# What is known about the functional square root of the Riemann zeta function?

Let us consider the Riemann zeta function $$\zeta(s)$$, where $$s$$ can take on values on the domain $$\mathbb{R}_{>1}$$:

$$\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}} .$$

I wonder what is known about the functional square root(s) of the Riemann zeta function defined on the aforementioned domain (*). In other words, I'm curious about the properties of the function(s) $$f$$ such that $$f(f(s)) = \zeta(s). \qquad \qquad (1)$$

Questions

1. Has a closed-form solution been found for $$f$$ in equation $$(1)$$ ?
2. If not (which I expect), have partial results been found for such a function? Properties like existence, (non)uniqueness, continuity, or results about the functional square root of the partial sums? $$f(f(s)) = \sum_{n=1}^{k} \frac{1}{n^{s}} \qquad \qquad k \in \mathbb{Z}_{>0}$$
3. If so, I would be grateful if you have some pointers to relevant articles or other sources.

(Cross-post from MSE.)

(*) Edit as per Gerald Edgar's answer, this condition should be changed. We must define $$f$$ on $$(0, \infty) \cup X$$ for some subset $$X \subset \mathbb{R} \setminus (0,\infty)$$. Then $$f$$ must map $$(1,\infty)$$ bijectively onto $$X$$, and $$X$$ iself onto $$(0,\infty)$$. Under these conditions, there is still a possibility that $$f$$ is both continuous and real-valued. I am interested in the properties of such an $$f$$.

• Are you sure this is the question you want to ask? The zeta function on the domain you consider takes values outside this domain, so it cannot have a functional square root. In fact, for z with real part greater than 1, $\zeta(z)$ may have negative real part. See e.g. arxiv.org/pdf/1112.4910.pdf. – Dmitry Vaintrob May 1 '20 at 23:37
• Well, it turns out that one needs to deal with fixpoints. I no longer have all the references on a website, but take a look at mathoverflow.net/questions/45608/… and my self-answers. I can see that there is a fixpoint in the reals between 1 and, say, 10. – Will Jagy May 2 '20 at 3:35
• @DmitryVaintrob I should also include that I'm only considering real $s>1$, so the values $\zeta(s)$ take are positive and real too – Max Muller May 2 '20 at 8:17
• For real $s > 1$, $\zeta(s)$ maps $(1,\infty)$ bijectively onto itself. It is continuous and decreasing. A functional square root can therefore not be real-valued and continuous. – Gerald Edgar May 2 '20 at 11:27
• @GeraldEdgar Could you please elaborate on that? I don't see why such a root can't be real-valued and continuous at once – Max Muller May 2 '20 at 22:00

Note that $$\zeta$$ maps $$(1,\infty)$$ bijectively onto itself, $$\zeta$$ is continuous, $$\zeta$$ is decreasing on $$(1,\infty)$$.
Suppose $$\zeta = f \circ f$$ where $$f$$ also maps $$(1,\infty)$$ onto itself. A continuous injective function $$f$$ from an interval to an interval is either increasing everywhere or decreasing everywhere. (This is from the intermediate value theorem.) But in either case, $$f\circ f$$ is increasing, so $$f \circ f \ne \zeta$$.
Now of course we can define $$f$$ on some larger set, say $$(1,\infty) \cup X$$, where $$X$$ has the power of the continuum. Let $$f$$ map $$(1,\infty)$$ bijectively onto $$X$$ and $$X$$ bijectively onto $$(1,\infty)$$. Easy.
• Ahh, I see. Thank you! I'll modify the question according to the answer you've given, because I'm still curious about the existence of such an $f$ defined on a larger set. – Max Muller May 3 '20 at 16:46