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The transformation of differentiation is a linear operator over the vector space of entire functions (call this space $\mathbb{C}^E.$) Is there any simple set of properties that uniquely determines this linear operator that uses only the field structure of the complex numbers and the vector space structure of $\mathbb{C}^E$ and how they "interact" with each other, rather than using an "absolute value" that cannot be defined in terms of the field structure alone?

mathlander
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    Basically the same question but with the real numbers: https://math.stackexchange.com/questions/4587371/is-there-any-simple-set-of-properties-that-uniquely-characterizes-differentiatio – mathlander Jan 17 '23 at 01:16
  • The proof given in that question adapts to this case, I think. – Qiaochu Yuan Jan 17 '23 at 01:47
  • The proof uses Taylor's theorem. Apart from that, I'm pretty sure everything adapts to this case too. – mathlander Jan 17 '23 at 01:49
  • Ah, I didn't notice the square. I haven't checked but I suspect that step can be replaced with writing $f(x) = f(a) + (x - a) f'(a) + (x - a)^2 h(x)$ which still works for analytic functions. – Qiaochu Yuan Jan 17 '23 at 01:56
  • Would Peetre's theorem be something helpful? In the simplest case, that a linear operator, without continuity assumptions, on test functions, that does not increase supports, must be a differential operator. ... ? – paul garrett Jan 17 '23 at 02:19
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    If you want it short and from the standpoint of a linear operator then it is: "Differentiation is a derivation." – Marius S.L. Jan 17 '23 at 02:28
  • Yes, but is is the unique derivation that can be applied on all of $\mathbb{C}^A$ to get another function in that space? – mathlander Jan 17 '23 at 02:31
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    @mathlander No, since $h(z)d/dz$ is also a derivation on entire functions, when $h$ is an entire function: so $3\ d/dz$, $e^z\ d/dz$ and so on. But $d/dz$ is the only derivation sending $z$ to $1$. – KCd Jan 17 '23 at 03:24
  • If you have a simple proof of the second statement, you can write that as an answer. – mathlander Jan 17 '23 at 03:43

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Let $D$ be a derivation on $\mathbb C^E$ such that $Dz = 1$. Since $Df = D(f \cdot 1) = Df + f D1$ for every $f$, we must have $D1 = 0$. Moreover $Dz^2 = D(z \cdot z) = 2 z Dz = 2 z$.

Now for any entire function $f(z)$ and $a \in \mathbb C$ we have $f(z) = f(a) + f'(a) (z-a) + (z-a)^2 h(z)$ where $h(z)$ is an entire function. Then $Df(z) = f'(a) + 2 (z-a) h(z) + (z-a)^2 Dh(z)$; substitute $z=a$ and this says $Df(a) = f'(a)$. But that is true for all $a\in C$, so $Df = f'$.

Of course this would also work for analytic functions in any open subset of $\mathbb C$.

Robert Israel
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