Does the product rule imply the chain rule?

Question

Let $\mathbb{F}$ be a field, and consider $\mathbb{F}^\mathbb{F}$ as an algebra over $\mathbb{F}$ with the standard function multiplication. Let $D$ be a derivation on a subalgebra of $\mathbb{F}^\mathbb{F}$ closed under function composition that takes the identity function to the function that sends everything to $1,$ as in the answers to these two questions. Does $D$ necessarily satisfy the chain rule for arbitrary $\mathbb{F}$?

Answer 1

I believe the answer is no, the product rule does not imply the chain rule, although counterexamples seem tedious to construct. Let $F = \mathbb{R}$ or $\mathbb{C}$ and let $A$ be the smallest $F$-subalgebra of $F^F$ closed under composition and containing the functions $x$ and $\exp(x)$. I believe that there exists a derivation $D$ on $A$ with the following properties:

• $D(x) = 1$
• $D(\exp(x)) = \exp(x)$
• $D(\exp^{\circ k}(x)) = 0$ for all $k \ge 2$.

If so then $D$ does not satisfy the chain rule. To motivate the existence of such a derivation let's see what constrains the possible values of $D$. Every element of $A$ is a sum of terms of the form $x^k \exp(a)$ where $a \in A$. By applying the sum-to-product exponent rule to $\exp(a)$ we can write every element of $A$ as a sum of products of copies of either $x$ or $\exp(a)$ where $a$ itself has the form $x^k \exp(a')$ for some $a' \in A$. $D$ being a derivation then means that the value of $D$ is determined by $D(x)$ and by the values $D(\exp(x^k \exp(a))$.

Now the point here is that being a derivation does not constrain the value of $D(\exp(x^k \exp(a))$ at all, except in the trivial case $D(\exp(c)) = 0$ where $c$ is a scalar. In fact we ought to be able to show that $A$ is basically a free algebra, albeit on a very large set of generators (if so, we can freely specify the value of $D$ on every generator). So we have a great deal of freedom to assign values to expressions of the form $D(\exp(-))$, although again it seems tedious to pin down whether we have enough freedom for the construction we want. Hopefully somebody else can do this!

What is true is that if $D$ is a derivation (so satisfies linearity and the product rule) then $D(p(f)) = p'(f) D(f)$ where $p$ is a polynomial and $p'$ is its ordinary derivative.