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Let $E_{\alpha}(x^{\alpha})$ be a Mittag-Leffler function, $\alpha \in (0,1)$. It is an eigenfunction for nonlocal fractional derivative, defined as a convolution with $$ \Phi_{\lambda}(x) = \frac{x_{+}^{\lambda-1}}{\Gamma(\lambda)} $$ when $\lambda = -\alpha$ (Gelfand, Shilov, "Generalized functions"). We can define a local fractional derivative of order $\alpha$ by $$ \tilde{D}^{\alpha}f(x) = \lim\limits_{\delta \to +0} \frac{\Gamma(\alpha+1)(f(x+\delta)-f(x))}{\delta^{\alpha}} $$ Then for any $x>0$ we will have $\tilde{D}^{\alpha}E_{\alpha}(x^{\alpha}) = 0$ and $E_{\alpha}(x^{\alpha})$ is not invariant under action of local fractional derivative. I would like to know which function is an eigenfunction of local fractional derivative. If this function is well-known, is there a representation of such function that depends on $x$ through $x^{\alpha}$ like the Mittag-Leffler function? I think that the main advantage of such function is that $f(x+a)$ will still be an eigenfunction for any $a$ because of locality of operator. This property doesn't hold for nonlocal fractional derivative defined above, so $E_{\alpha}((x+a)^{\alpha})$ is not an eigenfunction for nonlocal fractional derivative if $a > 0$.

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A consequence of your definition is that your local fractional derivative vanishes for all Lipschitz continuous functions, and even on Hölder continuous functions with exponent $>\alpha$. Huge kernel ! – Bazin Nov 7 '12 at 22:48
Yes, it may be some fractal function (Holder-continuous of order $\alpha$) that depends on $x$ via $x^{\alpha}$. – Nimza Nov 8 '12 at 7:53
From my point of view, your definition is strange, or at least it has a contra-intuitive feature: because $x$ and $\delta$ should have the same unit (e.g. some length unit), $\delta^\alpha$ will have the unit e.g. length$^\alpha$. Then $\tilde D^a f(a)$ will have the unit "meter$^{1-\alpha}$". But maybe this is OK? Please correct me if I'm wrong. – Petern Jan 26 '13 at 19:58

Nothing interesting is going to happen. Suppose $f$ is an eigenfunction, then it certainly must be continuous. Consider a closed interval on which $f$ is bounded away from $0$, then the fractional derivative of $f$, which is just a constant multiple of $f$, will also be bounded away from $0$. Say it is positive. Then $f - k x$ will also have positive fractional derivative for each constant $k$, since $kx$ has zero fractional derivative. A function of positive fractional derivative must be increasing, so $f-kx$ is an increasing function for all $k$. This is clearly impossible.

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I think that this type of fractional differentiation, say of order $\alpha$, maps a Holder $\mathcal{H}^\beta$ function $f(x)$ into $\mathcal{H}^{\beta - \alpha}$ function. Therefore, if the derivative does not diverge or does not vanish identically it will map a $\mathcal{H}^{\alpha}$ function into a $\mathcal{H}^{0}$ function, which is discontinuous. If you apply it to a $\mathcal{H}^{0}$ function you will get $\mathcal{H}^{-\alpha}$ function which will alternate between infinity and zero. Therefore, only zero remains, i.e. $f(x) \equiv 0$ $\tilde{D}^{\alpha}0= k \; 0$.

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