Proof of Proposition 3.1

We first prove (a). Let $(R_n)_n\subset [0,1)$ be such that $0<R_n<r_n<R_{n+1}<r_{n+1}<1$ , $n\in {\mathbb {N}}$ . Let us define, for every $0<r<1$ and $k\geq 0$ ,

(3.1) $$ \begin{align} H_k(r)=\sum_{j\geq k}h_j(r)\quad \mbox{ where }\quad h_j(r):=\sum_{i=j}^{+\infty}2ir^i. \end{align} $$

Observe that $H_k(r)\rightarrow 0$ as k tends to infinity. We choose $u_1>0$ such that

$$ \begin{align*}\max\left(H_{u_{1}}(r_{1}),\frac{(r_1/R_{2})^{u_1}}{1-r_1/R_{2}}\right)\leq \varepsilon_1.\end{align*} $$

By Mergelyan’s theorem, we find $P_1=\sum \limits _{i\geq u_1}a_{i,1}z^i$ so that

$$ \begin{align*}\sup_{\vert z\vert\leq R_1}\vert P_1(z)\vert\leq \varepsilon_1\quad \mbox{ and }\quad \sup_{z\in K_{\alpha(1)}}|P_1(r_1z)-\varphi_{\beta(1)}(z)|\leq \varepsilon_1.\end{align*} $$

Let us build by induction sequences $(P_n)_n$ and $(Q_n)_n$ of polynomials and an increasing sequence $(u_n)_n$ of integers as follows. Let $Q_0=0$ and suppose that $P_1,\dots ,P_{n-1}$ , $u_1,\dots ,u_{n-1}$ , and $Q_0,\dots ,Q_{n-2}$ have been chosen for $n\geq 2$ . We shall write these polynomials under the following form:

$$ \begin{align*}\mbox{for } k=2,\dots, n-2,\hspace{6pt} P_k=\sum_{i=u_k}^{u_{k+1}-1}a_{i,k}z^i,\hspace{6pt} Q_{k}=\sum_{i=u_{k}}^{u_{k+1}-1}b_{i,k}z^i,\hspace{6pt} \mbox{ with }u_k\geq\deg(P_{k-1})+1,\end{align*} $$

and

$$ \begin{align*}P_{n-1}=\sum_{i=u_{n-1}}^{\deg(P_{n-1})}a_{i,n-1}z^i,\mbox{ with }u_{n-1}\geq\deg(P_{n-2})+1,\end{align*} $$

where the coefficients $a_{i,j}$ and $b_{i,j}$ will satisfy suitable conditions that will be specified in the induction. We are going to construct $u_n$ , $Q_{n-1}$ , and $P_n$ in this order. First, choose $u_n\geq 1+ \deg (P_{n-1})$ satisfying

1. (1) $\max \left (H_{u_{n}}(r_{n}), \frac {(r_n/R_{n+1})^{u_n}}{1-r_n/R_{n+1}}\right )\leq \varepsilon _n$ .

Set $a_{i,n-1}=0$ for $i=\deg (P_{n-1})+1,\dots ,u_n-1$ , which implies that $P_{n-1}=\sum \limits _{i=u_{n-1}}^{u_n-1}a_{i,n-1}z^i.$ Let us consider $w_{n-2}(z)=\sum \limits _{j=1}^{n-2}(P_j+Q_j)(z)$ , $z\in {\mathbb {C}}$ . Then we define $Q_{n-1}=\sum \limits _{i= u_{n-1}}^{u_n-1}b_{i,n-1}z^i$ , where the coefficients $b_{i,n-1}$ for $u_{n-1}\leq i\leq u_n-1$ are built by induction as follows. We first set

$$ \begin{align*} b_{u_{n-1},n-1}:=\left\{\begin{array}{ll} 0, & \text{if } \vert \sum\limits_{l=1}^{u_{n-1}}S_l(w_{n-2})(1) + a_{u_{n-1},n-1}\vert \geq u_{n-1},\\ 2u_{n-1}, & \text{otherwise}. \end{array} \right. \end{align*} $$

Then, once $b_{u_{n-1},n-1},b_{u_{n-1}+1,n-1},\ldots , b_{k-1,n-1}$ have been built for some $u_{n-1}+1\leq k\leq u_n-1$ , we set:

• • $b_{k,n-1}=0$ if we have $\left \vert \sum \limits _{l=1}^{k}S_l(w_{n-2})(1) + \sum \limits _{l=u_{n-1}}^{k-1}\sum \limits _{i=u_{n-1}}^{l}(a_{i,n-1}+b_{i,n-1}) + \sum \limits _{i=u_{n-1}}^{k-1}(a_{i,n-1}+b_{i,n-1}) + a_{k,n-1}\right \vert \geq k,$

• • $b_{k,n-1}=2k$ otherwise.

Doing so, we obtain for all $n\geq 2$ and for all $k=u_{n-1},\dots ,u_n-1$ ,

1. (2) $\vert \sum _{l=1}^{k}S_l(\sum _{j=1}^{n-1}(P_j+Q_j))(1) \vert \geq k$ ;

2. (3) $|b_{k,n-1}|\leq 2k$ .

Then apply Mergelyan’s theorem to get $P_n=\sum \limits _{i\geq u_n}a_{i,n}z^i$ so that the following conditions hold:

1. (4) $\sup _{|z|\leq R_n} | P_n(z) | \leq \varepsilon _n$ ;

2. (5) $\sup _{z\in K_{\alpha (n)}}|P_n(r_nz)-\left (\varphi _{\beta (n)}(z)-\sum _{0\leq j\leq n-1}(P_j+Q_j)(r_nz)\right )|\leq \varepsilon _n$ .

Finally, we set $f=\sum \limits _{n\geq 1}(P_n+Q_n)$ and get from (3) and (4) that $f\in H({\mathbb {D}})$ . Moreover, it is clear from (2) that $|\sum \limits _{l=1}^{k}S_l(f)(1)|\geq k$ , for all $k\in {\mathbb {N}}$ , and therefore $f\not \in {\mathcal {U}}_{Ces}({\mathbb {D}},{\mathbb {T}})$ . The proof will be completed once we have proved that $f\in {\mathcal {U}}_A({\mathbb {D}},\varrho )$ . To do this, we will use the equality (3.1). Let us fix $n,m\in {\mathbb {N}}$ , and let $(n_l)_l\subset {\mathbb {N}}$ be such that for any $l\in {\mathbb {N}}$ , $\alpha (n_l)=n$ and $\beta (n_l)=m$ . By (5), we have, for any $\zeta \in K_n$ ,

$$ \begin{align*} \left|f(r_{n_l}\zeta)-\varphi_m(\zeta)\right| \leq \varepsilon_{n_l} + \left|Q_{n_l}(r_{n_l}\zeta)\right| + \left|\sum _{j\geq n_l+1} \left(P_j(r_{n_l}\zeta)+Q_j(r_{n_l}\zeta)\right)\right|\!. \end{align*} $$

