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For $\alpha > 0$, let $n =\lfloor \alpha \rfloor + 1$ and $f : (0,\infty)\to\mathbb{R}$ be continuous. Define $$D^\alpha f(x) = \frac{1}{\Gamma(n - \alpha)} \left(\frac{d}{dx}\right)^n \int_0^x \frac{f(t)dt}{(x - t)^{\alpha - n + 1}},$$ provided the RHS exist for $x\in(0,\infty)$.

Question: Which functions satisfy $D^\alpha f(x) = 0$?

The following is a corollary from Theory and Applications of Fractional Differential Equation by Kilbas, Srivastava, and Trujillo. They state it without proof, just by referring as a Corollary of the fact that $D^\alpha x^{\alpha - k} = 0$ for each $1 \leq k \leq \lfloor\alpha\rfloor + 1$

Corollary 2.1 The equality $D^\alpha f(x) = 0$ is valid if, and only if $$f(x) = \sum_{k = 1}^n c_k x^{\alpha - k}.$$

It is clear that, by direct computation, functions of the form $$f(x) = \sum_{k = 1}^n c_k x^{\alpha - k},$$ where $c_k$ are constants, indeed satisfy this equality. To see this, note that for any integer $1\leq k \leq \lfloor \alpha \rfloor + 1$ we have

\begin{align*} D^\alpha x^{\alpha - k} &= \frac{1}{\Gamma(n - \alpha)} \left(\frac{d}{dx}\right)^n \int_0^x \frac{t^{\alpha - k} dt}{(x - t)^{\alpha - n + 1}} \\ &= \frac{1}{\Gamma(n - \alpha)} \left(\frac{d}{dx}\right)^n \int_0^1 (ux)^{\alpha - k} (x - ux)^{n - \alpha - 1}(x\,du)\\ &= \frac{1}{\Gamma(n - \alpha)}\left( \int_0^1 u^{\alpha - k} (1 - u)^{n - \alpha - 1}\, du\right) \left(\frac{d}{dx}\right)^n x^{n - k}\\ &= 0. \end{align*} The conclusion follows from the linearity of $D^\alpha$.

But are these the only functions?

I have tried the following. Suppose $D^\alpha f(x) = 0$, then \begin{align*} \frac{1}{\Gamma(n - \alpha)} \left(\frac{d}{dx}\right)^n \int_0^x \frac{f(t)dt}{(x - t)^{\alpha - n + 1}} &= D^\alpha f(x) \\ \left(\frac{d}{dx}\right)^n \int_0^x \frac{f(t)dt}{(x - t)^{\alpha - n + 1}} &= 0\\ \int_0^x \frac{f(t)dt}{(x - t)^{\alpha - n + 1}} &= \sum_{k = 0}^{n - 1} c_k x^k. \end{align*}

This is where I got stuck. I could not extract the $f$ in the integral. I was about to use the Fundamental Theorem of Calculus but the denominator would be zero so it wouldn't work.

Update

I tried to substitute $u = t/x$ in the last integral and got

$$\int_0^1 f(ux)(1 - u)^{n - \alpha - 1} du = \sum_{k = 1}^n c_k x^{\alpha - k}$$ I don't know how to continue from here. Am I even on the right track? Any help would be greatly appreciated.

Update II I got the following remark in Fractional Differential Equations An Approach via Fractional Derivatives by Bangti Jin which gives the result with some additional assumptions.

In particular, for $\alpha \in (0, 1)$, we have $(x − a)^{\alpha - 1}$ belongs to the kernel of the operator $D^\alpha$ and plays the same role as a constant function for the first-order derivative. Generally, it implies that if $f, g \in L^1(D)$ with $I^{n - \alpha} f, I^{n - \alpha} g \in AC(D)$ and $n − 1 < α \leq n$, then $$D^\alpha f(x) = D^\alpha g(x) \quad\iff\quad f (x) = g(x) + \sum_{k = 1}^n c_kx^{\alpha - k},$$ where $c_j$, $j = 1, 2, . . ., n$, are arbitrary constants.

