Take the 2-minute tour ×
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.

Assume that $f$ is a differentiable function on $[0,\infty]$ and $f(0) = 0$. Furthermore, assume that for all $x \in (0,\infty)$ there is a $c \in (0,x)$ such that $$ 0 < f'(c) = \frac{f(x) - f(0)}{x} = \frac{f(x)}{x} $$ Must there exist a point $p \in (0,\infty)$ such that for all $x \in (0,p)$ we have $f'(x) > 0$?

This is not a book or homework problem, this is a a question of my own. So far I have tried a contradiction, assume that for all $p \in (0, \infty)$ there is at least one $x \in(0,p)$ such that $f'(x) \leq 0$. Consider such a point $x$, now, by Darboux's Theorem there should be a point $x_0 \in (0,x)$ such that $f'(x_0) = 0$. Let $p = x_0$, there must be another point $x$ such that $f'(x) \leq 0$ , and hence, another point $x_1$ such that $f'(x_1) = 0$ (it is of course trivial that $x_1 < x_0$). As this process keeps repeating, there should be a infinite sequence of points $x_0,x_1,x_2,...$ converging to $0$ (the sequence is bounded below by $0$ by definition so the Monotone Convergence Theorem should guarantee convergence to $0$ right?) such that $f'=0$ at any point in this sequence. And here I'm stuck. Any convergent sequence in $R$ is Cauchy could that help at all? Or is this the entirely wrong approach? This is my first semester of Advanced Calculus, so please try not to laugh too hard if this question is stupid. Thanks!

share|improve this question
    
If a sequence is decreasing and bounded from below by zero, it converges. but it does not have to converge to zero –  Amr Nov 22 '12 at 5:32
    
inf$\{x_n: n \in N\} = 0$ and the sequence is strictly decreasing, so the limit of the sequence should be $0$ correct? –  Eric Nov 22 '12 at 5:37

1 Answer 1

up vote 2 down vote accepted

A counterexample is $f(x)=x^2(2+\sin(1/x))$, when $x\neq 0$, and $f(0)=0$. Below is an image from Mathematica to help you believe that it works, but you can calculate the derivative to check that it works.

enter image description here

share|improve this answer
    
Is there always a point such the the derivative is greater than $0$? –  Eric Nov 22 '12 at 5:40
    
Yes there is, although I just realized I did not quite include your condition that $f(x)>0$ (I only have $f(x)\geq 0$), so I should adjust it. Edit: Done. (The old example was $x^2(1+\sin(1/x))$, and the $1$ was changed to $2$ so that $f$ is always strictly positive when $x\neq 0$.) –  Jonas Meyer Nov 22 '12 at 5:41
    
Awesome :D How ever I am wonder what happens if the function is uniformly continuous on $R$? –  Eric Nov 22 '12 at 5:49
    
@Eric: That causes no problems whatsoever. Only what happens near $0$ is important, and every continuous function on $[0,1]$ is uniformly continuous. Every differentiable function on $[0,1]$ can be extended to a uniformly continuous differentiable function on $[0,\infty)$, for example by extending the tangent line at $(1,f(1))$ to the right as the definition of $f$ on $(1,\infty)$. –  Jonas Meyer Nov 22 '12 at 5:51
    
Thank you so much. The question was bothering me for a long time! –  Eric Nov 22 '12 at 5:53

Your Answer

 
discard

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.