Moreover, it follows from (3), (3.1), and (1) that for any $\zeta \in K_{n}$ ,

$$ \begin{align*} \left|Q_{n_l}(r_{n_l}\zeta)\right| \leq \sum_{i\geq u_{n_l}}2ir_{n_l}^i = h_{u_{n_l}}(r_{n_l})\leq H_{u_{n_l}}(r_{n_l})\leq \varepsilon_{n_l}. \end{align*} $$

Now, let us observe by (4) and Cauchy’s inequalities that we have $\left |a_{i,j}\right |\leq \varepsilon _{j}R_{j}^{-i}$ for any $j\geq 1$ and any $u_j \leq i \leq u_{j+1}-1$ , whence by (1), (3.1), and (3),

$$ \begin{align*} \left|\sum_{j\geq n_l+1} \left(P_j(r_{n_l}\zeta)+Q_j(r_{n_l}\zeta)\right)\right| & \leq \sum_{j\geq n_l+1}\sum_{i\geq u_{j}}\varepsilon_{j}\left(\frac{r_{n_l}}{R_{j}}\right)^i+ \sum_{j\geq n_l+1}\sum_{i\geq u_{j}}2ir_{n_l}^i \\& \leq \sum_{j\geq n_l+1}\varepsilon_{j}\left(\frac{r_{n_l}}{R_{n_l+1}}\right)^{u_{n_l}}\frac{1}{1-r_{n_l}/R_{n_l+1}}+ \sum_{j\geq n_l+1} h_{u_j}(r_{n_l}) \\&\leq \sum_{j\geq n_l+1}\varepsilon_{j}+H_{u_{n_l}}(r_{n_l}) \\& \leq \sum_{j\geq n_l+1}\varepsilon_{j} + \varepsilon_{n_l}. \end{align*} $$

Altogether we get

$$ \begin{align*} \sup_{\zeta\in K_n}\left|f(r_{n_l}\zeta) -\varphi_m (\zeta) \right|\rightarrow 0\quad\text{as }l\to \infty, \end{align*} $$

and thus $f\in {\mathcal {U}}_A({\mathbb {D}},\varrho )$ .

Let us prove (b). Let $\varrho '=\left (r_n'\right )_n\subset [0,1)$ be an increasing sequence such that $r_n < r^{\prime }_n < r_{n+1}$ , $n\in {\mathbb {N}}$ . Using Runge’s theorem, we build by induction a sequence of polynomials $P_n$ such that:

1. (1) $\sup _{|z|\leq r^{\prime }_{n-1}}\vert P_n(z)\vert \leq \varepsilon _n$ ;

2. (2) $\vert P_n(r_n) + \sum _{0\leq j\leq n-1} P_j(r_n)\vert \leq \varepsilon _n$ ;

3. (3) $\sup _{\zeta \in K_{\alpha (n)}}\left \vert P_n(r^{\prime }_n\zeta ) - (\varphi _{\beta (n)}(\zeta ) - \sum _{0\leq j\leq n-1} P_j(r^{\prime }_n\zeta ))\right \vert \leq \varepsilon _n$ .

Then we set $f=\sum _{j\geq 0}P_j$ . We get from (1) that $f\in H({\mathbb {D}})$ . Let us now check that $f\in {\mathcal {U}}_A({\mathbb {D}})\setminus {\mathcal {U}}_A({\mathbb {D}},\varrho )$ . It is first clear from (1) and (2) that $f\not \in {\mathcal {U}}_A({\mathbb {D}},\varrho )$ since $|f(r_n)|\leq \sum _k\varepsilon _k$ . Now, let us fix $n,m\in {\mathbb {N}}$ and let $(n_l)_l\subset {\mathbb {N}}$ be increasing such that for any $l\in {\mathbb {N}}$ , $\alpha (n_l)=n$ and $\beta (n_l)=m$ for any $l\in {\mathbb {N}}$ . Then, by (1) and (3), for any $l\in {\mathbb {N}}$ and any $\zeta \in K_{n}$ ,

$$ \begin{align*} \left|f\left(r^{\prime}_{n_l}\zeta\right) - \varphi_m(\zeta) \right| \leq \varepsilon_{n_l} + \sum _{j\geq n_l+1}\left|P_j\left(r^{\prime}_{n_l}\zeta\right)\right| \leq \sum _{j\geq n_l}\varepsilon_j, \end{align*} $$

which goes to $0$ as $l\to \infty $ . Thus, $f\in {\mathcal {U}}_A({\mathbb {D}},\varrho ')\subset {\mathcal {U}}_A({\mathbb {D}})$ .

We turn to proving (c). For $\zeta \in {\mathbb {T}}$ and $A\subset [0,1)$ , we will denote by $A\zeta $ the set $\{r\zeta :\,r\in A\}$ . Let us fix $\zeta _1\neq \zeta _2$ in ${\mathbb {T}}$ , and let $(a_n)$ be an increasing sequence in $[0,1)$ , tending to $1$ . Note that for any $r\in [a_0,1)$ and any $K\subset {\mathbb {T}}$ containing $\zeta _1$ and $\zeta _2$ , we have

$$ \begin{align*} rK\cap \bigcup_n([a_{2n},a_{2n+1}]\zeta_1\cup [a_{2n+1},a_{2n+2}]\zeta_2) \neq \emptyset. \end{align*} $$

We also fix a sequence $(\eta _n)$ of positive real numbers such that

$$ \begin{align*} \eta_n < \min(a_{2n+2}-a_{2n+1},a_{2n+3}-a_{2n+2},|\zeta_1-\zeta_2|/(2\pi)),\quad n\in {\mathbb{N}}. \end{align*} $$

Remark that $\eta _n \to 0$ as $n\to \infty $ . Let us now consider, for $n\in {\mathbb {N}}$ , the compact set $L_n$ defined by

$$ \begin{align*} L_n=\overline{D(0,a_{2n+2}+\eta_n)}\setminus D((a_{2n+2}+\eta_n)\zeta_1,2\eta_n). \end{align*} $$

The sequence $(L_n)$ is clearly an exhaustion of ${\mathbb {D}}$ by compact sets with connected complement. Moreover, we have, for any $n\in {\mathbb {N}}$ and any $m>n$ ,

$$ \begin{align*} \text{int}(L_n) \supset [a_{2n},a_{2n+1}]\zeta_1 \cup [a_{2n+1},a_{2n+2}]\zeta_2\quad \end{align*} $$

and

$$ \begin{align*} L_n\cap ([a_{2m},a_{2m+1}]\zeta_1\cup [a_{2m+1},a_{2m+2}]\zeta_2)=\emptyset. \end{align*} $$

