# Is there a simpler proof of this fact in analysis?

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 $$0 . Then there must exist an $$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 ?

• 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. – Dole Dec 27 '18 at 21:15
• @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 $f(x)\geq0$. – Simon Dec 28 '18 at 1:31
• @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. – Simon Dec 28 '18 at 1:39
• @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 $f'(x_0)\leq0$. – Simon Dec 28 '18 at 2:00
• 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, $f'(x_0) \leqslant 0$. – RRL Dec 28 '18 at 2:32

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.

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

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

• 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 $f'(x_1)\gt0=f'(y)$? – bof Dec 28 '18 at 3:04
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 $$x_0$$.