Clearly f(0)=0 is minimum and unique.

We use Taylor's expansion in a neighbourhood of 0, to obtain

or

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.