Then, for any $I\subset {\mathbb {T}}$ , we denote by $C(I)$ the cone $\{r\zeta :\,r\in [0,1),\,\zeta \in I\}$ and by $C_n(I)$ the set $C(I)\cap \partial L_n$ . Thus, for any $n\geq 1$ , the sets $L_{n-1}$ , $[a_{2n},a_{2n+1}]\zeta _1\cup [a_{2n+1},a_{2n+2}]\zeta _2$ , and $C_n(K_{\alpha (n)})$ are pairwise disjoint (and with connected complement), so we can apply Runge’s theorem to build a sequence $(P_n)$ of polynomials (with $P_0=0$ ), satisfying for any $n\geq 1$ ,

1. (1) $\sup _{z\in L_{n-1}}\vert P_n(z)\vert \leq \varepsilon _n$ ;

2. (2) $\sup _{z\in [a_{2n},a_{2n+1}]\zeta _1\cup [a_{2n+1},a_{2n+2}]\zeta _2}\vert P_n (z) + \sum _{0\leq j\leq n-1} P_j(z)\vert \leq \varepsilon _n$ ;

3. (3) $\sup _{z\in C_n(K_{\alpha (n)})}\left \vert P_n(z) - (\varphi _{\beta (n)}(z/|z|) - \sum _{0\leq j\leq n-1} P_j(z))\right \vert \leq \varepsilon _n$ .

We set $f=\sum _{n\geq 0}P_n$ . By (1), $f\in H({\mathbb {D}})$ . Moreover, $f\not \in {\mathcal {U}}_A({\mathbb {D}})$ . Indeed, let us consider the compact set $K=\{\zeta _1\}\cup \{\zeta _2\}$ and fix $r\in [0,1)$ . Then there exists $n\in {\mathbb {N}}$ such that

$$ \begin{align*} rK\cap ([a_{2n},a_{2n+1}]\zeta_1\cup [a_{2n+1},a_{2n+2}]\zeta_2)\neq \emptyset. \end{align*} $$

So, by (1) and (2), there exists $i\in \{1,2\}$ such that $|f(r\zeta _i)|\leq \sum _k\varepsilon _k$ . This proves that $f\not \in {\mathcal {U}}_A({\mathbb {D}})$ .

It remains to check that $f\in {\mathcal {U}}_C({\mathbb {D}})$ . To do so, we will first prove that f has some universality property at the boundary, with respect to $(\partial L_n)_n$ . More precisely, let us fix $n,m\in {\mathbb {N}}$ and let $(n_l)_l\subset {\mathbb {N}}$ be increasing such that $\alpha (n_l)=n$ and $\beta (n_l)=m$ for any l. Then, by (1) and (3), for any $l\in {\mathbb {N}}$ and any $z \in C_{n_l}(K_n)$ ,

$$ \begin{align*} \left|f\left(z\right) - \varphi_m\left(\frac{z}{|z|}\right) \right| \leq \varepsilon_{n_l} + \sum _{j\geq n_l+1}\left|P_j(z)\right| \leq \sum _{j\geq n_l}\varepsilon_j, \end{align*} $$

which goes to $0$ as $l\to \infty $ . Now, let $\gamma $ be any continuous path to some $\zeta \in {\mathbb {T}}$ and let $c\in {\mathbb {C}}$ . By construction, if $I\subset {\mathbb {T}}$ with $\zeta \in I$ is an arc, then $\gamma $ intersects all but finitely many $C_k(I)$ . In particular, $\gamma $ intersects all $C_{n_l}(K_n)$ , $l\in {\mathbb {N}}$ , for some $n\in {\mathbb {N}}$ and some increasing sequence $(n_l)_l\subset {\mathbb {N}}$ so that

$$ \begin{align*} \sup_{z\in {\mathcal{C}}_{n_l}(K_n)}\left|f\left(z\right) - c \right| \to 0,\quad \text{ as } l\to \infty. \end{align*} $$

Thus, if $(z_{n_l})_l$ is any sequence so that $z_{n_l}\in \gamma \cap C_{n_l}(K_n)$ , we have $z_{n_l}\to \zeta $ and $f(z_{n_l})\to c$ as $l\to \infty $ . This completes the proof that $f\in {\mathcal {U}}_C({\mathbb {D}})$ .

Proof If $rK\cap [0,1)=\emptyset $ , it is enough to apply Mergelyan’s theorem on $rK\cup \overline {V}\cup \overline {D(0,r')}$ , where V is any neighborhood of $[0,1]$ with $\overline {V}\cap rK=\emptyset $ , and then Weierstrass’ theorem.

Let us now focus on the case where $rK\cap [0,1)=\{r\}$ (note that this intersection cannot contain more than $1$ point). For $\delta>0$ , let us denote $C_{\delta }:=\{z\in {\mathbb {C}}:\,\text {dist}(z,[r,1])\leq \delta \}$ . Since $r'<r$ and by uniform continuity of $\varphi $ on $rK$ , there exist $\delta>0$ such that $\overline {D(0,r'+\delta )}\cap C_\delta =\emptyset $ , and a function $\tilde {\varphi }$ continuous on $\overline {D(0,r'+\delta )}\cup rK\cup C_{\delta }$ , such that

(4.1) $$ \begin{align}\hspace{15pt} \sup_{z\in rK}|\varphi(z)-\tilde{\varphi}(z)|\leq \varepsilon/2 \quad \text{and} \quad \tilde{\varphi}(z)= \left\{\begin{array}{lll} \varphi(r), & \text{if } & z\in C_{\delta},\\ 0, & \text{if } & z\in \overline{D(0,r'+\delta)}. \end{array} \right. \end{align} $$

Note that $\overline {D(0,r'+\delta )}\cup rK\cup C_{\delta }$ has connected complement and that $\tilde {\varphi }$ is continuous on this set and holomorphic in its interior (since it is constant there). Now, by Mergelyan’s theorem, there exists a sequence $(P_n)_n$ of polynomials tending to $\tilde {\varphi }$ uniformly on $\overline {D(0,r'+\delta )}\cup rK\cup C_{\delta }$ . In particular, by Weierstrass’ theorem, $P^{(l)}_n\to 0$ uniformly on $[0,r']\cup [r,1]$ , whence we conclude that there exists $n_0\in {\mathbb {N}}$ such that $P_{n_0}$ satisfies (1)–(3) in the statement.

Proof of Theorem 4.1

Let us fix $l\geq 1$ . We let $\varrho =(r_n)_n$ and set $r_{-1}=0$ . Without loss of generality, we shall assume that $(r_n)_n$ is increasing. We recall that the notation $(\varepsilon _n)_n$ denotes a decreasing sequence of positive real numbers such that $\sum _n\varepsilon _n<1$ (see Section 2). Using Lemma 4.2, we build by induction polynomial $P_n$ , $n\in {\mathbb {N}}$ , such that:

1. (1) $\sup _{|z|\leq r_{n-1}}\vert P_n (z) \vert \leq \varepsilon _n$ ;

2. (2) $\sup _{z\in [0,r_{n-1}]\cup [r_n,1)}\vert P^{(l)}_n(z) \vert \leq \varepsilon _n$ ;

3. (3) $\sup _{z \in K_{\alpha (n)}}\left \vert P_n (r_nz)-(\varphi _{\beta (n)}(z) - \sum _{0\leq j\leq n-1} P_j(r_nz)) \right \vert \leq \varepsilon _n$ .

