Let $f$ be a $C^2$ function defined on an open interval $I \subseteq \mathbb{R}$ near $c \in I$. If $c$ is a local minumum, is it true that $f''(c)\geq0$? Why?
I'm learning semi-definiteness of a multi-variable function via approaching a point with different direction in $\mathbb{R}^n$. The above statement is used. How can I approach the question above? I've tried proving $f$ is convex near $c$, it does not work out. Or am I losing some condition to prove it?
Continuous differentiability is not needed here, we only need twice differentiability at the point in question. Suppose $f$ has a local minimum at $c$ and for the sake of contradiction, suppose $f''(c)<0$. Recall that by definition, this means \begin{align} \lim_{h\to 0}\frac{f'(c+h)-f'(c)}{h}=\lim_{h\to 0}\frac{f'(c+h)}{h}<0 \end{align} So, there is a $\delta>0$ such that for all $0<|h|<\delta$, we have $\frac{f'(c+h)}{h}<0$. So, $0<h<\delta$ implies $f'(c+h)<0$ while $-\delta<h<0$ implies $f'(c+h)>0$. In words, $f'$ is positive when we're slightly to the left of $c$ (hence by mean-value theorem, $f$ is strictly increasing here) and $f'$ is negative slightly to the right of $c$ (hence $f$ is strictly decreasing here). Thus, we must have that $c$ is a strict local maximum point.
But, we assumed that $c$ is a local minimum, from which it follows $f$ has to be constant in some neighborhood of $c$, but that would mean $f''(c)=0$, contradicting our assumption $f''(c)<0$.
In higher dimensions, an analogous theorem is true; the second derivative is a bilinear symmetric mapping $D^2f_c:\Bbb{R}^n\times\Bbb{R}^n\to\Bbb{R}$, so it is the positive/negative definiteness of this which determines the nature of the critical point.
