comparison property for fractional derivative I am studying fractional derivative. In the case of normal derivative, it is known that for $u,v \in C^1[0,1],$
if $u(0)=v(0)~\hbox{and} ~u'(0) < v'(0)$, then $u(t) < v(t)$ for all small $t>0$.
I was wondering if there exists a similar property for Riemann–Liouville fractional derivative $D_{0+}^{\alpha}$ with $\alpha \in (0,1)$ (see 1 for definition of Riemann-Liouville fractional derivative). Here, $$ D_{0+}^{\alpha} y(t):=\frac{1}{\Gamma(1-\alpha)}  \frac{d}{dt}  \left(\int_0^t \frac{y(s)}{(t-s)^{\alpha}} \, ds \right) \text{.}  $$
I would appreciate it if you could give any comments on my question.
 A: There is a comparison principle for this derivative. In fact, there are comparison principles for similar notions of fractional derivatives : they all roughly follow the same proof idea (even the proof for the usual derivative can be brought into this bracket).
Everything I'm writing here can be found in the THEORY OF FRACTIONAL DIFFERENTIAL INEQUALITIES AND
APPLICATIONS by Lakshmikantham and Vatsala. In fact, this is the first inequality considered in the given paper.
I'll take some liberty in emphasising the important parts of the proof and speaking about them. I think there are some errors in the presentation as well, so I'll make sure what I write is correct for sure!

The first result is quite interesting and is the heart of the matter. Let's take it to the usual derivative first : suppose that $f: [0,T] \to \mathbb R$ is differentiable, and satisfies $f(0)=f(T') = 0$ and $f(x)\leq 0$ on $[0,T']$ for some $T'<T$. You would expect $f'(T')\geq 0$, since the function is changing from a (possibly) negative value to the value $0$ as the argument approaches $T'$ from the left : so it's increasing near $T'$'s left. The result that $f'(T')\geq 0$ is quite easy to see and prove from the definition of the usual derivative.
This holds for the RL fractional derivative as well. First I'll explain how the proof works, then I'll explain how this goes on to establish the most basic comparison principle in the subject.

Lemma : Let $m : \mathbb R_+ \to \mathbb R$. Suppose that for some $t_1>0$ we have $m(t_1)=0$ and $m(t)\leq 0$ for $0 \leq t \leq t_1$. Suppose, furthermore, that $m$ is locally Hölder continuous of order $\lambda$ at the point $t_1$. Then, $D^\alpha m(t_1) \geq 0$ for any $\alpha \in (0,\lambda)$.

Proof : For the usual derivative, the proof goes via observing that the differential quotient along the left is non-negative for all small enough perturbations, and hence non-negative in the limit.
For this proof as well, we need to observe the differential quotient : not of $m$ itself, but of the integral appearing in the derivative.
So , by definition :
$$
D^{\alpha}m(t_1) = \frac 1{\Gamma(1-\alpha)} \frac{d}{dt}\Bigg|_{t = t_1}\left(\int_0^t \frac{m(s)}{(t-s)^\alpha}ds\right)
$$
we let $H(t) = \int_0^t \frac{m(s)}{(t-s)^\alpha}ds$. Naturally, we want the left hand derivative of $H$, like for the usual derivative, so for $h>0$, we do :
$$
H(t_1) - H(t_1-h) = \int_0^{t_1} \frac{m(s)}{(t_1-s)^\alpha}ds - \int_0^{t_1} \frac{m(s)}{(t_1-h-s)^\alpha}ds \\
= \int_0^{t_1-h} \left(\frac{1}{(t_1-s)^\alpha} -\frac{1}{(t_1-h-s)^\alpha} \right) m(s)ds + \int_{t_1-h}^{t_1} \frac{m(s)}{(t_1-s)^\alpha}ds
$$
The first of these integrals is non-negative. That's because on the interval $[0,t_1-h]$, we can easily see that $m(s) \leq 0$ and $\frac{1}{(t_1-s)^\alpha} -\frac{1}{(t_1-h-s)^\alpha} < 0$, so their product is non-negative. This leads to the inequality :
$$
H(t_1) - H(t_1-h) \geq \int_{t_1-h}^{t_1} \frac{m(s)}{(t_1-s)^\alpha}ds
$$
At this point, it is clear where the locally Hölder continuous hypothesis is going to come into play. Indeed, by local Hölder continuity, we know that there is a $K>0$ such that for all $s \in [t_1-h,t_1]$ we have $m(s) \geq -K (t_1-s)^{\lambda}$. Directly applying this bound gives us :$$
\int_{t_1-h}^{t_1} \frac{m(s)}{(t_1-s)^\alpha}ds \geq \int_{t_1-h}^{t_1} -K(t_1-s)^{\lambda-\alpha} ds = -K\frac{h^{\lambda-\alpha+1}}{\lambda-\alpha+1}
$$
Which therefore implies that :$$
H(t_1) - H(t_1-h) \geq -K\frac{h^{\lambda-\alpha+1}}{\lambda-\alpha+1} \implies \frac{H(t_1) - H(t_1-h)}{h} \geq -K\frac{h^{\lambda-\alpha}}{\lambda-\alpha+1}
$$
for all $h$ small enough. Letting $h$ go to zero sends the RHS of the above inequality to zero : so it follows that $H'(t_1) \geq 0$ and therefore that $$
D^{\alpha} m(t_1) = \frac{H'(t_1)}{\Gamma(1-\alpha)} \geq 0
$$
as desired. $\large\blacksquare$
We make the casual but important observation that if we set $\alpha = 0$ formally, then we are basically getting the usual derivative back. So the argument presented here overarches the argument for the usual derivative.

