Proving that an analytic function on a connected set which includes the upper half plane and satisfies a certain inequality is of a certain form

Question

Let $f$ be an analytic function on a connected open set $X$ which includes the closed upper half plane $\bar{\mathbb{H}}=\{z \in \mathbb{C};Im(z)\geq0\}$. Suppose there exist positive real numbers $M$ and $\delta$ s.t. $$|f(z)|\leq \frac{M}{(|z|+1)^{\delta}}$$ for all $z$ $\in \bar{H}$. Show that $$f(z)=\frac{1}{2\pi i}\int_{-\infty}^{\infty} \frac{f(t)}{t-z}dt$$ for all $z$ $\in \mathbb{H}=\{z \in \mathbb{C};Im(z)>0\}.$

How would I go about proving this? I'm quite stuck. I feel like $f$ must be a constant, and I would like to use the Maximum Principle to prove that, but I don't see how that's supposed to help me.

Any help would be greatly appreciated!

Answer 1

Here's a hint: Use the Cauchy integral theorem to write $$f(z) = \frac{1}{2\pi i} \oint_C \frac{f(t)}{t - z} \, dt$$ where $C$ is a closed semicircular contour of radius $R$ centered at the origin. Examine the behavior as $R \rightarrow \infty$.

Edit: To address your original idea, it is not true that $f$ will be constant. For example, consider $f = 1/(z + 2)$. It's true that $f$ decays at $\infty$, but since it is not entire it doesn't contradict Liouville's theorem.