Unique fixed point of imaginary part of Riemann Zeta function. Let, $a \in \Bbb R$ be any real number with $a>1$. Define, $f:\Bbb R\to \Bbb R$ by $$f(x)=-\sum_{n=1}^\infty\frac{\sin (x\log n)}{n^a},\forall x\in \Bbb R.$$ I have managed to find a way to show it has a fixed point in $\Bbb R$ by Brower's fixed point theorem [which is obviously zero]. 
I want to show it is unique. I am unable to proceed with the known fixed point theorems. After some regular homework on that series, I have concluded that if we are able to show that, "$f(x)$ attains its upper bound/maximum after $x=\zeta(a)$ where $\zeta$- is the Riemann Zeta function then the fixed point will be unique [Not verified yet!]."
Loosely speaking, the problem is directly related to the fact that the function $f_a(x)=\operatorname{Im}(\zeta(a+ix)),x\in\Bbb R$ has a unique fixed point [precisely zero] for every $a>1$.
How should I proceed? Any related paper/note/answer is highly appreciated.
Thanks in advance!
 A: Is $(1,\infty)$ the whole vanishing set of $\Im(\zeta(s) - s), \Re(s) > 1$ ?


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*For $\Re(s)$ large enough it follows from that $|\zeta'(s)| < 1$. 

*For $\Re(s) = 1+\epsilon$ use that $\zeta(s) = \frac{1}{s-1}+O(\log (2+|\Im(s)|))$ uniformly on $\Re(s) > 1$ which restricts the possible zeros of $\Im(\zeta(s) - s)$ to a bounded domain where you can do the numerical checks
A: If $x>0$, then $f(x)<0$ you add infinite negative numbers, if $x<0$, then $f(x)>0$ you add infinite positive numbers, that means you can't find an $x>0$, so that $f(x)>0$ or $x<0$ so that $f(x)<0$, besides $x=0$, where you get $f(0)=0$.
Im able to show and prove that your $f(x)=x$ ,can only happen: if and only if: $s=a+ix$ and the conjugate $s*=a-ix$ are the only fixed points with $ζ(s)=s=ζ(a+ix)=a+ix$ and $ζ(s*)=s*=ζ(a-ix)=a-ix$ , respectively.
(To determine the value of the fixed/critical/equilibrium point(s) , the system behavior has to be stable at $x=0$ or else we say its unstable!)
This is my proof: https://postimg.cc/68hQVYCs
(My apologies no time to right this in latex form)
If this is true the conjugate of $s$, the $s*$ produces a $g(x)=Im(ζ(a-ix))=-x$ , the same way i just proved $f(x)=x$, but you also want for $g(x)$ to find a way to show it has also a fixed point too, so we have $g(x)=x$ for the conjugate too, so by equating we have $-x=x =>2x=0 =>x=0$, the only way this can happen! Done!
In graphical terms, a fixed point $x$ means the point $(x,f(x))$ is on the line $y = x$, or in other words the graph of $f$ has a point in common with that line.
The graph of $g$ has only one point in common with the line $y=x$ and that is when $x=0$. It is unique for $g$ , since $g(x)=-x$, they only meet diagonally at $x=0$, therefore it is unique for $g$,
so equating the unique solution $x=0$ for both $f$ and $g$ we get $f(0)=0$ and $g(0)=0$ for their imaginary part of $s$ and the conjugate $s*$.
This unique solution $x=0$ , means that the fixed point of the zeta function can only be real since the imaginary parts have to be $0$, and must be for $α>1$.
With the same technic i just used with the conjugates of $s$ you can prove that there is an $α$ that gives $ζ(α)=α$ , when $α>1$ for the real part of the zeta function(the infinite summation of $cos$).
And thats because we indicated at first that the s and the conjugate of s are the only two fixed points that the zeta function has, which turns out it is only one unique point at (α,ζ(α)) with α>1.
