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I have a small questions concerning the fractional derivative of a test function. Is it true that if $u \in C^{\infty}_{c}(\mathbb{R})$ and we define the fractional derivative of this function as $(D^{\alpha}u)(x)=\mathcal{F}^{-1}( | \xi |^{\alpha} \hat{u}(\xi))(x)$ for $\alpha \in (0,2]$,

then $\int_{\mathbb{R}}{D^{\alpha}u(x) dx}=0$?

I somehow can't really prove it, any ideas would be nice!

Aga
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1 Answers1

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Note that $\int_{\mathbb{R}} v(x)\,\text{d}x = \hat{v}(0)$ by setting $\xi = 0$ in the definition of fourier transform of v. From the definition of fractional derivative you already know the fourier transform of $D^\alpha u$ so you can just plug in $\xi = 0$.

dioid
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