, an estimate on the number of (say) 4-tuples gives us a free estimate on the number of (say) 8-tuples.  If there are only this many 8-tuples, we say the set is NOT smoothing.  If there are many more, we say the set IS smoothing.  For example, consider a set of size M randomly distributed in a subspace of size M^{1+c} for some 0&lt;c&lt;1.  This is the additively smoothing example because M-M is all of the subspace, which has perfect additive structure.  An non-smoothing example is something like a bunch of (smaller) subspaces with no relation to each other.  Here the set of (popular) differences is just the set we started with, which has (trivially) no additional additive structure.  </p>
<p>Epsilon being small allows us to conclude that the spectrum is nonsmoothing.  The paper includes a structural theorem for nonsmoothing sets.  Large epsilon means smoothing is a possibility.  If there were an equally strong structural theorem for sets with additive smoothing, there might be some hope; but such a theorem would not be a trivial extension of our result.</p>
<p>Michael</p>
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		<title>By: gowers</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2834</link>
		<dc:creator><![CDATA[gowers]]></dc:creator>
		<pubDate>Mon, 14 Feb 2011 19:54:07 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2834</guid>
		<description><![CDATA[Tom Sanders has also asked this question. He is pretty sure there wouldn&#039;t be much problem (apart from a certain necessary technical slog) in improving any of the 1/2, 2/3 or 3/4 bound by the Bateman-Katz epsilon. It seems that that epsilon is necessarily very small if you stick closely to their argument. Whether one can be more precise and say that there is a certain barrier beyond which one cannot go without a new idea I do not know.]]></description>
		<content:encoded><![CDATA[<p>Tom Sanders has also asked this question. He is pretty sure there wouldn&#8217;t be much problem (apart from a certain necessary technical slog) in improving any of the 1/2, 2/3 or 3/4 bound by the Bateman-Katz epsilon. It seems that that epsilon is necessarily very small if you stick closely to their argument. Whether one can be more precise and say that there is a certain barrier beyond which one cannot go without a new idea I do not know.</p>
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		<title>By: Gil Kalai</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2830</link>
		<dc:creator><![CDATA[Gil Kalai]]></dc:creator>
		<pubDate>Mon, 14 Feb 2011 18:55:03 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2830</guid>
		<description><![CDATA[I have a question as an &quot;observer&quot;. Is it clear why the epsilon gained by Michael and Netz necessary small or perhaps we can hope also for a large epsilon? (If so maybe we can start with the 2/3 exponent, anyway, or even the loglogn Roth). A related question is: Is the argument of Michael and Netz allows adding an epsilon to Bourgain 2/3? (I am aware better results are known but still this seems interesting.)]]></description>
		<content:encoded><![CDATA[<p>I have a question as an &#8220;observer&#8221;. Is it clear why the epsilon gained by Michael and Netz necessary small or perhaps we can hope also for a large epsilon? (If so maybe we can start with the 2/3 exponent, anyway, or even the loglogn Roth). A related question is: Is the argument of Michael and Netz allows adding an epsilon to Bourgain 2/3? (I am aware better results are known but still this seems interesting.)</p>
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		<title>By: gowers</title>
		<link>http://polymathprojects.org/2011/02/13/can-bourgains-argument-be-usefully-modified/#comment-2829</link>
		<dc:creator><![CDATA[gowers]]></dc:creator>
		<pubDate>Mon, 14 Feb 2011 15:28:22 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=202#comment-2829</guid>
		<description><![CDATA[I think the kind of statement I&#039;m going to want is that the only counterexamples are fairly &quot;continuous&quot; functions, which will necessarily have large Fourier coefficients (and not just any old Fourier coefficients either).]]></description>
		<content:encoded><![CDATA[<p>I think the kind of statement I&#8217;m going to want is that the only counterexamples are fairly &#8220;continuous&#8221; functions, which will necessarily have large Fourier coefficients (and not just any old Fourier coefficients either).</p>
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