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Construct an example of a differentiable function such that $$ \forall r \in {\Bbb Q}\quad f(r) \in {\Bbb Q}\text{ but } f'(r) \notin {\Bbb Q} $$ this example is not trivial, in a paper they prove the existence, but they don't give an explicit example, but they said that other paper as an explicit example of that, but i could not find it.

EDITED: This is the paper:

Walter Rudin, Restrictions on the Values of Derivatives, The American Mathematical Monthly, Vol. 84, No. 9 (1977), pp. 722-723, MR480908.

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The link works for you...Its better to give the journal ref. and then links. –  Tapu Oct 13 '11 at 19:10
    
On MathSciNet there is the possibility to create a link to the review: right below the bibliographic info there are five red links PDF | Clipboard | Journal | Article | Make Link Click on Make Link, and you'll get a nice and clickable link, looking like this: http://www.ams.org/mathscinet-getitem?mr=480908. While you're at it, you can also click on "Article" (if it exists) and you'll get a stable link to the paper. Please paste both of these links in the future. –  t.b. Oct 13 '11 at 19:23
    
Thanks t.b! it looks so better –  August Oct 13 '11 at 19:24

1 Answer 1

up vote 6 down vote accepted

The problem posed is available here, but here it is in its entirety:

Hammer's problem

Here's a link to the paper you're looking for:

F. D. Hammer and William Knight, Solution to problem 5955, The American Mathematical Monthly Vol. 82, No. 4 (Apr., 1975), pp. 415-416. MR1537708.


The solution given in that paper is very simple and elegant:

Let $\tilde{g}: \left[-\frac{1}{2},\frac{1}{2}\right] \to \mathbb{R}$ be the function $\tilde{g}(x) = x(1-4x^2)$. Its $1$-periodic extension $g$ to all of $\mathbb{R}$ is a $C^1$-function that is zero at the integers and whose derivative at the integers is $1$.

Now let $$f(x) = \sum_{n=0}^{\infty} \frac{g(n!x)}{(n!)^2}.$$

It is straightforward to check that $f\in C^1$ [let $f_k \in C^1$ be the $k$th partial sum of the series. Then $f_k \to f$ pointwise and $f_{k}^\prime$ is uniformly Cauchy, so $f \in C^1$]. Now notice that for rational $x$ only finitely many summands of $f$ are non-zero, hence $f(\mathbb{Q}) \subset \mathbb{Q}$. On the other hand, for rational $x$ we have $$f'(x) - e = f'(x) - \sum_{n=1}^{\infty} \frac{1}{n!} = \text{finitely many rational terms} \in \mathbb{Q},$$ so $f'(x) \notin \mathbb{Q}$, as desired.

(Thanks to robjohn for pointing out a mistake in the previous version of this answer)

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$f'-e\in\mathbb{Q}$ on $\mathbb{Q}$. –  robjohn Oct 13 '11 at 23:25
    
Thanks! Fixed... –  t.b. Oct 13 '11 at 23:32
    
O= I Know the theorem that if the sequence of continuous functions, converges uniformly then the limit is also continuous. But there you are proving that f is $ C^1 $ what theorem is that? –  August Oct 14 '11 at 2:03
    
@August: I don't know its name. But here's the statement and proof: Let $f$ and $g$ be continuous, $f_n \in C^1$ on an interval. If $f_n \to f$ pointwise and $f_{n}^\prime \to g$ uniformly then $f \in C^1$ and $f^\prime = g$. To see this, write $$f_{n}(x) = f_{n}(a) + \int_{a}^{x} f_{n}^\prime(t)\,dt.$$ Passing to the limit we get $$f(x) = f(a) + \int_{a}^{x} g(t)\,dt$$ where interchange of integration and limit is justified by uniform convergence. By the fundamental theorem of calculus, the derivative of the right hand side exists and is $g(x)$, so $f' = g$ is continuous –  t.b. Oct 14 '11 at 8:19

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