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Suppose that $f:(0,1)\to\mathbb{R}$ is differentiable, and that $f(x_1)=f(x_2)=0$ and $f’(x_1)>0$ and $f’(x_2)>0$ for some $0 <x_1<x_2<1$. Then there must exist an $x_0\in(x_1,x_2)$ such that $f(x_0)=0$ and $f’(x_0)\leq0$, as follows: Let $A=\{x\in(x_1,x_2):f(x)\geq0\}$, and note that $A$ is nonempty, since the condition $f’(x_1)>0$ guarantees that there exist points that exceed $x_1$ by arbitrarily small amounts, at which $f$ is strictly positive. Also note that $A$ is bounded above, by $x_2$. Therefore, let $x_0=\sup A$. Note that $x_0 > x_1$. Since $f’(x_2)>0$, there exist points that $x_2$ exceeds by arbitrarily small positive amounts, at which $f$ is strictly negative. It follows that $x_1 < x_0 < x_2$. By continuity of $f$, we have that $f(x_0)=0$. Since $f(x)<0$ for all $x\in(x_0,x_2)$, it follows also that $f’(x_0)\leq0$, as required.

Is this proof correct, and is there a simpler proof, perhaps using a ready-made theorem such as the Intermediate Value Theorem ?

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    $\begingroup$ Quick thought: by the direction of the derivative there must exist points $f(x_3)>0$ and $f(x_4)<0$. Then apply intermediate value theorem? The derivative is such for at least one point as the function crosses over. $\endgroup$
    – Dole
    Commented Dec 27, 2018 at 21:15
  • $\begingroup$ @RRL, Please check my argument again - in fact I do not appeal to the IVT. Also the reason why $f(x)<0$ for all $x\in(x_0,x_2)$ is that $x_0$ is an upper bound for the set where $f(x)\geq0$. $\endgroup$
    – Simon
    Commented Dec 28, 2018 at 1:31
  • $\begingroup$ @Dole, thank you. This was my first thought, too. Trying to prove the last sentence in your comment is what lead me to the proof I gave in my original question. $\endgroup$
    – Simon
    Commented Dec 28, 2018 at 1:39
  • $\begingroup$ @RRL, in fact I am not arguing that there are points in a right neighbourhood of $x_0$ where $f$ is positive. I am arguing that in the interval $(x_0,x_2)$, $f$ must be negative. I am not using the IVT. I believe that I have shown rigorously, that $f(\sup A)=0$ (I am calling $\sup A$ $x_0$), and that $f'(x_0)\leq0$. $\endgroup$
    – Simon
    Commented Dec 28, 2018 at 2:00
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    $\begingroup$ OK I finished my proof. It gets to the same ultimate argument -- finding a smallest or largest zero $x_0$ bounded away from either $x_1$ or $x_2$ so that, consequently, $f'(x_0) \leqslant 0$. $\endgroup$
    – RRL
    Commented Dec 28, 2018 at 2:32

2 Answers 2

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By your argument there are points $x_1 < x' < x'' < x_2$ where $f(x') > 0$ and $f(x'') < 0$. By the IVT there is at least one point $y_1$ (and possibly more) where $x' < y_1 < x''$ and $f(y_1) = 0$.

If $f'(y_1) \leqslant 0$, then we are done. On the other hand, if $f'(y_1) > 0$, then we have the same problem with $y_1$ replacing $x_2$ and there exists a point $y_2$ between $x_1$ and $y_1$ such that $f(y_2) = 0$.

Continuing in this way we either find a zero where the derivative is less than or equal to $0$ or generate a sequence $y_n \in (x',x'')$ such that $f(y_n) = 0$ and $f'(y_n) > 0$.

However, it can be shown that if there are no zeros of a function that is differentiable on a closed interval where the derivative is also $0$, then the set of zeros is finite. Since $f$ is differentiable on the closed interval $[x_1,x_2]$ there exists only a finite set of zeros $\{y_1,y_2, \ldots, y_n\}$ between $x_1$ and $x_2$.

Armed with this, you can now show that $f'(y_n) \leqslant 0$ since $y_n$ must be the smallest number between $x'$ and $x''$ with $f(y_n) = 0$. If $f'(y_n) > 0$ then there would be another zero between $x'$ and $y_n$, a contradiction.

Addendum

Suppose $f$ is differentiable on $[a,b]$ and at no point $x \in [a,b]$ do we have $f(x) = f'(x) = 0$. Then the set of points in $[a,b]$ where $f(x) = 0$ is finite.

To prove this, assume otherwise. Then there is an infinite sequence of zeros and by compactness and continuity a subsequence $(x_n)$ converging to some point $c \in [a,b]$ such that $f(x_n) = f(c) = 0$. Since $f$ is differentiable

$$f'(c) = \lim_{n \to \infty} \frac{f(x_n) - f(c)}{x_n - c} = 0,$$

a contradiction.

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  • $\begingroup$ Thank you for your answer, RRL. Can you elaborate on how to prove the statement preceded by "it can be shown that" ? $\endgroup$
    – Simon
    Commented Dec 28, 2018 at 1:37
  • $\begingroup$ After the second paragraph why don't you just say: the sequence $y_1\gt y_2\gt y_3\gt\cdots$ is decreasing and bounded, so it converges to a limit $y=\lim_{n\to\infty}y_n$. Then $x_1\le y\lt x_2$, and $f(y)=0$ by continuity since $f(y_n)=0$, and $f'(y)=0$ since $$f'(y)=\lim_{n\to\infty}\frac{f(y_n)-f(y)}{y_n-y}=\lim_{n\to\infty}0=0,$$ so $x_1\lt y$ since $f'(x_1)\gt0=f'(y)$? $\endgroup$
    – bof
    Commented Dec 28, 2018 at 3:04
  • $\begingroup$ @bof: That is a great suggestion. On the other hand what I like about this site is the opportunity to pursue less than conventional lines of thought unlike the many more constrained aspects in my life. :-) $\endgroup$
    – RRL
    Commented Dec 28, 2018 at 3:45
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Your proof is correct (+1) and the key idea is that if $f$ is continuous on $[a, b] $ then the set $A=\{x\mid x\in[a, b], f(x) =k\} $ is closed (inverse image of a closed set under continuous map is closed, similar result holds for open sets also).

Here you choose $a$ near and to the right of $x_1$ so that $f(x) >0\,\forall x\in(x_1,a]$ and $b$ near and to the left of $x_2$ such that $f(x) <0\,\forall x\in[b, x_2)$. By IVT the set $$A=\{x\mid x\in[a, b], f(x) =0\} $$ is non empty and as noted above is closed. Since $A$ is obviously bounded and closed it has a minimum and a maximum. Both $\min A$ and $\max A$ (which can be same) work as the desired point $x_0$.

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  • $\begingroup$ Thank you Paramanand. $\endgroup$
    – Simon
    Commented Dec 28, 2018 at 10:33

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