# 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

• See also the postscript in my response. Good luck to your talk! – GH from MO May 31 '13 at 20:46
• @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 ). Of course this does not affect the main argument.

• 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 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

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

• 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

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 $A>B$.

• 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. – Sungjin Kim 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 $y_i>K$. – Sungjin Kim Jun 1 '13 at 5:52

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.