$f'(x)\geq 0$ only, but still strictly increasing

Question

It is a quick corollary of the Mean Value Theorem that for a real differentiable function $f:(a,b)\rightarrow\mathbb R$ that

• if $f'(x)\geq 0$ for all $x\in (a,b)$ then $f$ is (weakly) increasing
• if $f'(x)>0$ for all $x\in(a,b)$ then $f$ is strictly increasing

However, this result can be significantly strengthened. For example

• if $f'(x)\geq 0$ for all $x\in (a,b)$ and $Z(f')=\{x\in(a,b):f'(x)=0\}$ is a discrete subset of $(a,b)$, then $f$ is strictly increasing
• if $f'(x)\geq 0$ for all $x\in (a,b)$ and if the derived set of $Z(f')$ is discrete, then $f$ is strictly increasing

This pattern continues. Given $A\subseteq\mathbb R$, let $A^{(0)}=A$ and $A^{(n+1)}=(A^{(n)})'$ where $E'$ denotes the derived set of $E$. We then have this result:

• if $f'(x)\geq 0$ for all $x\in(a,b)$ and $(Z(f'))^{(n)}$ is discrete for some $n$, then $f$ is strictly increasing (note that $Z(f')$ must be countable in this case)

The next-simplest-yet-more-complicated example that I couldn't wrangle is when $Z(f')$ has an order type like $\omega^\omega+1$. Thus the question is

Let $f'(x)\geq 0$ for all $x\in (a,b)$ such that $Z(f')$ has order type $\omega^\omega+1$. Is $f$ necessarily strictly increasing on $(a,b)$? You may assume that $Z(f')$ is embedded as a closed subset of $(a,b)$ if necessary.

The proof of the three claims before-hand rely heavily on two facts:

• Strictly increasing is a local property. That is, if $f:(a,b)\rightarrow\mathbb R$ such that for all $x\in (a,b)$ there is an $\varepsilon>0$ such that $f$ is strictly increasing on $(x-\varepsilon,x+\varepsilon)$ then $f$ is strictly increasing on $(a,b)$
• if $f$ is strictly increasing on $(a,b)$ and extends continuously to $[a,b)$, then $f$ is strictly increasing on $[a,b)$. The dual statement is also true.

When $Z(f')$ had order type $\omega+1$, the real question is what happens around the 'top' of $Z(f')$. When $Z(f')$ had order type $\omega^2+1$, the question is what happens around $Z(f')^{(1)}$. And so forth. This argument can be extended to deal with $A$ such that $A^{(n)}$ is discrete for some $n$. This wraps up a lot of sets which are order-isomorphic to countable ordinals. But with a set with order type $\omega^\omega+1$, we have that every element, except one (the 'bottom'), is an accumulation point. And I can not apply the induction I was using to prove the third claim.

Answer 1

Suppose $f'(x)\geq 0$ for all $x\in (a,b)$. Then $f$ is strictly increasing on $(a,b)$ iff $\{x\in(a,b):f'(x)=0\}$ has empty interior. Here's an easy proof. Suppose $f$ is not strictly increasing, so there exists $c<d$ in $(a,b)$ such that $f(c)\not<f(d)$. Since $f$ is weakly increasing on $(a,b)$, this implies $f(c)=f(d)$, and in fact $f(c)=f(x)=f(d)$ for all $x\in (c,d)$ as well. It follows that $f'(x)=0$ for all $x\in(c,d)$, so $\{x\in(a,b):f'(x)=0\}$ has nonempty interior.

Conversely, if $\{x\in(a,b):f'(x)=0\}$ has nonempty interior, there is an open interval $(c,d)$ on which $f'(x)=0$, and then $f$ is constant (and thus not strictly increasing) on that interval.

Answer 2

I'm not quite sure what $\omega^\omega + 1$ is, but we can find large subsets where $f'=0$ and still have $f$ strictly increasing. Here's an example: Let $K\subset [0,1]$ be the usual Cantor set, and let $d(x,K)$ denote the distance from $x$ to $K.$ (Recall that $d(x,K)$ is a continuous function.) Then

$$f(x) = \int_0^x d(t,K)\,dt$$

is continuously differentiable function that is strictly increasing on $[0,1]$ (because $K$ has no interior), and $f'(x) = 0$ precisely on the set $K$ by the FTC.