# Prove that a function is holomorphic

I want to prove that for $f:\mathbb{R}\to\mathbb{R}$ a continuous function such that $\int_{\mathbb{R}}\frac{|f(t)|}{1+t^{2}}dt<\infty$ ,for $z=x+iy$, $$u(z)=\frac{y}{\pi}\int_{\mathbb{R}}\frac{f(t)}{(x-t)^{2}+y^{2}}dt$$ is harmonic in the upper half plane $U=\{z\in\mathbb{C}:Im(z)>0\}$. For this I want to see that $u(z)$ is the real part of a holomorphic function, I obtain that $$u(z)= Re\left(\frac{1}{\pi}\int_{\mathbb{R}}\frac{i}{z-t}f(t)dt\right).$$

How can I prove that $\int_{\mathbb{R}}\frac{i}{z-t}f(t)dt$ is holomorphic?

• What are the conditions on $f$?
– kobe
Commented Apr 3, 2018 at 3:37
• $f:\mathbb{R}\to\mathbb{R}$ a continuous function such that $\int_{\mathbb{R}}\frac{|f(t)|}{1+t^{2}}dt<\infty$ Commented Apr 3, 2018 at 6:21

Let $g(z)=\int_\mathbb R \frac 1 {z-t} f(t) dt$. The integral exists for $Im(z) >0$. [ Use the hypothesis and the fact that $\frac {1+t^{2}} {(x-t)^{2}+y^{2}}$ is a bounded function on $\mathbb R$. (Here x and y are the real and imaginary parts of z). Now consider $g(z+h)-g(z)= \int_\mathbb R (\frac 1 {z+h-t}-\frac 1 {z-t}) f(t) dt$ $= \int_\mathbb R (\frac {-h} {(z+h-t)(z+t)}f(t) dt$. Divide by $h$ and take the limit. Once again use the fact that $\frac {1+t^{2}} {(z+h-t)(z-t)}$ is uniformly bounded in $t$ for $|h| <y/2$ so Dominated Convergence Theorem can be applied to show that $g$ is differentaiable at each point of the upper half plane.