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



On the Universality of the Riemann zeta-function
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.


