3
$\begingroup$

Does there exist an operation $T: C^{\infty}(\mathbb{R},\mathbb{C}) \to C^{\infty}(\mathbb{R},\mathbb{C})$ satisfying $T^2=D$ where $D$ is the derivative operator?

The original question imposed the condition that $T$ is endomorphism of the vector space $C^{\infty}(\mathbb{R},\mathbb{C})$ over $\mathbb{R}$, in which case its quite easy to show that there cant exist such a $T$.

If we remove the requirement that $T$ be a linear transformation and now allow $T$ to be any function from $C^{\infty}(\mathbb{R},\mathbb{C})$ to $C^{\infty}(\mathbb{R},\mathbb{C})$, I believe the answer is yes, such an endomorphism exists. I have an idea for how to construct such a $T$, by partitioning $C^{\infty}(\mathbb{R},\mathbb{C})$ into equivalence classes defined by the equivalence relation $f \sim g$ iff there exist $(n,m) \in \mathbb{N^*}^2$ such that $f^{(n)}=g^{(m)}$. Defining $T(f)$ then determines (more or less) the image of the equivalence class of $f$ by $T$. Depending on the function $f$, the choice of $T(f)$ may be free or not (ex: $f=e^x$, then $T(f)$ its its own derivative). I can go more into the details, which I haven't yet fully worked out, tomorrow.

  • 0
    What condition exactly you want to remove? – 2017-02-19
  • 0
    @MarianoSuárez-Álvarez : See my edit. – 2017-02-19
  • 2
    Are you familiar with [fractional calculus](https://en.wikipedia.org/wiki/Fractional_calculus)? – 2017-02-19
  • 0
    @Chappers : No. maybe you can write an answer elaborated on that. I read the Wikipedia article on it and I guess the fractional derivative comes from generalizing the natural solution for monomials by using the gamma function. What I was trying to do here however was construct abstractly an "pathological" solution. – 2017-02-27

0 Answers 0