# Continuous function with 4-point mean value property is harmonic

Suppose a continuous function $$u : \mathbb{C} → \mathbb{R}$$ has the following property: $$u(x + iy) = \frac{1}{4}[(u(x + a + iy) + u(x − a + iy) + u(x + i(y + a)) + u(x + i(y − a)))]$$ for all $$a\in\mathbb{C}$$. Does it imply that u is harmonic?

I am inclined to believe so, but when I true to compute a mean value property for any $$z_0\in\mathbb{C}$$, I get: $$\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta}+z_0)d\theta=\frac{1}{2\pi}\sum_{k=0}^3\int_{\frac{k\pi}{4}}^{\frac{(k+1)\pi}{4}} u(re^{i\theta}+z_0)d\theta=\frac{1}{2\pi}\frac{1}{4}\int_{0}^{\frac{\pi}{4}}u(z_0)d\theta=\frac{u(z_0)}{2}\neq u(z_0)$$.

Where am I going wrong?

The idea is good, but you have to split the integration path into four segments of angle $$\pi/2$$, not $$\pi/4$$. Also the factor $$1/4$$ in the third expression should be $$4$$: $$\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta}+z_0)d\theta=\frac{1}{2\pi}\sum_{k=0}^3\int_{\frac{k\pi}{2}}^{\frac{(k+1)\pi}{2}} u(re^{i\theta}+z_0)d\theta=\frac{1}{2\pi} \int_{0}^{\frac{\pi}{2}}4 u(z_0)d\theta= u(z_0)$$