Here $I^\alpha$ denotes the fractional integral $$\frac{1}{\Gamma (\alpha)}\int_0^x (x - t)^{\alpha - 1} f(t) dt$$

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  • $\begingroup$ Why the functions that you mention satisfy it? If $\alpha<1$, I would say only constant functions do ... $\endgroup$
    – LL 3.14
    Aug 27, 2021 at 4:07
  • $\begingroup$ @LL3.14 I have added some explanation. In general, the $D^\alpha (constant)$ is not equal to zero if $\alpha$ is not an integer. $\endgroup$
    – Azlif
    Aug 27, 2021 at 5:33

1 Answer 1

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Let $f$ be smooth and rapidly decaying* as $x \to \infty$ with $D^\alpha f =0$ i.e $$ 0 = \frac{d^n}{dx^n} \int_0^x \frac{f(t)dt}{(x-t)^{\alpha-n+1}}.$$ Integrating backwards $n$ times gives $$ \int_0^x \frac{f(t)dt}{(x-t)^{\alpha-n+1}} = \sum_{k=0}^{n-1} \frac{a_k}{k!} x^k$$ for some constants $a_k$, $k=0,\dots,n-1$. Next, take the Laplace transform of both sides: \begin{align*}\mathcal L \bigg \{\int_0^x \frac{f(t)dt}{(x-t)^{\alpha-n+1}} \bigg \}(s) &= \mathcal L\{f\}(s) \cdot \frac{\Gamma(n-\alpha)}{s^{n-\alpha}} \end{align*} and \begin{align*} \mathcal L \bigg \{ \sum_{k=0}^{n-1} \frac{a_k}{k!} x^k \bigg \}(s) &= \sum_{k=0}^{n-1} \frac{a_k }{s^{k+1}}. \end{align*} Thus, \begin{align*} \mathcal L\{f\}(s) &= \sum_{k=0}^{n-1} \frac{a_k }{\Gamma(n-\alpha)} \cdot \frac1{s^{\alpha-n+k+1}}. \end{align*} It follows that \begin{align*} f(x) &= \sum_{k=0}^{n-1} \frac{a_k}{\Gamma(n-\alpha)\Gamma(\alpha-n+k+1)} x^{\alpha-n+k}. \end{align*} Making a change of dummy variable $\ell = n-k$ and letting $$ c_\ell = \frac{a_{n-\ell}}{\Gamma(n-\alpha)\Gamma(\alpha-\ell+1)} $$ gives $$f(x) = \sum_{\ell=1}^n c_\ell x^{\alpha-\ell}$$ as required.

*I know you only assumed $f$ is continuous, but I'm not entirely convinced that this is a strong enough assumption. I feel at the very minimum you need some find of integrability assumption on $f$ and probably also some differentiability assumption as well. I've made these assumptions here just for simplicity.

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  • $\begingroup$ If $D^{\alpha} f = 0$, then $D^{\beta} f = 0$ for all $\beta=\alpha+k, k>0$, so I think that this kind of bootstrapping should lead to $f$ being really regular : regular enough to apply your arguments, I would think. Continuity is not a strong enough assumption for $D^{\alpha} f$ to exist : looking at it's expression , one is differentiating $n$ times and integrating $f$ only once, therefore regularity on $f$ will be expected! Your assumptions , in this light, seem optimal to me. $\endgroup$ Aug 30, 2021 at 11:35
  • $\begingroup$ Thank you for the response. Yes, continuity is of course not enough to ensure it has derivative at all. But I think the condition $D^\alpha f(x) = 0$ somehow gives us implication to the existence of $f$ derivative at a certain degree somehow. Just like the condition $f^{(n)}(x) = 0$ for all $x$, which implies $f$ has $n$-th derivative. $\endgroup$
    – Azlif
    Aug 31, 2021 at 6:48
  • $\begingroup$ What you are referring to here is a question of regularity for your equation which is an entirely different problem which would have to be treated separately. I also think it would be much harder than the original problem. $\endgroup$
    – JackT
    Sep 5, 2021 at 13:13
  • $\begingroup$ I should also say I don't agree with your statement that "$f^{(n)}(x)=0$ implies $f$ has an $n$-th derivative". In order to make sense of the statement "$f^{(n)}(x)=0$" you need to assume $f$ is $n$ times differentiable or come up with an appropriate weak formulation of the problem. $\endgroup$
    – JackT
    Sep 5, 2021 at 13:16

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