On the Universality of the Riemann zeta-function

Question

Hi,

I have a question regarding the universality property of the Riemann zeta-function. I am no expert on this, so I'd be glad for any relevant reference.

First, recall Voronin's remarkable theorem on the Universality of the Riemann zeta-function :

Let $K$ be a compact subset with connected complement lying in the strip $\{1/2 < \operatorname{Re}(z)<1\}$, and let $f : K \rightarrow \mathbb{C}$ be continuous, holomorphic on the interior of $K$, and zero-free on $K$. Then for each $\epsilon>0$, there exists $t>0$ such that $$\max_{z \in K} |\zeta(z+it)-f(z)|<\epsilon.$$ Even more : the lower density of the set of such $t$'s is positive..!

Note that of course, the hypothesis that the complement of $K$ is connected is essential in the above theorem.

My question is the following :

Is there some sort of (modified) zeta-function universality-like result for compact sets $K$ with disconnected complements? For example, if $\mathbb{C}_\infty \setminus K$ has a finite number of components?

EDIT

Of course I know that a sequence of the form $f_n(z):=\zeta(z+it_n)$ won't work in the case when the complement of $K$ is disconnected (such a sequence cannot approximate uniformly say $1/z$ on an annulus centered at $0$). I'm asking wether there is some sequence of functions, involving the Riemann zeta-function, that could work in this case, and generalize Voronin's Theorem. Note that such functions will necessarily have poles in each component of the complement of $K$.

2nd EDIT

Let me explain what I was looking for here. Basically, I'd like to know if there exists a result of the following form :

Let $K$ be a compact subset whose complement has finitely many components lying in the strip $\{1/2 < \operatorname{Re}(z)<1\}$, and let $f : K \rightarrow \mathbb{C}$ be continuous, holomorphic on the interior of $K$, and zero-free on $K$. Then for each $\epsilon>0$, there exists...

Here insert some uniform approximation of $f$ on $K$ by a function involving the Riemann zeta-function

Furthermore, in the case when $K$ has connected complement, I would like the above result to reduce to Voronin's Theorem.

In summary, I want to know if there exists a generalization of Voronin's Theorem to compact sets whose complement have finitely many components.

Thank you, Malik

Answer 1

Here is a cheaper alternative depending on what you mean by a modification. Consider $L(z)$ any Dirichlet L function different from $\zeta$.

Joint universality theorem: Let $K$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$ with connected complement. For any two functions $f_1$ and $f_2$ holomorphic in the interior of $K$ (vanishing or not) and every $\epsilon>0$, we have that the limit $$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \{ t \leq T: \sup |f_1(z) - \log \zeta(z +i t)| + \sup |f_2(z) - \log L(z +i t)| < \epsilon\} $$ is positive for $\lambda$ being the Lebesgue measure.

From this, we can deduce:

Corollary: Let $K_0$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$. Let $f$ be a continuous function on $K_0$, which is holomorphic on an open set containing $K_0$. For every $\epsilon_0>0$, we have that the limit $$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \Big\{ t \leq T: \sup\limits_{z \in K_0} \left| f(z) - \frac{\log \zeta(z +i t)}{\log L(z+ it)}\right| < \epsilon_0\Big\} $$ is positive for $\lambda$ being the Lebesgue measure.

Proof: By Runge's theorem, it is sufficient to approximate rational functions, whose poles lie outside of $K_0$. Let $p(z)$ and $q(z)$ be polynomials such that $q$ does not vanish on $K_0$. Consider $\epsilon_0>0$ sufficiently small (to be made precise as we go on).

Let $K :=\mathbb{C}-O$, where $O$ is the unbounded, connected component of $\mathbb{C}-K_0$. Consider $\epsilon>0$ sufficiently small, then use the joint universality theorem for $f_1(z)=p(z)$ and $f_2(z) =q(z)$.

We want to show that $$\sup | f_1/f_2(z) - \frac{\log \zeta}{\log L}(z+i t) |< \epsilon_0.$$

We estimate the left-hand side: $$ \leq \sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | + \sup | \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)|.$$

The first summand is easy to estimate: $$\sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | \leq \sup_{z \in K_0} \left| f_2(z)^{-1} \right| \epsilon.$$ The second one is a little bit harder: $$ \sup \Big| \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)\Big| \leq $$ $$ \sup \Big| \frac{\log \zeta(z+i t)}{f_2(z)\log L(z+i t)} \Big| \sup | \log L(z+i t) -f_2(z) | < \sup \Big| \frac{\log \zeta(z+i t)}{f_2(z) \log L(z+i t)} \Big| \epsilon,$$ because we have to estimate $$ \sup | \frac{\log \zeta}{\log L}(z+i t) | $$ uniformly in $t$.

This is indeed possible, we have that $$\sup | f_2(z) | - \sup | \log L(z + i t) | < \epsilon$$ and $$ \sup | \log \zeta(z + i t) | - \sup | f_1(z) | < \epsilon$$ by the reversed triangle inequality. So for $\epsilon \leq \sup | f_2(z) |/2$ and $\epsilon \leq \sup | f_1(z) |$ , we have that $$ \sup | \log L(z + i t) | > \sup | f_2(z) |/2$$ and $$ \sup | \log \zeta(z + i t) | < 2 \sup | f_1(z) | .$$ So $$\epsilon_0 := \max\{ \frac{1}{2} \sup |f_2^{-1}| \epsilon, \frac{1}{2} 4* \sup |f_1f_2^{-2}| \epsilon \}$$ will do.

This finishes the proof of the corollary assuming the Joint universality theorem.

Answer 2

No, it cannot be true. Suppose the theorem holds for an appropriate translation of the annulus $K:=\{1/2 \leq z \leq 1\}$ and the function $f(z)=1/z$. Then there is a sequence $f_n$ of holomorphic functions on the closed unit disc, continuous on the boundary, such that $f_n (z) \to 1/z$ uniformly in $K$ (because $\zeta$ does not have a pole away from $1$). By the maximum principle, the convergence extends to the whole unit disc, contradicting the identity theorem, because $f$ is not holomorphic on the disc.