By (1), $f:=\sum _nP_n \in H({\mathbb {D}})$ and, by (2), for any $n\in {\mathbb {N}}$ and any $k\geq 0$ , $\vert P_n^{(l)}(r_k)\vert \leq \varepsilon _n$ , so $\vert f^{(l)}(r_k)\vert =\sum _n \vert P_n^{(l)}(r_k)\vert \leq \sum _n \varepsilon _n <1$ . In particular, $f^{(l)} \not \in {\mathcal {U}}_A({\mathbb {D}},\varrho )$ . The proof that $f\in {\mathcal {U}}_A({\mathbb {D}},\varrho )$ is now quite standard. We fix $n,m \in {\mathbb {N}}$ and let $(n_k)_k\subset {\mathbb {N}}$ such that $\alpha (n_k)=n$ and $\beta (n_k)=m$ for any $k\in {\mathbb {N}}$ . By construction, we have

$$ \begin{align*}\sup_{z\in K_n}|f(r_{n_k}z) - \varphi_m(z)| & \leq \varepsilon_{n_k} + \sum _{j>n_k} \vert P_j (r_{n_k}z)\vert \\ & \leq \varepsilon_{n_k} +\sum _{j>n_k}\varepsilon_j, \end{align*} $$

which goes to $0$ as $k\to \infty $ .

Proof Let us first prove the result for $l=2$ . Let $f=\sum _ka_kz^k \in {\mathcal {U}}({\mathbb {D}},{\mathbb {T}})$ . For any k, let the real number $\alpha _k$ be such that for any $k\geq 0$ ,

$$ \begin{align*} \Re((k+2)(k+1)(a_{k+2}+\alpha_{k+2}))=\max\{N\in {\mathbb{Z}}:\, N\leq \Re((k+2)(k+1)a_{k+2})\}. \end{align*} $$

Then, for any $k\geq 2$ ,

$$ \begin{align*} \vert \alpha_k\vert \leq (k(k-1))^{-1}. \end{align*} $$

In particular, $\sum _k \alpha _k\zeta ^k$ is uniformly convergent for any $\zeta \in {\mathbb {T}}$ , and thus the series $g:=\sum _k(a_k+\alpha _k)z^k$ still belongs to ${\mathcal {U}}({\mathbb {D}},{\mathbb {T}})$ . However, $g^{(2)}$ cannot be universal at $1$ , since $S_n(g^{(2)})(1)\in \mathbb {Z}$ for any $n\in {\mathbb {N}}$ .

We can proceed similarly for any $l\geq 3$ . The details are left to the reader. Now, if it were true that $f'\in {\mathcal {U}}({\mathbb {D}},{\mathbb {T}})$ whenever $f\in {\mathcal {U}}({\mathbb {D}},{\mathbb {T}})$ , then $f^{(2)}$ would also belong to ${\mathcal {U}}({\mathbb {D}},{\mathbb {T}})$ for any $f\in {\mathcal {U}}({\mathbb {D}},{\mathbb {T}})$ . That would contradict the first part of the proof.

Proof Let $\varepsilon $ , l, N, K, R, and $\varphi $ be fixed as in the lemma. Let $\eta>0$ be such that $0<\eta < \text {dist}({\mathbb {D}}_R,K)$ , and let $M\geq N$ be an integer such that $\frac {\varepsilon k^{l+2}R^k}{(R+\eta )^k}\leq 1$ for any $k\geq M$ . Then Mergelyan’s theorem gives the existence of a polynomial $P=\sum _{k\geq M}a_kz^k$ such that

$$ \begin{align*} \sup_{|z|\leq R+\eta} \vert P(z) \vert\leq \varepsilon \quad \text{and}\quad \sup_{z\in K}\vert P(z)-\varphi(z) \vert\leq \varepsilon. \end{align*} $$

By Cauchy’s inequalities, we deduce from the first inequality that for any $k\geq M$ ,

$$ \begin{align*} |a_k|\leq \frac{\varepsilon}{(R+\eta)^k}. \end{align*} $$

This gives the conclusion by the choice of M.

Proof of Theorem 4.4

Let us first fix $R\geq 1$ and $l\geq 2$ as in the statement. We keep the notations used in the previous proofs and let $(L_n)_n$ be an enumeration of compact sets in $(\overline {{\mathbb {D}}_R}\setminus {\mathbb {D}})\cup ({\mathbb {C}}\setminus \overline {{\mathbb {D}}_R})$ , with connected complement, such that for any compact sets $K_1\subset \overline {{\mathbb {D}}_R}\setminus {\mathbb {D}}$ and any $K_2\subset {\mathbb {C}}\setminus \overline {{\mathbb {D}}_R}$ , both with connected complement, there exists n such that $K_1\cup K_2 \subset L_n$ (the proof of the existence of such sequence is straightforward and omitted). We also let $\psi _1:{\mathbb {N}}\to {\mathbb {N}}$ , $\psi _2:{\mathbb {N}}\to {\mathbb {N}}$ and $\psi _3:{\mathbb {N}}\to {\mathbb {N}}$ , such that for any $n_1,n_2,m \in {\mathbb {N}}$ , there exist infinitely many $n\in {\mathbb {N}}$ such that $(\varphi _{n_1},\varphi _{n_2},L_m)=(\varphi _{\psi _1(n)},\varphi _{\psi _2(n)},L_{\psi _3(n)})$ . Let $Q_0=P_0=0$ and $R_0=0$ . We build by induction three sequences $(P_n=\sum a_k(n)z^k)_n$ , $(Q_n=\sum b_k(n)z^k)_n$ , and $(R_n)_n$ of polynomials (using Runge’s theorem for $(P_n)_n$ and Lemma 4.5 for $(Q_n)_n$ ), in the following order $Q_n\rightarrow P_n\rightarrow R_n$ , with $R_n=\sum \limits _{k=\text {val}(P_n)}^{\deg (P_n)}\alpha _k z^k$ , such that:

1. (1) $\text {val}(Q_n)> \text {deg}(P_{n-1})$ ;

2. (2) $\sup _{|z|\leq R}\vert Q_n(z) \vert \leq \varepsilon _n$ ;

3. (3) $\sup _{z\in L_{\psi _3(n)}\cap ({\mathbb {C}}\setminus \overline {{\mathbb {D}}_R})}\vert Q_n(z) - \left (\varphi _{\psi _2(n)}(z) - \sum _{k=0}^{n-1} (Q_k + P_k + R_k)(z) \right )\vert \leq \varepsilon _n$ ;

