Let $g(t,z)$ be a continuous function defined for $a \leq t \leq b$ and $z$ in a domain $D$, and suppose that $g(t,z)$ is a harmonic function of $z$ for each fixed $t$. Show that

$$G(z) = \int_{a}^{b} g(t,z)dt$$ is harmonic on $D$

I know I have to show this satisfies the Mean Value Property. The MVP I am given though is $$A(r) = \int_{0}^{2\pi} h(z_0 + re^{i\theta})\frac{d\theta}{2\pi}$$ which is said to be the average value of $h(z)$. This $h(z)$ is single valued, and I have the integral from $0$ to $2\pi$. Due to that, I am not sure how to show this applies to our case. Any help is appreciated!


1 Answer 1


i) The first thing to check is that $G$ is continuous in $D.$

ii) Having done i), it suffices to show

$$G(z_0) = \frac{1}{2\pi}\int_0^{2\pi}G(z_0+re^{is})\,ds$$

whenever $\overline {D(z_0,r)}\subset D.$

iii) Thus we want to show

$$G(z_0) = \frac{1}{2\pi}\int_0^{2\pi}\int_a^b g(t,z_0+re^{is})\, dt\,ds.$$

iv) Whenever faced with an iterated integral, reverse the order of integration (assuming it's legit), and then ask yourself why you did it.


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