TITLE: What are the obstructions to showing that $\zeta$ doesn't vanish on the strip $1- \varepsilon < {\rm Re}(s) \leq 1$
QUESTION [18 upvotes]: Most (if not all) of the proofs of the Prime Number Theorem that I have seen in the
literature rely on the fact that the Riemann zeta function, $\zeta(s)$, does not vanish
on the line ${\rm Re}(s) = 1$.  Since we have a lot of numerical evidence for the validity of RH, one would suspect that for small $\varepsilon$ (to start with) we have that $\zeta(s) \neq 0$ on the strip $1- \varepsilon < {\rm Re}(s) \leq 1$.  Does anyone know what are the main difficulties for proving such a result?

REPLY [2 votes]: The third approach looks at a known zero free region. The simplest such theorem is that there is some $A$ such that
$$
\zeta(s)\ne0\quad\text{for}\quad \sigma>1-A/\log(t).
$$
What prevents us from 'straightening it out' so the  boundary is a vertical line?
The proof begins like the proof that $\zeta(1+it)\ne0$ (see previous answer), but considers instead of $\log(\zeta(s))$, rather $\log(\zeta(s))^\prime=\zeta^\prime(s)/\zeta(s)$, again expanded as a sum over primes.  For $\sigma>1$, and any $t$, taking real parts and using the same trig identity shows that
$$
0\le -3\frac{\zeta^\prime}{\zeta}(\sigma)-4\text{Re} \frac{\zeta^\prime}{\zeta}(\sigma+it)-\text{Re}\frac{\zeta^\prime}{\zeta}(\sigma+2it).
$$
We will estimate each of the three terms on the left side.  The first is easy:
$$
-3\frac{\zeta^\prime}{\zeta}(\sigma)=\frac{3}{\sigma-1}-3C+O(\sigma-1)<\frac{3}{\sigma-1},
$$
where $C$ is the Euler constant.
For the other two, we need the expression of $\zeta(s)$ as a product over nontrivial zeros $\rho=\beta+i\gamma$.  Taking logarithmic derivatives gives
$$
-\frac{\zeta^\prime(s)}{\zeta(s)}=-\log(2\pi)+1+C+\frac{1}{s-1}+\frac12\frac{\Gamma^\prime(s/2+1)}{\Gamma(s/2+1)}-\sum_\rho\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right).
$$
Via Stirling's formula, for $s\to\infty$ in a vertical strip
$$
\frac{\Gamma^\prime(s)}{\Gamma(s)}\sim \log(s)\sim\log(t).
$$
This is where the $\log(t)$ in our zero free region will come from, but so far
$$
-\frac{\zeta^\prime(s)}{\zeta(s)}+\sum_\rho\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right)\sim\frac12\log(t/2).
$$
Taking real parts, the asymptotic relation above gives that
$$
-\text{Re}\frac{\zeta^\prime(s)}{\zeta(s)}+\sum_\rho\left(\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma)^2}+\frac{\beta}{\beta^2+\gamma^2}\right)<\log(t).
$$
Recalling that $\sigma>1$, every term in the sum over $\rho$ is positive, so we estimate the last of our three terms as
$$
-\text{Re}\frac{\zeta^\prime(\sigma+2it)}{\zeta(\sigma+2it)}<\log(2t).
$$
To get  a zero free region, we now fix a zero $\rho_0=\beta_0+i\gamma_0$, and set $t=\gamma_0$ in the $s$ parameter.  In estimating the middle of our three terms, we drop every term in the sum except that corresponding to $\rho=\rho_0$. This gives
$$
-4\text{Re}\frac{\zeta^\prime(\sigma+i\gamma_0)}{\zeta(\sigma+i\gamma_0)}<4\log(\gamma_0)-\frac{4}{\sigma-\beta_0}.
$$
Combining the three estimates, we now have
$$
0\le \frac{3}{\sigma-1}+4\log(\gamma_0)-\frac{4}{\sigma-\beta_0}+\log(2\gamma_0)\le\frac{3}{\sigma-1}-\frac{4}{\sigma-\beta_0}+5\log(\gamma_0).
$$
Thus
$$
\frac{3}{\sigma-1}+5\log(\gamma_0)\ge \frac{4}{\sigma-\beta_0}
$$
$$
\sigma-\beta_0\ge \frac{4(\sigma-1)}{3+5(\sigma-1)\log(\gamma_0)},
$$
and
$$
1-\beta_0\ge \frac{4(\sigma-1)}{3+5(\sigma-1)\log(\gamma_0)}+1-\sigma.
$$
Finally setting the parameter $\sigma=1+1/(10\log(\gamma_0))$, the right side of the inequality simplifies and we obtain
$$
1-\beta_0\ge\frac{1}{70\log(\gamma_0)}.
$$

How might we do better?  As mentioned above, the $\log(t)$ arises from the contribution of the logarithmic derivative of the $\Gamma$ function.  Except for the factor of $1/2$ we dropped, this is the best estimate possible.  But we lost a lot by first introducing all the zeros of $\zeta(s)$, and then throwing out all but one.  A better zero free region may be obtained by subtracting off from the logarithmic derivative only the contribution of the nearby zeros.  The best known bounds replace $1/\log(t)$ with $1/(\log(t)^{2/3}\log\log(t)^{1/3})$.