$f:\mathbb{R}^2\to\mathbb{R}$ is a smooth bounded function. In any unit circle, its value in the center of the circle is equal to the average value on the circumference.
Prove that $f$ is constant.
Attempt
I was trying to prove $f'(x,y)=0$ for all $x,y\in \mathbb{R}^2$. We know that $$f(x,y) = \int_{x-1}^{x+1} \frac{f(x_1,y+\sqrt{1-(x_1-x)^2})+f(x_1,y-\sqrt{1-(x_1-x)^2})}{2\pi}dx_1$$ Then $$f'(x,y) = \frac{f(x+1,y)-f(x-1,y)}{2\pi}=\frac{f(x,y+1)-f(x,y-1)}{2\pi},$$ so for all $x,y\in\mathbb{R}:f(x+1,y)+f(x,y+1)=f(x-1,y)+f(x,y-1)$, or $f(x+n,y+n+1)+f(x+n+1,y+n)=const \ \forall n\in\mathbb{Z}$
Because $f(x,y)$ is smooth and bounded, $\exists (x_M, y_m):f(x_M, y_M) = M$ and $f'(x_M, y_M)=0$.
That means that all points on the circumference with center in $(x_M, y_M)$ and radius $1$ are such that $f(x,y)=M$
So $f(x_M+1, y_M) = -f(x_M, y_M+1) = M = 0$.
How do I extend this result to all $(x,y)$?