A fractional derivative of distributions is usually introduced using definition of fractional integral as a convolution of two distributions. However there is another approach based on fractional integration by parts rule, suggested by Samko, Kilbas, Marichev. Let

$C_0^\infty(a,b)=\{ \phi \in C^\infty(a,b): \phi^{(k)}(a) = \phi^{(k)}(b) = 0, k=1,2,\ldots \}$

be a space of test functions. Linear functionals on $C_0^\infty(a,b)$ are distributions or generalized functions. Fractional derivative of a distribution is then defined in terms of adjoint fractional differentiation operators: $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle,\ 0 < \alpha < 1$.

However in this case Riemann--Liouville and regularized fractional derivatives are indistinguishable. Consider regularized (Caputo) derivative $D_{a+}^{(\alpha)} f(x) = D_{a+}^\alpha (f(x) - f(a))$. Obviously

$D_{a+}^{(\alpha)} \phi = D_{a+}^{\alpha} \phi\ \forall \phi \in C_0^\infty(a,b)$

and $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle = \langle D_{a+}^{(\alpha)} f,\phi \rangle$.

On the other hand, we know that the regularized derivative of a constant is zero while for the Riemann--Liouville derivative we have $D^\alpha_{a+}(x-a)^{\alpha - 1}=0$.

Hence a question: which distribution has zero fractional derivative in the sense described above? Is this a constant or $(x-a)^{\alpha - 1}$ or both?

Thank you.

A fractional derivative of distributions is usually introduced using definition of fractional integral as a convolution of two distributions. However there is another approach based on fractional integration by parts rule, suggested by Samko, Kilbas, Marichev. Let

$C_0^\infty(a,b)=\{ \phi \in C^\infty(a,b): \phi^{(k)}(a) = \phi^{(k)}(b) = 0, k=0, 1, 2,\ldots \}$

be a space of test functions. Linear functionals on $C_0^\infty(a,b)$ are distributions or generalized functions. Fractional derivative of a distribution is then defined in terms of adjoint fractional differentiation operators: $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle,\ 0 < \alpha < 1$.

However in this case Riemann--Liouville and regularized fractional derivatives are indistinguishable. Consider regularized (Caputo) derivative $D_{a+}^{(\alpha)} f(x) = D_{a+}^\alpha (f(x) - f(a))$. Obviously

$D_{a+}^{(\alpha)} \phi = D_{a+}^{\alpha} \phi\ \forall \phi \in C_0^\infty(a,b)$

and $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle = \langle D_{a+}^{(\alpha)} f,\phi \rangle$.

On the other hand, we know that the regularized derivative of a constant is zero while for the Riemann--Liouville derivative we have $D^\alpha_{a+}(x-a)^{\alpha - 1}=0$.

Hence a question: which distribution has zero fractional derivative in the sense described above? Is this a constant or $(x-a)^{\alpha - 1}$ or both?

Thank you.

A fractional derivative of distributions is usually introduced using definition of fractional integral as a convolution of two distributions. However there is another approach based on fractional integration by parts rule, suggested by Samko, Kilbas, Marichev. Let

$C_0^\infty(a,b)=\{ \phi \in C^\infty(a,b): \phi^{(k)}(a) = \phi^{(k)}(b) = 0, k=1,2,\ldots \}$

be a space of test functions. Linear functionals on $C_0^\infty(a,b)$ are distributions or generalized functions. Fractional derivative of a distribution is then defined in terms of adjoint fractional differentiation operators: $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle,\ 0 < \alpha < 1$.

However in this case Riemann--Liouville and regularized fractional derivatives are indistiguishable. Consider regularized (Caputo) derivative $D_{a+}^{(\alpha)} f(x) = D_{a+}^\alpha (f(x) - f(a))$. Obviously

$D_{a+}^{(\alpha)} \phi = D_{a+}^{\alpha} \phi\ \forall \phi \in C_0^\infty(a,b)$

and $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle = \langle D_{a+}^{(\alpha)} f,\phi \rangle$.

On the other hand, we know that the regularized derivative of a constant is zero while for the Riemann--Liouville derivative we have $D^\alpha_{a+}(x-a)^{\alpha - 1}=0$.

Hence a question: which distribution has zero fractional derivative in the sense described above? Is this a constant or $(x-a)^{\alpha - 1}$ or both?

A fractional derivative of distributions is usually introduced using definition of fractional integral as a convolution of two distributions. However there is another approach based on fractional integration by parts rule, suggested by Samko, Kilbas, Marichev. Let

$C_0^\infty(a,b)=\{ \phi \in C^\infty(a,b): \phi^{(k)}(a) = \phi^{(k)}(b) = 0, k=1,2,\ldots \}$

be a space of test functions. Linear functionals on $C_0^\infty(a,b)$ are distributions or generalized functions. Fractional derivative of a distribution is then defined in terms of adjoint fractional differentiation operators: $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle,\ 0 < \alpha < 1$.

However in this case Riemann--Liouville and regularized fractional derivatives are indistinguishable. Consider regularized (Caputo) derivative $D_{a+}^{(\alpha)} f(x) = D_{a+}^\alpha (f(x) - f(a))$. Obviously

$D_{a+}^{(\alpha)} \phi = D_{a+}^{\alpha} \phi\ \forall \phi \in C_0^\infty(a,b)$

and $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle = \langle D_{a+}^{(\alpha)} f,\phi \rangle$.

On the other hand, we know that the regularized derivative of a constant is zero while for the Riemann--Liouville derivative we have $D^\alpha_{a+}(x-a)^{\alpha - 1}=0$.

Hence a question: which distribution has zero fractional derivative in the sense described above? Is this a constant or $(x-a)^{\alpha - 1}$ or both?

Thank you.