A Lipschitz function must has a set of intervals s.t. $\cup_kI_k=\mathbb R$ and $f$ is either convex or concave on either $I_k$?

Question

Let $f:\mathbb R\to\mathbb R$ be a continuous function.

$f$ is said to be "regular" if there exists a set of intervals $\{I_k\}_{k\in K}$ (indexed with $k$ where $K$ is an arbitrary set), such that $\cup_kI_k=\mathbb R$ and $f$ is either convex or concave on $I_k$ $\forall k$.

Claim: $f$ is regular iff $f$ is absolute continuous.

If the claim is wrong, then what is the sufficient and necessary condition for the property “regular"?

It seems that Lipschitz is necessary and second differentiable is sufficient.

What theorems might be useful for me to explore the answer?

Answer 1

No, a regular function need not be Lipschitz. Consider $f(t)=t^2$. Of course that's "locally Lipschitz", meaning Lipschitz when restricted to any compact interval. If $f(t)=|t|^{1/2}$ then $f$ is regular but not even locally Lipschitz.

Presumably in the definition you mean that each interval contains more than one point, because if $\{a\}=[a,a]$ counts as an "interval" then every function is regular. Assuming so, then no, $f''$ continuous does not imply that $f$ is regular: Let $$g(t)=\begin{cases}t\sin(1/t),&(t\ne0), \\0,&(t=0).\end{cases}$$ Then $g$ is continuous, there exists $f$ with $f''=g$, but $f$ is not concave or convex on any interval containing the origin, because $f''$ changes sign on any such interval.

I doubt that the sort of characterization you want exists.

Answer 2

An absolutely continuous function is differentiable a.e. but only once. A convex function in $I_k$ has an increasing derivative at all but countably many points and so it is differentiable twice a.e. So a "regular" function must be differentiable at all but countably many points in the interior of the $I_k$ and the second derivative has to exist a.e. in the interior of the $I_k$. So the equivalence is false. I don't know what family of functions would be equivalent to regular. Definitely polynomials and very smooth functions, but it is not an iff.