jerrylicious/garden
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Added setup for the GKZ theorem.

Browse Source
This commit is contained in:
Bokuan Li
2026-06-02 19:11:59 -04:00
parent 44d8bbd4da
commit 54a518e9e2
5 changed files with 153 additions and 2 deletions
Download Patch File Download Diff File Expand all files Collapse all files

4
src/op/banach/fc.tex
View File

@@ -103,7 +103,7 @@
\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 $H(\sigma_A(x); \complex)$. 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}. Therefore by the homomorphism property,
\[
f(x) - f(\lambda) = (x - \lambda)h(x) = h(x)(x - \lambda)
\]
@@ -120,7 +120,7 @@
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 $g \in \complex(x) \cap H(f(\sigma_A(x)); \complex)$.
By \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} again and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$.
\end{proof}
\begin{proposition}