Theorem 13.5.6 (Fundamental Theorem of Calculus for Path Integrals).label 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 $f \in C^{1}_{\sigma}(U; E)$,
\[\int_{\gamma} D_{\sigma} f = f(\gamma(b)) - f(\gamma(a))\]
In particular, if $\gamma(a) = \gamma(b)$, then $\int_{\gamma} D_{\sigma} f = 0$.
Proof. Using Lemma 13.5.5, assume without loss of generality that $\gamma$ is piecewise smooth. By the Chain Rule, $f \circ \gamma \in C^{1}([a, b]; F)$ with $D(f \circ \gamma)(t) = Df(\gamma(t)) \cdot D\gamma(t)$. In which case, by Proposition 13.7.2 and the Fundamental Theorem of Calculus,
\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))\end{align*}
$\square$