From 56f3ae37f7844096fdf9bdf0e2f8a21199801440 Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Wed, 3 Jun 2026 14:37:13 -0400
Subject: [PATCH] Slight adjustments in known results.

---
 src/op/banach/fc.tex     | 13 +++++++++----
 src/op/example/hardy.tex |  3 +--
 2 files changed, 10 insertions(+), 6 deletions(-)

diff --git a/src/op/banach/fc.tex b/src/op/banach/fc.tex
index dd438d7..88eb925 100644
--- a/src/op/banach/fc.tex
+++ b/src/op/banach/fc.tex
@@ -23,12 +23,18 @@
         \]
 
         for all $f \in H(U; \complex)$. 
+        \item For each $U \subset \complex$ open and $f \in H(U; \complex)$, the mapping
+        \[
+        \bracs{x \in A| \sigma_A(x) \subset U} \to A \quad x \mapsto f(x)
+        \]
+
+        is continuous. 
     \end{enumerate}
 
     The mapping $f \mapsto f(x)$ is the \textbf{holomorphic functional calculus} of $x$. 
 \end{definition}
 \begin{proof}[Proof, {{\cite[Proposition I.2.7]{Takesaki1}}}. ]
-    (Definition): Let $U, V \in \cn_\complex(\sigma_A(x))$ such that $\ol V$ is a compact subset of $U$, then by \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_A(x)$ such that 
+    (Definition), (3): Let $U, V \in \cn_\complex(\sigma_A(x))$ such that $\ol V$ is a compact subset of $U$, then by \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_A(x)$ such that 
     \begin{enumerate}[label=(\alph*)]
         \item For all $f \in H(V; \complex)$ and $z_0 \in \sigma_A(x)$, 
         \[
@@ -89,7 +95,6 @@
     \]
 
     (Uniqueness): By (2), the homomorphism extends uniquely to $\complex(z) \cap H(\sigma_A(x); \complex)$. By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, it extends uniquely to $H(\sigma_A(x); \complex)$ by continuity. 
-
 \end{proof}
 
 \begin{theorem}[Spectral Mapping Theorem (Holomorphic)]
@@ -103,7 +108,7 @@
 \begin{proof}
     (1): Let $\lambda \in \complex \setminus f(\sigma_A(x))$, then $1/(\lambda - f) \in H(\sigma_A(x); \complex)$. Since the holomorphic functional calculus is a homomorphism, $[1/(\lambda - f)](x) = (\lambda - f(x))^{-1}$. Thus $\sigma_A(f(x)) \subset f(\sigma_A(x))$. 
 
-    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$ by \autoref{proposition:zero-finite-multiplicity}. Therefore by the homomorphism property, 
+    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$ by \autoref{proposition:zero-finite-multiplicity}. By the homomorphism property, 
     \[
     f(x) - f(\lambda) = (x - \lambda)h(x) = h(x)(x - \lambda)
     \]
@@ -120,7 +125,7 @@
     
     By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, there exists $h \in U \cap \complex(z)$. Thus $\norm{g(f(x)) - (g \circ f)(x)}_A < \eps$ as well. As this holds for all $\eps > 0$, $g(f(x)) = (g \circ f)(x)$ for any $f \in H(\sigma_A(x); \complex)$ and $g \in \complex(x) \cap H(f(\sigma_A(x)); \complex)$. 
     
-    Through a second application of \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} again and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
+    Through a second application of \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
 \end{proof}
 
 \begin{proposition}
diff --git a/src/op/example/hardy.tex b/src/op/example/hardy.tex
index e39c1b7..f4ba611 100644
--- a/src/op/example/hardy.tex
+++ b/src/op/example/hardy.tex
@@ -23,12 +23,11 @@
         \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$. 
         \item $\phi_{\fU}$ is a multiplicative linear functional on $H^\infty(D)$. 
     \end{enumerate}
-    
 \end{proposition}
 \begin{proof}
     Let $f \in H^\infty(D)$, then by \autoref{proposition:imagefilterbase}, $f(\fU)$ is an ultrafilter base. Since $f$ is bounded, $f(\fU)$ converges to exactly one element of $\complex$. Hence the limit is well-defined. 
 
-    (2): By \autoref{proposition:operator-space-completeness}. 
+    (2): By \autoref{definition:multiplicative-linear-functional-space}. 
 \end{proof}