$\nabla^2f(x)>0$ is a necessary or sufficient condition of the Newton direction $s=-[\nabla^2f(x)]^{-1}\nabla f(x)$ to be a descent direction? Let $f: \mathbb{R}^N \rightarrow \mathbb{R}$. The Newton direction $s=-[\nabla^2f(x)]^{-1}\nabla f(x)$ a descent direction if:
$\nabla f(x)^Ts<0 \implies -\nabla f(x)^T[\nabla^2f(x)]^{-1}\nabla f(x)<0 $
$\quad\quad\quad\quad\quad\quad\implies \nabla f(x)^T[\nabla^2f(x)]^{-1}\nabla f(x)>0$
$\quad\quad\quad\quad\quad\quad\implies [\nabla^2f(x)]^{-1} $ must be positive definite.
We know that the inverse of a matrix positive definite is also positive definite.
My question is: $\nabla^2f(x)>0$ is a necessary or sufficient condition? and why?
Thanks in advance for any help!
 A: $\nabla^2 f(x) \succ 0$ is equivalent to "$v^\top [\nabla^2 f(x)] v > 0$ for any nonzero vector $v$."
However, you are only asking that the latter condition hold for a specific vector, namely $v = \nabla f(x)$. So positive-definiteness of the Hessian is sufficient for the step to be a descent direction, but not necessary.
(Recall that you run this step many times in Newton's method for different values of $x$ that you do not know in advance. One usually assumes the Hessian is positive-definite everywhere so that we have this assumption that the step is a descent direction. Otherwise it is quite hard to certify whether or not a step will be a descent, and more complex analyses of $f$ would be required to assess how Newton's method will behave.)
A: Locally, you have:
$$f(x) \approx f(c) + \nabla f(c)^T(x-c) + 0.5(x-c)^T\nabla^2f(c)(x-c)$$
If you move from $c$ to $c-[\nabla^2f(c)]^{-1}\nabla f(c)$, the objective function changes from $f(c)$ to:
$$f\left(c-[\nabla^2f(c)]^{-1}\nabla f(c)\right) \approx f(c) - 0.5 \nabla f(c)^T [\nabla^2f(c)]^{-1}\nabla f(c).$$
The final term is negative, because if $\nabla^2f(c)$ is positive definite, then so is its inverse. So, if the local approximation is valid, there is a reduction in function value.
If the step size is large, the local approximation is not valid, and it is possible that the new iterate has a higher function value. To circumvent that, many algorithms do not take a full Newton step, but a damped step where the new iterate is $c-\alpha[\nabla^2f(c)]^{-1}\nabla f(c)$ for some parameter $0<\alpha\leq 1$. The process of finding $\alpha$ is called line search.
