I have been working on the following problem:
Let $f$ be a continuously differentiable function on the unit circle $\partial \mathbb{D} = \{ z \in \mathbb{C} : |z| =1 \}$. Let $\gamma (t) = e^{it}$, $0 \leq t \leq 2\pi$. Define $ g(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(w) dw}{w-z}, \quad z\in \mathbb{D}. $ Prove that $g$ is bounded in $\mathbb{D}$.
There is a hint with the problem which says to consider the point $w \in \partial \mathbb{D}$ that is closest to $z$.
It's easy to see that $ |g(z)| \leq \frac{M}{r}, $ where $\displaystyle M = \max_{w \in \partial \mathbb{D}} |f(w)|$ and $r$ is the minimum distance from the unit circle to the point $z$, but of course this bound approaches $\infty$ as $z$ nears the unit circle.
I've seen a proof of this in one of the texts I have, but it uses some heavy computations to show that $g$ is analytic, not just bounded, and it doesn't make any use of the closest point. I'm wondering if there's a simple way to show boundedness that's eluding me.
Thanks.
Edit: In an attempt to use the differentiability condition, I can rewrite the relevant integral as \begin{align*} \int_\gamma \frac{f(w) dw}{w-z} &= \int_\gamma \frac{f(w) - f(w_0)}{w-z} dw \quad + \quad f(w_0)\int_\gamma \frac{1}{w-z}dw \\ &= \int_\gamma \frac{f(w) - f(w_0)}{w-w_0} \cdot \frac{w-w_0}{w-z} dw \quad + \quad f(w_0)\int_\gamma \frac{1}{w-z} dw\\ &= \int_\gamma \left[ \frac{f(w) - f(w_0)}{w-w_0} -f'(w_0) \right] \cdot \frac{w-w_0}{w-z} dw \\ &\quad + \quad f'(w_0) \int_\gamma \frac{w-w_0}{w-z} dw \quad + \quad f(w_0) \int_\gamma \frac{1}{w-z}dw \end{align*} where $w_0$ is the point on the unit circle which is closest to $z$. The last two integrals in the final expression will be bounded for any $z$ and corresponding $w_0$ since $f$ and $f'$ are bounded. I still can't see whether I can use the continuous differentiability to get a bound on the first integral, though.