Closing the loophole in this real analysis paper In this paper, Theorem 1 states, given $F$ an arbitrary ordered subfield of $\mathbb{R}$, $F$ is complete iff every continuous function defined on a closed and bounded interval has a uniformly differentiable anti-derivative. The authors mention that they have not been able to show whether or not the "uniformly differentiable" hypothesis can be dropped without penalty. So, my question is, has anyone done further research to show whether or not this loophole can be closed? Can the uniform differentiability hypothesis be dropped without affecting the theorem?
 A: If $K$ is a countable subfield of $\mathbb R,$ and $f$ is any function $K\to \mathbb R,$ then there is an anti-derivative $F:K\to K$ in the sense that for each $k\in K$ the limit of the quotient $(F(x)-F(k))/(x-k)$ as $x\to k$ (restricted to $x\in K\setminus\{k\}$) exists and equals $f(k).$
I don't know what happens for uncountable $K.$
Let $x_1,x_2,\dots$ be an enumeration of $K.$ We will pick quadratic conditions inductively.
Suppose that we have chosen pairs $(y_1,a_1),\dots,(y_k,a_k)\in K\times \mathbb R_{>0}$ such that for each real $x$ there exists a real $y$ such that for each $1\leq i\leq k$ with $x_i\neq x,$
$$|(y-y_i)-(x-x_i)f(x_i)|< a_i(x-x_i)^2\tag{*}$$
and furthermore $(x,y)=(x_j,y_j)$ satisfies (*) whenever $1\leq j\leq k$ with $i\neq j.$
The values $y$ satisfying (*) for $x=x_{k+1}$ and for all $1\leq i\leq k$ form a non-empty open set in $\mathbb R,$ which must contain an element of $K.$ Pick one and call it $y_{k+1}.$
If I give you terms $b,c$ of a quadratic $ax^2+bx+c$ with $c>0,$ you can always pick $a$ large enough that the quadratic is strictly positive everywhere (algebraically: take $b^2-4ac<0$). By translating in the $x$ direction this means that for all sufficiently large positive $a$:

*

*the convex parabola $(y-y_{k+1})-(x-x_{k+1})f(x_{k+1})= a(x-x_{k+1})^2$ lies strictly above the $k$ concave parabolas $((y-y_i)-(x-x_i)f(x_i))= -a_i(x-x_i)^2,$ $1\leq i\leq k$

*the concave parabola $(y-y_{k+1})-(x-x_{k+1})f(x_{k+1})= -a(x-x_{k+1})^2$ lies strictly below the $k$ concave parabolas $((y-y_i)-(x-x_i)f(x_i))= a_i(x-x_i)^2,$ $1\leq i\leq k$
Pick one such value of $a$ and call it $a_{k+1}.$ By induction we can pick values $(y_i,a_i)$ for all $i$ such that (*) holds at $(x,y)=(x_j,y_j)$ for $j\neq i.$ Then the function $F:K\to K$ defined by $F(x_i)=y_i$ is an anti-derivative of $f.$
