# Is there a differentiable function such that $f(\mathbb Q) \subseteq \mathbb Q$ but $f'(\mathbb Q) \not \subseteq \mathbb Q$?

Is there a differentiable function $$f:\mathbb R \rightarrow \mathbb R$$ such that $$f(\mathbb Q) \subseteq \mathbb Q$$, but $$f'(\mathbb Q) \not \subseteq \mathbb Q$$? A friend of mine asserted this without giving any examples. I seriously doubt it, but I had hard time trying to disprove it since analysis isn't really my thing. I can't even think of any class of differentiable functions with $$f(\mathbb Q) \subseteq \mathbb Q$$ other than the rational functions.

• Just an observation :- If $f'(x)$ could be made a constant irrational , then $f(x)$ maps rationals to irrationals Aug 12, 2020 at 12:18
• Yeah, but that can only happen if $f(x)$ is linear, no? Aug 12, 2020 at 12:20
• Yes, if $f'(x)$ is constant irrational then $f$ is of the form $f(x) =ax+b$ with $a \in \Bbb{R}\setminus \Bbb{Q}$ Aug 12, 2020 at 12:21
• $g(t)=\int_a^{ta} m(x)dx\quad \text{where}\quad a\in \mathbb R/\mathbb Q$ Functions of this figure may be realized Aug 12, 2020 at 12:43
• Pretty much any non-rational analytic function that maps rationals to rationals would do that, see Functions that take rationals to rationals Aug 12, 2020 at 12:46

Nice question.

Here is my solution. It would be nice to know if there is a simpler example.

This is basically a "spline" example. First, if $$A,B,C,D,r,s$$ are all rational numbers, and $$r < s$$, then there exist rational numbers $$a,b,c,d$$ so that the cubic polynomial $$f(x) = ax^3+bx^2+cx+d$$ satisfies $$f(r)=A,f'(r)=B, f(s)=C, f'(s)=D$$. The proof is: write down the system of equations and solve for $$a,b,c,d$$. As long as $$r \ne s$$ the solution is rational in $$A,B,C,D,r,s$$.

Remark 1. Computation shows that if $$B=D=0$$, then the graph of $$f$$ on $$[r,s]$$ lies inside the rectangle with opposite corners $$(r,A)$$ and $$(s,C)$$. [Differentiate the cubic, then factor to find that the derivative vanishes at the two endpoints $$r,s$$.]

The construction First, choose your favorite irrational number, let's say $$\pi$$. Consider the two curves $$\phi_1(x)=\pi x$$ and $$\phi_2(x) = \pi x + x^3$$. Of course, any function $$f$$ with graph between these two has $$f'(0)=\pi$$. Define such a function by using a sequence of rational "knots" $$r_n \searrow 0$$ and $$A_n$$ so that $$\phi_1(r_n) < A_n < \phi_2(r_n)$$ Then fill in with the splines as explained, to get a function $$f$$ with $$f(r_n) = A_n$$ and $$f'(r_n) = 0$$ for all $$n$$. Do the same thing on the negative side. Finallly let $$f(0)=0$$.

This gives us $$f : \mathbb R \to \mathbb R$$ such that $$f(\mathbb Q) \subseteq \mathbb Q$$, $$f$$ is differentiable except possibly at $$0$$, and $$\lim_n \frac{f(r_n)-f(0)}{r_n-0} = \pi$$ is irrational.

What remains: we need to choose the sequences $$r_n$$ and $$A_n$$ so that the rectangles with opposite vertices $$(r_n,A_n)$$ and $$(r_{n+1},A_{n+1})$$ lie entirely between $$\phi_1$$ and $$\phi_2$$. Then by Remark 1, $$f$$ remains between $$\phi_1$$ and $$\phi_2$$, so $$f'(0)=\pi$$.

