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> 



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


