Is there any simple set of properties which uniquely characterizes differentiation?

Question

The transformation of differentiation is a linear operator over $C^\infty(\mathbb{R}),$ the vector space of smooth functions over $\mathbb{R}.$ Is there any simple set of properties that uniquely determines this linear operator other than the standard definition?

Answer 1

Claim Let $D : C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ be a nonzero linear operator such that $D(fg) = D(f)g + fD(g)$ and $D(f \circ g) = (D(f) \circ g)D(g)$ for all $f,g \in C^\infty(\mathbb{R})$ (i.e. $D$ satisfies the product rule and the chain rule). Then $D$ is the derivative operator.

The proof comes in a few steps. I will use $1$ to denote the constant function $t \mapsto 1 : \mathbb{R} \to \mathbb{R}$ and $x$ to denote the identity function $t \mapsto t : \mathbb{R} \to \mathbb{R}$.

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Lemma 1: Any operator on $C^\infty(\mathbb{R})$ which satisfies the product rule sends $1$ to $0$.

Let $T : C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ be an operator satisfying the product rule. Then $$T(1) = T(1 \cdot 1) = T(1) \cdot 1 + 1 \cdot T(1) = T(1) + T(1)$$ implies that $T(1) = 0$.

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Lemma 2: If $T_1, T_2 : C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ are linear operators satisfying the product rule and $T_1(x) = T_2(x)$, then $T_1 = T_2$.

Let $f \in C^\infty(\mathbb{R})$ and $a \in \mathbb{R}$ be arbitrary. Since $f$ is smooth, by Taylor's theorem we have that $$f = f(a)1 + f'(a)(x-a1) + h(x-a1)^2$$ for some $h \in C^\infty(\mathbb{R})$.

By Lemma 1 and linearity, $$T_i(f(a)1) = f(a)T_i(1) = 0$$ for $i \in \{1,2\}$.

By linearity, we have $$T_i(f'(a)(x-a1)) = f'(a)(T_i(x)-aT_i(1)) = f'(a)T_i(x)$$ for $i \in \{1,2\}$.

By the product rule, we have $$T_i(h(x-a1)^2) = T_i(h)(x-a1)^2 + hT_i((x-a1)^2) = T_i(h)(x-a1)^2 + 2h(x-a1)T_i(x-a1)$$ for $i \in \{1,2\}$. When evaluated at $a$, this yields $$T_i(hx^2)(a) = (T_i(h)(x-a1)^2 + 2h(x-a1)T_i(x-a1))(a) = 0 + 0 = 0$$ for $i \in \{1,2\}$ (because both terms have a factor of $(x-a1)$).

Now by linearity, $$T_1(f)(a) = f'(a)T_1(x)(a) = f'(a)T_2(x)(a) = T_2(f)(a),$$ as desired.

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Lemma 3: If $D$ is a nonzero linear operator on $C^\infty(\mathbb{R})$ which satisfies the product rule and the chain rule, then $D(x) = 1$.

By the chain rule, $D(x) = D(x \circ x) = (D(x) \circ x) D(x) = D(x)^2$, so for all $t \in \mathbb{R}$ we have $D(x)(t) \in \{0,1\}$. Since $D(x)$ is continuous, this means that $D(x) = 0$ or $D(x) = 1$. Since the zero operator on $C^\infty(\mathbb{R})$ satisfies the product rule, and we've assumed that $D$ is not the zero operator, Lemma 2 tells us that $D(x) \neq 0$. Thus, $D(x) = 1$.

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Proof: Lemma 3 tells us that $D(x) = 1$. Since the derivative operator satisfies the product rule, Lemma 2 now implies that $D$ is the derivative operator. Since the derivative operator satisfies the chain rule, it is valid. $\square$

Since the chain rule isn't used anywhere other than Lemma 3, replacing the chain rule with Lemma 3 would also work, as user7530 suspected.