4. (4) $|b_k(n)|\leq R^{-k}k^{-(l+2)}$ for any $n,k$ ;

5. (5) $\sup _{|z|\leq 1-1/n}\vert P_n(z) \vert \leq \varepsilon _n$ ;

6. (6) $\sup _{z\in L_{\psi _3(n)}\cap (\overline {{\mathbb {D}}_R}\setminus {\mathbb {D}})}\vert P_n(z) - (\varphi _{\psi _1(n)}(z) - \sum _{k=0}^{n-1} P_k(z))\vert \leq \varepsilon _n$ ;

7. (7) $\text {val}(P_n)> \text {deg}(Q_n)$ ;

8. (8) $\Re (k(k-1)\ldots (k-l+2)(k-l+1)R^{k-l}(a_{k}(n)+\alpha _{k}))$ is the integer part of $\Re (k(k-1)\ldots (k-l+2)(k-l+1)R^{k-l}a_{k}(n))$ for $k=\text {val}(P_n),\dots ,\deg (P_n)$ .

We set $f=\sum _n Q_n +\sum _n P_n + \sum _{n} R_n$ . Because of (2), (5), and (8), $f \in H({\mathbb {D}})$ and $\sum _{n} R_n$ is absolutely convergent on $\overline {{\mathbb {D}}_R}$ (indeed, (8) implies $|\alpha _k|\leq R^{-k}(k(k-1))^{-1}$ ).

Moreover, (6) implies that $\{S_j(\sum _n P_n)|_K:\,j\in {\mathbb {N}}\}$ is dense in $A(K)$ for any $K\subset \overline {{\mathbb {D}}_R}\setminus {\mathbb {D}}$ with connected complement. By (8) and (4), respectively, $\sum _{n} R_n$ and the Taylor expansion at $0$ of $\sum _n Q_n$ are absolutely convergent on $\overline {{\mathbb {D}}_R}$ , so $\{S_j(f)|_K:\,j\in {\mathbb {N}}\}$ is also dense in $A(K)$ for any $K\subset \overline {{\mathbb {D}}_R}\setminus {\mathbb {D}}$ with connected complement.

Now, (3) and (7) imply that $\{S_j(f)|_K:\,j\in {\mathbb {N}}\}$ is dense in $A(K)$ for any $K\subset {\mathbb {C}}\setminus \overline {{\mathbb {D}}_R}$ with connected complement.

Last, let us observe that by (8) the set $\{S_j\left ((\sum _n P_n+\sum _n R_n)^{(l)}\right )(R):\,j\in {\mathbb {N}}\}$ is contained in ${\mathbb {Z}}$ . Now, since by (4) the Taylor series of the lth derivative of $\sum Q_n$ is absolutely convergent on $\overline {{\mathbb {D}}_R}$ , the set $\{S_j(f^{(l)})(R):\,j\in {\mathbb {N}}\}$ cannot be dense in ${\mathbb {C}}$ . In particular, $f^{(l)}\not \in {\mathcal {U}}({\mathbb {D}},R)$ and, by induction, there must be some $f\in {\mathcal {U}}({\mathbb {D}},R)$ such that $f'\not \in {\mathcal {U}}({\mathbb {D}},R)$ .

Question 4.6 Does the set ${\mathcal {U}}({\mathbb {D}})$ coincide with ${\mathcal {U}}({\mathbb {D}},R)$ for some $R\geq 1$ ?

Question 5.1 If $\{f_1,\dots ,f_l\}$ is $\varrho $ -Abel multiuniversal (resp. Abel multiuniversal), does there exist $i\in \{1,\dots ,l\}$ such that $f_i$ is a $\varrho $ -Abel universal functions (resp. Abel universal functions)?

Proof We assume that $\sum _n\varepsilon _n< 1/4$ . Let us set, for any $n\in {\mathbb {N}}$ , $I_n=K_n \cup \{1\}$ . Note that $I_n$ is different from ${\mathbb {T}}$ for any $n\in {\mathbb {N}}$ and that any compact subset of ${\mathbb {T}}$ containing $1$ and different from ${\mathbb {T}}$ is contained in some $I_n$ . Let us also denote by $(\tilde {\varphi }_n)_n$ an enumeration of all polynomials with rational real and imaginary parts coefficients with value at $1$ off $D(a,1)$ . Note that every continuous function on ${\mathbb {T}}$ with value at $1$ off $\overline {D(a,1)}$ can be uniformly approximated on every compact subset of ${\mathbb {T}}$ containing $1$ and different from ${\mathbb {T}}$ , by a subsequence of $(\tilde {\varphi }_n)_n$ .

Set $f_1(z)=\tilde {\varphi }_{\alpha (1)}\left (\frac {z}{r_{1}}\right )$ . Observe that $f_1(r_1)\notin D(a,1)$ . Moreover, we can assume that, for all $k\geq 2$ , $\varepsilon _k<d(\tilde {\varphi }_{\alpha (k)}(1),D(a,1))$ , where d is the usual distance on $\mathbb {C}$ . Using Mergelyan’s theorem, we build by induction a sequence of polynomials $f_k$ , $k\geq 2$ , such that, for any $k\geq 2$ ,

1. (1) $\sup _{z\in r_{k}I_{\beta (k)}}\left \vert f_{k}(z)-\tilde {\varphi }_{\alpha (k)}\left (\frac {z}{r_k}\right ) \right \vert <\varepsilon _k$ ;

2. (2) $\sup _{\vert z\vert \leq r_{k-1}}\left \vert f_{k}(z)-f_{k-1}(z) \right \vert <\varepsilon _k$ ;

3. (3) $\sup _{r\in [r_{k-1},r_k]}\vert f_{k}(r)-h_k(r)\vert <\varepsilon _k$ , where $h_k:[r_{k-1},r_k]\rightarrow \mathbb {C}$ is a continuous functions such that $h_k(r_{k-1})=f_{k-1}(r_{k-1})$ , $h_{k}(r_k)=\tilde {\varphi }_{\alpha (k)}(1)$ and for all $r_{k-1}\leq r\leq r_{k}$ , $h_k(r)\notin D(a,1)$ .

Observe that the choice of the sequence $(\varepsilon _k)_k$ and the property (1) ensure that $f_k(r_k)\notin D(a,1)$ .

