Let $D:\mathbb{R}(x)\to\mathbb{R}(x)$ be the algebraic(!) differential operator such that $D(x)=1$. Is it true that $D(c)=0$ for all $c\in\mathbb{R}$?

Question

Define the field of rational functions $\mathbb{R}(x)$ with algebraic differential operator $D:\mathbb{R}(x)\to\mathbb{R}(x)$ such that $D(x)=1$. By algebraic differential operator, we mean a mapping such that the additive rule and product rule apply for all $f,g\in\mathbb{R}(x)$: $$D(f+g)=D(f)+D(g),$$ $$D(fg)=fD(g)+gD(f).$$ Also, with this definition of the derivative, the following are easy to prove:

1. $D(0)=D(1)=0$;
2. $D(-f)=-D(f)$;
3. $D\left(\frac{f}{g}\right)=\frac{gD(f)-fD(g)}{g^2}$ for $g\neq0$;
4. $D(f^n)=nf^{n-1}D(f)$ for every $n\in\mathbb{Z}$ and $f\neq0$;
5. If $c\in\mathbb{R}(x)$ is such that $D(c)=0$, then $D(cf)=cD(f)$.

I want to prove that, for all $c\in\mathbb{R}$, we have $D(c)=0$. I've already proved that this is true if $c\in\mathbb{Q}$. I wanted to use an argument that involved sequences, like using the fact that for every $c\in\mathbb{R}$ we could define a sequence $(c_n)\subseteq\mathbb{Q}$ such that $c_n\to c$. However, this doesn't work, as the differential operator is (usually) not continuous, and we can't just do $$D(c)=D\left(\lim_{n\to\infty}c_n\right)=\lim_{n\to\infty}D(c_n)=\lim_{n\to\infty}0=0.$$ Hence, I'm stumped. For now I've been defining the field of constants of $\mathbb{R}(x)$ as $\mathbb{R}$, but I would like to know if this result can be proved just using algebra and non-differential tools from analysis.

Answer 1

As the commenters above have already pointed out, the answer is no, not without additional conditions on $D$. The reason is that any transcendental real number, for example, $\pi$, acts like an indeterminate with respect to algebraic conditions. So, you can start with $D$ defined on $\mathbb{Q}(x)$ by $D(x)=1$ and then define $D(\pi)$ to be 1, $3\pi^2+7x$, or whatever you like. Then (to put it briefly) as long as some new real number $r$ is available which is algebraically independent of the numbers you've already defined $D$ on, you again have complete freedom in choosing $D(r)$. When no new algebraically independent real numbers are available, any new real number $s$ must be algebraic over the field $F$ which you've already defined $D$ on. Then, if $f$ is the minimal polynomial of $s$ over $F$, $$f(s)=0$$ so you must have also $$D(f(s)) = f_D(s) + f'(s) (Ds) = 0$$ where $f_D$ is given by coefficient-by-coefficient differentiation of the cofficients of $f$ and $f'$ is the formal derivative of $f$. So, the only way that you can possibly extend $D$ is by setting $$ Ds := -\frac{f_D(s)}{f'(s)}. $$ It can be shown that this does give a consistent extension of $D$ to $\mathbb{R}(x)$.