Is $F$ an isometry between $l^1(\mathbb(N))$ and $C(D)$?

Question

Let:

$D=${$z \in \mathbb{C}: |z| \leq 1$}

$l^1(\mathbb{N})=${$ (a_n)_{n \in \mathbb{N}}:a_n \in \mathbb{C} \ \ \forall n \geq 0 \ ; \ \sum_{n=0}^{\infty}|a_n| < \infty $}, endowed with the norm $||(a_n)_{n \in \mathbb{N}}||= \sum_{n=0}^{\infty}|a_n|$.

$C(D) =${$f:D \rightarrow \mathbb{C}: f$ is continuous$ $}, endowed with the norm $||f||=$sup$_{z \in D}|f(z)|$.

Now consider $F:l^1( \mathbb{N}) \rightarrow C(D)$ defined by:

$F((a_n)_{n})= \sum_{n=0}^{\infty}a_n z^n$.

I'm wondering if $F$ is an isometry.

I guess it is not. My first attempt at a solution was try to truly find an element of $l^1(\mathbb{N})$ such that $||F((a_n)_{n \in \mathbb{N}})|| \neq ||(a_n)_{n \in \mathbb{N}}||$, what I couldn't do.

Next, I thought about $F$ as an homomorphism between unital algebras, with the usual products. It is not hard to show that $F$ is not surjective. So if I find an subalgebra $\mathbb{A}$, with Range$(f) \supset \mathbb{A}$, of $C(D)$ such that:

(1) $\mathbb{A}$ separates points of $\mathbb{C}$

(2) $1_{l^{1}(\mathbb{N})} \in \mathbb{A}$

(3)$f \in \mathbb{A} \Rightarrow \bar{f} \in \mathbb{A}$

Then $\mathbb{A}$ is dense in $C(D)$ by the Stone Weierstrass theorem, therefore Range$(F)= C(D)$, contradition. (An isometry between Banach spaces must be closed). But that does not work too, because all the subalgebras I thought dont satisfy the third required property.

Any clues?

Answer 1

Hint: consider the subspace

$$V = \{ ( a_0, a_1, a_2, 0, 0, 0, \ldots ) : a_0, a_1, a_2 \in \mathbb{C} \} \subseteq \ell^1(\mathbb{N}).$$

Show that $F : V \to C(D)$ is not an isometry. So the goal is to find a function $f(z) = a_0 + a_1 z + a_2 z^2$ such that $|f(z)| < |a_1| + |a_2| + |a_3|$ for all $z \in D$.

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Moreover, let $S^1 = \{ z \in \mathbb{C} : |z| = 1 \}$ and for $\alpha \in S^1$ define $R(\alpha) = \{ t \cdot \alpha : t \in [0, \infty) \}$. Then for all $a \in \ell^1(\mathbb{N})$:

$\| F( a ) \| = \| a \|$ if and only if exist $\alpha, \omega \in S^1$ such that $a_n \in R(\alpha \cdot \omega^n)$ for all $n \in \mathbb{N}.$

Of course this doesn't happen very often, because it implies, loosely speaking, that arguments of $a_n$ form an arithmetic progression.