Let us set

$$ \begin{align*} f(z)=f_1(z)+\sum_{k=1}^{+\infty}(f_{k+1}(z)-f_k(z)). \end{align*} $$

From (2), we deduce that $f\in H(\mathbb {D})$ . Now, by (1) and (2), we have, for any $n\in {\mathbb {N}}$ and any $z\in I_{\beta (n)}$ ,

(5.1) $$ \begin{align} \begin{array}{lll} \left\vert f(r_{n}z)-\tilde{\varphi}_{\alpha(n)}\left(z\right) \right\vert & \leq & \displaystyle \left\vert f_n(r_{n}z)-\tilde{\varphi}_{\alpha(n)}\left(z\right)\right\vert+ \sum_{k=n}^{+\infty}\left\vert f_{k+1}(r_{n}z)-f_k(r_{n}z)\right\vert\\ & \leq & \displaystyle \left\vert f_n(r_{n}z)-\tilde{\varphi}_{\alpha(n)}\left(z\right)\right\vert+ \sum_{k=n}^{+\infty}\sup_{\vert z\vert \leq r_{k}}\left\vert f_{k+1}(z)-f_k(z)\right\vert\\ & < & \displaystyle \sum_{k=n}^{+\infty}\varepsilon_k.\end{array} \end{align} $$

This yields the point (2) of the proposition. Indeed, if $n,m\in {\mathbb {N}}$ are fixed and $(n_l)_l\subset {\mathbb {N}}$ is an increasing sequence such that $\alpha (n_l)=n$ and $\beta (n_l)=m$ for any $l\in {\mathbb {N}}$ , then by (5.1) $f_{r_{n_l}}$ tends to $\tilde {\varphi }_{n}$ uniformly on $I_m$ as $l\to \infty $ .

For the proof of (1), let us fix $r\in [r_1,1)$ and let $n\geq 2$ be an integer such that $r_{n-1} \leq r\leq r_n$ . Since $\sum _k\varepsilon _k< 1/4$ , we deduce from (3) that

$$ \begin{align*} \begin{array}{lll}\vert f(r)-h_n(r)\vert &\leq &\vert f(r)-f_n(r)\vert+\vert f_n(r)-h_n(r)\vert\\&\leq& \displaystyle\sum_{k\geq n}\vert f_{k+1}(r)-f_k(r)\vert +\vert f_n(r)-h_n(r)\vert\\ &<&\displaystyle\frac{1}{4} + \frac{1}{4}=\frac{1}{2}.\end{array} \end{align*} $$

We conclude the proof thanks to the fact that $h_n(r)\in {\mathbb {C}}\setminus D(a,1)$ for any $r_{n-1}\leq r\leq r_{n}$ .

Proof Let $a_1,a_2\in \mathbb {C}$ such that $\vert a_1- a_2\vert>2$ . Let $f_1$ and $f_2$ in $H({\mathbb {D}})$ be given by Proposition 5.2 applied, respectively, with $a=a_1$ and $a=a_2$ . Clearly, neither $f_1$ nor $f_2$ is Abel universal. Now, since $\overline {D(a_1,1)}\cap \overline {D(a_2,1)}=\emptyset $ , it is also clear that the family $\{f_1,f_2\}$ is $\varrho $ -Abel multiuniversal. To complete the proof, it is enough to observe that any $\varrho $ -Abel multiuniversal set is Abel multiuniversal.

Proof By Mergelyan’s theorem, we may and shall assume that for any $l,n\in {\mathbb {N}}$ , $V_n(l)$ is centered at some entire function $g_{n,l}$ and has radius $\delta _{n,l}$ , where $(\delta _{n,l})_n$ , $l\in \mathbb {N}$ , are some sequences of positive numbers. By uniform continuity, for any $l,n\in {\mathbb {N}}$ , we can define $r(l,n)\in [0,1)$ such that

(6.1) $$ \begin{align} \sup_{r\in [r(l,n),1]}\sup_{\zeta\in K_l}|g_{n,l}\circ \phi_r(\zeta)-g_{n,l}(\zeta)| < \frac{\delta_{n,l}}{4}. \end{align} $$

We shall also assume that $r(l,n)$ is larger than the radius of D for any $l,n\in {\mathbb {N}}$ . For any $l,n,m\in {\mathbb {N}}$ , let us denote by $\tilde {\Gamma }(l,n,m)$ the set $\Gamma (l,n,m)\cap [r(l,n),1)$ and by $\tilde {\Gamma }(l,n)$ the set

$$ \begin{align*} \tilde{\Gamma}(l,n):=\Gamma(l,n)\cap[r(l,n),1) = \bigcup_{m\in {\mathbb{N}}}\tilde{\Gamma}(l,n,m). \end{align*} $$

It may happen that the family $\{\tilde {\Gamma }(l,n,m):\, l,n,m\in {\mathbb {N}}\}$ is finite. In that case, we choose $r(n,l)$ larger than the largest element in these sets, so that nothing further needs to be proved. Let us then assume that the family $\{\tilde {\Gamma }(l,n,m):\, l,n,m\in {\mathbb {N}}\}$ is infinite. Furthermore, upon removing all the empty sets $\tilde {\Gamma }(l,n,m)$ and reordering the remaining nonempty ones, we may and shall assume that every $\tilde {\Gamma }(l,n,m)$ , $l,n,m \in {\mathbb {N}}$ , is nonempty.

Since $\{\tilde {\Gamma }(l,n,m):\, l,n,m\in {\mathbb {N}}\}$ is locally finite, we can then find an ordering $(\tilde {\Gamma }_i)_{i\in {\mathbb {N}}}$ of it and an increasing sequence $(s_i)_{i\geq 1}$ in $(0,1)$ , tending to $1$ , such that for any $i\geq 1$ ,

$$ \begin{align*} \max(\tilde{\Gamma}_{i-1}) < s_i < \min(\tilde{\Gamma}_{i}). \end{align*} $$

Note that for any $i\in {\mathbb {N}}$ , there exists a unique pair $(l,n)$ such that $\tilde {\Gamma }_{i}\subset \tilde {\Gamma }(l,n)$ . Let us write $D_0=D$ and denote by $D_i$ , $i\geq 1$ , the closed disk centered at $0$ , with radius $s_i$ . By induction, let us build a sequence $(P_i)_i$ of polynomials as follows. If $D_0=\emptyset $ , set $P_0=0$ . If not, observe that by construction the real number $\min (\tilde {\Gamma }_{0})$ is larger than the radius of D and let us apply Runge’s theorem to define $P_0$ as any polynomial satisfying $\sup _{D_0} |P_0-g| < \varepsilon /2$ and

$$ \begin{align*} \sup_{\zeta\in K_l}|P_0\circ \phi_r(\zeta)-g_{n,l}\circ \phi_r(\zeta)| < \frac{\delta_{n,l}}{4} \end{align*} $$

for any $r\in \tilde {\Gamma }_0$ , where $(l,n)$ is the only pair such that $\tilde {\Gamma }_{0}\subset \tilde {\Gamma }(l,n)$ . Then we build by induction positive real numbers $\eta _k$ , $k\geq 1$ , and by applying Runge’s theorem, polynomials $P_k$ , $k\geq 1$ , such that for any $j\geq 1$ and any $0\leq i\leq j-1$ :

1. (1) $\sup _{K_l}\left \vert \sum _{k=0}^i P_k\circ \phi _r - g_{n,l}\circ \phi _r \right \vert + \sum _{k=i+1}^j \eta _k < \frac {\delta _{n,l}}{4}$ for any $r \in \tilde {\Gamma }_{i}$ where $(l,n)$ is the only pair for which $\tilde {\Gamma }_{i}\subset \tilde {\Gamma }(l,n)$ ;

2. (2) $\eta _j < \varepsilon /2^{j+1}$ ;

3. (3) $\sup _{K_l}\left \vert \sum _{k=0}^j P_k\circ \phi _r - g_{n,l}\circ \phi _r \right \vert < \frac {\delta _{n,l}}{4}$ for any $r \in \tilde {\Gamma }_{j}$ where $(l,n)$ is the only pair for which $\tilde {\Gamma }_{j}\subset \tilde {\Gamma }(l,n)$ ;

4. (4) $\sup _{D_{j}}|P_j|< \eta _{j}$ .

