A contour integral of a jointly continuous, parameter-holomorphic integrand is holomorphic

The parameter interval $[a,b]$ is compact by [L4] and $\gamma$ is continuous, so the trace ${\gamma }^{\ast }$ is compact by [L6].

For each $z\in \mathrm{\Omega }$ the map $\zeta \mapsto \phi (\zeta ,z)$ is continuous on ${\gamma }^{\ast }$, so $F(z)$ exists by [L8].

Write $P_N$ for the uniform partition of $[a,b]$ into $N$ parts, with points $t_i^N=a+i(b-a)\mathrm{/}N$, and set $S_N(z)={\sum }_{i<N}\phi (\gamma (t_i^N),z)(\gamma (t_{i+1}^N)-\gamma (t_i^N))$ for $z\in \mathrm{\Omega }$, assuming $a<b$. Each summand is a constant multiple of a function holomorphic on $\mathrm{\Omega }$, so $S_N$ is holomorphic on $\mathrm{\Omega }$ by [L7].

Fix $z_0\in \mathrm{\Omega }$. By [L11] choose $s>0$ with $B(z_0,s)\subseteq \mathrm{\Omega }$ and put $\rho =s\mathrm{/}2>0$, so $\bar{B}(z_0,\rho )\subseteq B(z_0,s)\subseteq \mathrm{\Omega }$; put $K=\bar{B}(z_0,\rho )$, which is closed and bounded, hence compact by [L4]. By [L5] and step 1.1 both ${\gamma }^{\ast }$ and $K$ are closed and bounded, and their coordinate images have these properties by the given isometry. Their product is bounded because squared product distance is the sum of the two squared distances. It is closed because a point outside the product has a coordinate outside one closed factor; a ball in the open complement of that factor lifts, by the $1$-Lipschitz product projection, to a product-metric ball missing the product. Hence $J[{\gamma }^{\ast }\times K]$ is closed and bounded in ${\mathbb{R}}^4$ and compact by [L4]; the distance identity for $J$ transfers this compactness back by carrying open covers and finite subcovers in both directions; since $\phi$ is continuous there, [L3] makes it uniformly continuous on ${\gamma }^{\ast }\times K$.

Let $\epsilon >0$. Step 2.1 gives $\eta >0$ such that $\mathrm{\mid }\phi (u,z)-\phi (v,z)\mathrm{\mid }\le \epsilon$ whenever $u,v\in {\gamma }^{\ast }$ satisfy $\mathrm{\mid }u-v\mathrm{\mid }<\eta$ and $z\in K$. The interval $[a,b]$ is compact by [L4], so $\gamma$ is uniformly continuous on it by [L3]: there is $\delta >0$ with $\mathrm{\mid }\gamma (t)-\gamma (s)\mathrm{\mid }<\eta$ whenever $\mathrm{\mid }t-s\mathrm{\mid }<\delta$. By [L10] applied to $\delta \mathrm{/}(b-a)$ there is a natural $n\ge 1$ with $(b-a)\mathrm{/}n<\delta$.

Let $N\ge n$ and $z\in K$. Every two parameters in a subinterval of $P_N$ differ by at most $(b-a)\mathrm{/}N\le (b-a)\mathrm{/}n<\delta$, so any two points of $\gamma ([t_i^N,t_{i+1}^N])$ are within $\eta$ of each other and step 3.1 bounds $\mathrm{\mid }\phi (u,z)-\phi (v,z)\mathrm{\mid }$ by $\epsilon$ for such points. Applying [L1] to $f=\phi (\cdot ,z)$ on each subarc, with $\omega =\epsilon$ on that subarc, and summing the subarc bounds gives $\mathrm{\mid }F(z)-S_N(z)\mathrm{\mid }\le \epsilon L(\gamma )$.

Since $\epsilon >0$ was arbitrary and $L(\gamma )$ is a fixed nonnegative real by [L12], step 4.1 says $S_N\to F$ uniformly on $K$, hence uniformly on the open neighbourhood $B(z_0,\rho )$ of $z_0$; as $z_0\in \mathrm{\Omega }$ was arbitrary, $S_N\to F$ locally uniformly on $\mathrm{\Omega }$, and [L2] with step 1.3 makes $F$ holomorphic on $\mathrm{\Omega }$. If instead $a=b$ then ${\int }_{\gamma }fdz=0$ for every continuous $f$, so $F$ is identically $0$ and holomorphic by [L7].