Exercise 6.8.19

Here we use the noation in the proof of Theorem 6.37. Also, the equation

is helpful.

1.

Since $0<\mathrm{rank}<!--\; FUNCTION\; APPLICATION\; -->(A)<n$ and $A$ has no negative eigenvalues, we could find a positive eigenvalue ${\lambda }_i$ of $A$. Then take $x=s_i{v}_i$. Then we’ll have that

$$f(x)>(\frac{1}{2}{\lambda }_i-𝜖)s_i^2>0=f(0).$$

We may pick $s_i$ arbitrarily small such that $\parallel x-p\parallel$ could be arbitrarily small. Hence $f$ has no local maximum at $p$.

2.

Since $0<\mathrm{rank}<!--\; FUNCTION\; APPLICATION\; -->(A)<n$ and $A$ has no positive eigenvalues, we could find a negative eigenvalue ${\lambda }_i$ of $A$. Then take $x=s_i{v}_i$. Then we’ll have that

$$f(x)<(\frac{1}{2}{\lambda }_i+𝜖)s_i^2<0=f(0).$$

We may pick $s_i$ arbitrarily small such that $\parallel x-p\parallel$ could be arbitrarily small. Hence $f$ has no local minimum at $p$.