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Let $f : \mathbb{R}\to \mathbb{R}$ be a differentiable function such that $f$ and its derivative have no common zero in the closed interval $[0, 1]$. Show that f cannot have infinitely many zeroes in $[0, 1]$.

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marked as duplicate by tomasz, Dominic Michaelis, Stefan Hansen, Micah, Davide Giraudo Mar 17 '13 at 16:12

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Beware, there is a trap in this question. $f$ differentiable over $[0,1]$ doesn't imply $f'$ is continuous over $[0,1]$. –  achille hui Mar 17 '13 at 12:39

3 Answers 3

Since $[0,1]$ is compact, if $f$ had infinitely many zeroes there, they’d have to lie non-discretely, i.e. the zeroes of $f$ have a limit point in $[0,1]$. Because $f$ is continuous, that limit point would be a zero of $f$, too. Taking the derivative of $f$ at that limit point by choosing the zeros as approximating sequence for the difference quotients, one sees that the derivative of $f$ at that limit point would be zero as well. This is a contradiction. Therefore, $f$ cannot have infinitely many zeros in $[0,1]$.

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I need a little more clarification for "Taking the derivative of $f$ at that limit point by choosing the zeros as approximating sequence for the difference quotients." –  Sugata Adhya Mar 17 '13 at 13:24
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Let $x$ be a limit point of the zeroes of $f$ and let $(x_n)_{n ∈ ℕ}$ be a sequence of zeroes of $f$ different from $x$, but converging to $x$. Now $f(x) = \lim_{n→∞} f(x_n) = 0$, so $f'(x) = \lim_{n→∞} \tfrac{f(x_n) - f(x)}{x_n - x} = 0$. –  k.stm Mar 17 '13 at 13:29
    
Thanks so much. –  Sugata Adhya Mar 17 '13 at 13:32

Hints:

Suppose $\,\{x_n\}_{n\in\Bbb N}\,$ is an infinite sequence of zeros of $\,f\,$

1) Show there exists a subsequence $\,\{x_{n_k}\}\subset\{x_n\}\,$ s.t. that

$$\lim_{k\to\infty}x_{n_k}=x_0\;,\;\;x_0\in [0,1]$$

2) Show $\,f(x_0)=0\,$

3) Show that between any two different zeros of $\,f\,$ there exists a zero of $\,f'\,$ (M.V.T.)

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Yup, my answer above clearly goes for using $\,f'\,$ continuous. I must have confused this with other question where it was given $\,f\in C^1(0,1)\,$...oh, well. –  DonAntonio Mar 17 '13 at 12:45

I assumed at first that the derivative is continuous which is clearly wrong, sry for that.

Maybe you enjoy the example that it is not enough to have a function $f:[0,1]\to \mathbb{R}$ continuous in $[0,1]$ and continuous differentiable in $(0,1)$ (in fact it is $C^{\infty}$ in $(0,1)$

$$f(x)=\begin{cases} x\cdot \sin\left(\frac{1}{x}\right) & x \neq 0\\ 0& x=0 \end{cases}$$

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You mean an accumulation point of the zeroes of the derivative, right? Also, to conclude, don’t you assume continuity of the derivative? –  k.stm Mar 17 '13 at 12:25
    
mh i thought at first to get it over the darboux property but i don't know if it works i will change it –  Dominic Michaelis Mar 17 '13 at 12:36

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