Part a) Find $v^{+}=\min_{y \in D} \max_{x \in C}$ $f(x,y)$ and
$v^{-}=\max_{x \in C}\min_{y \in D} $ $f(x,y)$
I'm sure I got this: $v^{+} = (-1)^2 - 1^2 = 0$ and
$v^{-} = 1^2-(-1)^2 = 0$
Part b wants a pure saddle point which I find by fixing $y=0$ $ \forall x$ and then fixing $x=0$ $\forall y$? That would be $f(0,0)$, since $f(x,0) <f(0,0) < f(0,y)$
If my reasoning is off, please give me a hand.
Thanks!
You are correct, but let's make it clear. A saddle point $(x_0,y_0)$ of a function $f(x,y)$ of two variables is a critical point of $f$ such that $$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-[f_{xy}(x_0,y_0)]^2\le 0$$ Here $f_{xx}\equiv -2$, $f_{yy}\equiv2$ and $f_{xy}\equiv0$, so any critical point is a saddle point. A critical point is given by $$f_x(x_0,y_0)=-2x_0\overset{!}=0\\f_y(x_0,y_0)=2y_0\overset{!}=0$$ so, indeed $(x_0,y_0)=(0,0)$ is the only saddle point.
But when you have an exercise with parts a) and b) it is 101% sure that in part b) you should part a). So, you should use that if $$\max\min f(x,y)=f(x_0,y_0)=\min\max f(x,y)$$ then $(x_0,y_0)$ is a saddle point of $f$. The reason you missed it, is that you calculated correctly $v^+$ and $v^-$ but you did not observe that $v^+$ and $v^-$ are simultaneously obtained at point $(0,0)$ and not only on the boundaries as you have it.