# How does Yitang Zhang use Cauchy's inequality and Theorem 2 to obtain the error term coming from the $S_2$ sum

I have been reading Yitang Zhang's paper now for one and a half weeks and also volunteered to give a popular talk on the paper next week at Stockholm University.

Today I found a detail in the proof that seems outright wrong, and I am starting to worry that it is a serious problem (in particular since I promised to give a talk about the result next week).

Can anyone explain the last three lines on page 22, or if there indeed is a mistake on those lines give an alternative argument for obtaining the error term that we want to have? Zhang has the following sum $$\mathcal E_i= \sum_{ d < D^2, d | \mathcal P } \tau_3(d) \rho_2(d) \sum_{ c \in \mathcal C_i ( d ) } | \Delta( \theta,d,c) | .$$ Then he says: By Cauchy's inequality and Theorem 2 we have $$\mathcal E_i \ll x \mathcal L^{-A}.$$ Here $\mathcal L$ denotes $\log x$. I do not see this. The reason is that Theorem 2 which is given in the following way: For $1 \leq i \leq k_0$ we have $$\sum_{ d < D^2 , d | \mathcal P } \sum_{ c \in \mathcal C_i(d) } | \Delta (\theta,d,c)|\ll x \mathcal L^{-A},$$

is given in $L^1$-norm, Cauchy's inequality would need something in $L^2$-norm. The natural inequality to use would be $$\| f g \|_1 \leq \| f \| _\infty \| g \| _1,$$ where the first function would be the divisor function $\tau_3(d) \rho_2(d)$ and the second would be the sum in $c$. Thus it seems that Theorem 2 in its current form should not really give anything better than

$$\mathcal E_i \ll \left( \max_{ d < D^2, d | \mathcal P} \tau_3(d) \rho_2(d) \right) \sum_{ d < D^2, d | \mathcal P } \sum_{ c \in \mathcal C_i(d) } | \Delta( \theta,d,c ) | \ll \left( \max_{ d < D^2, d | \mathcal P } \tau_3(d) \rho_2(d) \right) x \mathcal L^{-B}$$ for any $B>0$. Here $D^2$ is a little more than $\sqrt x$ (to be precise $x^{ 1 / 2 + 1 / 584 }$), and $\mathcal P$ is the product of all primes up to a small power of $x$.

Now the sup norm for the divisor function $\tau_3(n)$ certainly grows faster than any power of $\log x$ (even on square free numbers, e.g on primorials), even if it on the average grows like a power of $\log x$). The function $\rho_2(d)$ is defined on page 7 is multiplicative, has support on square free numbers and defined to be $v_p-1$ on the primes, where $v_p$ are the number of residue classes of $\mathcal H$ mod p. With the exception of finitely many $p$ this will be $k_0-1 =3.5\cdot 10^6-1$, i.e really large and contribute much more than the divisor function. Thus this should not get what we need, i.e. a bound of the form $x \mathcal L^{-B}$.

If my concerns are correct, I guess that you can start looking closely at the proof of theorem 2 and see if the same proof holds with the divisor functions thrown in.

If I have made som simple error in the above reasoning I would appreciate your help to understand it, since I would like to understand at least all details on how Theorem 1 implies Theorem 2 before my talk next week (Of course as much as possible of the proof of Theorem 2 also).

Reference: Yitang Zhang: Bounded gaps between primes http://annals.math.princeton.edu/articles/7954

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See also the postscript in my response. Good luck to your talk! – GH from MO May 31 '13 at 20:46
This to me is nearly a perfect showcase example for MathOverflow. (I would change "outright wrong" to "wrong to me" and "som" to "some".) If people saw this first before submitting questions, it might serve as an effective filter. Even better would be a short but representative list of quality questions and responses for people to browse before submitting. Any moderators willing to set up a tour link? Gerhard "Ask Me About Community Relations" Paseman, 2013.05.31 – Gerhard Paseman May 31 '13 at 21:51
@Johan: There is a typo in your arXiv preprint in the display before (8): $\tau_3(d)$ and $\rho_2(d)$ should be squared as in my response below. Thanks for the footnote. – GH from MO Jun 4 '13 at 15:21

That also puzzled me first, but I think it is ok. By Cauchy-Schwarz,

$$\mathcal E_i \leq \left(\sum_{ d < D^2, d | \mathcal P }\sum_{ c \in \mathcal C_i ( d ) } \tau_3^2(d) \rho_2^2(d) | \Delta( \theta,d,c) |\right)^{1/2}\left(\sum_{ d < D^2 , d | \mathcal P } \sum_{ c \in \mathcal C_i(d) } | \Delta (\theta,d,c)|\right)^{1/2}.$$

In the first parenthesis, $| \Delta( \theta,d,c) |\ll x\mathcal{L}/d$ by trivial estimation (for $d < x$), hence the first parenthesis is $\ll x\mathcal{L}^B$ for some fixed $B>0$. The second parenthesis, on the other hand, is $\ll x\mathcal{L}^{-A}$ for any $A>0$. Combining these, $\mathcal E_i \ll x\mathcal{L}^{-C}$ for any $C>0$, and this is sufficient.

