if a is local minimum of f show that $\forall h \in \Bbb{R^n} < H_f(a)\cdot h,h> \ge 0 $

Question

Let $ U \in \Bbb{R^n}$ be a domain, and $f: U \to \Bbb{R^m}$ ,$f \in C^2(U)$

show that if $a \in U $ is a local min of $f$ then

$\forall h \in \Bbb{R^n} < H_f(a)\cdot h,h> \ge 0 $

where $H_f(a)$ is the Hessian matrix

I tried to define $g(h) = < H_f(a)\cdot h,Ih>$ (where I is the unit matrix) and find its deritative using the deritative show that $h=0$ is a local minimum of $g(h)$ and that it equals to 0, so for every other $h \in \Bbb{R^n} g(h) > 0$

I'm not sure I'm calculating the deritative correctly: $\dot g(h) = < \dot H_f(a), Ih> + < H_f(a), \dot Ih> = 2< H_f(a), h> $ if It is the deritative then its easy to show that $\dot g(h) = 0 $ iff $h=0$

but is it that correct? and if so, how to show it is a minimum?

Answer 1

Hint:

Use Taylor's formula at order $2$. It can be written as $$f(a+h)=f(a)+\nabla f(a)\cdot h + \langle H_f(a)h,h\rangle+o\bigl(\lVert h\rVert\bigr).$$ At a critical point, $\nabla f(a)=0$, and we obtain $$f(a+h)-f(a)=\langle H_f(a)h,h\rangle+o\bigl(\lVert h\rVert\bigr).$$ If $\lVert h\rVert$ is small enough, the sign of the r. h. s. is the sign of the inner product. Thus if $f(a)$ is a local minimum, this sign is positive or $0$, which means the quadratic form $\;\langle H_f(a)h,h\rangle$ is semi-definite positive.