[proofplan] We reduce the vector-valued statement to the classical scalar Cauchy theorem on cycles via duality. Given any $\varphi \in X^*$, the composition $\varphi \circ f : U \to \mathbb{C}$ is scalar-valued and analytic on $U$, so the scalar Cauchy theorem applied to $\varphi \circ f$ on the cycle $\Gamma$ (with the same winding-number hypothesis) gives $\int_\Gamma (\varphi \circ f)(z)\, dz = 0$. The commutation property [Functionals Commute with Vector-Valued Integrals](/theorems/2682) lets us pull $\varphi$ through the vector integral, so $\varphi(\int_\Gamma f\, dz) = 0$ for all $\varphi \in X^*$. The [Hahn-Banach Theorem (Normed Space Version)](/theorems/2629)(ii) then forces $\int_\Gamma f\, dz = 0_X$ in $X$. [/proofplan] [step:Reduce to scalar functions via duality] Let $\varphi \in X^*$ be arbitrary. Define \begin{align*} g_\varphi : U &\to \mathbb{C} \\ z &\mapsto \varphi(f(z)). \end{align*} We claim $g_\varphi$ is analytic on $U$. Since $f : U \to X$ is analytic and $\varphi : X \to \mathbb{C}$ is bounded (in particular $\mathbb{C}$-linear and continuous), at every $z_0 \in U$ the limit \begin{align*} g_\varphi'(z_0) = \lim_{h \to 0} \frac{g_\varphi(z_0 + h) - g_\varphi(z_0)}{h} = \lim_{h \to 0} \varphi\!\left(\frac{f(z_0 + h) - f(z_0)}{h}\right) = \varphi(f'(z_0)) \end{align*} exists, the second equality using linearity of $\varphi$ and the third using continuity of $\varphi$ at $f'(z_0) \in X$ (the difference quotient converges in norm to $f'(z_0)$ because $f$ is analytic at $z_0$, so $\varphi$ converts norm convergence into scalar convergence). Hence $g_\varphi$ is complex-differentiable at every point of $U$, i.e., holomorphic. [/step] [step:Apply the scalar Cauchy theorem on cycles to $g_\varphi$] The hypotheses of the scalar **Cauchy Theorem on Cycles** (the homological form: for any cycle $\Gamma$ in an open $U \subseteq \mathbb{C}$ with $n(\Gamma, \omega) = 0$ for all $\omega \in \mathbb{C} \setminus U$, and any holomorphic $g : U \to \mathbb{C}$, $\int_\Gamma g(z)\, dz = 0$) are met: - $\Gamma$ is a cycle in $U$, by hypothesis; - the winding-number condition $n(\Gamma, \omega) = 0$ for all $\omega \in \mathbb{C} \setminus U$ is given by hypothesis; - $g_\varphi : U \to \mathbb{C}$ is holomorphic on $U$, by the previous step. Therefore \begin{align*} \int_\Gamma g_\varphi(z)\, dz = \int_\Gamma \varphi(f(z))\, dz = 0. \tag{$\ast$} \end{align*} [/step] [step:Pass $\varphi$ through the vector-valued integral] Recall that the cycle integral $\int_\Gamma f(z)\, dz$ is defined as a finite $\mathbb{Z}$-linear combination of piecewise-$C^1$ contour integrals, each of which is in turn defined by parametrising a piecewise-$C^1$ closed curve $\gamma : [a, b] \to U$ as \begin{align*} \int_\gamma f(z)\, dz := \int_a^b f(\gamma(t))\, \gamma'(t)\, d\mathcal{L}^1(t) \in X, \end{align*} the right-hand side being the vector-valued integral of the continuous map $t \mapsto f(\gamma(t))\, \gamma'(t)$ on the compact interval $[a, b]$. Apply [Functionals Commute with Vector-Valued Integrals](/theorems/2682). Theorem 2682 requires (i) a Banach space (we have $X$), (ii) a continuous map $[a, b] \to X$ (we have $t \mapsto f(\gamma(t))\, \gamma'(t)$, continuous because $f$ is continuous on $U$, $\gamma$ is continuous on $[a, b]$, $\gamma'$ is piecewise-continuous, and the scalar–vector multiplication is jointly continuous), and (iii) a bounded linear functional $\varphi \in X^*$ (given). Both hypotheses (i)–(iii) hold, so theorem 2682 applies and gives \begin{align*} \varphi\!\left(\int_a^b f(\gamma(t))\, \gamma'(t)\, d\mathcal{L}^1(t)\right) = \int_a^b \varphi(f(\gamma(t)))\, \gamma'(t)\, d\mathcal{L}^1(t), \end{align*} where the scalar $\gamma'(t)$ has been pulled out of $\varphi$ by linearity. Re-interpreting the right-hand side as a contour integral of $g_\varphi = \varphi \circ f$ along $\gamma$ gives \begin{align*} \varphi\!