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Let $f:(0,1)\to \Bbb R$ be a differentiable function. Which of the following statements is true?

(a)If $f'$ is bounded, then $\lim_{x\rightarrow 0^{+}}f(x)<\infty$

(b)If $f$ is uniformly continuous, then $f'$ is bounded.

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Have you tried something? And I believe the notation $(0,1)$ means the interval is open? –  Patrick Da Silva Jan 7 '13 at 17:31
3  
It's usually not a great idea to post three unrelated questions, and it's even worse where you show no attempt to solve the problem yourself. Please give us some idea of what part of the theory you need help with; preferably, ask about that instead of this question directly. –  Ben Millwood Jan 7 '13 at 17:34
    
(a) is a true statement. –  emka Jan 7 '13 at 22:17
    
(b) is a false statement. –  sos440 Jan 8 '13 at 0:07
    
The converse of (b) is true. –  emka Jan 8 '13 at 0:54

2 Answers 2

up vote 1 down vote accepted

(a) True: Since $f'$ is bounded, if follows the Lipschitz condition, i. e. we can find an M, $|f(x)-f(x_0)| \le M|x-x_0|$. Therefore the function is uniformly continuous and therefore the function is bounded.

(b) False: A counter example is $f(x)=\sqrt x $. As $x \to 0$, $f'(x) \to -\infty $.

If you are having problems with these, you should try reviewing the Litpschitz condition. It should be the first instinct whenever you are dealing with bounded derivatives.

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Just a few helpful (hopefully) remarks:

Fix $x_0$

Since $f'$ is bounded we know $\frac{f(x)-f(x_0)}{x-x_0}\leq M$ for some $M$.

Since $f$ is differentiable: for every $\epsilon > 0$ there is a $\delta > 0$ such that $|f(x)-f(x_0)|< \epsilon$ whenever $|x-x_0|<\delta$.

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