0

I don't know how to prove this theorem (please help):

Let $f$ be a twice differentiable function on an open set $U \subset E$ (normed vector space). Let $x$ be a critical point of $f$, we suppose there exists an open set $B \subset U$ such that $x \in B$ and $v^tHessf(y)v \geq 0$ $\forall y \in B$, $\forall v \in E$. Then $x$ is an optimal minimum of f.

I proved this theorem for the case $z^tHessf(y)v >0$ is positive, but I don't know how to prove it for the case $v^tHessf(y)v \geq 0$...

array.02
  • 25

1 Answers1

0

Given a fixed unit vector $u$, let $g: \mathbb{R} \to \mathbb{R}$ be defined by $g(t) = f(x + tu)$.

Note that $g'(t) = (\nabla f(x + tu))^\top u$ and $g''(t) = u^\top (Hf(x+tu)) u$. Because $x$ is a critical point of $f$, note that $g'(0)=0$.

By the mean value theorem, $$f(x+tu) = g(t) = g(0) + g'(0) t + \frac{1}{2} g''(\xi) t^2 = f(x) + 0 + (u^\top Hf(x+\xi u) u)\frac{t^2}{2}$$ where $\xi$ lies between $t$ and $0$.

In particular, we have $$f(x+tu) - f(x) = (u^\top Hf(x+\xi u) u) \frac{t^2}{2}.$$ If $x+tu \in B$, then $x + \xi u \in B$ too, so $u^\top Hf(x+\xi u) u \ge 0$.

angryavian
  • 89,882