Let $X$ be a set and $\mathcal{F}\subseteq {\mathbb{R}^X}$ an arbitrary family of functions. The superposition closure of $\mathcal{F}$ is defined as $$ \overline{\mathcal{F}}=\{ H\circ(f_1\times\cdots f_i)\mid H\in C^\infty(\mathbb{R}^i),\ f_1,\ldots,f_i\in \mathcal{F},\ i\in\mathbb{N}\}\, , $$ where $(f_1\times\cdots f_i)(x)=(f_1(x),\ldots, f_i(x))$, $\forall x\in X$, and it is a commutative unitary sub-algebra of ${\mathbb{R}^X}$ containing $\mathcal{F}$. For instance, if $X=\mathbb{R}^2$ and $\mathcal{F}=C^\infty(\mathbb{R})\otimes_{\mathbb{R}} C^\infty(\mathbb{R})$, then $\overline{\mathcal{F}}=C^\infty(\mathbb{R}^2)$ (basically, this is the Kolmogorov superposition theorem discussed in the post Kolmogorov superposition for smooth functions).

Suppose now that $\mathcal{F}$ is already an unitary sub-algebra of ${\mathbb{R}^X}$. Then, we have a sequence of algebra inclusions $$ {\mathcal{F}}\subseteq \overline{\mathcal{F}}\subseteq \mathbb{R}^X $$

QUESTION: is it true that any derivation of ${\mathcal{F}}$ can be uniquely extended to a derivation of $\overline{\mathcal{F}}$?

By derivation I simply mean an $\mathbb{R}$-linear map $\Delta:{\mathcal{F}}\to {\mathcal{F}}$ satisfying the Leibniz rule.

I've been trying for many years to answer this question: all the examples I could think of gave a positive answer, but I could not lay down a general proof. All the experts I contacted gave me unsatisfactory answers (basically, one uses the chain rule for derivatives and tries to prove that the result is independent on the choice of the $H$ and the $f$'s above). Finding a counterexample would be really nice!


The counterexample proposed in the answer below is correct, and based on the fact that there can be a subalgebra $\mathcal{F}$ of $C^\infty(\mathbb{R})$ whose (real) spectrum is $\mathbb{R}^2$: it suffices to take two algebraically independent smooth functions. Then the inclusion $\mathcal{F}\subseteq C^\infty(\mathbb{R})$ should correspond, at the level of spectra, to a map $\mathbb{R}\to \mathbb{R}^2$, i.e., a "curve". I guess that, in this case, the ambiguity boils down to the fact that there are many ways to extend a vector field on a curve to the ambient space, loosely speaking. What I'm really looking for is a counterexample where $\mathcal{F}$ and its superposition closure $\overline{\mathcal{F}}$ share the same spectrum.


1 Answer 1


Not every derivation can be extended to the closure: consider the subalgebra $\mathcal{F}\subset C^\infty(\mathbb{R})$ generated by functions $x$ and $e^x$. Since these are algebraically independent this algebra is isomorphic to a polynomial algebra in two variables and we can define a derivation $\Delta$ on generators by $$\Delta(x)=1 \quad \text{and} \quad \Delta(e^x)=x.$$ Now the closure of $\mathcal{F}$ is $C^\infty(\mathbb{R})$ but no derivation on $C^\infty(\mathbb{R})$ satisfies both of these equations.

On the other hand, if there is an extension, it is unique by the following

Lemma: For any derivation $D$ on the closure $\overline{\mathcal{F}}$ of a function subalgebra $\mathcal{F}\subset \mathbb{R}^X$ we have

$$ D\left(H(f_1,\ldots,f_n)\right)=\sum_{i=1}^n \partial_i H (f_1,\ldots,f_n)\cdot D(f_i) $$

for any $H\in C^\infty(\mathbb{R}^n)$ and $f_i \in \mathcal{F}$.

Proof: Fix any point $p\in X$ and apply Hadamard's Lemma on $H$ to write

$$ H(f_1,\ldots,f_n)=H\left(f_1(p),\ldots,f_n(p)\right)+ \sum_{i=1}^n \partial_i H \left(f_1(p),\ldots,f_n(p)\right)\cdot\left(f_i-f_i(p)\right)+\sum_{|\sigma|=2}G_{\sigma}(f)\cdot(f-f(p))^\sigma $$ Now apply $D$ on both sides and then evaluate the result at $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.