Theorem 27.1.2 (Cauchy).label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $f \in C^{1}(U; E)$, and $\gamma, \mu \in C([a, b]; U)$ be closed rectifiable paths. If $\gamma$ and $\mu$ are homotopic, then
\[\int_{\gamma} f = \int_{\mu} f\]
Proof of smooth case. Let $\Gamma \in C^{\infty}([0, 1] \times [a, b]; U)$ be a smooth homotopy of loops from $\gamma$ to $\mu$, and
\[F: [0, 1] \to E \quad t \mapsto \int_{\Gamma (t, \cdot)}f = \int_{a}^{b} (f \circ \Gamma)(t, s) \Gamma(t, ds)\]
then for any $t \in [0, 1]$, by the change of variables formula,
\begin{align*}F(t)&= \int_{a}^{b} (f \circ \Gamma)(t, s) \Gamma(t, ds) \\&= \int_{a}^{b} (f \circ \Gamma)(t, s) \frac{\partial \Gamma}{\partial s}(t, s) ds\end{align*}
Now, by Proposition 26.9.2,
\[\frac{dF}{dt}(t) = \int_{a}^{b} \frac{\partial}{\partial t}\braks{(f \circ \Gamma)(t, s) \frac{\partial \Gamma}{\partial s}(t, s)}ds\]
Under the identification that $\complex = \real^{2}$, by the power rule and the chain rule,
\[\frac{\partial }{\partial t}\braks{(f \circ \Gamma) \frac{\partial \Gamma}{\partial s}}= (Df \circ \Gamma)\paren{\frac{\partial\Gamma}{\partial t}}\frac{\partial \Gamma}{\partial s}+ (f \circ \Gamma) \frac{\partial^{2}\Gamma}{\partial t \partial s}\]
Now, since $f \in C^{1}(U; E)$ satisfies the Cauchy-Riemann equations,
\begin{align*}(Df \circ \Gamma)\paren{\frac{\partial\Gamma}{\partial t}}\frac{\partial \Gamma}{\partial s}&= (Df \circ \Gamma)\frac{\partial\Gamma}{\partial t}\frac{\partial \Gamma}{\partial s}= (Df \circ \Gamma)\paren{\frac{\partial \Gamma}{\partial s}}\frac{\partial\Gamma}{\partial t}\end{align*}
so
\begin{align*}\frac{\partial }{\partial t}\braks{(f \circ \Gamma) \frac{\partial \Gamma}{\partial s}}&= (Df \circ \Gamma)\paren{\frac{\partial\Gamma}{\partial s}}\frac{\partial \Gamma}{\partial t}+ (f \circ \Gamma) \frac{\partial^{2}\Gamma}{\partial s \partial t}\\&= \frac{\partial }{\partial s}\braks{(f \circ \Gamma) \frac{\partial \Gamma}{\partial t}}\end{align*}
Hence by the Fundamental Theorem of Calculus,
\begin{align*}\frac{dF}{dt}(t)&= \int_{a}^{b} \frac{\partial}{\partial s}\braks{(f \circ \Gamma)(t, s) \frac{\partial \Gamma}{\partial t}(t, s)}ds \\&= (f \circ \Gamma)(t, b)\frac{\partial \Gamma}{\partial t}(t, b) - (f \circ \Gamma)(t, a)\frac{\partial \Gamma}{\partial t}(t, a)\end{align*}
Since $\Gamma(t, a) = \Gamma(t, b)$ for all $t \in [0, 1]$, the above expression evaluates to $0$, so
\[\int_{\gamma} f = F(0) = F(1) = \int_{\mu} f\]
by Proposition 26.3.6.$\square$
Proof of general case. Let $\Gamma \in C([0, 1] \times [a, b]; \complex)$ be a homotopy of loops from $\gamma$ to $\mu$. By augmenting $\Gamma$ and using Lemma 13.5.5, assume without loss of generality that:
(a)
$\mu$, $\gamma$ are piecewise linear.
Furthermore, by passing through a reparametrisation, assume without loss of generality that:
(b)
For each $t \in [0, \eps)$, $\Gamma(t, \cdot) = \gamma$.
(c)
For each $t \in (1 - \eps, 1]$, $\Gamma(t, \cdot) = \mu$.
(d)
For each $t \in [0, 1]$, $\Gamma$ is constant on $\bracs{t}\times ([a, a + \eps] \cup [b - \eps, b])$.
Extend $\Gamma$ to $[0, 1] \times \real$ by
\[\Gamma_{0}: \real^{2} \to \complex \quad (t, s) \mapsto \begin{cases}\Gamma(t, s) &t \in k(b-a) + [a, b], k \in \integer \\\end{cases}\]
then extend $\Gamma_{0}$ to $\real^{2}$ by
\[\ol \Gamma: \real^{2} \to \complex \quad (t, s) \mapsto \begin{cases}\Gamma(t, s) &t \in [0, 1] \\ \Gamma(1, s) &t \ge 1 \\ \Gamma(0, s) &t \le 0\end{cases}\]
Let $\varphi \in C_{c}^{\infty}(\real^{2}; \real)$ with $\int_{\real^2}\varphi = 1$. For each $\delta \ge 0$, let
\[\Gamma_{\delta}: [0, 1] \times [a, b] \to \complex \quad (t, s) \mapsto \frac{1}{\delta^{2}}\int_{\real^2}\Gamma(y) \varphi\paren{\frac{(t, s) - y}{\delta}}dy\]
Since for each $k \in \integer$ and $(t, s) \in \real^{2}$, $\Gamma(t, s + k(b - a)) = \Gamma(t, s)$, $\Gamma_{\delta}(t, a) = \Gamma_{\delta}(t, b)$ for all $t \in [0, 1]$. Therefore $\Gamma_{\delta}$ is a homotopy of loops. Since $\Gamma$ is continuous, $\Gamma([0, 1] \times [a, b])$ is compact, so $\Gamma_{\delta}$ lies in $U$ for sufficiently small
By assumptions (b) and (c), for sufficiently small $\delta$, there exists $\psi \in C_{c}^{\infty}(\real; \real)$ with $\int_{\real}\psi = 1$ such that
\[\Gamma_{\delta}(0, s) = \frac{1}{\delta}\int_{\real^2}\Gamma(0, y) \psi\paren{\frac{s - y}{\delta}}dy\]
and
\[\Gamma_{\delta}(1, s) = \frac{1}{\delta}\int_{\real^2}\Gamma(1, y) \psi\paren{\frac{s - y}{\delta}}dy\]
By assumption (a), (d), and Lemma 13.5.6,
\[\int_{\gamma} f = \lim_{\delta \downto 0}\int_{\Gamma_\delta(0, \cdot)}f = \lim_{\delta \downto 0}\int_{\Gamma_\delta(1, \cdot)}f = \int_{\mu} f\]
$\square$
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