P.S. The first display on page 5 requires a small correction: $\mathcal{E}$ should be multiplied by $\mathcal{L}^{2k_0+2l_0}$, because $\lambda(n)^2$ in (2.2) is not bounded (cf. (9.7) in [6]). Of course this does not affect the main argument.

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Thanks, yes I understand the argument now. It is correct, but possibly a case where Zhang could have improved his exposition (since we are at least two persons who got puzzled by it). – Johan Andersson May 31 '13 at 20:44
I agree. I think Zhang understood several years ago that this estimate was the one to focus on, and later he did not bother or forgot to explain it so well. – GH from MO May 31 '13 at 20:46
GH, I acknowledged yours and Denis help in understanding this detail in my recent arxiv paper arxiv.org/abs/1306.0511 "Bounded prime gaps in short intervals" (footnote 3) where I improve on Zhang's result in a certain direction. BTW thanks for the "good luck" my talk is tomorrow, so I am preparing at the moment. – Johan Andersson Jun 4 '13 at 7:48
@GH, Thanks I will fix that error in my next version (I do not have time to do it right now but probably within a week's time). – Johan Andersson Jun 4 '13 at 15:42
There is now an updated version of my paper on arxiv where I fix this as well as some other errors. – Johan Andersson Jun 7 '13 at 9:07

The incorporation of the factor of divisor function, $\tau_?(d)$, in the Bombieri-Vinogradov Theorem is familiar feature in sieve application. For example, this occurs in GPY and Graham's paper "Small gaps between product of two primes" page 748.

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After a conversation with Prof. Terry Tao, he told me to use weighted C-S or split the sum, However, I realized that this can be done by splitting the sum and using the trivial bound, and without appealing to C-S at all.

First sum is $$\sum_{f(d) < L^B} f(d) |\Delta(d)|\ll_A xL^BL^{-A}$$

Second sum is the remaining ones $$\sum_{f(d)>L^B} f(d) |\Delta(d)|\ll\sum_d \frac{f(d)^2}{L^B} \frac{xL}{d}\ll\frac{xL^C}{L^B}$$ for some absolute constant $C$.

Then choose $B>C$, and $A>B$.

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This is nice, but I think that secretly you reproduced the proof of Cauchy-Schwarz. At least one proof goes like this: $2\sum x_iy_i\leq\sum (x_i^2k^{-2}+y_i^2k^2)$. Minimmizing the right hand side in $k$ yields Cauchy-Schwarz. – GH from MO May 31 '13 at 22:37
@GH: Good point! I tried reproducing proof of C-S regarding your comment, but misses a factor of 2. So, this method overestimates than C-S by a factor of 2. – i707107 May 31 '13 at 22:50
The method I outlined gives C-S precisely (note the factor 2 on the left). – GH from MO Jun 1 '13 at 1:01
Yes, your outline precisely is C-S. But mine is truncation of sums, so I would have $\sum x_iy_i\leq K\sum x_i+K^{-1}\sum x_iy_i^2$, since the first sum on the RHS is upper bound for the sum of terms with $y_i\leq K$, and the second sum on the RHS is for $y_i>K$. – i707107 Jun 1 '13 at 5:52
@i707107: Got it, thanks! – GH from MO Jun 1 '13 at 14:25

The argument is OK (in fact it appears already -- probably as sketchily -- in Goldston, Pintz, Yildirim and certainly many other papers involving the Selberg sieve for instance. The point is to use Cauchy-Schwarz with the square root of the modulus of the error term $\Delta$, and one uses a trivial bound on the error term (it could be of size $D(\log D)^{A}$ for some $A$) in one factor, and Theorem 2 in the second: in other words, one writes

$\sum_d f(d) |\Delta(d)|\leq (\sum_d f(d)^2|\Delta(d)|)^{1/2} (\sum_d |\Delta(d)|)^{1/2}.$

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OK. I think I got it now. The trival bound in this case will however not be $D(\log D)^A$. That was what confused me (again removed my previous comments). The trivial bound in this case for the term $\Delta(d)$ will be $x/\phi(d)$. When summing over $d$ and the arithmetical function $f(d)^2$ we will get that after taking square roots that the first term will be of form $(\log x)^A \sqrt{x}$. Then we apply theorem 2 with sufficiently large constant $B$ for the second term. It is a good argument, although I still think that Zhang could have spelled it out less sketchily. – Johan Andersson May 31 '13 at 20:35
Denis, I acknowledged yours and GH's help in understanding this detail in my recent arxiv paper arxiv.org/abs/1306.0511 "Bounded prime gaps in short intervals" (footnote 3) where I improve on Zhang's result in a certain direction. Thanks for your help. – Johan Andersson Jun 4 '13 at 7:50