[proofplan]
The proof reduces the Suita inequality to the sharp one-point $L^2$ [extension theorem](/theorems/59) of Błocki and Guan-Zhou. That theorem produces, for each fixed point $z \in \Omega$, a [holomorphic function](/page/Holomorphic%20Function) with prescribed value $1$ at $z$ and squared $L^2$ norm bounded by the sharp constant
\begin{align*}
\frac{\pi}{c_\Omega(z)^2}.
\end{align*}
The Bergman kernel is the reciprocal of the minimal squared $L^2$ norm among holomorphic functions taking value $1$ at $z$, so the extension estimate immediately gives $\pi K_\Omega(z) \ge c_\Omega(z)^2$. The equality statement is then exactly the equality statement in this reciprocal variational formulation.
[/proofplan]
[step:Define the one-point extremal norm]
Fix $z \in \Omega$. Define the [Hilbert space](/page/Hilbert%20Space)
\begin{align*}
A^2(\Omega) := \left\{f \in \mathcal{O}(\Omega) : \int_\Omega |f(\zeta)|^2\, d\mathcal{L}^2(\zeta) < \infty\right\},
\end{align*}
with norm
\begin{align*}
\|f\|_{A^2(\Omega)}^2 := \int_\Omega |f(\zeta)|^2\, d\mathcal{L}^2(\zeta).
\end{align*}
Define the one-point extremal quantity
\begin{align*}
M_\Omega(z) :=
\inf\left\{\|f\|_{A^2(\Omega)}^2 : f \in A^2(\Omega),\ f(z)=1\right\}.
\end{align*}
Since $\Omega$ is bounded, the constant function $1:\Omega \to \mathbb{C}$ belongs to $A^2(\Omega)$ and satisfies $1(z)=1$, so the admissible class is non-empty and $M_\Omega(z)<\infty$.
We also record that $M_\Omega(z)>0$. Choose $r_z>0$ such that the open Euclidean ball
\begin{align*}
B(z,r_z):=\{\zeta\in\mathbb{C}: |\zeta-z|<r_z\}
\end{align*}
is contained in $\Omega$. For any $f\in A^2(\Omega)$, the function $|f|^2:\Omega\to [0,\infty)$ is subharmonic, so the sub-[mean value inequality](/theorems/328) on the disk $B(z,r_z)$ gives
\begin{align*}
|f(z)|^2
\le
\frac{1}{\pi r_z^2}\int_{B(z,r_z)} |f(\zeta)|^2\,d\mathcal{L}^2(\zeta)
\le
\frac{1}{\pi r_z^2}\int_\Omega |f(\zeta)|^2\,d\mathcal{L}^2(\zeta).
\end{align*}
If $f(z)=1$, this yields $\|f\|_{A^2(\Omega)}^2\ge \pi r_z^2$. Taking the infimum over the admissible class gives $M_\Omega(z)\ge \pi r_z^2>0$.
[/step]
[step:Identify the Bergman kernel with the reciprocal extremal norm]
We claim that
\begin{align*}
K_\Omega(z)=\frac{1}{M_\Omega(z)}.
\end{align*}
Indeed, if $f \in A^2(\Omega)$ satisfies $f(z)=1$, then $g := f/\|f\|_{A^2(\Omega)}$ satisfies $\|g\|_{A^2(\Omega)}=1$ and
\begin{align*}
K_\Omega(z) \ge |g(z)|^2 = \frac{1}{\|f\|_{A^2(\Omega)}^2}.
\end{align*}
Taking the supremum over all such $f$ gives $K_\Omega(z) \ge 1/M_\Omega(z)$.
Conversely, if $h \in A^2(\Omega)$ and $\|h\|_{A^2(\Omega)} \le 1$, then either $h(z)=0$, in which case $|h(z)|^2 \le 1/M_\Omega(z)$, or $h(z)\ne 0$, in which case $f := h/h(z)$ satisfies $f(z)=1$. Hence
\begin{align*}
M_\Omega(z) \le \|f\|_{A^2(\Omega)}^2
=
\frac{\|h\|_{A^2(\Omega)}^2}{|h(z)|^2}
\le
\frac{1}{|h(z)|^2}.
