I have been asked to show that: $$\frac1 N \sum_{i=1}^Ng(x_i,y_i)=g(\bar x,\bar y)$$ Where $$\bar g=\frac1 N \sum_{i=1}^Ng(x_i,y_i)$$
A clue given in the problem sheet is:
$$g(x_i,y_i)-g(\bar x,\bar y)=(\frac{\partial g}{\partial x})_y(x_i-\bar x)+(\frac{\partial g}{\partial y})_x(y_i-\bar y)$$
Clearly this rearranges to give: $$g(x_i,y_i)=(\frac{\partial g}{\partial x})_y(x_i-\bar x)+(\frac{\partial g}{\partial y})_x(y_i-\bar y)+g(\bar x,\bar y)$$ which may be plugged into the first equation: $$\frac1 N \sum_{i=1}^N[(\frac{\partial g}{\partial x})_y(x_i-\bar x)+(\frac{\partial g}{\partial y})_x(y_i-\bar y)+g(\bar x,\bar y)]$$
$$\frac1 N \sum_{i=1}^N(\frac{\partial g}{\partial x})_y(x_i-\bar x)+\frac1 N \sum_{i=1}^N(\frac{\partial g}{\partial y})_x(y_i-\bar y)+\frac1 N \sum_{i=1}^Ng(\bar x,\bar y)$$
$g(x_i,y_i)$ is contant so:
$$\frac1 N \sum_{i=1}^N(\frac{\partial g}{\partial x})_y(x_i-\bar x)+\frac1 N \sum_{i=1}^N(\frac{\partial g}{\partial y})_x(y_i-\bar y)+g(\bar x,\bar y)$$
But I don't know where to go from here...
The proposition is certainly false. We can resort to $g(x)$ by letting $g(x, \cdot)$ constant. Take $g(x) = x^2, x_1 = 1, x_2 = -1$ then $$\bar g = \frac12 \sum_{i=1}^2 g(x_i) = \frac12 \cdot (1+1) = 1 \ne 0 = g(0) = g(\frac12 \sum_{i=1}^2 x_i) = g(\bar x)$$
Actually the proposition implies that $g$ is more or less affine: $$\frac12 (g(x) + g(x')) = g(\frac12(x+x')) \qquad\forall\ x,x' \\ \stackrel{x'=0}\Rightarrow \frac12 g(x) + \frac12 g(0) = g(\frac12 x) \qquad \forall\ x$$