## Saturday, October 31, 2009

### On general derivatives

Long time ago I remember one of my former professors back home talking about a way of generalize the order of derivatives to real and then complex numbers. At the moment I was maybe in introduction to analysis or so, and I only did understand the meaning of the $n$th derivative, or at least, I knew how to calculate them.

Then, I came to graduate school and people where talking about fractional calculus, which in some sense is a generalization for the usual derivatives, providing a way of calculating the $\frac{p}{q}$ derivative of a function.

The purpose is to calculate $\frac{d^\alpha}{dt^\alpha}f(x)$ for $\alpha\in\mathbb{R}^+,\mathbb{R},\mathbb{C}$.

I was thinking of some sort of an easy and non-elaborated way of generalizing this idea, and it came to my mind the spectral theorem. If we define an operator $T=\frac{d}{dx}$ to be the derivative operator for a space of functions $H$ to itself, one would like to calculate $T^\alpha$ of a function, and a nice way of solving this would be to use functional calculus by means of the spectral theorem, which would say that

$T^\alpha(f)(x)=\int_{\sigma(T)}\lambda^\alpha dE(\lambda)(f)(x)$

where $\sigma(T)$ is the spectrum of the operator $T$ and $E(\lambda)$ is a partition of unity. This would provide a quick and dirty way of accomplish our mission, but we have one problem: $T$ is not a bounded operator, and we cannot quite use this same result.

Anyhow, trying to go further with this first approach, one has some classical results when applying this idea.

It is well known that the eigenfunctions of $T$ are exponentials given by the solutions of

$Tf-\lambda f=0$

subject to some boundary/initial conditions. Suppose that we work in the space $H=L^2$ with the usual inner product and the condition for the eigenfunctions to be $f(0)=1$. Then this result is the same as taking the Fourier transform of the function and passing the derivative along the integration sign

$T^\alpha (f)(x)=\int_{-\infty}^\infty \hat{f}(t) (2\pi i t)^\alpha e^{2\pi i xt}dt$

for $\alpha=0$ we have $T^0(f)(x)=f(x)$. When $\alpha=n\in\mathbb{N}$ we have the old result from Fourier analysis and in fact, we have that $T^n(f)(x)=\frac{d^n}{dx^n}f(x)$.

Also, for negative integers $\alpha=-n\in\mathbb{Z}^-$, this realization for the generalized derivative agrees with the result od Fourier transforms, having that $T^{-n}(f)(x)=F^{(n)}(x)$, where $F^{(n)}(x)$ is the $n$th antiderivative of $f(x)$.

But this method is not so true, since as we pointed out before, $T$ is not bounded in $L^2$, so the idea is to use some kind of spectral theorem for symmetric unbounded operators. This can be done by finding a self adjoint extension of $T$ and apply the spectral theorem to it or equivalently, to use the symmetric unbounded version of this result to $T$.

By this means, one can calculate the derivative of any complex order $\alpha$ of (in general) any $L^2$ function.