\left(\int_\gamma f(z)\, dz\right) = \int_\gamma \varphi(f(z))\, dz = \int_\gamma g_\varphi(z)\, dz. \end{align*} Summing over the (finitely many) curves making up the cycle $\Gamma$ with the same integer multiplicities (linearity of $\varphi$ commutes with finite sums): \begin{align*} \varphi\!\left(\int_\Gamma f(z)\, dz\right) = \int_\Gamma g_\varphi(z)\, dz \overset{(\ast)}{=} 0. \end{align*} [guided] The whole strategy hinges on reducing a vector-valued statement to a scalar one. We have a scalar version of the Cauchy theorem at our disposal, and we need a way to extract scalar information from a vector. Bounded linear functionals are the natural tool: in a Banach space, scalar-valued probes $\varphi \in X^*$ resolve every vector (this is the content of the Hahn-Banach norming-functional result). So we test the integral against an arbitrary $\varphi \in X^*$ and check that the scalar result is $0$. **Why does $\varphi$ pass through the integral?** Because the vector integral is, by construction, a norm-limit of Riemann sums, and $\varphi$ is linear and continuous; both properties are needed. The packaged result is theorem 2682, which we apply on each smooth segment of $\Gamma$. The hypothesis check: - $X$ is Banach (given). - $t \mapsto f(\gamma(t))\, \gamma'(t)$ is continuous on each smooth piece of $\gamma$: $f$ is continuous on $U$ (it is analytic, hence in particular continuous), $\gamma$ is continuous on $[a, b]$, and $\gamma'$ is continuous on each smooth piece of $[a, b]$. - $\varphi \in X^*$ is bounded linear (given). So 2682 applies on each smooth piece, and summing over the finitely many pieces and integer multiplicities of $\Gamma$ gives \begin{align*} \varphi\!\left(\int_\Gamma f(z)\, dz\right) = \int_\Gamma \varphi(f(z))\, dz. \end{align*} **Why is $\varphi \circ f$ holomorphic?** This is the second place the linearity-and-continuity of $\varphi$ pays off. The complex difference quotient $h^{-1}(f(z_0 + h) - f(z_0))$ converges in $X$-norm to $f'(z_0)$ (definition of $f$ analytic at $z_0$). Linearity lets us pull $\varphi$ inside the difference quotient; continuity lets us pull $\varphi$ through the limit. The result is a complex difference quotient of $\varphi \circ f$ converging to $\varphi(f'(z_0))$, i.e., $\varphi \circ f$ is complex-differentiable at $z_0$ — hence holomorphic on $U$. **Why does the scalar Cauchy theorem apply?** It applies to any holomorphic function on $U$ and any cycle in $U$ whose winding number vanishes outside $U$. Both conditions are inherited: the cycle $\Gamma$ is the same one given by hypothesis, and its winding-number property is stated for **every** $\omega \in \mathbb{C} \setminus U$, independent of the function. The function $\varphi \circ f$ is holomorphic on $U$ by the previous paragraph. So the scalar theorem gives $\int_\Gamma \varphi \circ f\, dz = 0$. Combining, \begin{align*} \varphi\!\left(\int_\Gamma f(z)\, dz\right) = \int_\Gamma \varphi(f(z))\, dz = 0. \end{align*} [/guided] [/step] [step:Use Hahn-Banach to conclude the vector integral vanishes] Set $J := \int_\Gamma f(z)\, dz \in X$. The previous step showed $\varphi(J) = 0$ for **every** $\varphi \in X^*$. Suppose for contradiction $J \neq 0_X$. By the [Hahn-Banach Theorem (Normed Space Version)](/theorems/2629)(ii), applied to the non-zero element $J \in X$, there exists $\varphi_0 \in X^*$ with $\|\varphi_0\|_{X^*} = 1$ and $\varphi_0(J) = \|J\|_X > 0$. The hypothesis of 2629(ii) — that $J$ is a non-zero element of the normed space $X$ — is the standing assumption. But this contradicts $\varphi_0(J) = 0$. Hence $J = 0_X$, i.e., $\int_\Gamma f(z)\, dz = 0_X$. [/step]