I know practically nothing about fractional calculus so I apologize in advance if the following is a silly question. I already tried on math.stackexchange.


I just wanted to ask if there is a notion of fractional derivative that is linear and satisfy the following property $D^u((f)^n) = \alpha D^u(f)f^{(n-1)}$ where $\alpha$ is a scalar. In the case of standard derivatives we would have $\alpha = n$.


Thank you very much.



A question on fractional derivatives
There are basically no interesting solutions to this equation beyond first and zeroth order operators, even if one only imposes the stated constraint for $n=2$.


First, we can [depolarise](https://en.wikipedia.org/wiki/Polarization_identity) the hypothesis
$$ D^u(f^2) = \alpha\_2 D^u(f) f \quad (1)$$
by replacing $f$ with $f+g, f-g$ for arbitrary functions $f,g$ and subtracting (and then dividing by $4$) to obtain the more flexible Leibniz type identity
$$ D^u(fg) = \frac{\alpha\_2}{2}( D^u(f) g + f D^u(g) ). \quad (2)$$


There are now three cases, depending on the value of $\alpha\_2$:


1. $\alpha\_2 \neq 1,2$. Applying (2) with $f=g=1$ we then conclude that $D^u(1)=0$, and then applying (2) again with just $g=1$ we get $D^u(f)=0$. So we have the trivial solution $D^u=0$ in this case.
2. $\alpha\_2=2$. Then $D^u$ is a [derivation](https://en.wikipedia.org/wiki/Derivation_(differential_algebra)) and by induction we have $D^u(f^n) = n D^u(f) f^{n-1}$, just as with the ordinary derivative, so we just have $\alpha\_n=n$ for all $n$ with no fractional behavior.
3. $\alpha\_2=1$. Applying (2) with $g=1$ we obtain (after a little bit of algebra) $D^u(f) = mf$ where $m := D^u(1)$. Thus $D^u$ is just a multiplier operator, which obeys $D^u(f^n) = D^u(f) f^{n-1}$, thus $\alpha\_n=1$ for all $n$.


Thus there are no linear solutions to your equation other than the usual derivations (e.g., $D^u(f) = a(x) \frac{d}{dx} f$ for any smooth symbol $a$) and multiplier operators $D^u(f) = mf$, i.e., first order and zeroth order operators.


On the other hand, fractional derivatives $D^u$ tend to obey a "fractional chain rule"
$$ D^u( F(f) ) = D^u(f) F'(f) + E$$
for various smooth functions $F,f$, where the error $E$ obeys better estimates in various Sobolev spaces than the other two terms in this equation. In particular, for $F(t) = t^n$, we would have
$$ D^u(f^n) = n D^u(f) f^{n-1} + E$$
for a "good" error term $E$. For instance, taking $u=n=2$ with $D$ the usual derivative, we have
$$ D^2(f^2) = 2 D^2(f) f + E \quad (3)$$
with $E$ the "[carré du champ](https://mathscinet.ams.org/mathscinet-getitem?mr=521767)" operator
$$ E := 2 (Df)^2.$$
Note that the error $E$ is controlled uniformly by the $C^1$ norm of $f$ but the other two terms in (3) are not. See my previous MathOverflow answer at <https://mathoverflow.net/a/94039/766> for some references and further discussion.


