Clearly f(0)=0 is minimum and unique.
We use Taylor's expansion in a neighbourhood of 0, to obtain
with the obvious notation.
We also have . One of the principal curvatures at 0 must be negative, say . Since this is an eigenvalue of
there is a direction on the plane, of length , such that . Along this direction, we will have
a smooth function.
Therefore, there is an such that , a contradiction.