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 $$\Phi_{\lambda}(x) = \frac{x^{\lambda-1}}{\Gamma(\lambda)} frac{x_{+}^{\lambda-1}}{\Gamma(\lambda)}$$ 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 $a > 0$.
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 $a > 0$.