MHB Can an analytic function on the half plane be represented by an integral?

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Here is this week's POTW:

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Let $f$ be an analytic function on the half plane $\Omega := \{z\in \Bbb C : \operatorname{Im}(z) \ge 0\}$ such that for some $\alpha > 0$ and $M > 0$, $\lvert z^\alpha f(z)\rvert < M$ for all $z\in \Omega$. Prove that $f$ has integral representation

$$f(z) = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{f(t)}{t - z}\, dt\quad (\operatorname{Im}(z) > 0)$$

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This week's problem was solved correctly by Ackbach. You can read his solution below.

Spoiler

Let $W=\{z\in\mathbb{C}:\operatorname{Im}(z)>0\}$ be the strict upper half plane. Since $f$ is analytic on $\Omega,$ it is analytic on $W$. Note that $W$ is an open subset of $\mathbb{C}.$ Let $z\in W$ be arbitrary, and given by $z=x+iy,\; y>0.$ Let $D$ be the closed disk $D=\{w:|w-z|\le y/2\},$ and let $\gamma$ be its boundary, oriented counterclockwise. Note that $D$ is completely contained in $W,$ and hence by the Cauchy Integral Formula we have that
$$f(z)=\frac{1}{2\pi i}\oint_{\gamma}\frac{f(t)}{t-z}. $$
Let $R>0$ such that $R^2>x^2+y^2,$ and consider the contour $\gamma'$ consisting of the straight line segment from $-R$ to $R,$ followed by the half-circle from $R$ back to $-R$ in the upper-half-plane. Because $f$ is analytic on $\Omega,$ and the only singularity of $f(t)/(t-z)$ is at $z,$ it follows from homotopy that
$$\oint_{\gamma}\frac{f(t)}{t-z}=\oint_{\gamma'}\frac{f(t)}{t-z}=\int_{-R}^{R}\frac{f(t)}{t-z}\,dt+\underbrace{\int_{0}^{\pi}\frac{f(Re^{i\theta})}{Re^{i\theta}-z}\,i\,R e^{i\theta}\,d\theta}_{t=Re^{i\theta},\; dt=iRe^{i\theta}\,d\theta}.$$
We examine this second integral, the $d\theta$ integral, and use the $ML$ estimate:
\begin{align*}
\left|\int_{0}^{\pi}\frac{f(Re^{i\theta})}{Re^{i\theta}-z}\,i\,R e^{i\theta}\,d\theta\right|&\le
\int_{0}^{\pi}\left|\frac{f(Re^{i\theta})}{Re^{i\theta}-z}\,i\,R e^{i\theta}\right|d\theta \\
&\le \int_{0}^{\pi}\left|\frac{f(Re^{i\theta})}{Re^{i\theta}-z}\,R\, \right|d\theta \\
&\le \int_{0}^{\pi}\left|\frac{M/R^{\alpha}}{e^{i\theta}-z/R}\, \right|d\theta \\
&\le \int_{0}^{\pi}\frac{M/R^{\alpha}}{1-|z|/R}\,d\theta \\
&=\pi\cdot\frac{M/R^{\alpha}}{1-|z|/R}.
\end{align*}
As $R\to\infty,$ the denominator approaches $1\not=0;$ in fact, we can bound it away from zero in magnitude. The numerator goes to zero as $R\to\infty.$

Hence, we would like to write
$$f(z)=\frac{1}{2\pi i}\lim_{R\to\infty}\int_{-R}^{R}\frac{f(t)}{t-z}\,dt=\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{f(t)}{t-z}\,dt.$$
We just need to make sure this integral converges. But this must happen because of our previous formula:
\begin{align*}
\oint_{\gamma}\frac{f(t)}{t-z}&=\int_{-R}^{R}\frac{f(t)}{t-z}\,dt+\int_{0}^{\pi}\frac{f(Re^{i\theta})}{Re^{i\theta}-z}\,i\,R e^{i\theta}\,d\theta \\
\lim_{R\to\infty}\oint_{\gamma}\frac{f(t)}{t-z}&=\lim_{R\to\infty}\int_{-R}^{R}\frac{f(t)}{t-z}\,dt+\lim_{R\to\infty}\int_{0}^{\pi}\frac{f(Re^{i\theta})}{Re^{i\theta}-z}\,i\,R e^{i\theta}\,d\theta \\
\oint_{\gamma}\frac{f(t)}{t-z}&=\lim_{R\to\infty}\int_{-R}^{R}\frac{f(t)}{t-z}\,dt \\
\oint_{\gamma}\frac{f(t)}{t-z}&=\int_{-\infty}^{\infty}\frac{f(t)}{t-z}\,dt,
\end{align*}
and the theorem is proved.