Background: Bombieri and Lagarias showed that a function $f$ with roots $\rho=x+iy$ satisfies has all its roots lying on $x=\frac12$ if and only if

$$\lambda_n :=\sum_\rho 1-\left(1-\frac{1}{\rho}\right)^n$$

is nonnegative for all $n$.

When $f$ is the Riemann zeta restricted to the critical strip, this gives an equivalent statement for the Riemann Hypothesis: find a (sharp, I think) lower bound for $\lambda_n$.

Question: Is there a simple function that serves as an upper bound for $\lambda_n$ for all $n$?

Motivation: I was trying to answer this question, and I suspected there would be a nice way to say that the question was hard because it implies the Riemann Hypothesis. I managed to find another way to show the implication, but I would still like to complete this approach.

This formulation of $\lambda_n$ suggests an easy continuous extension to $\lambda:\Bbb R_+\to\Bbb R$. However, for the purposes of the equivalence, I wanted a function which never went above 1, so if we had some upper bound $\lambda\leq B$, then I could use $f(x)=\lambda(x)/B(x)$ [and so $\lfloor f\rfloor$ in the question would be nonzero iff RH.]

However, I'm a completely new to analytic number theory and so I have absolutely no intuition about these numbers. I used WolframAlpha to plot the first two terms in the partial sum, and I played around with it a little bit, so I am pretty sure that $\lambda(x)$ is real. But that's about it.

  • $\begingroup$ if these sums converge, they are real simply because the roots of $\zeta(s)$ are symmetric around the real axis and $\overline{(1-1/\rho)^{\,n}} = (1-1/\overline{\rho})^n$ $\endgroup$ – reuns Mar 26 '16 at 21:52
  • $\begingroup$ and for an upper bound, you can use that $Im(\rho)$ is distributed like $\frac{1}{2\pi} \log |Im(\rho)|$, see fr.wikipedia.org/wiki/… where $N(T)$ is the number of non-trivial zeros with $|Im(\rho)| \le T$ $\endgroup$ – reuns Mar 26 '16 at 21:57
  • $\begingroup$ @user1952009: Why don't you post this as an answer? $\endgroup$ – Eric Stucky Mar 27 '16 at 4:51
  • $\begingroup$ because it is your question, your problem, if you want a rigorous answer, try yourself if everything works fine $\endgroup$ – reuns Mar 27 '16 at 16:22

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