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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?

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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
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
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 $|\frac{f(c+s) - f(c) - sA}{s}| = \lim_{n\to \infty} |\frac{f(c+h_n)-f(c) -h_nA}{h_n}| \le \epsilon$ .

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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

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