# Is the mapping from $l^1(\mathbb{N})$ to the space of holomorphic functions on the unit disc a homeomorphism?

The space $l^1(\mathbb{N})$ has the topology of a Banach space, and we can endow the set of all holomorphic functions on the disc with the topology of uniform convergence on compact sets.

Is the map $(a_n) \mapsto f(z) = \sum\limits_{n \in \mathbb{N}} a_nz^n$ a homeomorphism with respect to these topologies?

Injectivity comes from the fact that a function is identically zero iff its Taylor coefficients are all zero. Surjectivity comes from the fact that every holomorphic function on the disc has a power series expansion at the origin. And finally $l^1$ convergence of power series coefficients means the corresponding functions become arbitrarily close in sup norm when restricted to a compact set.

Does this mean I can translate between absolutely convergent series and holomorphic functions on the disc?

Edit: I see this is not surjective, but what if we consider the Hardy space of functions on the disc instead of all holomorphic functions?

• The mapping is not surjective. $\frac{1}{1-z} = \sum_{n=0}^\infty 1z^n$ is holomorphic in the unit disk, but $(1, 1, 1, \ldots) \not \in l^1$. Also (unless I am misunderstanding your notation) $l^1(\mathbb{N})$ is the space of all summable integer sequences, i.e. that would not even cover $f(z) = z/2$. – Martin R Dec 24 '17 at 23:38
• I see. I thought of this question as I was reading about Hardy spaces, so what if we restrict to bounded holomorphic functions on the disc? – dinstruction Dec 24 '17 at 23:41
• The space where the collection of functions on the circle are uniformly bounded with the $L^1(S^1)$ norm. – dinstruction Dec 24 '17 at 23:48
• I have no idea what you mean by "The space where the collection of functions on the circle are uniformly bounded with the $L^1(S^1)$ norm. " A reasonable conjecture might be that the map maps $\ell_1$ onto $H^\infty$, but that's clearly not so, since an $H^\infty$ function need not be continuous on the closed disk. It doesn't map onto "the disk algebra" (the space of holomorphic functions continuous on the closure) either, altough that's not so obvious. – David C. Ullrich Dec 24 '17 at 23:54
• @MartinR No, $\ell^1(\mathbb N)$ is not integer sequences, it is complex-valued sequences indexed by $\mathbb N$. – Robert Israel Dec 25 '17 at 0:03

## 1 Answer

Let's say $S(a)=\sum a_nz^n$ for $a\in\ell_1$.

$S$ does not map $\ell_1$ onto any of the spaces one might reasonably hope it might. We can still ask whether $S$ is a homeomorphism onto $S(\ell_1)$, with the topology of uniform convergence on compact sets.

But no, the inverse of $S$ is not continuous on $S(\ell_1)$. For example $S(e_n)=z^n\to0$ although $||e_n||_1=1$.

Regarding $S$ not being surjective: Since $S(a)$ is continuous on the closed disk the only really reasonable conjecture is that $S(\ell_1)=A$, where $A$ is "the disk algebra", namely the space of functions continuous on the closed disk and holomorphic in the interior. But it's not so.

If $S(\ell_1)=A$ then the Open Mapping Theorem would show that the inverse was bounded: $$||a||_1\le c||S(a)||_\infty.$$ But for example the Rudin-Shapiro polynomials $P_n$ satisfy $P_n=S(a_n),$where $$||P_n||_\infty\sim 2^{n/2},\quad ||a_n||_1=2^n.$$