• as you say, there is probably a shorter argument. But yours is the only one among those proposed that I understand, and a nice idea. Nice job. Aug 12, 2020 at 12:51
• I wish I could upvote that several times. Very nice answer! Aug 12, 2020 at 13:16
• @GEdgar Please clarify one question . Why is $f(\mathbb{Q}) \subseteq \mathbb{Q}$ true? Is it because $f(r_n)=A_n$ ? But then, are all the positive rationals included in the decreasing sequence $\{r_n\}$ ?Thanks, sir. Aug 12, 2020 at 15:08
• @user710290 ... The function $f$ is made up piecewise of segments where it is a cubic spline polynomial with rational coefficients. This is explained at the beginning. Not all rationals are included, the sequence $r_n$ converges to $0$. Aug 12, 2020 at 16:05
• @GEdgar Ok , got it. Since the coeffiecients are rationals , $f$ cannot map rationals to irrationals. Thanks ! Aug 12, 2020 at 16:26

This answer shares the use of cubic splines with that of GEdgar. Consider the following polynomial function: $$p(x)=2(x-1)^3-3(x-1)^2+1$$ $$p$$ satisfies $$p'(1)=p'(2)=0$$. Studying $$p$$ in the interval $$[1,2]$$ we see that in this interval $$p$$ is non negative and strictly decreasing, with $$p(1)=1$$ its maximum and $$p(2)=0$$ its minimum. Therefore the function: $$h(x)= \begin{cases} 1& |x|\leq 1\\ p(|x|)& 1<|x| \leq 2\\ 0 &|x|>2 \end{cases}$$ is in $$C^1(\mathbb{R})$$. Now set: $$g_n(x)=h(nx)/n!$$ and consider $$g(x):=\sum_{n=0}^{+\infty} g_n(x)$$.

What we are going to prove is that $$f(x):=xg(x)$$ is a differentiable function and satisfies $$f(\mathbb{Q})\subseteq \mathbb{Q}$$ and $$f'(0)=e$$.

Let's first discuss the values taken by $$f$$ on the rationals. The function $$g_n$$ takes only rational values in every rational point and for $$n>0$$ we have $$g_n$$ is zero outside $$[-2/n,2/n]$$. This means that for every $$q \in \mathbb{Q}-\{0\}$$ the sequence $$(g_n(q))$$ is eventually zero, so that $$g(q)$$ is in fact a finite sum of rationals and $$f(q)=qg(q)$$ is also rational. We also have $$g(0)=e$$ but clearly $$f(0)=0\cdot e=0 \in \mathbb{Q}$$.

To show $$f$$ is differentiable, it is enough to show $$g$$ is. Since $$g'_n(x)= h'(nx)/(n-1)!$$ for $$n>1$$, we have $$\sum g'_n$$ converges uniformly on $$\mathbb{R}$$ and since $$\sum g_n(0)$$ converges we deduce that also $$g$$ is differentiable, with $$g'=\sum g_n'$$.

Finally, $$f'(0)= g(0)=e$$. This concludes the proof.

• +1 I like it... Aug 12, 2020 at 16:12
• @GEdgar Thank you! Also, we get $f \in C^1(\mathbb{R})$ (even if not required,though) Aug 12, 2020 at 17:01

Here is another answer. Define $$h:[0,+\infty)\to \mathbb{R}$$ as follows: $$h(x):= \frac{1}{(n+1)!}(n+1-n(x-n)) \quad \text{for } x\in [n,n+1)$$ The function $$h$$ is continuous and satisfies $$h(k)=1/k!$$ for every $$k$$ integer in its domain. Moreover: $$\int _0^{q}h(t)dt \in \mathbb{Q} \quad \forall q \in \mathbb{Q}$$ since on every interval of the type $$[n,n+1)$$ the function $$h$$ is an affine function with rational coefficients. A direct computation shows: $$\int _0^{\infty}h(t)dt = \sum_{k=0}^{+\infty} \frac{1}{2}\big( \frac{1}{k!} + \frac{1}{(k+1)!}\big) = e - \frac{1}{2}$$ Define now $$f:\mathbb{R}\to \mathbb{R}$$ as follows: $$f(x):= \begin{cases} 0 & x=0\\ x \int _0^{1/|x|}h(t)dt & x\neq 0 \end{cases}$$ Then $$f$$ satisfies $$f(\mathbb{Q})\subseteq \mathbb{Q}$$ and it is differentiable in $$x\neq 0$$ for the fundamental theorem of integral calculus. What about $$x=0$$? We have: $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{x \int _0^{1/|x|}h(t)dt}{x}= \int _0^{+\infty}h(t)dt = e - \frac{1}{2}$$ Hence $$f$$ is differentiable on the whole real line and $$f'(0)= e-1/2 \notin \mathbb{Q}$$