I have seen that many online sources (including other math stack exchange questions) say that the following are equivalent definitions of a point of inflection:
If $f$ is differentiable on $I$ we say that $f$ has a point of inflection at $a$ if $f'$ has an isolated local extremum at $a$.
We say that $f$ has a point of inflection at $a$ if $f$ changes concavity at $a$. This definition is somewhat ambiguous to me, so I decided that it would mean that $\exists \delta>0$ such that $f$ is strictly convex on $[a-\delta,a]$ and strictly concave $[a,a+\delta]$ or $f$ is strictly concave on $[a-\delta,a]$ and strictly convex $[a,a+\delta]$.
I want to prove that these definitions are equivalent (I know that this might require clarifying definition 2. in a different way than I have).
I was able to prove that 2. implies 1. This is my proof:
- Suppose that $\exists \delta>0$ such that $f$ is strictly convex on $[a-\delta,a]$ and strictly concave $[a,a+\delta]$. Then $f'$ is strictly increasing on $[a-\delta,a]$ and $f'$ is strictly decreasing on $[a,a+\delta]$, so $f'$ has an isolated local maximum at $a$.
The case when $f$ is strictly concave on $[a-\delta,a]$ and strictly convex $[a,a+\delta]$ is similar.
Now I want to show that 1. implies 2.
- Suppose that $f'$ has a isolated relative extremum. Without loss of generality, suppose that $f'$ has a isolated relative minimum. I need to show that $f'$ is strictly increasing on $[a-\delta,a]$ and $f'$ is strictly decreasing on $[a,a+\delta]$ (which is equivalent to showing that $f$ is convex on $[a-\delta,a]$ and concave on $[a,a+\delta]$).
I know that this is not true for a general function $g$ that if $g$ has an isolated extremum at $g$, $g$ is strictly monotone on either side of $a$. (for example, if the g has a jump discontinuity, which I know is not possible for $f'$). Why is true for $f'$?
$1$ does not imply $2$ !
Take $f(x) = \int_{0}^{x} g(t) dt$, where $g(t)= 2| t | +t \sin \frac{1}{t} $ , with $g(0) = 0 $. Note that $g $ is continuous, so $f $ is differentiable and $ f' = g $.
it is easy to check that $$ |x | \leq g(x) \leq 3 |x| $$ for all $x \in \Bbb{R} $.
This implies $ g $ gets an isolated minimum at $ x = 0 $ but clearly in not monotone function around $ 0 $.