If $f(x)$ is convex function , is Newton step be a descent direction?
In the optimization, the Newton step is $-\nabla^2f(x)^{-1}\nabla f(x)$ and $d$ is a descent direction if $ d.\nabla f(x) < 0 $
I know when $f$ is a convex function, we have always $\nabla^2f(x) >0$ but how I can show when $d=-\nabla^2f(x)^{-1}\nabla f(x)$ then $ d.\nabla f(x) < 0 $ ?
I read this statement :
Now, since $\nabla^2f(x)$ is positive definite, $\nabla^2f(x)^{-1}$ is also positive definite and therefore $$ \forall \nabla f(x) \; \; \; \nabla^2f(x)^{-1}\nabla f(x) \cdot \nabla f(x)>0.$$
I can't understand it. Why for all $y$ the inequality hold?
since $\nabla^2f(x)$ is positive definite, $\nabla^2f(x)^{-1}$ is also positive definite and therefore $$ \forall y, \; \; \; \nabla^2f(x)^{-1}y\cdot y>0.$$ this is true for case of $\nabla^2f(x)^{-1} >0$ How can show for case of $\nabla^2f(x)^{-1} =0$
f must be strictly convex ,then $\nabla^2f(x)$ is positive definite, $ \nabla^2 f > 0$ . Now, since $\nabla^2f(x)$ is positive definite, $\nabla^2f(x)^{-1}$ is also positive definite and therefore $$ \forall \nabla f(x) \; \; \; \nabla^2f(x)^{-1}\nabla f(x) \cdot \nabla f(x)>0.$$
When f be a convex and not strictly , maybe determinant equal zero and $ \nabla^ 2 f^ {-1}$ is not exist