Pustyl’nikov’s equivalent criteria for RH

Question

In this paper (Pustyl’snikov 1999 Russ. Math. Surv. 54 262), Pustyl’nikov proved the following two theorems

Theorem 1 All the even derivative of $\xi(s) $ at point $s=1/2$ are strictly positive.

Theorem 2 If at least one of those even derivatives were not positive, then RH would be false: in this case there would exist a complex zero $s_1$ ( and it’s mirror image $1-s_1$) that does not lie on the line $\mathrm{Re}(s)$=1/2.

As you can see that this is similar to Li’s criteria.

Question What is the gap between these two theorems and a proof of RH?

Answer 1

Theorem 1 is trivial $$\xi^{(2n)}(1/2) = 2^{2-2n}\int_1^\infty f(x)(\log x)^{2n}dx$$ with $f(x)= x^{1/4} \sum_{n\ge 1}\pi n^2 e^{-\pi n^2 x}(x\pi n^2-\frac32 ) \ge 0$ on $[1,\infty)$

So the theorem 2 doesn't mean anything. When replacing $\zeta(s)$ by the Dirichlet L-function of a quadratic character we'll get that $\xi_\chi$ has similar properties except that $\xi_\chi(1)=0$ so its derivatives can't be all $\ge 0$, therefore I hardly see what the theorem 2 means.

Answer 2

This is not an answer. It just allows me to easily expand my comment because I am typing this on a tiny cell phone screen. I tested two polynomials (even in z) to see if the coefficients are alternating in sign. $$f(z)=(z^2-1)(z^2-4)(z^2-9)(z^2-25)$$ $$=z^8-39z^6+399z^4-1261z^2+900\tag{1}$$ $$g(z)=(z^2-4)(z^2-9)(z-(1+3i))(z-(1-3i))(z+(1+3i))(z+(1-3i))$$ $$=z^8+3z^6-72z^4-724^2+3600\tag{2}$$.

I do see the sign pattern that Pustyl’nikov implied in his theorem 2 ( and I guessed). Update I did find a counter example to theorem 2. $$ h(z)=(z^2-4)(z^2-9)(z-(3+i))(z-(3-i))(z+(3+i))(z-(3-i))$$ $$=z^8-29z^6+344z^4-1876z^2+3600\tag{3}$$. The sign pattern is alternating.