Proof of Corona problem for $\mathbb{D}$ Denote by $H^\infty$ all analytic and bounded functions on the unit disk $\mathbb{D}:=\{x\in\mathbb{C}:|x|<1\}$. I want to show that $\Delta_0:= \{[f\to f(w)]:w\in\mathbb{D}\}\subseteq\Delta(H^\infty)$ is dense in $\Delta(H^\infty)$ (Gelfand spectrum) if and only if for every $f_1,\dots,f_n\in H^\infty$ with $\inf_{z\in\mathbb{D}}\sum_{k=1}^n |f_k(z)|>0$ there exist $g_1,\dots,g_n\in H^\infty$ with $\sum_{k=1}^n f_k g_k=1$.
$\Longrightarrow$: The existence of such $g_k$'s is equivalent to $1\in I(f_1,\ldots,f_n)$, where $I$ is the smallest ideal that contains all $f_k$'s. This means that $I$ is no proper ideal, i.e. $I=H^\infty$. How can I show this? I know that this is true in the space $C(\mathbb{D})$, since $I$ could otherwise be extended to a maximal ideal $\tilde{I}$ and then there would exist an $x\in \mathbb{D}$ such that $\tilde{I} = \{f\in C(\mathbb{D}):f(x)=0\}$, which contradicts $\inf_{z\in\mathbb{D}}\sum_{k=1}^n |f_k(z)|>0$. But is it possible to extend $I\subseteq H^\infty$ to a proper ideal in $C(\mathbb{D})$?
$\Longleftarrow$: I tried taking a $\phi\in\Delta(H^\infty)\backslash \Delta_0$ and a neighborhood $U$ with $U\cap \Delta_0=\emptyset$ and then tried to characterize $U$ by the neighborhoods in weak-* topologies, but I did not arrive at anything useful.
 A: $\newcommand\spec{\Delta (H^\infty )}
\newcommand\D{\mathbb D}$ This question has been without an answer for eight months, perhaps due to a comment by user @Conrad  discouraging
pedestrian attempts.  It is true that the Corona Theorem is a very hard problem,   first stated by
Kakutani in 1941 and only settled by Carleson in 1962.  Nevertheless the statement of the question is definitely not asking for a
solution of the whole problem, but only for a proof of the equivalence between two well known formulations.
In the hope that there is still interest, here is an answer which is in fact not too hard to give.

($\Leftarrow$).  Assuming the condition given, let us prove that $\Delta _0$ is dense in $\spec$.
Arguing by contradiction
suppose that  $\varphi $ is a point of $\spec$ which is not in the closure of $\Delta _0$.  Therefore there exists an open subset
$U\subseteq \spec$ containing  $\varphi $ and which is disjoint from $\Delta_0$.
By the definition of the topology of $\spec$ (pointwise convergence) there are $h_1,h_2,\ldots ,h_n\in H^\infty $, and $\varepsilon >0$, such that
$$
  V:= \{\psi \in \spec: |\psi (h_i)-\varphi (h_i)| <\varepsilon \} \subseteq  U.
  $$
For each $w\in \D$, let $\delta _w$ be the character of $H^\infty $ given by $\delta _w(f)=f(w)$, so $\delta _w$ lies in $\Delta _0$ and consequently
not in $V$.  This implies that there exist some $i$ such that
$$
  |\delta _w(h_i)-\varphi (h_i)| =  |h_i(w)-\varphi (h_i)| \geq \varepsilon .
  $$
For each $j$, let $f_j\in H^\infty $ be defined by $f_j(z)=h_j(z)-\varphi (h_j)$, so the above implies that
$$
  \sum_{j=1}^n |f_j(w)|\geq |f_i(w)|\geq \varepsilon >0.
  $$
Since $w$ is arbitrary we conclude that
$$
  \inf_{z\in \D}\ \sum_{j=1}^n |f_j(z)|\geq \varepsilon >0,
  $$
and hence the hypothesis yields $g_1,g_2,\ldots ,g_n$ in $H^\infty $ such that $\sum_{j=1}^n f_jg_j=1$.  So
$$
  1 = \varphi (1) = \varphi \Big (\sum_{j=1}^n f_jg_j\Big ) = \sum_{j=1}^n \varphi (h_j-\varphi (h_j))\varphi (g_j) =0,
  $$
a contradiction.  This shows that $\Delta _0$ is dense in $\spec$.
($\Rightarrow$).  Assuming that  $\Delta _0$ is dense in $\spec$, let $f_1,f_2,\ldots ,f_n\in H^\infty $ be such that
$$
  \inf_{z\in \D}\
  \sum_{i=1}^n |f_i(z)|>0. \tag{$\dagger$}
  $$
Letting $J$ be the ideal of $H^\infty $ generated by the $f_i$, our goal is to show that $J=H^\infty $.  Again arguing by
contradiction, suppose that $J$ is proper, so Zorn's Lemma implies that there exists a maximal ideal $I\supseteq J$.  By
the Gelfand-Mazur theorem there is $\varphi $ in $\spec$ whose kernel coincides with $I$, and hence $\varphi $ vanishes on all of the
$f_i$'s.
By hypothesis $\varphi $ lies in the closure of $\Delta _0$, so there is a net $\{w_\lambda \}_{\lambda \in \Lambda }$ in $\D$ such that $\delta _{w_\lambda }\to \varphi $.  In
particular this means that, for every $i$, one has that
$$
  0=\varphi (f_i) =
  \lim_\lambda  \delta _{w_\lambda }(f_i) =
  \lim_\lambda  f_i(w_\lambda ),
  $$
hence also
$$
  \lim_\lambda   \sum_{i=1}^n |f_i(w_\lambda )|=0,
  $$
but this contradicts ($\dagger$),  hence showing that $J=H^\infty $, as desired.
