Question: Let $f\colon [0,1] \to \mathbb{C}, f(t) = u(t) + i v(t)$ be such that both $u$ and $v$ are of bounded variation on $[0,1]$ with $f(0) = f(1)$. Let $g(e^{2 \pi i t}) = f(t)$. Show that the function $h\colon D_1(0) \to \mathbb{C}$ defined by $$h(z) := \int_{\gamma} \frac{g(\xi)}{\xi - z}\, d \xi$$ is analytic in $D_1(0)$ where $\gamma$ is the circle $|z| = 1$.
My attempt: It suffices to prove that $h$ is holomorphic in $D_1(0)$, i.e. $h$ is complex differentiable on $D_1(0)$. Let $z_0 \in D_1(0)$ be given and define $r = (1 - |z_0|)/2$ then the closed ball $\overline{D_r(z_0)}$ is contained in $D_1(0)$. Let $(k_n) \subset \mathbb{C} - \{0\}$ be a sequence that converges to $0$. Assuming without loss of generality that $(z_0 + k_n)_n \subset \overline{D_r(z_0)}$.
by using the parameterisation of $\gamma$ as $\xi = \xi(t) \colon [0,1] \to (|z|= 1)$ as $\xi(t) = e^{i 2 \pi t}$. For each $n$ define $g_n \colon [0,1] \to \mathbb{C}$ as
\begin{align*}
g_n(t) := \frac{f(t)}{(e^{i 2\pi t} - z_0 - k_n)(e^{i 2\pi t} - z_0)}(2 \pi i) e^{i 2 \pi t}
\end{align*}
Given any $t \in [0,1]$ we have $|e^{i 2\pi t} - z_0 - k_n|\geq 1 - |z_0| - |k_n| \geq r$ and similarly $|e^{i 2 \pi t} - z_0| \geq 1 - |z_0| = 2r$ and as functions of bounded variation, $u,v$ are bounded on $[0,1]$, so there exists $M > 0$ so that $|u(t)|, |v(t)| \leq M$ for all $t \in [0,1]$. Thus,
\begin{align*}
|g_n(t)| \leq \frac{\sqrt{2} \pi M}{r^2}
\end{align*}
where $\frac{\sqrt{2} \pi M}{r^2}$ is independent of $n$ and integrable over $[0,1]$ so by Lebesgue's dominated convergence theorem we have
$$\lim_{n\to \infty} h_n(z_0) = \int_0^1 \frac{f(t)}{(e^{i 2\pi t} - z_0)^2}(2 \pi i)e^{i 2\pi t}\, dt = \int_\gamma \frac{g(\xi)}{(\xi - z_0)^2}\, d \xi = h'(z_0)$$
so $h$ is complex differentiable at $z_0$ and since $z_0$ and $(h_n)_n$ are arbitrary $h$ is complex differentiable in $D_1(0)$ so $h$ is analytic in $D_1(0)$.
May I check if there are any gaps in my proof? As I did not use the hypothesis that $f(0) = f(1)$. Any hints/solutions are appreciated.
