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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?

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    $\begingroup$ The product rule will do most of the heavy lifting here. I guess you also need something like $Lx = 1$. $\endgroup$
    – user7530
    Commented Nov 29, 2022 at 1:42
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    $\begingroup$ I wonder if this question could be modified to ask about a different space, since $C^\infty(\mathbb R)$ is defined in terms of the linear operator we're trying to describe, which seems circular somehow. Or maybe the space and the operator could be paired together, and both must be determined by some simple properties. $\endgroup$
    – mr_e_man
    Commented Nov 30, 2022 at 1:03
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    $\begingroup$ I know. The question makes sense, but it feels slightly unnatural. $\endgroup$
    – mr_e_man
    Commented Nov 30, 2022 at 1:06
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    $\begingroup$ Well, maybe I shouldn't have said "modified"; I meant making a new question similar to this one. $\endgroup$
    – mr_e_man
    Commented Nov 30, 2022 at 1:10
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    $\begingroup$ Relevant: en.wikipedia.org/wiki/Differential_algebra $\endgroup$ Commented Jan 13, 2023 at 21:13

1 Answer 1

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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 $t \mapsto t : \mathbb{R} \to \mathbb{R}$.


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 $T(1) = 0$.


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.


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, $D(x) = 1$.


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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Commented Dec 3, 2022 at 8:10
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    $\begingroup$ what is $C^{\infty}(\mathbb{R})$? Can we define this set without knowing differentiability? $\endgroup$
    – whoisit
    Commented Dec 31, 2022 at 23:25
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    $\begingroup$ See my linked question. math.stackexchange.com/questions/4588139/… $\endgroup$
    – mathlander
    Commented Jan 2, 2023 at 3:11
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    $\begingroup$ See also related mathoverflow.net/questions/44774/… (product rule, and sending the identity function to the constant function 1 also suffice) $\endgroup$
    – D.R.
    Commented Jan 31, 2023 at 23:14
  • $\begingroup$ This is basically what user7530 stated. $\endgroup$
    – mathlander
    Commented Feb 2, 2023 at 4:56

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