Let $B_n \subset \mathbb{C}^n$ be the $n$-dimensional open ball of radius $1$ centered at the origin. Let $A$ be the set consisting of holomorphic functions in $B_n$ which are continuous in $\overline{B_n}$, then $A$ is a Banach space with the sup-norm and pointwise operations. Let $M$ be the maximal ideal space of $A$, that is, $M$ is the set of non-trivial linear multiplicative functionals in $A$.
I want to prove that $M$ and $\overline{B_n}$ are homeomorphic via point-evaluations. In other words, I want to prove that the map $F: \overline{B_n} \longrightarrow M$ defined by $F(x)(\varphi)=\varphi(x)$ is an homeomorphism. Of course, in order to do so the first step is to prove that $F$ is a bijective function. Clearly $F$ is injective. My idea to prove surjectivity was the following: take any $\varphi \in M$ and let $p(z)=\sum_{|\alpha|=0}^{n}{c_{\alpha} z^{\alpha}}$ be a polynomial (here we're using the multi-index notation). For each $i, 1 \leq i \leq n$, let $f_i:\overline{B_n} \longrightarrow \mathbb{C}$ be the map defined by $f_{i}(z)=z_i$, then we can write $p=\sum_{|\alpha|=0}^{n}{c_{\alpha} f^{\alpha}}$, where $f=(f_1,\ldots,f_n)$. Therefore, since $\varphi \in M$ we have that $$ \varphi(p)=\sum_{|\alpha|=0}^{n}{c_{\alpha} \varphi(f)^{\alpha}}=p(\varphi(f)) $$ therefore, if I could prove that $\varphi(f)=(\varphi(f_1),\ldots,\varphi(f_n))$ is in $\overline{B_n}$ I would be done proving the surjectivity (since polynomials are dense in $A$). Of course, I can see that $|\varphi(f_i)| \leq 1$ but I don't see why would we have that $|\varphi(f)| \leq 1$.
Any suggestion in order to prove that $|\varphi(f)| \leq 1$? Or, if my approach is wrong I would appreciate any hint.
In advance thank you very much.
Let's take $\phi \in M, \phi(f_j)=w_j$ where $f_j(z)=z_j$ and assume $w=(w_j)$ is not in $\overline{B_n}$ so $\|w\|_2^2=c > 1$ and clearly $$ \phi\left(\sum \bar w_j f_j\right)=\|w\|_2^2=c. $$ But $$ \left|\phi\left(\sum \bar w_j f_j\right)\right| \le \max_{\|z\|_2 \le 1} \left|\sum (\bar w_j f_j)(z)\right|=\max_{\|z\|_2 \le 1} |\sum \bar w_j z_j| \le (\|w\|_2^2)^{1/2} $$ by Cauchy-Schwarz, hence $c \le \sqrt c$ which contradicts $c>1$.
This shows that for every $\phi \in M$ we have $(\phi(f_j))=(w_j) \in \overline{B_n}$ and from here it easily follows that $\phi$ is the evaluation at $w=(w_j)$ and we are done!