The kind of comparison principle that has been discussed in the question would have this natural interpretation : If $v(0)<w(0)$ and $D^{\alpha }v(0)< D^{\alpha} w(0)$ then there should exist a $t>0$ such that $v(x)<w(x)$ for $x \in [0,t]$.
I will be proving a similar comparison principle below, but I will briefly talk about why I don't expect the above to be true. The point is, that $D^{\alpha}w(t) >0$ implies that the derivative of the function $t \to \int_0^t \frac{w(s)}{(t-s)^{\alpha}}ds$ is positive. Put $\alpha=0$ and this is basically (following a use of FTC) reducing to $w$, so for the usual derivative, we obtain a direct connection between the function values and the derivative's sign.
However, I don't expect it to be true that if the function $t \to \int_0^t \frac{w(s)}{(t-s)^{\alpha}}ds$ is increasing for a particular $\alpha$, that $w(s)$ must be increasing as well. It's quite an arbitrary-looking condition (given that, in greater generality, there is nothing special about the function $\frac 1{(t-s)^{\alpha}}$), and counterexamples should be easily generated for this claim. SUch counterexamples would refute the possibility of any such "local" comparison principle.

What is true is this, though : on an entire interval, if one derivative dominates the other, then domination occurs over this interval. This is a modification of the very general Theorem 2.3 from the same paper.

Comparison principle for the $\alpha$ RL fractional derivative : Suppose that $v,w : [0,T] \to \mathbb R$ are locally Hölder continuous of order $\lambda$ everywhere. Suppose for some $\alpha<\lambda$ that $D^{\alpha}v(x)<D^{\alpha}w(x)$ for every $x \in [0,T]$. Then, $v(0)<w(0)$ implies $v(t)<w(t)$ on $[0,T]$.

Proof : Set $m(t) = v(t)- w(t)$. We know that $m$ is a continuous function. Suppose that $m(t)$ doesn't have the same sign through the interval : then, consider the point where $m$ first changes sign, which will be the point $t_1 = \inf\{t : m(t) = 0\}$. The fact that $t_1 \neq 0$ can be shown by continuity and the fact that $m(0)<0$.
But then, $m(t) \leq 0$ for $0 \leq t\leq t_1$, and $D^{\alpha}m(t) \leq 0$ (because $D^{\alpha}$ is linear , just like the usual derivative) and $\alpha<\lambda$. It follows that $D^{\alpha} m_1(t_1) \geq 0$, but then that contradicts the assumption that $D^{\alpha}v(t_1)<D^{\alpha}w(t_1)$.
It follows that $t_1$ doesn't exist i.e. that $m$ never changes sign. This is precisely the conclusion desired. $\blacksquare$

These comparison principles, having been established very early on in the paper, then go on to contribute to the theory of existence-uniqueness for such equations.  Similar computations for other forms of the fractional derivative can be found here.
It is also good to read the book titled "Theory and Applications of Fractional Differential Equations" by Kilbas, Srivastava, and Trujillo, where existence and uniqueness for FDEs are not proved via the method of Lakshmikantham, but rather (similar to the usual "classical" versus "distributional" contrast in PDEs) using abstract fixed-point methods and working with the fractional differential operator as a linear operator on a Sobolev space. This book is a nice, if tough, read, and will be very useful to get a look into the subject and its nuances better.