Then we set $f=\sum _{i\geq 0} P_i$ . By (b) and (d) and the choice of $P_0$ , we get $f\in H({\mathbb {D}})$ and $\sup _{D} |f - g| < \varepsilon $ . Let us now check that f satisfies $f\circ \phi _r \in V_n(l)$ for any $l,n\in {\mathbb {N}}$ and any $r\in \tilde {\Gamma }(l,n)$ . Fix $l,n\in {\mathbb {N}}$ and $r\in \tilde {\Gamma }(l,n)$ . There exists a unique $i\in {\mathbb {N}}$ such that $r\in \tilde {\Gamma }_i$ . We deduce from (a), (c), and (6.1) that

$$ \begin{align*} f\circ \phi_r = \sum_{k=0}^i P_k\circ \phi_r + \sum _{k\geq i+1}P_k\circ \phi_r \in V_n(l). \end{align*} $$

This gives the desired conclusion.

Proof The assertion (1) is contained in [Reference Grosse-Erdmann and Peris Manguillot29, Lemma 9.5]. Let us prove (3). According to [Reference Grosse-Erdmann and Peris Manguillot29, Lemma 9.5], there exist pairwise disjoint subsets $A(l,n)$ , $l,n\geq 1$ , of ${\mathbb {N}}$ , with positive lower density, such that for any $j\in A(l,n)$ and $j'\in A(l',n')$ , one has

(6.2) $$ \begin{align}j\geq n\quad \text{and}\quad |j-j'|\geq n+n'\quad\text{if}\quad j\neq j'. \end{align} $$

For any $l,n\geq 1$ , let us write

$$ \begin{align*} A(l,n)=\{j_m^{l,n},\ m\geq 1\}, \end{align*} $$

where $(j_m^{l,n})_m$ is increasing. Now, for $l,n,m\geq 1$ , we set

$$ \begin{align*} \Gamma(l,n,m):=\left[1-\frac{1}{j_m^{l,n}}-\frac{1}{3}\left(\frac{1}{j_m^{l,n}-1}-\frac{1}{j_m^{l,n}}\right), 1-\frac{1}{j_m^{l,n}}+\frac{1}{3}\left(\frac{1}{j_m^{l,n}}-\frac{1}{j_m^{l,n}+1}\right) \right] \end{align*} $$

and for $l,n\geq 1$ ,

$$ \begin{align*} \Gamma(l,n):=\bigcup_{m\geq 1}\Gamma(l,n,m). \end{align*} $$

Using (6.2), it is not difficult to check that the sets $\Gamma (l,n)$ , $l,n\geq 1$ , are pairwise disjoint and that $r\geq 1-\frac {1}{n-1}$ for any $r\in \Gamma (l,n)$ . This shows that the family $\{\Gamma (l,n,m):\,l,n,m\in {\mathbb {N}}\}$ is locally finite. Now, let us fix $l,n\geq 1$ and simply denote $j_m^{l,n}=j_m$ , $m\geq 1$ . An easy calculation leads to

$$ \begin{align*} |\Gamma(l,n,m)|=\frac{2}{3}\frac{1}{j_m^2-1}. \end{align*} $$

Now, taking into account that $A(l,n)$ has positive lower density, there exist $M>1$ such that $m\leq j_m\leq Mm$ , $m\geq 1$ . Therefore, for any integer $N\geq 1$ , we get

$$ \begin{align*} N\left|\Gamma(l,n)\cap [1-1/N,1)\right| & \geq \frac{2N}{3}\sum_{\substack{m\geq 1\\j_m\geq N}}\frac{1}{j_m^2-1}\\ &\geq \frac{2N}{3}\sum_{m\geq N}\frac{1}{M^2m^2}\\ &\geq \frac{2N}{3M^2N}=\frac{2}{3M^2}. \end{align*} $$

This finishes the proof of (3).

The proof of (2) is standard and similar to that of (4), so we only check (4). By induction, one can build two increasing sequences $(N_i)_i$ and $(M_i)_i$ of positive integers such that $2^{N_i-M_i}\to 0$ as $i\to \infty $ and such that $N_{i+1}>M_i$ , $i\in {\mathbb {N}}$ . Then we set $\Gamma _i=[1-2^{-N_i},1-2^{-M_i}]$ and fix increasing sequences $(u_{m}(l,n))_m$ , $l,n\in {\mathbb {N}}$ , such that $u_m(l,n)\neq u_{m'}(l',n')$ whenever $(m,l,n)\neq (m',l',n')$ . It is now easily checked that the family $\{\Gamma _{u_m(l,n)}:\,l,n,m\in {\mathbb {N}}\}$ is locally finite and that the sets

$$ \begin{align*} \Gamma(n,l):=\bigcup_{m\in {\mathbb{N}}}\Gamma_{u_m(l,n)},\quad l,n\in {\mathbb{N}} \end{align*} $$

have upper density equal to $1$ .

Proof Setting $\Gamma (l,n,m)=\{r_{j_m}\}$ , $l,n,m\in \mathbb {N}$ , where $(j_m)_m$ is the increasing sequence of elements of the set $A(l,n)$ given by point (2) of Lemma 6.4, and applying Proposition 6.1, we get that the set $\mathcal {FU}^{u=1}_A(\mathbb {D},\varrho )$ consisting of those $f\in H({\mathbb {D}})$ such that $\overline {d}(N_f(V,K,\varrho ))=1$ for any compact set $K\subset {\mathbb {T}}$ different from ${\mathbb {T}}$ and any $V\in {\mathcal {V}}(K)$ , is dense in $H({\mathbb {D}})$ . Similarly, if the sets $\Gamma (l,n,m)$ , $l,n,m\in {\mathbb {N}}$ , are instead given by point (4), then Proposition 6.1 gives us that the set $\mathcal {FU}^{u=1}_A(\mathbb {D})$ consisting of those $f\in H({\mathbb {D}})$ such that $\overline {d_u}(N_f(V,K))=1$ for any compact set $K\subset {\mathbb {T}}$ different from ${\mathbb {T}}$ and any $V\in {\mathcal {V}}(K)$ , is dense in $H({\mathbb {D}})$ .

