The following is part of an assignment so please no full answers.
I am trying to prove the following theorem. In the Dogleg method we consider
$$ p^u = - \frac{g^Tg}{g^TBg}g $$
and
$$ p(\tau) = \tau p^u $$
for $0\leq \tau \leq 1$.
Now, let $B$ be positive definite:
(1) $||p(\tau)||$ is a increasing function of $\tau$
(2) $m(p(\tau))$ is a decreasing function of $\tau$, where $m$ is simply the quadratic model.
The following is what I came up with.
(1) $||p(\tau)||$
Consider the function $h(\tau) = \frac{1}{2}||\tau p^u||^2$:
$$ h(\tau) = \frac{1}{2} \tau ^2 \left\Vert \frac{g^Tg}{g^TBg}g \right\Vert ^2 .$$
If we take the derivative we obtain:
$$ h'(\tau) = \tau \left\Vert \frac{||g||^2}{||g||_B^2}g \right\Vert ^2 = C||g||^2.$$
With $C$ being a positive constant. Therefore the derivative is positive. But, and this is the same problem I have with the second part, this quantity can be $0$, since $\tau$ can be $0$.
(2) $m(p(\tau))$
And this case I study the following expression:
$$ m(\tau p^u) = \frac{1}{2} (\tau p^u)^TB\tau p^u + \tau g^Tp^u .$$
With some computations, I obtain:
$$ m(\tau p^u) = \frac{1}{2}\tau ^2 \frac{||g||^4}{||g||_B^2} - \tau \frac{||g||^4}{||g||_B^2} .$$
Again, we take the derivative and obtain:
$$ m'(\tau p^u) = \frac{||g||^4}{||g||_B^2} \left( \tau -1 \right) .$$
In this case, if $\tau = 0$ we obtain a negative quantity, but when $\tau=1$ we obtain $0$. Am I missing something? Decreasing function or increasing function means also derivative equal to $0$?
