POTW Contour Integral Representation of a Function

[Euge]

Suppose $f$ is holomorphic in an open neighborhood of the closed unit disk $\overline{\mathbb{D}} = \{z\in \mathbb{C}\mid |z| \le 1\}$. Derive the integral representation $$f(z) = \frac{1}{2\pi i}\oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\,\frac{w + z}{w - z}\, dw + i\operatorname{Im}(f(0))$$ for $|z| < 1$.

 

[anuttarasammyak]

Suppose $f$ is holomorphic in an open neighborhood of the closed unit disk $\overline{\mathbb{D}} = \{z\in \mathbb{C}\mid |z| \le 1\}$. Derive the integral representation $$f(z) = \frac{1}{2\pi i}\oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\,\frac{w + z}{w - z}\, dw + i\operatorname{Im}(f(0))$$ for $|z| < 1$.

By residue theorem may we say
\frac{1}{2\pi i}\oint_{|w| = 1} \frac{Re(f(w))}{w}\,\frac{w + z}{w - z}\, dw
=\frac{1}{2\pi i}\oint_{|w| = 1} [ \frac{2Re(f(w))}{w-z}-\frac{Re(f(w))}{w} ]\, dw
=2Re(f(z))-Re(f(0))
I have no idea for going further.

 

[julian]

Not sure you can use Cauchy's integral formula with $Re (f(w)) = \frac{1}{2} (f(w) + \overline{f (w)})$ because $\overline{f (w)}$ is not holomorphic.

 

[julian]

I have partial result, in that I might have proved the integral for functions specified in the Schwarz Reflection Principle.

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Schwarz Reflection Principle:​

Let $D$ be an open domain in the complex plane who's reflection about the real axis is symmetric. Define $D^+ = \{ z \in D : Im(z) >0 \}$. Suppose that $f$ is an analytic function which is defined in $D^+$. Further suppose that $f$ extends to a continuous function on the real axis, and takes on real values on the real axis. Then $f$ can be extended to an analytic function on $D$ by the formula

$$
f(\overline{z}) = \overline{f(z)} .
$$

and the values for $z$ reflected across the real axis are the reflections of $f(z)$ across the real axis.
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We assume that $D$ contains the disk $\overline{\mathbb{D}} = \{ z \in \mathbb{C} : |z| \leq 1 \}$.

In the following we only consider functions as described above. First note: $Im (f(0)) =0$.

Now

\begin{align}
\frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+z}{w-z} dw & = \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{1}{2} [f (e^{i \theta}) + \overline{f (e^{i \theta})}] \dfrac{e^{i \theta} + z}{e^{i \theta} - z} i d \theta
\nonumber \\
& = \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{1}{2} [f (e^{i \theta}) + f (e^{-i \theta})] \dfrac{e^{i \theta} + z}{e^{i \theta} - z} i d \theta
\nonumber \\
& = \frac{1}{4 \pi i} \int_0^{2 \pi} \left[ f (e^{i \theta}) \dfrac{e^{i \theta} + z}{e^{i \theta} - z} + f (e^{-i \theta}) \dfrac{e^{i \theta} + z}{e^{i \theta} - z} \right] i d \theta
\nonumber \\
& = \frac{1}{4 \pi i} \int_0^{2 \pi} f (e^{i \theta}) \left[ \dfrac{e^{i \theta} + z}{e^{i \theta} - z} + \dfrac{e^{-i \theta} + z}{e^{-i \theta} - z} \right] i d \theta
\nonumber \\
& = \frac{1}{4 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} \left[ \dfrac{w+z}{w-z} + \dfrac{1+wz}{1-wz} \right] dw \quad (*)
\nonumber
\end{align}

First, using $(*)$ when we have $z=0$:

$$
\frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+0}{w-0} dw = \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} dw = f(0) .
$$

Now take $z \not=0$. Using $(*)$,

\begin{align}
\frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+z}{w-z} dw & = \frac{1}{4 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} \left[ \dfrac{w+z}{w-z} - \dfrac{w+z^{-1}}{w-z^{-1}} \right] dw
\nonumber \\
& = \frac{1}{4 \pi i} \oint_{|w|=1} f(w) \left[ \dfrac{2}{w-z} - \frac{1}{w} - \dfrac{2}{w-z^{-1}} + \frac{1}{w} \right] dw
\nonumber \\
& = \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w-z}dw = f(z) .
\nonumber
\end{align}

 

[Euge]

Spoiler

Let $|z| < 1$. Cauchy's integral formula gives $f(z) = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{f(w)}{w - z}\, dw$. In particular, $f(0) = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{f(w)}{w}\, dw$. For all $w$ on the unit circle, $w\overline{w} = 1$. Hence $\overline{f(0)} = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\overline{f(w)}}{w} dw$, and
\begin{align*}
f(z) + \overline{f(0)} &= \frac{1}{2\pi i} \oint_{|w| = 1} \frac{w(f(w) + \overline{f(w)}) - z\overline{f(w)}}{w(w-z)} \, dw\\
&= \frac{2}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w - z}\, dw - \frac{z}{2\pi i} \oint_{|w| = 1} \frac{\overline{f(w)}}{(w - z)}\, \frac{dw}{w}\\
&= \frac{2}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w - z}\, dw + \frac{z}{2\pi i}\overline{ \oint_{|w| = 1} \frac{f(w)}{(\bar{w} - z)}\, \frac{dw}{w}}\\
&=\frac{2}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w - z}\, dw + \frac{z}{2\pi i}\overline{\oint_{|w| = 1} \frac{f(w)}{(1 - zw)}\, dw}\\
&= \frac{2}{2\pi i}\oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w-z}\, dw
\end{align*}
where the last equality follows from Cauchy's theorem (the function $w\mapsto f(w)/(1 - zw)$ is holomorphic inside and on the unit circle for $|z| < 1$). Since $z$ was arbitrary it follows that $$\operatorname{Re}(f(0)) = \frac{f(0) + \overline{f(0)}}{2} = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\, dw$$yielding $$f(z) - i\operatorname{Im}(f(0)) = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\left[\frac{2w}{w - z} - 1\right]\, dw = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w} \frac{w + z}{w - z}\, dw$$which is equivalent to the desired result.