Ytukyg

Following the scheme given by mercio in the answer to this question, one can construct a function $f: mathbb Q rightarrow mathbb Q$, differentiable everywhere, with derivative function
$$
f'(x) = begin{cases}
0, & (x neq 0),\
1, & (x = 0).
end{cases}tag{1}label{one}
$$
Such function cannot be uniformly continuous, because otherwise, in any closed interval containing $0$, $f$ could be extended to a continuous function on $mathbb R$. Such extended function would give
begin{equation}
lim_{h rightarrow 0} frac{f(h)-f(0)}{h} = 1
end{equation}
and, at the same time,
begin{equation}
lim_{xrightarrow 0}f'(x) = 0,
end{equation}
which would contradict de l'Hôpital's rule.

My question is whether it is possible or not to demonstrate that $f$ is not uniformly continuous by using only condition eqref{one} and the definition of uniform continuity, i.e. (for the sake of completeness)
begin{equation}
forall epsilon, exists delta: forall x, y in mathbb Q, |x-y|<delta Rightarrow |f(x)-f(y)|<epsilon .
end{equation}

EDIT: Wojowu pointed out to me that the function may not be differentiable once extended on $mathbb R$ (thus not contradicting de l'Hôpital's rule) . So either $f$ is not uniformly continuous or its extension to the reals fails to be differentiable. So the question now is whether this can be estabished only by means of eqref{one} or not, and how.

Actually $f$ can be uniformly continuous.

Edit: A hugely simplified and cleaned up version of what I posted earlier:

Oops: I just realized that what's below is not quite a counterexample to the assertion in the OP. We construct a uniformly continuous $f:Bbb RtoBbb R$ such that $f'(0)=1$ and $f'(r)=0$ for every non-zero rational $r$, but there's no reason to think that $f(Bbb Q)subsetBbb Q$. I'm leaving this here anyway because (i) the construction is quite simple, (ii) it does serve to show that the OP's argument purporting to show that $f$ cannot be uniformly continous is wrong, since that argument would work just as well for $f:Bbb QtoBbb R$, and (iii) come to think of it it's not hard to fix it to give a literal counterexample (see Fix below).

Say $r_1,r_2,dots$ is an enumeration of the non-zero rationals. For each $n$ choose $delta_nin(0,|r_n|/2)$ in such a way that if $I_n=(r_n-delta_n,r_n+delta_n)$ and $$E=Bbb Rsetminusbigcup I_n$$then $$lim_{rhoto0}frac1rho m(Ecap(0,rho))
=lim_{rhoto0}frac1rho m(Ecap(-rho,0))
=1.$$(It's clear that happens if $delta_nto0$ fast enough; Details below if it's not clear). Let $$f(x)=int_0^xchi_E(t),dt.$$Then $f:Bbb RtoBbb R$ is Lipshitz, hence uniformly continuous, $f'(r_n)=0$ because $f$ is constant on $I_n$, and the condition on density of $E$ at the origin implies that $f'(0)=1$.

Details: We will show that if $delta_nto0$ fast enough then $f'(0)=1$ (that being what we care about - it's equivalent to the statement that $0$ is a point of density for $E$.) In fact

If $sumdelta_n/|r_n|<infty$ then $f'(0)=1$.

Note first that since $0<delta_n<|r_n|/2$ that condition is equivalent to $sumfrac{delta_n}{|r_n|-delta_n}<infty$.

Let $$g(x)=x-f(x)$$ and $$G(x)=sum g_n(x),$$ where $$g_n(x)=int_0^xchi_{I_n}.$$ We will avoid worrying about $int_0^x$ for $x<0$ by showing only that the right-hand derivative of $g$ at the origin vanishes; of course the same applies to the left-hand derivative. Since $$0lefrac{g(x)-g(0)}{x}lefrac{G(x)}{x}quad(x>0)$$it's enough to show that $$lim_{xto0^+}frac{G(x)}{x}=0.$$Define $D_n(x)$ for $xge0$ by $$D_n(x)=begin{cases}frac{g_n(x)}{x},&(x>0),
\0,&(x=0).end{cases}$$Since $g_n'(0)=0$ it follows that $D_n$ is continuous on $[0,infty)$. It's clear that $$D_n(x)=0quad(0le xle|r_n|-delta_n)$$ and $$|D_n(x)|lefrac{2delta_n}{|r_n|-delta_n}quad(x>|r_n|-delta_n).$$So the series $sum D_n$ converges uniformly on $[0,infty)$, and hence $$D=sum D_n$$ is continuous on $[0,infty)$. But $$D(x)=begin{cases}frac{G(x)}{x},&(x>0),\0,&(x=0);end{cases}$$hence $lim_{xto0^+}G(x)/x=0$.

Fix: A patch for the problem mentioned in Oops above:

Lemma. Suppose $(y_n)$ is a sequence of non-zero reals. There exists a uniformly continuous $psi:Bbb RtoBbb R$ such that $psi(0)=0$, $psi'(0)=1$, and $psi(y_n)$ is rational for every $n$.

That seems quite clear, since we're not requiring more than continuity except at one point. Anyway:

Proof: Suppose $y_nne y_m$ for $nne m$. For each $n$ choose an open interval $I_n$ with $0notin I_n$, $y_nin I_n$ and $y_knotin I_n$, $1le k<n$. Let $phi_n$ be a continuous function supported in $I_n$ with $phi_n(y_n)ne0$. We will let $$phi=sum c_nphi_n$$for suitable constants $c_n$.

First, if $c_nto0$ fast enough then the series converges uniformly, hence $phi$ is uniformly continuous, since the partial sums are uniformly continuous.

Second, if $c_nto0$ fast enough then $$|phi(x)|le x^2.$$(For example, if $c_n$ is small enough then $|c_nphi_n(x)|le x^2/2^n$ for all $x$.)

Finally, it's clear that we may choose the $c_n$ one by one, "small enough" as in the previous two paragraphs, so that $$y_n+sum_{j=1}^n c_jphi_j(y_n)inBbb Q$$ for all $n$. Since $y_nnotin I_j$ for $j>n$ this implies that $$y_n+phi(y_n)=y_n+sum_{j=1}^n c_jphi_j(y_n)inBbb Q.$$

Now let $psi(x)=x+phi(x)$. QED.

Now to fix the construction above: Let $E$ and $f$ be as above. Say the connected components of $E$ are $J_1,J_2,dots$. Note that $f$ is constant on each $J_n$; let $y_n$ be this constant. Choose $psi$ as in the lemma and let $F=psicirc f$.

Then $F'(0)=1$. Also $F(0)=0inBbb Q$, while if $r$ is a non-zero rational then there exists $n$ with $rin J_n$; hence $F'(r)=0$, since $F$ is constant on $J_n$, and also $F(r)=psi(y_n)inBbb Q$.