6 replaces L by zeta; edited body; added 10 characters in body

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 $$\sup | \log L(z \zeta(z + i t) | - \sup | f_1(z) | < \epsilon$$ 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 $$\sup | \log L(z \zeta(z + i t) | < 2 \sup | f_1(z) | .$$ 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.

5 added 29 characters in body

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: $$\sup \Big| \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)\Big|$$ t)\Big| \leq $$=$$ \sup \Big| \frac{\log \zeta}{\log L}(z+i t) zeta(z+i t)}{f_2(z)\log L(z+i t)} \Big| \sup | \log L(z+i t) -f_2(z) | \leq < \sup \Big| \frac{\log \zeta}{\log L}(z+i tzeta(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 L(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 L(z + i t) | < 2 \sup | f_1(z) | .$$This finishes the proof of the corollary assuming the Joint universality theorem. 4 More rigorous statements and proof. Joint universality theorem: Let K be a compact set in the right half of the critical stripe 1/2< \Re s<1 with connected complement. For any two functions f_1 and f_2 holomorphic in the interior of K (vanishing or not) and every \epsilon>, \epsilon>0, we have that the limit$$ \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\} $$This is From this, we can deduce: Corollary: Let K_0 be a so called joint universailty theorem. We work with compact set in the right half of the critical stripe \log, since you wanted to remove 1/2< \Re s<1. Let f be a continuous function on K_0, which is holomorphic on an open set containing K_0. For every \epsilon_0>0, we have that the vanishing assumptionlimit$$ \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\} $$is positive for \lambda being the Lebesgue measure. Then we write our meromorphic function as Proof:By Runge's theorem, it is sufficient to approximate rational functions, whose poles lie outside of g = f_1/f_2 K_0. Let p(z) and approximate with \frac{\log \zeta}{\log L}(z) on q(z) be polynomials such that K_0 compact q does not containing vanish on K_0. Consider \epsilon_0>0 sufficiently small (to be made precise as we go on). Let K :=\mathbb{C}-O, where O is the poles unbounded, connected component of g. Be careful \mathbb{C}-K_0. Consider \epsilon>0 sufficiently small, then use the joint universality theorem for f_1(z)=p(z) and f_2(z) =q(z). We want to show that < \epsilon_0.$$

We estimate the left-hand side:is needed in my proof.t)|.$$The first summand is easy to estimate, the :$$\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)|= \sup | \Big| \frac{\log \zeta}{\log L}(z+i t) | \Big| \sup | \log L(z+i t) -f_2(z) | ,$$\leq \sup \Big| \frac{\log \zeta}{\log L}(z+i t) \Big| \epsilon,$$$$\sup | \frac{\log \zeta}{\log L}(z+i t) | .$$$$uniformly in t. This is indeed possible, because f_2 must we have roughly locally the same number of zeros thatL(z+it), i.e., none in \sup | f_2(z) | - \sup | \log L(z + i t) | < \epsilon$$$K_0$ for sufficiently small $\epsilon$ \sup | \log L(z + i t) | - \sup | f_1(z) | < \epsilon$$by Hurwitz's theoremthe reversed triangle inequality.So we can estimatefor \epsilon \leq \sup | f_2(z) |/2 and \epsilon \leq \sup | f_1(z) | , we have that$$ \frac{\log sup | \zeta}{\log L}(z+i log L(z + i t) | \leq 3000 > \sup | f_1| * f_2(z) |/2 \sup|f_2^{-1}|.$$This also gives you positive lower density:sup | \log L(z + i t) | < 2 \sup | f_1(z) | .$$

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

3 K0 vs K
2 added 53 characters in body