If $f\in L^p(\mathbb{R})$ and $h_{\lambda}$ is Poisson Kernel of the upper half plane, then $f*h_{\lambda}(x)$ is harmonic

Question

I want to prove that if $f\in L^p(\mathbb{R})$ and $h_{\lambda}$ is Poisson Kernel of the upper half plane, then $f*h_{\lambda}(x)$ is a harmonic function of $x+i\lambda$ in the upper half plane $h_{\lambda}(x)=\sqrt{\frac{2}{\pi}}\frac{\lambda}{\lambda^2+x^2}$. Since $C(\mathbb{R})$ is dense in $L^p(\mathbb{R})$, there exists a continuous function $g$ such that $f=g$ a.e. Since $g$ is continuous on $\mathbb{R}$, $g*h_{\lambda}(=h_{\lambda}[g])$ is harmonic in the upper half plane. This is where I am stuck and I need help proving $f*h_{\lambda}(x)$ is harmonic. Thanks!

Answer 1

I will spare the details.

Let \begin{equation*} \begin{aligned} u(z) = (f*h_{\lambda})(x) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x-y) \sqrt{\frac{2}{\pi}}\frac{\lambda}{\lambda^2 + y^2}\,dy = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{f(t)\lambda}{\lambda^2 + (x-t)^2}\,dt \\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} \varphi(z,t) f(t) \,dt \\ \end{aligned} \end{equation*} It is easy to show that $\varphi$ is harmonic and continuous, hence it satisfies the mean value property in the upper half plane.

One can also show that $\varphi \in L^p(\mathbb{R})$ for $p \in [1, \infty]$.

Then by Lebesgue's Dominated Convergence Theorem, we have that for some $\alpha$ and $\{\alpha_n\}$ in the upper half plane that \begin{equation*} \begin{aligned} \lim_{n\to\infty}u(\alpha_n) &= \lim_{n\to\infty} \frac{1}{\pi}\int_{-\infty}^{\infty} \varphi(\alpha_n,t)f(t)\,dt = \frac{1}{\pi}\int_{-\infty}^{\infty} \varphi(\alpha,t)f(t)\,dt = u(\alpha) \end{aligned} \end{equation*} Hence $u$ is continuous at $\alpha$, hence $u$ is continuous in the upper half plane.

Now consider some $z$ in the upper half plane and some $r>0$ such that $\overline{D(z;r)}$ is contained in the upper half plane. Hence by Fubini Theorem, since $\varphi$ satisfies the mean value property, and by definition, we have that \begin{equation*} \begin{aligned} \frac{1}{2\pi} \int_{-\pi}^{\pi} u(z+re^{it})\,dt &= \frac{1}{2\pi^2} \int_{-\pi}^{\pi} \int_{-\infty}^{\infty} \varphi(z + re^{it}, \theta)f(\theta)\,d\theta\,dt \\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} \left( \frac{1}{2\pi} \int_{-\pi}^{\pi} \varphi(z + re^{it}, \theta)\,dt \right) f(\theta)\,d\theta \\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} \varphi(z,\theta) f(\theta)\,d\theta = u(z) \end{aligned} \end{equation*} Hence $u$ satisfies the mean value property.

Hence $u$ is harmonic in the upper half plane.