$f:\mathbb{R}\rightarrow\mathbb{R}$ continuous, $a\in\mathbb{R}$. Suppose that there exists $L\in\mathbb{R}$ such that for every $\varepsilon>0$ there exists $r(\varepsilon)>0$ such that $|\frac{f(x)-f(a)}{x-a}-L|<\varepsilon$ for every $x\in\mathbb{Q}$ and $|x-a|
If the condition of differentiability holds for the rationals then the function is differentiable?
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1@Srivatsan: It looks like $L$ is independent of $x$ but not of $a$. By the way,$a$function whose derivative is constant at all rational points, doesn't necessarily have constant derivative _everywhere_. [Minkowski's question mark function](http://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function) is a counterexample. – 2011-10-26
2 Answers
Fix a point $x\neq a$ such that $0 < |x - a| < r(\epsilon/2)$. Then, at the point x, the function $g(y) = \frac{f(y) - f(a)}{y - a}$ is continuous. Now, pick a point $p \in Q$ near $x$ such that $0< |p-a| < r(\epsilon/2)$ and $|g(p) - g(x)| < \epsilon/2$. This is possible by continuity. Now, use your condition to conclude via the triangle inequality, that $ \left|\frac{f(x) - f(a)}{x - a} - L\right| = |g(x) - L| \le |g(x) - g(p)| + |g(p) - L| < \epsilon/2 + \epsilon/2 = \epsilon.$
It follows by definition that $f$ is differentiable at $a$ with f'(a) = L.
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0The function $g$ is defined and continuous at all points $y\neq a$, in particular, at the point$x$I picked. – 2011-10-26
Since the variable name $x$ is already "taken" as ranging over the rationals, we use $z$ to range over the reals. Note that $\frac{f(z)-f(a)}{z-a}-L= \frac{(f(x)-f(a))-(f(x)-f(z))}{z-a}-L.$ The right-hand side above is equal to $\left(\frac{f(x)-f(a)}{x-a}\frac{x-a}{z-a}-L\right)-\frac{f(x)-f(z)}{z- a}.$
Let $z_1, z_2, z_3\dots$ be any sequence of reals with limit $a$, where the distance of $z_n$ from $a$ decreases monotonically. If we can find a sequence $(x_n)$ of rationals with limit $a$ such that $\lim_{n\to\infty}\frac{x_n-a}{z_n-a}=1 \qquad \text{and}\qquad \lim_{n\to\infty}\frac{f(x_n)-f(z_n)}{z_n-a}=0,$ the result will follow.
So $x_n$ has to be chosen far nearer to $z_n$ than $z_n$ is to $a$. We look at the second limit, because it is a little harder to achieve. By the continuity of $f$, and the fact that the rationals are dense in the reals, we can find for each $z_n$ a rational $x_n$ such that $|f(x_n)-f(z_n)|<(1/n)|z_n-a|$.
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0@Didier Piau: Thank you. I had fixed it while you were typing. – 2011-10-26