Existence of half derivative of 1

Question

I'm trying to find general properties for a half derivative operator, yet I encountered what seems to be a paradox. Let $H$ be a linear operator such that $H^2f = Df$ where $D$ is the first derivative. Then we have the following consequences

$$H0=0$$ $$H^31=DH1=HD1 \Rightarrow DH1 = H0=0 \Rightarrow H1 = c $$ $$H^2 x = 1 \Rightarrow H^3 x = c \Rightarrow Hx = cx + d \tag{1}\label{eq1}$$ where $c$ and $d$ are some constant functions. Consequently $$1 = H(Hx) = cHx + cd = c\left(Hx + d\right)\tag{2}\label{eq2}$$ $$\Rightarrow Hx = \frac{1}{c}-d\tag{3}\label{eq3}$$ By \eqref{eq3}, $c \neq 0$. However, by \eqref{eq1} and \eqref{eq3}, $c = 0$, which is contradiction.

So does this mean that $H1$ does not exist? Or is there something wrong with my calculations?

Answer 1

My understanding is that $H1=0$. But I don't think my explanation here is directly related to what you truely want to do.

The formula $H^2=D$ is a correct definition. However the domain of $H$ can be tricky.

If we consider the Fourier transform $(Df)^\wedge(\xi)=i\xi\hat f(\xi)$ (or $2\pi i\hat f(\xi)$ depending on the conventions). A standard fractional derivative is $(Hf)^\wedge(\xi)=\sqrt{i\xi}\hat f(\xi)$ where you can choose a branch of the square root for $\xi>0$ and $\xi<0$.

In this setting $\hat 1=\delta_0$ is the Dirac measure, it makes sense to say $\sqrt{|\xi|}\cdot\delta_0(\xi)=0$ as if you take any smooth approximation $f_\epsilon(\xi)\to\sqrt{|\xi|}$ and $g_\epsilon\to\delta_0$ (in the sense of measures) we always have $f_\epsilon(\xi)g_\epsilon(\xi)\to0$ in the sense of distributions.

On the other hand $\hat x=\delta_0'$ is the derivative of Dirac measure, and $\sqrt{i\xi}\cdot\delta_0'(\xi)$ is not defined because the similar approximation fails.

If you consider this way, that is to say $x\notin\operatorname{Dom}H$. Basically you cannot use the function $x$ in the argument.

In general if one define "function spaces with fractional derivative", we tend not to use such $H$, instead we use something like $(1-D^2)^{1/4}$ because its Fourier transform is smooth at 0. This is called Bessel potential, which works on all polynomials. By contrast, the Riesz potential $(-D)^{1/2}$ (not $D^{1/2}$) may have a domain problem.

Answer 2

After searching for a while, I've found the answer in this paper. $H(1)$ does indeed not exist. And this is not just the case for $H^2=D$ but for any linear operator $H$ that satisfies $H^a = D$ for $a > 1$. A fractional derivative with index law simply doesn't work on constant functions.

Answer 3

Problems

1. There are multiple problems and one of them is that $H$ has many difrent definitions.

2. We call $\operatorname{H}$ a semi derivative and $\operatorname{H}\left( \text{constant} \right)$ does not have to be a constant (e.g. see Riemann-Liouville operator of a constant).

3. Problem $3$ is that not all fractional derivatives are linear (you can quickly see this if you apply them to the series expansions of functions).

4. $\operatorname{D}_{x}^{\mu}\left[ \operatorname{D}_{x}^{\nu}\left[ f\left( x \right) \right] \right] = \operatorname{D}_{x}^{\nu}\left[ \operatorname{D}_{x}^{\mu}\left[ f\left( x \right) \right] \right]$ is not always true for fractional derivatives.

