Are the second derivatives of a function of several variables strictly non-negative at a local minimum?

Question

Let $f({\bf x})\equiv f(x_1,x_2,\ldots,x_n)$ is a function of $n$ real variables ${\bf x}=(x_1,x_2,\ldots,x_n)$. The Taylor expansion of $f({\bf x})$ is about the point ${\bf x}^0=(x_1^0,x_2^0,\ldots,x_n^0)$, reads, $$f({\bf x})=f({\bf x}^0)+\sum_i\left(\frac{\partial f}{\partial x_i}\right)_{{\bf x}={\bf x}^0}(x_i-x_i^0)+\frac{1}{2!}\sum_i\sum_j\left(\frac{\partial^2 f}{\partial x_i\partial x_j}\right)_{{\bf x}={\bf x}^0}(x_i-x_i^0)(x_j-x_j^0)+\ldots$$ If ${\bf x}^0$ is a local minimum, $$\partial f/\partial x_i=0, \forall i .$$ This is true.

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However, is it also true that at ${\bf x}^0$, $$H_{ij}\equiv\left(\frac{\partial^2 f}{\partial x_i\partial x_j}\right)_{{\bf x}={\bf x}^0}\geq 0$$ as claimed here in this Quantum Field Theory Textbook by Lewis Ryder (Eq. 8.24)? See here. If so, can you offer proof of this?

Answer 1

You can see the positivity from taking a limit. Note that for a function $g$ with second derivative at $0$ we have

$$g''(0) = \lim_{t\to 0} \frac{g(t) + g(-t) - 2 g(0)}{t^2}$$

and so, if moreover $0$ is a local minimum, we have $g''(0) \ge 0$.

Now take the case of your problem. Let $h\in \mathbb{R}^n$. The functions

$$t \mapsto g(t) \colon = f(x_0 + t h)$$

has a minimum at $t=0$, so $g''(0)\ge 0$. But we have

$$g''(0) = \sum_{i,j} \frac{\partial f}{\partial x_i \partial x_j}_{x=x_0} h_i h_j $$

Answer 2

Taylor theorem assert that $$ f(x + h) = f(x) + f'(x) \cdot h + f''(x) \cdot (h,h) + o(\|h\|^2), $$ this being true for $f$ twice differentiable in a neighbourhood of $x,$ everything in this equation taking place in normed spaces. Then, for any vector $v,$ let $h = \lambda v.$ Suppose $f(x) = \max f(B_x)$ and that $f(y) < f(x)$ for all other $y \in B_x,$ where $B_x$ is some ball centred at $x.$ Then, $f'(x) = 0$ as it is well-known, and $$ f(x + h) - f(x) = \lambda^2 f''(x) \cdot (v,v) + o(\lambda^2), $$ the right hand side is $< 0$ and of the form $c \lambda^2 + \lambda^2 o(1).$ If $\lambda$ is small enough, the sign of this will be that sign of $c = f''(x) \cdot (v,v),$ assuming $c \neq 0,$ therefore $f''(x) \cdot (v,v) < 0$ assuming $c \neq 0.$ Thus, $f''(x)$ is negative semidefinite (i.e., a quadratic opening down).