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Suppose $f:[0,1]\to[0,1)$ is a continuous function such that $f(0)=0$ and $f(x)>0$ for $x\neq 0$. Under what natural conditions there exists $\epsilon>0$ such that $f$ is strictly increasing over $[0,\epsilon]$?

For example, if $f$ is differentiable over [0,1] and $\partial_+ f(0)\neq 0$ (the right-derivative is non-zero), then such $\epsilon$ must exist, by here. Could we make the same conclusion if the differentiability assumption is removed (so, instead of differentiability, we have that $\partial_+ f(0)$ is defined and is positive).

afshi7n
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  • You seem to be saying that if the right derivative of $f$ is $0$ at $0$, then $f$ is increasing in some interval $[0,c]$. That is not true. See here: https://math.stackexchange.com/questions/807537/function-which-derivative-at-0-is-1-but-is-not-monotonic-increasing – user49640 Jun 14 '17 at 05:24
  • So, from the linked question, do we need that $f'$ is continuous on some interval $[0,c]$? Or, is that a sufficient condition, anyway? – G Tony Jacobs Jun 14 '17 at 05:29
  • @user49640 the right derivative of $f$ is positive at 0. But even so, I think you're saying that in addition to that, the derivative needs to be positive at a whole interval $(0,\epsilon]$? Wouldn't the condition that $f(x)>0$ for $x>0$ buy us anything here? – afshi7n Jun 14 '17 at 05:41
  • @Tony From looking at the proof, I think continuity of $f'$ is needed only at the limit point. – afshi7n Jun 14 '17 at 05:43
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    @afshi7n I don't think so. Imagine a function that essentially follows the line $y = x$, with small but sudden, violent oscillations, increasing in intensity as $x$ moves towards zero. It will be possible to keep these oscillations above the line $y = 0$. To answer your original question, continuity of $f_+'$ at $0$ is enough. Of course, the condition $f'(x) \geq 0$ is sufficient if it is also known that $f'(x)$ is not zero on any interval. – user49640 Jun 14 '17 at 06:22

2 Answers2

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If $f$ is not strictly increasing on $[0,\epsilon]$, there are $a,b$ with $0 \le a < b \le \epsilon$ and $f(a) \ge f(b)$. The maximum value of $f$ on $[a,b]$ is attained at some $c < b$, and then the upper right-hand Dini derivative $$f'_+(c) = \limsup_{x \to c+} \frac{f(x) - f(c)}{x-c} \le 0$$
Thus a sufficient condition for $f$ to be strictly increasing on $[0,\epsilon)$ is that $f'_+(x) > 0$ for all $x \in [0, \epsilon)$. On the other hand, a necessary condition for $f$ to be increasing on $[0,\epsilon)$ is that $f'_-(x) \ge 0$ for all $x \in [0,\epsilon)$, where $f'_-$ is the lower right-hand Dini derivative $$f'_-(x) = \liminf_{t \to x+} \frac{f(t) - f(x)}{t - x}$$

Robert Israel
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Take $f(x) = \frac{1}{2}x +x^2 \sin{\frac{1}{2x}} $ and $f(0)=0$

From the inequality $\sin{\frac{1}{2x}} \leq \frac{1}{2x} $ we can conclude that $f \ge 0$ one $[0,1]$. Also $f' (0+) = \frac{1}{2}$ and $f' (x) = \frac{1}{2} - \cos{\frac{1}{2x}} + 2x \sin{\frac{1}{2x}} $ for all $x >0$.

Observe that $f'$ oscillates with positive and negative values as $x \to 0^+.$ This shows $f$ can not be increasing on any interval $[0, \epsilon].$

This function satisfies all conditions in question but not the claim!

Red shoes
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  • Thanks, this example is very helpful. It shows that some sort of continuity of $f'$ at 0 is needed. Now, if it's given that $f'$ is continuous over (0,1), and that the right limit $\lim_{x\to 0^+} f'(x)$ exists and is finite, does it also imply that $\lim_{x\to 0^+} f'(x)=\partial_+ f(0)$? This will solve my problem.. thanks! – afshi7n Jun 14 '17 at 17:29
  • I think the answer to the last question is positive. That seems to be just a consequence of L'hospital's rule. – afshi7n Jun 14 '17 at 20:53
  • Yes, If $\lim_{x\to 0^+} f'(x)$ exists , it has to be equal to $\partial_+ f(0)$

    @afshi7n

     – Red shoes Jun 15 '17 at 00:35