Prove that this function is strictly increasing.

Question

I read about this function which is strictly increasing on [0,1] but with infinitely many critical points in [0,1].

$$f(x)=\begin{cases} x\left(2-\cos\log x - \sin\log x\right) &, 0<x\leq1,\\ 0&,x=0. \end{cases}$$

It's easy to show that this function has infinitely many critical points, but I am finding it difficult to show that it's strictly increasing using simple inequalities.

I think I'm close.. but here is my progress so far:

I know I need to show that if $a<b$, then $f(a)<f(b)$.

I have shown that if $x>0$, then $f(x)>0$. (To do this, I write, for $x\neq0$, $f(x)=x\left(2-\sqrt2\sin(\log x+\pi/4)\right)$).

It may help to note that $\int2 \cos \log t dt=x(\sin\log x+\cos\log x)$, but I wonder if this can be proved without using integrals.

Any suggestions would be appreciated. Thank you.

@OlivierRoche That's sort of the point -- that derivative sure is non-negative, but it cancels infinitely many times. — Clement C., Apr 27 '20 at 05:59
The derivative is zero at a countable number of points. Since the derivative is positive almost everywhere, its integral over any interval must be positive. — Angela Pretorius, Apr 27 '20 at 06:05
@AngelaPretorius I know, but that requires an argument, not just a "compute the derivative" (which clearly the OP must have done to prove there were infinitely many critical points). — Clement C., Apr 27 '20 at 06:07
Even more strange behavior can occur. See https://math.stackexchange.com/q/1845927/72031 — Paramanand Singh, Apr 27 '20 at 10:07

Hanul Jeon · Answer 1 · 2020-04-28T07:04:19.670

Observe that $f'(x)=2-2\cos\log x$ and $f'(x)>0$ for all but countably many points of $x\in(0,1]$. Moreover, every critical point of $f$ is isolated, in the sense that for any critical point $p$ we can find $a<b$ such that $a<p<b$ and no other critical point $q$ satisfy $a<q<b$.

In fact, we can do a bit more: $f'(x)=0$ iff $x=e^{-2n\pi}$ for some $n=0,1,2,\cdots$. For each $n$, choose $a_n=(e^{-2n\pi}+e^{-2(n+1)\pi})/2$ and $b_n=(e^{-2n\pi}+e^{-2(n-1)\pi})/2$. Then $a_n<e^{-2n\pi} < b_n$ and no other critical point other than $e^{-2n\pi}$ lies on $(a_n,b_n)$.

Let $x<y$. If there is no critical point of $f$ between $x$ and $y$, then $f(y)-f(x)=\int_x^y f'(t)dt>0$. If not, we may assume that there is only one critical point: if not, choose $x=t_0<t_1<\cdots<t_m=y$ such that each $(t_i,t_{i+1})$ isolates each of the critical point, and show that $f(t_i)<f(t_{i+1})$ for each $i$.

If $p=e^{-2n\pi}$ be the unique critical point between $x$ and $y$. Choose a small $\varepsilon>0$ and $\delta>0$ such that

$\delta<(y-x)/2$ and
if $x<z<y$ and $|p-z|>\delta$ then $f'(x )>\varepsilon$.

(Detail: Choose any $\varepsilon>0$. Since our $f'$ is continuous, we can find $\delta>0$ which satisfies the two conditions.) Then we have $f(y)-f(x)\ge (y-x-2\delta)\varepsilon>0$.

In fact, we can show the following general statement: if $f$ is absolutely continuous (this holds when $f$ has a continuous derivate), $f'(x)\ge0$ and the set of critical points has measure 0, then $f$ is strictly increasing.

Prove that this function is strictly increasing.

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