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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.

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How do you find "the closest point to any point" on the complex plane?? –  DonAntonio Jul 11 '12 at 2:28
    
I'm not sure what you're asking. Here, for a point $z = Re^{it}$ in the disk, the closest point on the unit circle is just $w = e^{it}$ (unless $z=0$). –  ec92 Jul 11 '12 at 2:50
    
You're right: didn't pay attention that for you it seems to be that $\,\Bbb D:=\{z\;\:\;|z|<1\}\,$...got it. –  DonAntonio Jul 11 '12 at 2:53
    
Since differentiability of $f$ is not necessary for $g$ to be holomorphic, my guess is that this assumption makes things easier...maybe an integration by parts could help? –  Jose27 Jul 11 '12 at 4:25
    
@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. –  ec92 Jul 11 '12 at 20:42
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1 Answer

up vote 3 down vote accepted

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 $\gamma$.

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Thanks; I think I see. To get a bound $ \displaystyle \left| \frac{f(w)-f(w_0)}{w-w_0} \right| \leq C$ for all $w$, $w_0$ on the unit circle, we use that $f'$ continuous on a compact set implies $f$ is uniformly differentiable there, correct? –  ec92 Jul 12 '12 at 17:25
    
@ecg2 Since $|f'|$ is continuous on the circle $\gamma$ it attains its maximum somewhere on $\gamma$. –  Andrew Jul 12 '12 at 17:37
    
Yes, but I don't see why $|f'| \leq C$ implies that $ \displaystyle \quad \left| \frac{f(w) -f(w_0)}{w-w_o} \right| \leq C \quad$ for all $w$, $W_0$. –  ec92 Jul 12 '12 at 17:56
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@ec92 it's a corollary of the mean value theorem. –  Andrew Jul 12 '12 at 18:00
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Oh, and I can apply the MVT here because $f$ is basically a function of one variable, the angle. Thanks. –  ec92 Jul 12 '12 at 18:11
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