2
$\begingroup$

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.

  • 0
    @Jose27 Integration by parts doesn't seem to help; the functions you get aren't holomorphic in the unit disk, so you can't use the fundamental theorem of calculus to simplify the results. I've tried writing $\int \frac{f(w) dw}{w-z} = \int \frac{f(w) - f(w_0)}{w-z} dw + f(w_0) \int \frac{1}{w-z} = \int \frac{f(w) - f(w_0)}{w-z} dw + f(w_0) \cdot 2 \pi i$ for $w_0$ on the unit circle, but that doesn't seem to be leading anywhere either.2012-07-11

1 Answers 1

3

Denote $C=\max_\gamma |f'|<\infty\;$. Let $z=rw_0$, $w_0\in \gamma$. Then $ \left|\int_\gamma \frac{f(w) - f(w_0)}{w-z}\, dw\right|\le \int_\gamma \left|\frac{f(w) - f(w_0)}{w-z}\right|\, dw\le $ $ C\int_\gamma \left|\frac{w - w_0}{w-rw_0}\right|\, dw\le C\int_\gamma \frac{|w - rw_0|+|rw_0 - w_0|}{|w-rw_0|}\, dw= $ $ C\int_\gamma \left(1+\left|\frac{1 - r}{w-r}\right|\right)\, dw\le C\int_\gamma 2\, dw= 4C\pi. $ In the last line it was taken into account what the point $1\in\gamma$ is the nearest to the point $r$ on the cirlce $\gamma$.

  • 1
    Oh, and I can apply the MVT here because $f$ is basically a function of one variable, the angle. Thanks.2012-07-12