Can a Local Fractional Differential Operator exist?

98 Views Asked by At

Let $f: \mathbb{R} \rightarrow \mathbb{R}$. The derivative of $f$ is defined pointwise, and we say that $f$ is differentiable if the derivative exists in each point. Higher order derivatives are defined recursively.

There are some common ways to extend the "order" to real numbers. This can be done abstractly by saying that an operator $D$ is a fractional differential operator of order 1/n if $D^nf(x)=\frac{d}{dx}f(x)$. Another definition is made by using the gamma function, and a third one using fourier analysis: \begin{equation} f^{(\alpha)}(x)= \mathcal{F}^{-1}((i\xi)^\alpha (\mathcal{F}f)(\xi)) \end{equation}

Now, however, the fractional derivative defined by the last two definitions is no longer local, in the sense that we can change the value of the derivative by changing the function $f$ far away from $z$. Can there be another definition of the fractional derivative that is local? Or can we show that this is not possible,for example by using the first, operator-theoretic, definition?