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Let $A$ be the set of all real-valued functions having their domain a subset of $\Bbb R$ which are at least differentiable on an open set, and for $f \in A$, let $U_f$ be the largest open set on which $f$ is differentiable. Let $B = \{ f \in A: \text{ Domain}(f) = U_f\}$, $C$ be the set of all real-valued functions having their domain a subset of $\Bbb R$, and $D: B \to C$ be given by $Df= f'$ (i.e. the derivative map of $f$).

Let $T: B \to C$ be a non-zero linear mapping which satisfies:

  • If $g(U_g) \subset U_f$, then $T(f \circ g) = (Tf \circ g) Tg$

  • $T(fg) = (Tf)g + f(Tg)$

Is it true that $T = D$?

From the assumptions on $T$, we can prove that for any polynomial $p$, $Tp = Dp$, and consequently that for any rational function $q$, $Tq = Dq$. I don't know what can be done at this point. If we were considering the maps $T$ and $D$ as maps from $C^1(\Bbb R)$ to $C(\Bbb R)$, for instance, then we could've concluded that $T = D$ using density and continuity.

  • How do your assumptions rule out taking Tf = 0 for all f? – Peter LeFanu Lumsdaine Jan 13 '17 at 23:49
  • @PeterLeFanuLumsdaine they don't. Sorry, I'll add another assumption. –  Jan 13 '17 at 23:50
  • I take it that "open sets" here are open subsets of $\mathbb{R}$? – Todd Trimble Jan 14 '17 at 00:13
  • @ToddTrimble yes, I'll add that as well. –  Jan 14 '17 at 00:13
  • Well, given that you took it as far as polynomials, it's very simple. See here: http://mathoverflow.net/questions/44774/do-these-properties-characterize-differentiation/44778#44778 – Todd Trimble Jan 14 '17 at 00:36
  • @ToddTrimble I saw that question and another related one, but the situation here is quite different. The set $B$ contains functions which may not even be continuously differentiable. –  Jan 14 '17 at 00:38
  • Oh sorry; you're right. – Todd Trimble Jan 14 '17 at 00:41
  • One more question, maybe nitpicky. Why do you write $\mathbb{R}^\mathbb{R}$ (presumably the set of functions $\mathbb{R} \to \mathbb{R}$) when $f'$ seems to be defined on only part of $\mathbb{R}$, like $U_f$? – Todd Trimble Jan 14 '17 at 01:14
  • @ToddTrimble sorry, I'll fix that. –  Jan 14 '17 at 01:15
  • How do you show that Tp = Dp when p is a polynomial? I can see how the "chain rule" implies Tx = (Tx)^2, but am not sure how to deduce that Tx=1 instead of Tx=0. – Graham Cox Jan 22 '17 at 00:19
  • @GrahamCox the key is that for each $x_0 \in \Bbb R$ one can construct $g\in B$ and $y_0$ such that $g(y_0) = x_0$ and $Tg(x_0) \neq 0$. –  Jan 22 '17 at 00:36
  • Personally I think without continuity one cannot assume this operator is the derivative mapping. It is possible to construct totally discontinuous functions, defined in odd manners, and taking what would be the equivalent of derivatives, except it isn't the derivative (I definitely know how to do this on $\mathbb{Q}$, I also think you can do it on $\mathbb{R}$ too). This leads me to doubt the possibility. Granted I can't give you a rigorous answer, this is just what's ringing in my head. –  Jan 23 '17 at 05:19
  • Nevermind, you are assuming continuity, just not a continuous derivative, this idea seems much more likely now. –  Jan 23 '17 at 05:33

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