[proofplan]
The reproducing property identifies evaluation at $a$ with the inner product against the kernel vector $k_a$. The Hilbert-space [Cauchy-Schwarz inequality](/theorems/432) gives the upper bound by $\|k_a\|_{A^2(\Omega)}^2$, and evaluating the reproducing identity on $k_a$ shows that this norm equals $K_\Omega(a,a)$. The reverse inequality is obtained by testing the supremum on the normalized kernel vector. For bounded pseudoconvex domains, an $L^2$ holomorphic jet realisation result supplies a square-integrable [holomorphic function](/page/Holomorphic%20Function) nonzero at $a$, forcing the extremal value and hence $K_\Omega(a,a)$ to be positive.
[/proofplan]
[step:Use the reproducing formula to bound every value at $a$ by the diagonal kernel]
Let $a \in \Omega$ satisfy $K_\Omega(a,a)>0$. By definition of the Bergman kernel, the function
\begin{align*}
k_a: \Omega &\to \mathbb{C} \\
z &\mapsto K_\Omega(z,a)
\end{align*}
belongs to $A^2(\Omega)$ and satisfies the reproducing identity
\begin{align*}
f(a) = (f,k_a)_{A^2(\Omega)}
\end{align*}
for every $f \in A^2(\Omega)$, where
\begin{align*}
(f,g)_{A^2(\Omega)} := \int_\Omega f(z)\overline{g(z)}\, d\mathcal{L}^{2n}(z).
\end{align*}
Applying the Hilbert-space [Cauchy-Schwarz inequality](/theorems/432) in $A^2(\Omega)$ to the pair $f,k_a \in A^2(\Omega)$ gives
\begin{align*}
|f(a)|^2
=
|(f,k_a)_{A^2(\Omega)}|^2
\le
\|f\|_{A^2(\Omega)}^2 \|k_a\|_{A^2(\Omega)}^2.
\end{align*}
Taking $f=k_a$ in the reproducing identity gives
\begin{align*}
K_\Omega(a,a)
=
k_a(a)
=
(k_a,k_a)_{A^2(\Omega)}
=
\|k_a\|_{A^2(\Omega)}^2.
\end{align*}
Therefore, for every nonzero $f \in A^2(\Omega)$,
\begin{align*}
\frac{|f(a)|^2}{\|f\|_{A^2(\Omega)}^2}
\le
K_\Omega(a,a).
\end{align*}
[guided]
The role of the kernel vector $k_a$ is to turn point evaluation into an inner product. Explicitly, define
\begin{align*}
k_a: \Omega &\to \mathbb{C} \\
z &\mapsto K_\Omega(z,a).
\end{align*}
The defining reproducing property of the Bergman kernel says that $k_a \in A^2(\Omega)$ and that every $f \in A^2(\Omega)$ satisfies
\begin{align*}
f(a) = (f,k_a)_{A^2(\Omega)}.
\end{align*}
Here the Bergman-space inner product is
\begin{align*}
(f,g)_{A^2(\Omega)} := \int_\Omega f(z)\overline{g(z)}\, d\mathcal{L}^{2n}(z),
\end{align*}
so the associated norm is $\|f\|_{A^2(\Omega)}^2=(f,f)_{A^2(\Omega)}$.
Now apply the Hilbert-space [Cauchy-Schwarz inequality](/theorems/432) to the two vectors $f$ and $k_a$ in the [Hilbert space](/page/Hilbert%20Space) $A^2(\Omega)$. This gives
\begin{align*}
|f(a)|^2
=
|(f,k_a)_{A^2(\Omega)}|^2
\le
\|f\|_{A^2(\Omega)}^2 \|k_a\|_{A^2(\Omega)}^2.
\end{align*}
The remaining point is to identify $\|k_a\|_{A^2(\Omega)}^2$ with the diagonal value $K_\Omega(a,a)$. We do this by applying the same reproducing identity to the particular function $f=k_a$, which is allowed because $k_a \in A^2(\Omega)$:
\begin{align*}
K_\Omega(a,a)
=
k_a(a)
=
(k_a,k_a)_{A^2(\Omega)}
=
\|k_a\|_{A^2(\Omega)}^2.
\end{align*}
Substituting this identity into the Cauchy-Schwarz estimate yields, for every nonzero $f \in A^2(\Omega)$,
\begin{align*}
\frac{|f(a)|^2}{\|f\|_{A^2(\Omega)}^2}
\le
K_\Omega(a,a).
