Fixed mistakes in FTC for path integrals.
All checks were successful
Compile Project / Compile (push) Successful in 27s

This commit is contained in:
Bokuan Li
2026-05-13 16:29:09 -04:00
parent 9f3c8a2e81
commit c1a9e11dbb
3 changed files with 32 additions and 8 deletions

View File

@@ -83,9 +83,7 @@
\begin{theorem}[Fundamental Theorem of Calculus for Path Integrals]
\label{theorem:ftc-path-integrals}
Let $[a, b] \subset \real$, $E, F$ be separated locally convex spaces, $\gamma \in C([a, b]; F)$ be a rectifiable path, $U \in \cn_F(\gamma([a, b]))$.
Let $\sigma \subset \mathfrak{B}(F)$ be an ideal containing all compact sets, then for any $f \in C^1_\sigma(U; E)$,
Let $[a, b] \subset \real$, $E, F$ be separated locally convex spaces, $\sigma \subset \mathfrak{B}(F)$ be an ideal containing all compact sets, $\gamma \in C([a, b]; F)$ be a rectifiable path, and $U \in \cn_F(\gamma([a, b]))$, then for any $f \in C^1_\sigma(U; E)$,
\[
\int_\gamma D_\sigma f = f(\gamma(b)) - f(\gamma(a))
\]
@@ -93,7 +91,7 @@
In particular, if $\gamma(a) = \gamma(b)$, then $\int_\gamma D_\sigma f = 0$.
\end{theorem}
\begin{proof}
Using \autoref{lemma:rectifiable-piecewise-linear}, assume without loss of generality that $\gamma$ is piecewise smooth. By the \hyperref[Chain Rule]{proposition:chain-rule-sets}, $f \circ \gamma \in C^1([a, b]; F)$ with $D(f \circ \gamma)(t) = Df(\gamma(t)) \cdot D\gamma(t)$. In which case, by \autoref{proposition:lebesgue-stieltjes-differentiable} and the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann},
Using \autoref{lemma:rectifiable-piecewise-linear}, assume without loss of generality that $\gamma$ is piecewise smooth. By the \hyperref[Chain Rule]{proposition:chain-rule-sets-conditions}, $f \circ \gamma$ is piecewise $C^1$ with $D(f \circ \gamma)(t) = Df(\gamma(t)) \cdot D\gamma(t)$ on all but finitely many points. In which case, by \hyperref[change of variables formula]{theorem:rs-change-of-variables} and the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann},
\begin{align*}
\int_\gamma D_\sigma f &= \int_a^b D_\sigma f (\gamma(t)) \cdot D\gamma(t)dt \\
&= \int_a^b D(f \circ \gamma)(t) dt = f(\gamma(b)) - f(\gamma(a))