\end{align*}
Thus $|h(z)|^2 \le 1/M_\Omega(z)$. Taking the supremum over all $h$ with $\|h\|_{A^2(\Omega)} \le 1$ gives $K_\Omega(z) \le 1/M_\Omega(z)$.
[/step]
[step:Apply the sharp one-point $L^2$ extension theorem]
Let $G_\Omega(\cdot,z):\Omega\setminus\{z\}\to (-\infty,0)$ denote the Green function with pole at $z$, normalized so that
\begin{align*}
G_\Omega(\zeta,z)-\log|\zeta-z| \to \log c_\Omega(z)
\end{align*}
as $\zeta\to z$. The sharp one-point $L^2$ [extension theorem](/theorems/59) of Błocki and Guan-Zhou, in its one-dimensional Suita form, states that for a bounded planar domain admitting such a Green function, for each pole $z\in\Omega$, and for each datum $a\in\mathbb{C}$, there exists a [holomorphic function](/page/Holomorphic%20Function) $F:\Omega\to\mathbb{C}$ satisfying $F(z)=a$ and
\begin{align*}
\int_\Omega |F(\zeta)|^2\,d\mathcal{L}^2(\zeta)
\le
\frac{\pi |a|^2}{c_\Omega(z)^2}.
\end{align*}
The hypotheses are satisfied here because $\Omega\subset\mathbb{C}$ is bounded and admits a Green function by assumption, and the normalization above is exactly the normalization defining $c_\Omega(z)$. Applying the theorem with $a=1$ gives a [holomorphic function](/page/Holomorphic%20Function)
\begin{align*}
F_z:\Omega &\to \mathbb{C}
\end{align*}
such that $F_z(z)=1$ and
\begin{align*}
\int_\Omega |F_z(\zeta)|^2\,d\mathcal{L}^2(\zeta)
\le
\frac{\pi}{c_\Omega(z)^2}.
\end{align*}
Therefore the definition of $M_\Omega(z)$ gives
\begin{align*}
M_\Omega(z) \le \frac{\pi}{c_\Omega(z)^2}.
\end{align*}
[/step]
[step:Convert the extension estimate into Suita's inequality]
Using the reciprocal identity $K_\Omega(z)=1/M_\Omega(z)$ and the estimate for $M_\Omega(z)$, we obtain
\begin{align*}
K_\Omega(z)
=
\frac{1}{M_\Omega(z)}
\ge
\frac{c_\Omega(z)^2}{\pi}.
\end{align*}
Multiplying both sides by $\pi$ gives
\begin{align*}
\pi K_\Omega(z) \ge c_\Omega(z)^2.
\end{align*}
Since $z \in \Omega$ was arbitrary, the inequality holds for every $z \in \Omega$.
[/step]
[step:Characterize equality through the extremal extension problem]
The preceding argument shows that equality
\begin{align*}
\pi K_\Omega(z)=c_\Omega(z)^2
\end{align*}
holds if and only if
\begin{align*}
K_\Omega(z)=\frac{c_\Omega(z)^2}{\pi}.
\end{align*}
Using $K_\Omega(z)=1/M_\Omega(z)$, this is equivalent to
\begin{align*}
M_\Omega(z)=\frac{\pi}{c_\Omega(z)^2}.
\end{align*}
By the definition of $M_\Omega(z)$, this says exactly that the best possible one-point $L^2$ extension constant at $z$ equals the sharp value $\pi/c_\Omega(z)^2$ for the datum $1\in\mathbb{C}$. Hence equality in Suita's inequality is precisely equality of the corresponding extremal infimum in the sharp $L^2$ [extension theorem](/theorems/59), without requiring an additional assertion that the infimum is attained.
[/step]
