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Let $\mathcal{D}(\mathbb{R})$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$ which are differentiable at all points, and let $\mathcal{F}(\mathbb{R})$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$.

Let's denote by $D$ the derivative operator, i.e.

$$D:\mathcal{D}(\mathbb{R})\to\mathcal{F}(\mathbb{R})$$

is such that

$$D(f)=f^{\prime}$$

It's clear that $\mathcal{F}(\mathbb{R})$ is a real vector space and that $\mathcal{D}(\mathbb{R})$ is a subspace of it. We also know that $D$ is linear and that it obeys the product rule

$$D(f\cdot g)=f\cdot D(g)+g\cdot D(f)$$

The question is: is $D$ the only such linear transformation, i.e., the only linear transformation that obeys product rule?

Of course, the $0$ operator and any real multiple of $D$, $\alpha\cdot D$ are trivial examples. But I'm interested in non-trivial examples.

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    $\begingroup$ You are looking for derivations. I believe there are no other examples in this case. $\endgroup$ – Omnomnomnom Mar 4 '16 at 1:06
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    $\begingroup$ are you requiring continuity? if not I think you could take different values of $\alpha$ for different subspaces. eg 1 for polynomials and 2 for non=polynomials $\endgroup$ – Mark Joshi Mar 4 '16 at 1:17
  • $\begingroup$ @MarkJoshi that's interesting. But continuity is always a good feature, I'll edit the question. $\endgroup$ – Larara Mar 4 '16 at 1:21
  • $\begingroup$ @Mark you need to be a bit more subtle than that; we'd need some axiom of choice there. Also, what should "continuous" mean here? $\endgroup$ – Omnomnomnom Mar 4 '16 at 1:25
  • $\begingroup$ Relative to any norm, a derivation operator over the polynomials must be unbounded $\endgroup$ – Omnomnomnom Mar 4 '16 at 1:29
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The differentiable functions are a subset of the continuous functions. The polynomials are dense in the continuous functions on closed intervals. (This is the Weierstrass approximation theorem.) It is sufficient, therefore to consider what this operator does to polynomials. It's action on non-polynomial differentiable functions or differentiable functions without compact support can be found by taking limits of polynomials. (For instance, this limit could be of a sequence of polynomials,$(p_n(x))_{n \in \Bbb{N}}$, approximating a given differentiable function to within $1/n$ on the closed interval $[-n,n]$. This is the same idea as is normally used for showing the simple functions are dense in the continuous functions.)

What happens when we put $1$ in for both functions? $$ D(1) = D(1 \cdot 1) = D(1) \cdot 1 + 1 \cdot D(1) = 2D(1) \text{.} $$ This means $D(1) = 0$. Since $D$ is linear, for any constant, $c$, $D(c) = 0$.

What happens to $x^2$ and $x^n$ (for $n \in \Bbb{N}$)? \begin{align} D(x^2) &= D(x \cdot x) = D(x) \cdot x + x \cdot D(x) = 2 x D(x), \text{ and} \\ D(x^n) &= D(x \cdot x^{n-1}) = \dots = n x^{n-1} D(x) \text{.} \end{align} So, for any polynomial, $f$, $D(f) = f' D(x)$. (Where we use the prime to denote "the usual derivative". You introduced an alternative usage in your Question, but since you didn't subsequently use it, I'm stealing it to use here.)

What if we decide $D(x) = x^2$? If $f$ and $g$ are polynomials, $$ D(f \cdot g) = D(f) \cdot g + f \cdot D(g) = f' x^2 g + f g' x^2 = (fg)' D(x) \text{.} $$ This example is warm-up for: what if we decide $D(x) = \phi(x)$ for any function $\phi$? If $f$ and $g$ are polynomials, $$ D(f \cdot g) = D(f) \cdot g + f \cdot D(g) = f' \phi g + f g' \phi = (fg)' D(x) \text{.} $$

So it appears that you can use any $\phi \in \mathcal{F}(\Bbb{R})$ as the image of $x$. The image of $\mathcal{D}(\Bbb{R})$ will just be a "scaled and rotated" version of the normal derivative operator where the preimage of $0$ is the constants (and everything else, if you chose $\phi=0$) and the preimage of $\phi$ is $x$ and the rest of the image of $\mathcal{D}(\Bbb{R})$ falls out of this one choice.

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    $\begingroup$ Why is it sufficient to determine $D$ on a dense set? Why should $D$ be uniformly continuous? $\endgroup$ – Omnomnomnom Mar 4 '16 at 1:48
  • $\begingroup$ @Omnomnomnom : My knee-jerk reaction is "Hahn-Banach". But now I'm divided over whether that's actually sufficient. $\endgroup$ – Eric Towers Mar 4 '16 at 3:38
  • $\begingroup$ @Omnomnomnom : On further thought, I'm more convinced. The LCH Stone-Weierstrass gives that the polynomials are dense in $\mathcal{C}_0(\Bbb{R}, \Bbb{R})$. This isn't $\mathcal{C}(\Bbb{R}, \Bbb{R})$, but I'm stumped to think of an element of $\mathcal{D}(\Bbb{R})$ that isn't in $(\text{the polynomials} \oplus \mathcal{C}_0(\Bbb{R}, \Bbb{R}))$. Maybe you can think of one that is eluding me? $\endgroup$ – Eric Towers Mar 4 '16 at 3:53
  • $\begingroup$ again: why does the density of the polynomials matter? All we know about $D$ is that it's linear. Can't there (assuming choice) be a map $D$ that is the usual derivative on polynomials but satisfies $D(e^x)=0$? $\endgroup$ – Omnomnomnom Mar 4 '16 at 12:04
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    $\begingroup$ @Omnomnomnom : Having slept for an hour or so I might see what you're getting at. For the usual derivative, this can't happen because the normal derivative is a densely defined, closed linear operator (on the space $\mathcal{C}^1(\Bbb{R})$). This means that any sequence of polynomials converging to (for example) $\mathrm{e}^x$ have all their derivatives converging to the same thing. I'm not 100% convinced that we can't break closedness. $\endgroup$ – Eric Towers Mar 4 '16 at 18:15

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