Moreover, in the same way, Proposition 6.1 and points (1) and (3) of Lemma 6.4 immediately tell us that the sets $\mathcal {FU}_A(\mathbb {D},\varrho )$ and $\mathcal {FU}_A(\mathbb {D})$ are dense in $H({\mathbb {D}})$ . Since $\underline {d}(E)=1-\overline {d}({\mathbb {N}}\setminus E)$ , $E\in {\mathbb {N}}$ (resp. $\underline {d_u}(\Gamma )=1-\overline {d_u}([0,1)\setminus \Gamma )$ , $\Gamma \subset [0,1)$ ), the corollary is proved once we have shown that the sets $\mathcal {FU}^{u=1}_A(\mathbb {D},\varrho )$ and $\mathcal {FU}^{u=1}_A(\mathbb {D})$ are $G_{\delta }$ in $H({\mathbb {D}})$ . The proof for both sets being very similar, we only deal with $\mathcal {FU}^{u=1}_A(\mathbb {D})$ .

To do so, note that

$$ \begin{align*} \begin{array}{l}\displaystyle\mathcal{F{\kern1pt}U}^{u=1}_A(\mathbb{D})=\\ \displaystyle\bigcap_{l\in {\mathbb{N}}}\bigcap_{n\in {\mathbb{N}}}\bigcap_{m\geq 1}\bigcap_{k\geq 1}\bigcup_{N\geq k}\left\{f\in H({\mathbb{D}}):\,\left|N_f(V_n(l),K_l)\cap [1-1/N,1)]\right|> \left(1-\frac{1}{m}\right)N\right\}\!.\end{array} \end{align*} $$

Since each set $\left \{f\in H({\mathbb {D}}):\,\left |N_f(V_n(l),K_l)\cap [1-1/N,1)]\right |> \left (1-\frac {1}{m}\right )N\right \}$ is easily seen to be open, the proof is complete.

Proof The proofs of the two equalities being analogous to that of [Reference Bonilla and Grosse-Erdmann9, Theorem 5.9], we only sketch the proof of the first equality. Let h be in $H(\mathbb {D})$ , and let $A(l,n)$ , $l,n\in {\mathbb {N}}$ , be given by point (1) of Lemma 6.4. We split each $A(l,n)$ into two disjoint sets $A^1(l,n),A^2(l,n)$ with positive lower density. Proceeding as in the proof of Theorem 6.1, we build $f\in H({\mathbb {D}})$ such that for any $k\in A^1(l,n)$ , $f\circ \phi _{r_k}\in V_n(l)$ and for any $k\in A^2(l,n)$ , $f\circ \phi _{r_k} \in h\circ \phi ^{-1}_{r_k}- V_n(l)$ , where $h\circ \phi ^{-1}_{r_k}- V_n(l):=\{h\circ \phi ^{-1}_{r_k} - g:\,g\in V_n(l)\}$ . Then f and $h-f$ belong to $\mathcal {FU}_A(\mathbb {D},\varrho )$ and we obviously have $h=f+(h-f)$ .

Proof Let $(I_m)_m$ be a sequence of compact sets whose union is $\Lambda $ . Clearly, we have

$$ \begin{align*} \bigcap_{\lambda\in\Lambda}\mathcal{U}_A(\mathbb{D}, \varrho(\lambda))=\bigcap_{j,l,m,s\geq 1}E(j,l,m,s), \end{align*} $$

where

$$ \begin{align*} E(j,l,m,s)= \left\{f\in H(\mathbb{D}):\,\forall\lambda\in I_m,\, \exists n\geq 0,\, \sup_{\zeta\in K_l}\vert f(r_{n}(\lambda)\zeta)-\varphi_j(\zeta)\vert<1/s\right\}\!. \end{align*} $$

By compactness of $I_m$ and continuity of $\lambda \mapsto r_n(\lambda )$ on $\Lambda $ , $n\in {\mathbb {N}}$ , each set $E(j,l,m,s)$ , $j,l,m,s\geq 1$ , is open. By the Baire Category Theorem, it is enough to prove that $E(j,l,m,s)$ is dense in $H(\mathbb {D})$ for any $j,l,m,s\geq 1$ . Let us then fix $j,l,m,s\geq 1$ and $g\in H(\mathbb {D})$ , $0<r<1$ and $\varepsilon>0$ . Let us set $r<\tilde {r}<1$ . Now, observe that the assumptions and Dini’s theorem imply that $(r_n(\cdot ))_n$ tends to $1$ uniformly on each compact set $I_m$ . Therefore, there exists $c_{n,I_m}>0$ with $c_{n,I_m}\rightarrow 0$ as $n\rightarrow +\infty $ so that $\vert r_n(\lambda )-1\vert \leq c_{n,I_m}$ for any $\lambda \in I_m$ , whence one can find $N_1\geq 1$ such that

$$ \begin{align*} \forall\lambda\in I_m,\ \forall n\geq N_1,\ r_n(\lambda)\geq \tilde{r}. \end{align*} $$

By Mergelyan’s theorem, there exists a polynomial f such that

$$ \begin{align*} \sup_{\vert z\vert\leq r}\vert f(z)-g(z)\vert<\varepsilon\mbox{ and } \sup_{z\in[\tilde{r},1]K_l}\vert f(z)-\varphi_j(z)\vert<1/2s. \end{align*} $$

We get, for any $\lambda \in I_m$ and any $n\geq N_1$ ,

$$ \begin{align*} \displaystyle\sup_{\zeta\in K_l}\vert f(r_{n}(\lambda)\zeta)-\varphi_j(\zeta)\vert&\leq \displaystyle \sup_{\zeta\in K_l}\vert f(r_{n}(\lambda)\zeta)-\varphi_j(r_{n}(\lambda)\zeta)\vert+\sup_{\zeta\in K_l}\vert \varphi_j(r_{n}(\lambda)\zeta)-\varphi_j(\zeta)\vert\\&\leq\displaystyle \frac{1}{2s}+\sup_{\zeta\in K_l}\vert \varphi_j(r_{n}(\lambda)\zeta)-\varphi_j(\zeta)\vert. \end{align*} $$

Using again the property $\vert r_n(\lambda )-1\vert \leq c_{n,I_m}$ with $c_{n,I_m}\rightarrow 0$ , one can find $N_2\geq N_1$ such that for all $\lambda \in I_m$ and $n\geq N_2$ , $\sup _{\zeta \in K_l}\vert \varphi _j(r_{n}(\lambda )\zeta )-\varphi _j(\zeta )\vert <1/2s.$ This completes the proof.