\end{align*}
[/guided]
[/step]
[step:Test the supremum on the normalized kernel vector]
Since $K_\Omega(a,a)>0$ and $K_\Omega(a,a)=\|k_a\|_{A^2(\Omega)}^2$, the function $k_a$ is nonzero. Define
\begin{align*}
e_a: \Omega &\to \mathbb{C} \\
z &\mapsto \frac{k_a(z)}{\|k_a\|_{A^2(\Omega)}}.
\end{align*}
Then $e_a \in A^2(\Omega)$ and $\|e_a\|_{A^2(\Omega)}=1$. Using $k_a(a)=K_\Omega(a,a)=\|k_a\|_{A^2(\Omega)}^2$, we compute
\begin{align*}
|e_a(a)|^2
=
\left|\frac{k_a(a)}{\|k_a\|_{A^2(\Omega)}}\right|^2
=
\frac{K_\Omega(a,a)^2}{K_\Omega(a,a)}
=
K_\Omega(a,a).
\end{align*}
Thus
\begin{align*}
\sup_{\|f\|_{A^2(\Omega)}\le 1}|f(a)|^2
\ge
K_\Omega(a,a).
\end{align*}
Combining this lower bound with the upper bound from the previous step gives
\begin{align*}
\sup_{\|f\|_{A^2(\Omega)}\le 1}|f(a)|^2
=
K_\Omega(a,a).
\end{align*}
[/step]
[step:Identify the homogeneous and unit-ball suprema]
For every nonzero $f\in A^2(\Omega)$, define
\begin{align*}
u_f: \Omega &\to \mathbb{C} \\
z &\mapsto \frac{f(z)}{\|f\|_{A^2(\Omega)}}.
\end{align*}
Then $u_f \in A^2(\Omega)$ and $\|u_f\|_{A^2(\Omega)}=1$, so
\begin{align*}
\frac{|f(a)|^2}{\|f\|_{A^2(\Omega)}^2}
=
|u_f(a)|^2
\le
\sup_{\|g\|_{A^2(\Omega)}\le 1}|g(a)|^2.
\end{align*}
Conversely, if $g \in A^2(\Omega)$ satisfies $\|g\|_{A^2(\Omega)}\le 1$ and $g\ne 0$, then
\begin{align*}
|g(a)|^2
=
\frac{|g(a)|^2}{\|g\|_{A^2(\Omega)}^2}\,\|g\|_{A^2(\Omega)}^2
\le
\frac{|g(a)|^2}{\|g\|_{A^2(\Omega)}^2}
\le
\sup_{\substack{f\in A^2(\Omega)\\ f\ne 0}}
\frac{|f(a)|^2}{\|f\|_{A^2(\Omega)}^2}.
\end{align*}
The case $g=0$ contributes $|g(a)|^2=0$ and does not change the supremum. Hence
\begin{align*}
\sup_{\substack{f\in A^2(\Omega)\\ f\ne 0}}
\frac{|f(a)|^2}{\|f\|_{A^2(\Omega)}^2}
=
\sup_{\|f\|_{A^2(\Omega)}\le 1}|f(a)|^2.
\end{align*}
Together with the previous step, this proves both extremal identities.
[/step]
[step:Use holomorphic jet realisation to prove positivity on bounded pseudoconvex domains]
Assume now that $\Omega$ is bounded and pseudoconvex, and fix $a\in\Omega$. By the $L^2$ holomorphic jet realisation theorem on bounded pseudoconvex domains (citing a result not yet in the wiki: $L^2$ holomorphic jet realisation on bounded pseudoconvex domains), there exists a function
\begin{align*}
h: \Omega &\to \mathbb{C}
\end{align*}
such that $h \in A^2(\Omega)$ and $h(a)=1$. Therefore
\begin{align*}
\sup_{\substack{f\in A^2(\Omega)\\ f\ne 0}}
\frac{|f(a)|^2}{\|f\|_{A^2(\Omega)}^2}
\ge
\frac{|h(a)|^2}{\|h\|_{A^2(\Omega)}^2}
=
\frac{1}{\|h\|_{A^2(\Omega)}^2}
>
0.
\end{align*}
The first part of the proof identifies this supremum with $K_\Omega(a,a)$ whenever the diagonal is positive; independently, the same Cauchy-Schwarz argument gives the general upper bound by $K_\Omega(a,a)$ for the Bergman kernel vector, and the existence of $h$ makes the evaluation functional at $a$ nonzero. Hence the representing vector $k_a$ is nonzero, so
\begin{align*}
K_\Omega(a,a)
=
\|k_a\|_{A^2(\Omega)}^2
>
0.
\end{align*}
Since $a\in\Omega$ was arbitrary, $K_\Omega(a,a)>0$ for every $a\in\Omega$.
[/step]