Problem $2$

Let's use the Riemann-Liouville operator. According it's definition (if $-b \in \mathbb{C} \setminus \mathbb{N}$): \begin{align*} \operatorname{^{RL}D}_{x}^{\alpha}\left[ a \cdot x^{b} \right] &= a \cdot \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - \alpha + 1 \right)} \cdot x^{b - \alpha}\\ \operatorname{^{RL}D}_{x}^{\alpha}\left[ \text{constant} \right] &= \lim\limits_{b \to 0}\left[ \operatorname{^{RL}D}_{x}^{\alpha}\left[ \text{constant} \cdot x^{b} \right] \right] = \lim\limits_{b \to 0}\left[ \text{constant} \cdot \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - \alpha + 1 \right)} \cdot x^{b - \alpha} \right]\\ \end{align*}

So: \begin{align*} H\left( \text{constant} \right) &= \lim\limits_{b \to 0}\left[ \text{constant} \cdot \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - \frac{1}{2} + 1 \right)} \cdot x^{b - \frac{1}{2}} \right]\\ H\left( \text{constant} \right) &= \frac{\text{constant}}{\Gamma\left( \frac{1}{2} \right)} \cdot x^{-\frac{1}{2}} = \frac{\text{constant}}{\sqrt{\pi \cdot x}}\\ \end{align*}

Is this ture? \begin{align*} H\left( H\left( \text{constant} \right) \right) &= \operatorname{^{RL}D}_{x}^{1}\left[ \text{constant} \right]\\ H\left( \frac{\text{constant}}{\Gamma\left( \frac{1}{2} \right)} \cdot x^{-\frac{1}{2}} \right) &= 0\\ \frac{\text{constant}}{\Gamma\left( \frac{1}{2} \right)} \cdot H\left( x^{-\frac{1}{2}} \right) &= 0\\ \frac{\text{constant}}{\Gamma\left( \frac{1}{2} \right)} \cdot \frac{\Gamma\left( -\frac{1}{2} + 1 \right)}{\Gamma\left( -\frac{1}{2} - \frac{1}{2} + 1 \right)} \cdot x^{-\frac{1}{2} - \frac{1}{2}} &= 0\\ 0 &= 0\\ \end{align*}

Yes (because $\frac{1}{\Gamma\left( 0 \right)} = 0$). Aka this a possible solution...

But it can also be a constant (e.g. using Caputo derivative: $H\left( \text{constant} \right) = 0$). They are not the same but satisfy the same equation.

Problem $3$

Let's say $D^{\alpha}$ is linear then $D_{x}^{\alpha}(f(x) + g(x)) = D_{x}^{\alpha}(f(x)) + D_{x}^{\alpha}(g(x))$ and $D_{x}^{1/2} = a \cdot \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - \alpha + 1 \right)} \cdot x^{b - \alpha}$. We know $D_{x}^{\alpha}( e^{x} ) = e^{x}$ but: \begin{align*} D_{x}^{\alpha}\sum_{n=0}^\infty\frac{x^n}{n!}=D_{x}^{\alpha}\sum_{n=0}^\infty e^{x}=\sum_{n=0}^\infty D_{x}^{\alpha}\frac{x^n}{n!}=\sum_{n=0}^\infty\frac{x^{n-\alpha}}{\Gamma(n-\alpha+1)}\neq\sum_{n=0}^\infty\frac{x^n}{\Gamma(n+1)} &= e^x\\ \end{align*}

The reason it is wrong is that the integration limits of $D_{x}^{\alpha}$ (in integral representation) are not all defined. Simple example $\alpha = -1$: \begin{align*} \int\limits_{0}^{x} u^{b}\, \operatorname{d}u &= \frac{1}{b + 1} \cdot x^{b + 1} = D_{x}^{-1}(x^{b})\\ \int\limits_{-\infty}^{x} e^{u}\, \operatorname{d}u = e^{x} &\wedge D_{x}^{-1}(e^{x}) = \int\limits_{0}^{x} e^{u}\, \operatorname{d}u = e^{x} - 1\\ \end{align*}

Problem $4$

Let's say that $H(f(x)) = \operatorname{^{RL}D}_{x}^{\frac{1}{2}}\left[ f\left( x \right) \right]$. If $f\left( x \right) = \text{constant}$ then \begin{align*} H( \frac{\operatorname{d} \text{constant}}{\operatorname{d}x} ) = H(0) = 0 \ne -\frac{\text{constant}}{2 \cdot \sqrt{\pi \cdot x^{3}}} = \frac{\operatorname{d} \frac{\text{constant}}{\sqrt{\pi \cdot x}}}{\operatorname{d}x} = \frac{\operatorname{d} H( \text{constant} )}{\operatorname{d}x}\\ \end{align*}