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Let the vector space of functions $f: \mathbb{R} \rightarrow \mathbb{R}$ be called $\mathbb{F}$. What are the linear operators on $\mathbb{F}$ that satisfy the product rule? It easily follows from this question that such a linear operator must be the derivative operator if $T(x) = 1$.

mathlander
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    The word you're looking for is derivations. Unless I'm missing something, the link you provided describes a proper subset of those derivations which satisfy the chain rule. – Dustan Levenstein Nov 30 '22 at 19:45
  • I was asking about the product rule, and I wanted something for all functions, not just smooth functions. – mathlander Nov 30 '22 at 19:47
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    I was replying to "If we restrict ourselves to smooth functions, then it easily follows from this question that such a linear operator must be a multiple of the derivative operator." – Dustan Levenstein Nov 30 '22 at 20:01
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    It follows that such a linear operator $T$ must be a multiple of the derivative operator... only if $T(x)$ is a constant function. (Here $x$ is the identity function.) What if $T(x)$ is not constant? – mr_e_man Dec 01 '22 at 00:40
  • Thanks for correcting me. – mathlander Dec 01 '22 at 00:41
  • Related: https://math.stackexchange.com/questions/4588886/interesting-properties-of-derivative-spaces – mathlander Dec 01 '22 at 02:20

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Let $D$ be a derivation on $\mathbb F$ (a derivation is simply a linear operator on an algebra which satisfies the product rule). For $S \subseteq \mathbb R$, let $\mathcal I_S$ denote the indicator function of $S$, and observe that $\mathcal I_S^2 = \mathcal I_S$.

So $D(\mathcal I_S) = D(\mathcal I_S^2) = 2\mathcal I_S D(\mathcal I_S)$ by the product rule. It readily follows that $D(\mathcal I_S)$ vanishes on both $\mathbb R \setminus S$ and on $S$, therefore $D(\mathcal I_S)=0$. Since $D$ is linear, it follows that all functions with finite range are in the kernel of $D$.

Now consider $\mathcal I_S g$ for arbitrary $g \in \mathbb F$. The product rule tells you $D(\mathcal I_S g) = \mathcal I_S D(g)$. But consider the case when $g$ vanishes outside of $S$, and therefore $g = \mathcal I_S g$. Then you get $D(g) = \mathcal I_S D(g)$, i.e., $D(g)$ vanishes everywhere $g$ vanishes.

Of course, functions don't vanish everywhere, but they take some value everywhere. Let $c \in g(\mathbb R)$ be any member of the range of $g$, and let $T$ be the fiber above $c$. By linearity, we have $D(g) = D(g)-0 = D(g)-D(c) = D(g-c)$ vanishes on $T$. By varying $c$, we can conclude that $D(g)$ vanishes everywhere. So $D$ is the zero operator.

Dustan Levenstein
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