I am having trouble with the following problem and would appreciate some help. It is a piece of a proof of Cauchy's integral formula for a circle that appears in Dieudonne's Treatise on Analysis, Vol. I. I am interested in this proof because it proceeds directly from the assumption that $f$ is complex differentiable in an open neighborhood of the disk. All other proofs that I have seen take a more indirect route, often going through some form of Cauchy's theorem first.
Let $f$ be complex differentiable on an open neighbourhood of the closed unit disc. Assume that, for each given $\lambda \in (0,1)$ and $\epsilon > 0$, there are points $t_0 = 0 < t_1 < \cdots < t_m = 2\pi$, a number $\delta > 0$, and in each interval $[t_k,t_{k+1}]$ a point $\theta_k$ such that $$ | f(\zeta_k + x) - f(\zeta_k) - f^{\prime}(\zeta_k)x | \leq \epsilon |x| $$ whenever $|h| \leq \delta$ and $t_k \leq t \leq t_{k+1}$. Here $\zeta_k = z + \lambda(e^{i\theta_k} - z)$ and $\zeta_k + x = z + (\lambda+h)(e^{it} - z)$.
Define $$ g(\lambda) = \int_{0}^{2\pi} \frac{ f(z + \lambda(e^{it} - z)) - f(z)}{e^{it} - z} e^{it} dt. $$
By comparing each integral $$ A_k = \int_{t_k}^{t_{k+1}} \frac{ f( z + (\lambda + h)(e^{it} - z) ) - f( z + \lambda(e^{it} - z) ) }{e^{it} - z} e^{it} dt $$ to the corresponding expression $$ B_k = \frac{h}{i \lambda} ( f( z + \lambda(e^{i t_{k+1}} - z) ) - f( z + \lambda(e^{i t_k} - z) ) ) $$ show that, for each given $\lambda \in (0,1)$ and $\epsilon > 0$, there is a $\delta > 0$ such that $$ | g(\lambda + h) - g(\lambda) - 0 \cdot h | \leq \epsilon |h| $$ for $|h| \leq \delta$. That is, show that $g^{\prime}(\lambda) = 0$ for $\lambda \in (0,1)$.
Note $$ \sum_{k=0}^{m-1} (A_k - B_k) = g(\lambda + h) - g(\lambda) - 0 \cdot h. $$
