It is well known that $\xi(s)=\xi(1-s)$ is a verified functional equation for all complex $s$, where $\xi(s) = s(s-1) \pi^{s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s)$.
let $k(s)=\xi(1-s)$ and $s=\alpha+\beta i$ be a complex variable , $\alpha , \beta $: are real numbers . and $\beta \neq 0$ defferent from zero
-Is there someone show me if this equivalence is true :
$(\xi \circ k)(s)=0 $ $\implies$ $k(s)=\xi(s)=0 $ is true if and only if the Riemann Hypothesis (RH) is false?
I see this claim help us to proof or disproof RH.
note : i edited the question which has same goal with the precedent.
I would be interest for any replies or any comments. Thank you.
Consider the following propositions: $$P(z): "\xi(k(z))=k(\xi(z))=0"$$ $$Q(z): "k(z)=\zeta(z)=0"$$ $$R(z): "\text{RH is false}"$$ So your problem can be stated as if the following equivalence relation is true: $$(P(z)\to Q(z))\leftrightarrow R(z)$$ Now the above biconditional statement is true if and only if $R(z)\equiv T$ and $P(z)\to Q(z)\equiv T$ or $R(z)\equiv F$ and $R(z)\equiv F$ where $T$ and $F$ stand for "true" and "false" respectively.
First take $R(z)\equiv T$ then there exists some $z_0$ with $\Re(z)\neq 1/2$ such that $\zeta(z_0)=0$. It is well known that zeros of $\xi(z)$ are precisely the non-trivial zeros of zeta function $\zeta(z)$. Therefore $\xi(z_0)=0$. By your definition $k(z)=\xi(1-z)$ and using the symmetry $\xi(z)=\xi(1-z)$ you just get $\xi(k(z))=k(\xi(z))=\xi(\xi(z))$. At $z_0$ we have $\xi(\xi(z_0))=\xi(0)=1/2$ a result you get from the functional definition of $\xi(z)$ without assuming any additional hypothesis. So $P(z)\equiv F$ and whatever the truth value of $Q(z)$ is one has $P(z)\to Q(z)\equiv T$. Therefore we have shown so far \begin{equation} R(z)\to (P(z)\to Q(z)) \end{equation} To check for the other direction notice that if $\xi(\xi(z))=0$ implies $\xi(z)=w_0$ where $w_0$ is a non-trivial zero of zeta function on the critical strip. Since there is no non-trivial zero on the interval $(0,1)$ then the statement $\xi(z)=\zeta(z)=w_0=0$ is false. Therefore the compound proposition $P(z)\to Q(z)$ is false. So for $P(z)\to Q(z)$ to be true you need now necessarily $P(z)\equiv F$ which also leaves $Q(z)$ free to take any of the two truth values because if $P(z)\equiv F$ then $P(z)\to Q(z)\equiv T$ whatever the truth value of $Q(z)$ is. But $P(z)\equiv F$ means $$\xi(\xi(z))\neq 0$$ implying that $\xi(z)\neq w_0$ where $w_0$ is any non-trivial zero of zeta function. However this condition does not pin down any thing on the validity of Riemann hypothesis. Therefore $$(P(z)\to Q(z))\to R(z)$$ is not assured to be true. As a result your biconditional statement is not assured to hold true. But of course you have that the following is true: \begin{equation} R(z)\to (P(z)\to Q(z)) \end{equation}