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I don't have any idea how to start solving this problem. Any help please?

Problem:

Suppose that a differentiable function $f:\mathbb R \to \mathbb R$ and its derivative $f'$ have no common zeros. Prove that $f$ has only finitely many zeros in $[0,1]$.

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Please start accepting answers to your earlier questions. In Math.SE, this is considered important feedback for answerers. You can accept an answer by clicking the green tick/check mark under it. –  Srivatsan Nov 10 '11 at 23:30
    
One can prove something slightly different but cute: if a (infinitely differentiable) function $\mathbb R\to\mathbb R$ has infinitely many zeroes in a bounded interval, then there is a common zero of $f$ and all its derivatives. –  Mariano Suárez-Alvarez Nov 10 '11 at 23:49
    
I accepted my previous questions. I am new to this forum, so I don't know how things work here. Thanks for letting me know. –  M.Krov Nov 11 '11 at 0:53
    
@Mariano: Isn't that strictly stronger than the OP's statement? –  jprete Nov 11 '11 at 3:10
    
I don't think it is comparable, really: that's why I said it is different. –  Mariano Suárez-Alvarez Nov 11 '11 at 3:12
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Suppose $(x_n)_{n\geq1}$ is a sequence of infinitely many distinct zeroes of $f$ in $[0,1]$. Since $[0,1]$ is compact, by eventually replacing this sequence with one of its subsequences, we can assume that there is a point $y\in[0,1]$ such that $\lim_{n\to\infty}x_n=y$. Since $f$ is continuous, this implies that $0=\lim_{n\to\infty}f(x_n)=f(y)$, so $f$ vanishes at $y$ also.

Now, since $x_n\to y$, we have $$f'(y)=\lim_{z\to y}\frac{f(z)-f(y)}{z-y}=\lim_{n\to\infty}\frac{f(x_n)-f(y)}{x_n-y}=0.$$ It follows that $y$ is a common zero of $f$ and $f'$.

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A different argument.

Suppose $f$ and $f'$ have no common zeroes. If $x\in\mathbb R$ be a zero of $f$, then $f'(x)\neq0$ and $f$ is therefore injective in a small neighborhood $I=(x-\varepsilon,x+\varepsilon)$ of $x$: in particular, the only zero of $f$ in $I$ is $x$ itself. This shows that the set of zeroes of $f$ is discrete in $\mathbb R$ and, therefore, intersects every bounded closed interval in a finite set.

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I don't think it's true that $f'(x)\ne 0$ implies $f$ is injective in a small neighborhood of $x$, although I think it is true that $f(y)$ differs from $f(x)$ for $y$ in a small neighborhood of $x$. For example, if $f(x)=x+x^2\sin(1/x)$ for $x\ne 0$ and $f(0)=0$, then $f'(0)=1$ and $f$ has no other zeros near $0$, but $f$ is not injective in any open neighborhood of $0$, no matter how small. So your argument has an error but it's not fatal. –  Michael Hardy Nov 11 '11 at 0:00
    
@Micheal: this is part of the content of the inverse function theorem, no? –  Mariano Suárez-Alvarez Nov 11 '11 at 0:01
    
I think you might have to assume the derivative is continuous in a neighborhood of that point. –  Michael Hardy Nov 11 '11 at 0:02
    
I always do that. Functions with non continuous derivatives donot interest me much. –  Mariano Suárez-Alvarez Nov 11 '11 at 0:02
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The question said $f$ is differentiable but not that the derivative is continuous. But your argument still survives with only a small modification. –  Michael Hardy Nov 11 '11 at 0:05
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If the set of zeros of $f$ in $[0,1]$ is infinite, then it has a limit point in $[0,1]$. Since $f$ is continuous, the limit point $x_0$ must be a zero of $f$. There is a sequence $\{x_n\}_{n=1}^\infty$ of zeros of $f$ that approaches $x_0$. So $$ f'(x_0) = \lim_{n\to\infty} \frac{f(x_n)-f(x_0)}{x_n-x_0} = \lim_{n\to\infty} \frac{0-0}{x_n-x_0} = 0. $$ So $x_0$ is a common zero of $f$ and $f'$.

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Prove the contrapositive. Suppose $f$ is differentiable and it has infinitely many zeros in $[0,1]$. Now $[0,1]$ is compact, hence there exists a subsequence of zeros $x_{n_k}$ such that $f(x_{n_k}) = 0$. It also converges to a point where $f(x) = 0$ by continuity, and $x \in [0,1]$. This means that $$ f'(x) = \lim_{k \to \infty} \frac{f(x_{n_k}) - f(x)}{x_{n_k} - x} = 0. $$ Thus $x$ is a common zero of $f$ and $f'$.

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(This is not the converse) –  Mariano Suárez-Alvarez Nov 10 '11 at 23:53
    
My english is wrong... I meant the contrapositive. Right? –  Patrick Da Silva Nov 10 '11 at 23:57
    
Just to make sure I understand your proof well: you are saying tha since (xn) belongs to a compact interval, then it is bounded) and then there is a subsequence that converges to a point where f(x)=0. Are you using the Bolzano Weierstarss theorem saying that every bounded sequence has a convergent subsequence? Please let me know if this true? –  M.Krov Nov 11 '11 at 0:50
    
Okay so, to go in the details of which theorems I use, there is a sequence $x_n$ of zeros of $f$. That sequence is in $[0,1]$, thus yes, you can use Bolzano-Weierstrass's theorem, but my way of seeing things is more general, i.e. in any metric space, a subset of the metric space is compact iff any sequence inside the compact admits a convergent subsequence. Since we are in a metric space, this applies, hence there exists $x_{n_k}$ which converges to $x \in [0,1]$. Now $f$ is differentiable, hence continuous, thus $0 = f(x_{n_k}) \to f(x)$, hence $f(x) = 0$. The rest is explained above. –  Patrick Da Silva Nov 11 '11 at 4:19
    
The property that a subset is compact (in the sense with open sets, not sequential compacity) iff any sequence admits a convergent subsequence is not true in general topological spaces, but convergent sequences are usually met in the context where a metric is around, so this is general enough for most human minds. =) –  Patrick Da Silva Nov 11 '11 at 4:21
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