Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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 $[0,1]$.

share|cite|improve this question
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
up vote 4 down vote accepted

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 $f'$.

share|cite|improve this answer

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.

share|cite|improve this answer
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
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

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 $f'$.

share|cite|improve this answer

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 $f'$.

share|cite|improve this answer
(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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.