The classical one-dimensional case is: for $\alpha\in\langle0,1\rangle$ and $1<p<q<\infty$ such that $1/q=1/p-\alpha$, there is a constant $C\_p>0$ depending on $p$ such that
$$\|I\_\alpha f\|\_{L^q}\leq C\_p \|f\|\_{L^p}\, .$$


My question is: does a similar result hold for Fourier multiplier operator with symbol $\left(i\xi\right)^{-\alpha}$ (i.e. ${\cal A}\_{\left(i\xi\right)^{-\alpha}}(u) = {\cal F}^{-1}\left( (i\xi)^{-\alpha}{\cal F}(u) \right)$)?


In the case of Riesz potential, we have absolute values in the symbol, but maybe the analogous result holds in this special one-dimensional case? I could not find it.



A variant of the Hardy-Littlewood-Sobolev inequality in one dimensional case
Yes. If $J\_\alpha$ is the operator with Fourier symbol $(i \xi)^{-\alpha}$, $\alpha \in (0, 1)$, then the convolution kernel of $J\_\alpha$ is $v\_\alpha(x) = c\_\alpha x^{\alpha-1} \mathbb{1}\_{(0, \infty)}(x)$ (or $v\_\alpha(-x)$, depending on your choice of parameters for the Fourier transform), and so $j\_\alpha$ is bounded by a constant times the convolution kernel $u\_\alpha(x) = c\_\alpha |x|^{\alpha-1}$ of the Riesz potential operator. Thus $$\bigl\|J\_\alpha f\bigr\|\_q \leqslant \bigl\|J\_\alpha |f|\bigr\|\_q \leqslant c\_\alpha \bigl\|I\_\alpha |f|\bigr\|\_q \leqslant c\_{\alpha,p} \bigl\||f|\bigr\|\_p = c\_{\alpha,p} \bigl\|f\bigr\|\_p .$$ The same is true for more general non-symmetric homogeneous kernels, also in higher dimensions, as long as they are bounded pointwise by the Riesz kernel.


