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.

If I know a function is continuous on $[0, 1]$, and that $\lim_{h \in \mathbb Q \to 0} {{f(c+h)-f(c)}\over h }$ exists, can I say $f'(c)$ exists?

I feel like this is an obvious yes, since by definition of derivative, $f'(c) = \lim_{h \to 0} {{f(c+h)-f(c)}\over h }$. Hence if that limit exists and is finite, the function at $c$ is differentiable, and so $f(c)$ exists.

Does the fact that $h \in \mathbb Q$ change anything?

And if so, where does $f(x)$ being continuous on $[0,1]$ come in?

share|improve this question
1  
What do you mean by Q{0}? –  Erick Wong Dec 9 '12 at 17:48
    
Sorry, it formatted that weirdly. Should be Q\{0}. –  GradStudent Dec 9 '12 at 17:52
1  
I kind of doubt it. Probably a counterexample of shape $f(x)=\sum a_n \sin(n! x)$. –  André Nicolas Dec 9 '12 at 17:59
    
@GradStudent: Maybe you should indicate that by $Q$ you mean the rationals, or use ${\mathbb Q}$. –  Christian Blatter Dec 9 '12 at 18:52
    
I probably should, but I do not know how to make the Q that represents rational numbers here. Can you show me? –  GradStudent Dec 9 '12 at 19:46

1 Answer 1

up vote 1 down vote accepted

To answer your last question, $f$ being continuous at a point is merely a consequence of it being differentiable. The reverse is not true.

Edit: Oh, I see.

It does not hold that just because $\lim_{h \in \mathbb{Q}: h \to 0} \frac{f(c+h) - f(c)}{h}$ exists that $f$ is differentiable at a point $c$. Consider the indicator function of the rationals defined by $\chi_{\mathbb{Q}}(x) = 1$ if $x \in \mathbb{Q}$ and 0 otherwise. Then we have that $\chi_{\mathbb{Q}}(h) - \chi_{\mathbb{Q}}(0) = 0$ for every rational $h$, but the function is clearly not differentiable at $0$, since it's not even continuous.

We'll show however that if $f$ is continuous and the above limit exists, then $f$ is differentiable at $c$. Let $A = \lim_{h \in \mathbb{Q}: h \to 0} \frac{f(c +h) - f(c)}{h}$. Fix an $\epsilon > 0 $. Find a $\delta > 0$ so that if $h$ is a nonzero rational which satisfies $|h| < \delta$ then $|\frac{f(c+h) - f(c) - hA}{h}| < \epsilon$. Let $s$ be any nonzero real satisfying $|s| < \delta$. Find a sequence $\{h_n\}_{n=1}^{\infty}$ of nonzero rationals so that $h_n \to s$ and $|h_n| < \delta$ for every $n$. By continuity of $f$, we have $|\frac{f(c+s) - f(c) - sA}{s}| = \lim_{n\to \infty} |\frac{f(c+h_n)-f(c) -h_nA}{h_n}| \le \epsilon$ .

share|improve this answer
    
Right, but then why is it mentioned in the first place? –  GradStudent Dec 9 '12 at 17:53
    
Edit: Oh, I see. Editing my original response. –  anonymous Dec 9 '12 at 17:58

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.