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According to wikipedia:

'In mathematics, the Riesz function is an entire function defined by Marcel Riesz in connection with the Riemann hypothesis, by means of the power series

$$ {\rm {Riesz}}(x)=-\sum _{k=1}^{\infty }{\frac {(-x)^{k}}{(k-1)!\zeta (2k)}}. $$

It can be shown that

$$ \operatorname {Riesz} (x)=O(x^{e})\qquad ({\text{as }}x\to \infty ) $$

Riesz showed that the Riemann hypothesis is equivalent to the claim that the above is true for any e larger than 1/4.'

Query1 : As i understand K is a positive Integer. The doubt is regarding the input 'x'. Is the domain of 'x' - positive real numbers ? Or it is complex numbers too?

Query2 : If real numbers, proving Riemann is equivalent to proving that for all real values of 'x', the Big O limit of 1/4 holds true?

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I would think the answer to both questions can be extracted from the Wikipedia entry:.

Q1: the Riesz function is a holomorphic function of $x$ over the whole complex plane.

Q2: the Riemann hypothesis holds if $|{\rm Riesz}\,(x)|\leq Mx^{e}$, with $e>1/4$ and some real $M$, for all real $x$ greater than some $x_0$. This follows from the identity

$${\rm Riesz}\,(x)=\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s+1)}{\zeta(-2s)}x^{-s}\,ds$$

which holds for $-1<c<-1/4$ if the Riemann hypothesis is true.

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  • $\begingroup$ Much thanks for the answer. Q1. Understood the domain of 'x' in Riesz in general. Q2. My confusion isn't really clear to its relation to Riemann. So, just to be clear, in its relation to Riemann 1. To prove Riemann, we have to prove there exists a real value (not Complex) of 'x' ( x0), such that for all real x beyond it, the value of Riesz(x), are bounded by the asymptotic as you pointed out. Thus, in this case with its relation to Riemann, we need to consider only the real number line (and not the complex plane). Apologies in advance, i am a computer science student. $\endgroup$
    – TheoryQuest1
    Commented Jan 3, 2016 at 17:22
  • $\begingroup$ added an identity to clarify the relation between the fourth-root rate of growth of the Riesz function along the positive real axis, and the absence of zeroes of $\zeta(z)$ in the strip $1/2<{\rm Re}\,z<2$. $\endgroup$
    – Carlo Beenakker
    Commented Jan 3, 2016 at 18:40
  • $\begingroup$ Thanks again. I do get that the original identity follows from the one updated now. (x = Real, s = the complex values over integral). But, i would be grateful if you could point out if my interpretation of the requirement for Proving Riemann are correct (mentioned in 1. w.r.t. original Riesz definition). An amateur still struggling to learn. $\endgroup$
    – TheoryQuest1
    Commented Jan 3, 2016 at 18:57
  • $\begingroup$ yes, your statement in 1. is correct $\endgroup$
    – Carlo Beenakker
    Commented Jan 3, 2016 at 19:03

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