Theorem 14.5.7 (Fundamental Theorem of Calculus for Path Integrals).label 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 $f \in C^{1}_{\sigma}(U; E)$,

\[\int_{\gamma} D_{\sigma} f = f(\gamma(b)) - f(\gamma(a))\]

In particular, if $\gamma(a) = \gamma(b)$, then $\int_{\gamma} D_{\sigma} f = 0$.

Proof. Using Lemma 14.5.5, assume without loss of generality that $\gamma$ is piecewise smooth. By the Chain Rule, $f \circ \gamma$ 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 change of variables formula 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*}

$\square$

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