$f>0$, $f'\geq 0$ but not continuous.

Question

This question is continuation of this question. (And last of this type)

Let $f: [a,b]\rightarrow \mathbb{R}$, where $f(x)$ is monotonically non-decreasing positive continuous function and differentiable on $(a,b)$ or simply $f'(x)\geq 0$ for $x\in (a,b)$.

can we find such function for which $f'(x)$ is not continuous?

I could not find any,obviously. Is there such function even exists!

The positivity assumption on $f$ is useless. $f$ is bounded on $[a,b]$ by continuity, so if you find a counterexample $f$ which is not positive, you can make it so by adding a constant, namely $\lVert f\rVert_\infty+1$. — Clement C., Aug 17 '17 at 14:20
BTW, Monotonously increasing and $f'\ge 0$, aren't the same thing. $f'\ge 0$ implies non-decreasing function. — Jaideep Khare, Aug 17 '17 at 14:20
@JaideepKhare are you saying I should write Monotonously non-decreasing? — user467365, Aug 17 '17 at 14:23

score 2 · Accepted Answer · answered Aug 17 '17 at 14:22

Take $x\in(-1,1)$ $$ f(x) =100000+ 100x+ x^2\sin\frac{1}{x} \qquad\text{with $f(0)=100000$}$$ which is differentiable, and $$ f'(x) = 100+ 2x\sin\frac{1}{x}-\cos\frac{1}{x} \qquad\text{with $f'(0)=100$}$$ $f'$ is not continuous a $x=0$ and you might choose the interval $[a,b]$ around $x=0$ as it suite to you. Indeed

$$ f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100+ 2x\sin\frac{1}{x}-1$$

Since $|\sin a| \le |a|$ and $-1\le-\cos a\le 1$ then $|2x\sin\frac{1}{x}|\le 2$ i.e $$2x\sin\frac{1}{x}\ge -2 $$

therefore

$$ f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100-2-1= 97>0$$ for every $x\in \mathbb R$. so $f $ is increasing and $f'$ is not continuous at $x=0$.

Also $|x^2\sin\frac{1}{x}|\le 1$ for all $x\in (-1,1)$ so that $$f(x) =100000+ 100x+ x^2\sin\frac{1}{x} > 100000-100 -1 >0$$ for every $x\in (-1,1).$

I don't want to be overly critical, but what is the point of reproducing pretty much exactly the answer to the previous question with all the analysis, while it suffices to explain "take the same function and add a constant, since all that was missing was $f>0$"? — Clement C., Aug 17 '17 at 14:30

José Carlos Santos · Answer 2 · 2017-08-17T14:48:54.057

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Define $f(x)=1+3x+x^2\sin\left(\frac1x\right)$ if $x\in(0,1]$ and $f(0)=1$. It satisfies your conditions, but $f'$ is discountinuous at $0$.

edited Aug 17 '17 at 14:48

answered Aug 17 '17 at 14:21

José Carlos Santos

440,053

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do we have $f'\ge 0$? – Guy Fsone Aug 17 '17 at 14:29
@GuyFabrice Not necessarily. I will edit my answer. – José Carlos Santos Aug 17 '17 at 14:31
I think $3x$ is more secure. – Guy Fsone Aug 17 '17 at 14:45
@GuyFabrice Right you are! Of I course, I could also put an interval shorter than $(0,1]$, but I will follow your suggestion. – José Carlos Santos Aug 17 '17 at 14:48

score 0 · Answer 3 · answered Aug 17 '17 at 14:23

Take the function $f$ from the question you link. It is already a monotonically increasing continuous function and differentiable on $[a,b]$, yet $f'$ is not continuous. It is only missing positivity.

Make it positive by adding $\lVert f\rVert_\infty+1$ (possible since $f$ is continuous on $[a,b]$, so bounded). This does not change the derivative.

You then have an answer to your new question.

$f>0$, $f'\geq 0$ but not continuous.

3 Answers3