Let $U$ be the open unit ball in $\mathbb{C}^N$, let $A(U)$ be the algebra of functions analytic on $U$ and continuous on $\bar U$, and let $u=(1,0,\ldots,0)$. Let $\mathcal{B}=\{f\in H^\infty (U): \|f\|_\infty\leq 1\}$, and endow $\mathcal{B}$ with the compact-open topology. In 1979, Chee proved the following.
There is a sequence $A_n$ of functions in $A(U)$ that is dense in $\mathcal{B}$, and satisfies $\Vert A_n \Vert_\infty \leq 1$, $A_n(u)=1$.
Chee, P. S., Universal functions in several complex variables, J. Aust. Math. Soc., Ser. A 28, 189-196 (1979). ZBL0423.32002.
My question is, can Chee's result be strengthened to say that each of the $A_n$ is also Lipschitz on $\bar U$? My motivation for this question is the same as Chee's: to study the universality of composition operators on $\mathcal{B}$.
Chee's proof was constructive, and to mention what I have tried, I will need to mention the construction (which is a bit technical). The following is the argument in Lemma 2 of Bayart and Gorkin in
Bayart, Frédéric; Gorkin, Pamela, How to get universal inner functions, Math. Ann. 337, No. 4, 875-886 (2007). ZBL1129.47027.
To prove Chee's result, it suffices to establish, for any polynomial $P$ with $\Vert P\Vert_{C(\bar U)}\leq r<1$, for any compact subset $K$ of $U$, and for any $\epsilon>0$, the existence of a function $f\in A(U)$ with $\Vert f\Vert_\infty=1$, $\Vert f-P\Vert_{C(K)}<\epsilon$ and $f(u)=1$.
The function $g(u)=\frac 12 (1+\langle z,u\rangle)$ peaks at $u$: $g\in A(U)$, $g(u)=1$, and $|g|<1$ on $\bar U\setminus \{u\}$. Setting $M=2r+1$ and $r<c<1$, given any compact set $F$ of $\bar U$ with $u\not \in F$, we note that there exists a function $h\in A(U)$ such that $\Vert h\Vert_{C(\bar U)}\leq M, h(u)=1, \Vert h-P\Vert_{C(K)}<\epsilon$ and $\Vert h\Vert_{C(F)}<c$: indeed, we may take $h=P+(1-P(u)) g^n$, for $n$ sufficiently large. The proof now follows that in Theorem 11.1 of Gamelin's book Uniform Algebras.
Let $s$ be sufficiently close to 1 with $0<s<1$ so that $M-1+s(c-M)<0$. Choose a sequence $\epsilon_n$ decreasing to 0 so that $$\epsilon_{n-1}(1-s^n)+s^n(M_1+s(c-M))<0, n\geq 1.$$ Let $F_n$ be a sequence of closed subsets of $\bar U$ such that $\cup_n F_n = \bar U\setminus \{u\}$. We choose a sequence $f_n$ by induction as follows. Take $f_0$ such that $\Vert f_0\Vert_{C(\bar U)}\leq M, f_0(u)=1, \Vert f_0-P\Vert_{C(K)}<\epsilon$.
Supposing $f_0,\ldots,f_n$ have been chosen, let $$W_n=\left\{z : \max_{0\leq j\leq n} |f_j(z)|\geq 1+\epsilon_n\right\}.$$ Now take $f_{n+1}$ in $A(U)$ such that $\Vert f_{n+1}\Vert_{C(\bar U)}\leq M, f_{n+1}(u)=1, \Vert f_{n+1}-P\Vert_{C(K)}<\epsilon$ and $\Vert f_{n+1}\Vert_{C(F_n\cup W_n)}< c$. Now define $$f=(1-s)\sum_{j=0}^\infty s^j f_j.$$ Then one can show (as in Gamelin) that $f\in A(U), \Vert f\Vert_{C(\bar U)}\leq 1, f(u)=1$, and $\Vert f-P\Vert_{C(K)}\leq \epsilon$, which concludes the proof.
My hope was to reconstruct this proof while keeping track of the Lipschitz constants $L_n$ of the $f_n$, with the hope that $\sum_{j=0}^\infty s^j L_j<\infty$. The function $\langle z,u\rangle$ has a Lipschitz constant of 1 by the Cauchy-Schwartz inequality. So if $L$ is the Lipschitz constant of $P$ above, then the function $h=P+(1-P(u))g^j$ above has a Lipschitz constant of $2L+1+j$ (maybe better). But during the induction step, the value of $j$ needed to make a function in the form of $h$ satisfy $\Vert h\Vert_{C(F_n\cup W_n)}< c$ is incredibly large - much larger than $1/s^n$.
I am looking for new ideas, suggestions, or potentially relevant references. My several variable knowledge and skills are at a beginner's level. The single variable case $N=1$ has a positive answer, so I am only concerned with considering $N\geq 2$.
