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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

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There is an approximation theorem of Arakelian, which is essentially an iff-statement. So there is no extension of Mergelyan in the sense you search for. the domain has to be contractible in any case. –  plusepsilon.de Jan 21 '13 at 21:35
    
contractible to a point, I meant to say. –  plusepsilon.de Jan 21 '13 at 21:37
1  
@MarcPalm : There is definitely a confusion here. I am talking about Mergelyan's theorem on uniform approximation by rational functions. Of course, uniform approximation by polynomials is only possible in the "connected complement" case! I'll edit the question for more clarity. –  Malik Younsi Jan 21 '13 at 22:45
    
Is this question purely out of curiosity, or do have something concrete in mind? –  plusepsilon.de Jan 24 '13 at 10:41
    
You misunderstood me. Observe the difference of K and K0 in my second post. –  plusepsilon.de Jan 25 '13 at 12:47
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2 Answers 2

up vote 2 down vote accepted

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.

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1  
This joint universality result is interesting, but what I'm looking for is really a result for compact sets with disconnected complements. Also, just to clarify : I don't want to remove the non-vanishing assumption. I'm fine with it! –  Malik Younsi Jan 24 '13 at 15:41
    
See my new edit. I hope it clarifies what I'm looking for here. –  Malik Younsi Jan 24 '13 at 15:50
    
I think I provide exactly what you want. K_0 is assumed to have holes at the zeros of f_2. Of course K is not K_0. –  plusepsilon.de Jan 25 '13 at 12:44
    
Oh I see, indeed I misread $K_0$ for $K$, sorry about that! But I don't understand why you work with $\log$..? –  Malik Younsi Jan 25 '13 at 20:03
    
it doesn't work without log in that way. sorry if that is not in the sense you wanted it to be. –  plusepsilon.de Jan 26 '13 at 10:50
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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.

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Yes, of course, the theorem, as stated, does not hold if you remove the hypothesis that the complement of $K$ is connected. I'm asking if there exists a (different) universality-like result that would work for compact sets with disconnected complements. Sorry if that wasn't clear, I'll edit the question... –  Malik Younsi Jan 21 '13 at 19:17
    
@JohannesEbert : Do you think the question is clear now? –  Malik Younsi Jan 21 '13 at 19:20
    
Yes, now the question is clear, I was misunderstanding it. –  Johannes Ebert Jan 21 '13 at 21:09
    
@MarcPalm : See my comment to the question... There is a confusion here. –  Malik Younsi Jan 21 '13 at 22:43
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