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Suppose $\nu > 0$ and $n$ is such that $\lceil\nu\rceil = n$. The Riemann-Liouville definition of the fractional derivative would be

$$f^{(\nu)}(x) = \frac{1}{\Gamma(1-\nu)}\frac{d^{n}}{dx^{n}}\int_0^x(x-t)^{n-\nu-1}f(t)\,dt.$$

Based on the Riemann-Liouville definition of the fractional derivative, it would seem to me that any smooth function has a fractional derivative of all orders but I am still somewhat skeptical of this. Is there an easy way to argue this? The differentiation under the integral sign seems a bit formidable.

Enrico M.
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Cameron Williams
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1 Answers1

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Not exactly unless you re-define smooth:

Smooth functions are considered to be those function with continuous derivatives of all orders

I am not sure if the function necessarily must have its (integral) derivative exist at all points (correct me if i'm wrong)

but that can mean a function such as the classic

$$ \begin{pmatrix} \frac{2x^{\frac{1}{2}}}{\sqrt{\pi}} x \ge 0 \\ -\frac{2(-x)^{\frac{1}{2}}}{\sqrt{\pi}} x < 0\end{pmatrix} $$

is smooth.

Since any integer derivative will also result in a function too (which will have singularities at the point 0 but these functions are still continuous!)

Interestingly if one takes a $\frac{3}{2}$ derivative of this function you will obtain the function 0 which will continue to be the function if you keep taking any sort of fractional derivative.

The lesson, a function can be smooth for all integral derivatives but not be smooth for all rational derivatives.

On the other hand smooth's original definition was "function with derivative of all orders" and I just assumed they meant integral since that was what was available at the time when the definition was made.

Now it could very well be that there was a specific reason for that ambiguity in the definition and that with this invention of the frational derivative it could mean smooth should also be smooth for fractional (And real) derivatives. It's not apparently clear if that was the intentionn

Sidharth Ghoshal
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  • Interesting. My question is a little less well-posed than I thought since "existence" was pretty vague in my post. Existence would naturally mean that $f^{(\nu)}(x)$ exists for all $\nu > 0$ and $x\in\Bbb R$. Clearly smooth as most people know it is not sufficient. Taking $f(x) = x$ and $\nu = \frac{3}{2}$, we would get a function proportional to $x^{-1/2}$ and this is singular at $0$. This then makes me wonder how common it is for a fractional derivative of all orders to suggest. If even monomials fail, it might be that the only function that has derivative all orders is the zero function. – Cameron Williams Aug 07 '14 at 04:06
  • Maybe existence could be relaxed to being just in the almost everywhere sense which naively seems to work (at least for polynomials), but I wouldn't know how to appropriately incorporate measure-theoretic arguments into this. The appearance of the integral might be masking the almost everywhere behavior but this is something I'd have to think about for a while. – Cameron Williams Aug 07 '14 at 04:07