On the Universality of the Riemann zeta-function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:55:57Z http://mathoverflow.net/feeds/question/119499 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function On the Universality of the Riemann zeta-function Malik Younsi 2013-01-21T18:14:45Z 2013-01-28T10:46:09Z <p>Hi,</p> <p>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.</p> <p>First, recall Voronin's remarkable theorem on the Universality of the Riemann zeta-function :</p> <p>Let $K$ be a compact subset with connected complement lying in the strip ${1/2 &lt; \operatorname{Re}(z)&lt;1}$, 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 $$\max_{z \in K} |\zeta(z+it)-f(z)|&lt;\epsilon.$$ Even more : the lower density of the set of such $t$'s is positive..!</p> <p>Note that of course, the hypothesis that the complement of $K$ is connected is essential in the above theorem.</p> <p>My question is the following :</p> <p><strong>Is there some sort of (modified) zeta-function universality-like result for compact sets $K$ with <em>disconnected</em> complements? For example, if $\mathbb{C}_\infty \setminus K$ has a finite number of components?</strong></p> <p><strong>EDIT</strong></p> <p>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 <strong>some</strong> 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 $K$. </p> <p><strong>2nd EDIT</strong></p> <p>Let me explain what I was looking for here. Basically, I'd like to know if there exists a result of the following form :</p> <p>Let $K$ be a compact subset <strong>whose complement has finitely many components</strong> lying in the strip ${1/2 &lt; \operatorname{Re}(z)&lt;1}$, 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...</p> <p><em>Here insert some uniform approximation of $f$ on $K$ by a function involving the Riemann zeta-function</em></p> <p>Furthermore, in the case when $K$ has <strong>connected</strong> complement, I would like the above result to reduce to Voronin's Theorem.</p> <p>In summary, I want to know if there exists a generalization of Voronin's Theorem to compact sets whose complement have finitely many components.</p> <p>Thank you, Malik</p> http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119502#119502 Answer by Johannes Ebert for On the Universality of the Riemann zeta-function Johannes Ebert 2013-01-21T19:12:02Z 2013-01-21T19:12:02Z <p>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.</p> http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119759#119759 Answer by Marc Palm for On the Universality of the Riemann zeta-function Marc Palm 2013-01-24T13:52:32Z 2013-01-28T10:46:09Z <p>Here is a cheaper alternative depending on what you mean by a modification. Consider $L(z)$ any Dirichlet L function different from $\zeta$.</p> <blockquote> <p><strong>Joint universality theorem</strong>: Let $K$ be a compact set in the right half of the critical stripe $1/2&lt; \Re s&lt;1$ 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 $$\inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \{ t \leq T: \sup |f_1(z) - \log \zeta(z +i t)| + \sup |f_2(z) - \log L(z +i t)| &lt; \epsilon\}$$ is positive for $\lambda$ being the Lebesgue measure.</p> </blockquote> <p>From this, we can deduce:</p> <blockquote> <p><strong>Corollary</strong>: Let $K_0$ be a compact set in the right half of the critical stripe $1/2&lt; \Re s&lt;1$. 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 $$\inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \Big\{ t \leq T: \sup\limits_{z \in K_0} \left| f(z) - \frac{\log \zeta(z +i t)}{\log L(z+ it)}\right| &lt; \epsilon_0\Big\}$$ is positive for $\lambda$ being the Lebesgue measure.</p> </blockquote> <p><em>Proof</em>: 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).</p> <p>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 $f_2(z) =q(z)$.</p> <p>We want to show that $$\sup | f_1/f_2(z) - \frac{\log \zeta}{\log L}(z+i t) |&lt; \epsilon_0.$$</p> <p>We estimate the left-hand side: $$\leq \sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | + \sup | \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)|.$$</p> <p>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: $$\sup \Big| \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)\Big| \leq$$ $$\sup \Big| \frac{\log \zeta(z+i t)}{f_2(z)\log L(z+i t)} \Big| \sup | \log L(z+i t) -f_2(z) | &lt; \sup \Big| \frac{\log \zeta(z+i t)}{f_2(z) \log L(z+i t)} \Big| \epsilon,$$ because we have to estimate $$\sup | \frac{\log \zeta}{\log L}(z+i t) |$$ uniformly in $t$.</p> <p>This is indeed possible, we have that $$\sup | f_2(z) | - \sup | \log L(z + i t) | &lt; \epsilon$$ and $$\sup | \log \zeta(z + i t) | - \sup | f_1(z) | &lt; \epsilon$$ 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 $$\sup | \log \zeta(z + i t) | &lt; 2 \sup | f_1(z) | .$$ So $$\epsilon_0 := \max\{ \frac{1}{2} \sup |f_2^{-1}| \epsilon, \frac{1}{2} 4* \sup |f_1f_2^{-2}| \epsilon \}$$ will do. </p> <p>This finishes the proof of the corollary assuming the Joint universality theorem.</p>