Character space of the disc algebra

$A(\mathbb{D})$ is a complex vector space closed under pointwise multiplication, with unit the constant function $1$ of norm one, and the supremum norm is submultiplicative; the constant functions and the coordinate function $z$ show that it is nonzero.

For every $a\in \bar{\mathbb{D}}$ the evaluation ${\mathrm{e}\mathrm{v}}_a$ is a character: it is nonzero, complex-linear and multiplicative, and ${\mathrm{e}\mathrm{v}}_a(1)=1$.

Let $\chi$ be a character of $A(\mathbb{D})$ and put $a:=\chi (z)$, where $z$ denotes the coordinate function. By [L1] $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, so $a\in \bar{\mathbb{D}}$; and for every polynomial $p$ one has $\chi (p)=p(a)$ by linearity, multiplicativity and $\chi (1)=1$.

Polynomials are uniformly dense in $A(\mathbb{D})$: for $f\in A(\mathbb{D})$ and $0<r<1$ the function $f_r(z):=f(rz)$ is holomorphic on the disc $\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}r$ and agrees with the sum of its Taylor series there, and the partial sums converge uniformly on the compact set $\bar{\mathbb{D}}$ by [L2]; moreover $f_r\to f$ uniformly on $\bar{\mathbb{D}}$ as $r\uparrow 1$ by uniform continuity of $f$ on the compact disc; hence $f$ is a uniform limit of polynomials.

$A(\mathbb{D})$ is complete: if $(f_n)\subseteq A(\mathbb{D})$ is Cauchy in the supremum norm, then it is uniformly Cauchy on $\bar{\mathbb{D}}$, so [L3] supplies a uniform limit $f$, and [L3] also makes $f$ continuous. For every filled triangle $\mathrm{\Delta }\subseteq \mathbb{D}$, [L4] gives ${\int }_{\mathrm{\partial }\mathrm{\Delta }}{f}_n=0$ because $f_n$ is holomorphic. Passing to the uniform limit in the contour integral by [L3] gives ${\int }_{\mathrm{\partial }\mathrm{\Delta }}f=0$, and Morera's direction of [L4] makes $f$ holomorphic on $\mathbb{D}$. Hence $f\in A(\mathbb{D})$ and $\mathrm{\parallel }{f}_n-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$.

Consequently $\chi (f)=f(a)$ for every $f\in A(\mathbb{D})$: by [step 1.4] take polynomials $p_n\to f$ uniformly and use continuity of $\chi$ from [L1] together with $\chi (p_n)=p_n(a)$ from [step 1.3]; hence $\chi ={\mathrm{e}\mathrm{v}}_a$ with $a=\chi (z)\in \bar{\mathbb{D}}$, so every character is an evaluation at a point of the disc, and the point is unique because ${\mathrm{e}\mathrm{v}}_a={\mathrm{e}\mathrm{v}}_b$ forces $a={\mathrm{e}\mathrm{v}}_a(z)={\mathrm{e}\mathrm{v}}_b(z)=b$.

The map $a\mapsto {\mathrm{e}\mathrm{v}}_a$ is continuous because every coordinate $a\mapsto {\mathrm{e}\mathrm{v}}_a(f)=f(a)$ is continuous; it is a bijection by [step 2.2] and [step 1.2]. Its domain $\bar{\mathbb{D}}$ is compact and its target is Hausdorff by [L6], so [L6] makes it a homeomorphism.

The boundary restriction is isometric: by [L5] applied to the bounded domain $\mathbb{D}$ and the function $f$, continuous on the closure, one has ${\sup }_{\bar{\mathbb{D}}}\mathrm{\mid }f\mathrm{\mid }={\max }_{\mathrm{\partial }\mathbb{D}}\mathrm{\mid }f\mathrm{\mid }$, so $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }f{\mathrm{\mid }}_{\mathrm{\partial }\mathbb{D}}{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $R$ preserves norms; and the character space is $\mathrm{\Delta }(A(\mathbb{D}))\cong \bar{\mathbb{D}}$, which contains points not on the boundary, so it is not the circle alone.