Let $0<\alpha<1$ and define


$$Tf(x):=\int e^{\dot{\imath} x y} \frac{f(y)}{|x-y|^{\alpha}}dy.$$


The Hardy-Littlewood-Sobolev inequality characterizes $L^p-L^q$ boundedness of
$Hf(x):=\int \frac{f(y)}{|x-y|^{\alpha}}dy$ which majorizes $|Tf(x)|$ by the triangle inequality, when $f>0$. It asserts that
$$ \|Tf\|\_{q}\leq C \|f\|\_{p}\qquad (1)$$
where $C$ is a constant independent of $f$,
if and only if
$$1<p,q<\infty\quad\frac{1}{p}-\frac{1}{q}=1-\alpha\qquad (2)$$
The necessity of (2) for (1) can be easily deduced by scaling.


My question:


Is (2) necessary for the inequality
$$ \|Tf\|\_{q}\leq C\_{1} \|f\|\_{p}\qquad (3)\qquad ?$$


It seems that $T$ does not have any scaling properties. I cannot come up with counterexamples that show that (3)
is false outside the range of exponents described by (2).
Is the boundedness of $T$ known ?



$L^p-L^q$ boundedness of this simple singular oscillatory integral operator
Let me consider instead the operator
$$Sf(x):=\int e^{-i x y} \frac{f(y)}{|x-y|^{\alpha}}dy$$
which has the same properties as $T$ since $Tf(x)=Sf(-x)$. Writing
$$e^{-i x y}=e^{i|x-y|^{2}/2}e^{-i|x|^{2}/2}e^{-i|y|^{2}/2}$$
the operator $S$ can be written
$$Sf(x)=e^{-i|x|^{2}/2}\int \frac{e^{i|x-y|^{2}/2}}{|x-y|^{a}}
 e^{-i|y|^{2}/2}f(y)dy.$$
Thus $T:L^{q}\to L^{p}$ is bounded iff convolution with the oscillating kernel $|x|^{-a}e^{i|x|^{2}/2}$ is bounded. This kind of kernel has been studied extensively, for instance in: P.Sjolin, Convolution with Oscillating Kernels, Indiana University Mathematics Journal Vol. 30, No. 1 (January–February, 1981), pp. 47-55.
The range of values of $p,q$ is larger than (2), e.g. Sjolin proves boundedness $L^{p}\to L^{p}$ for $p\in[p\_{0},p\_{0}']$ where $p\_{0}=\frac{2n}{n+a}$, provided $0\le a<n$. Thus the answer to your question seems to be: no, condition (2) is not necessary for boundendess of $T$.


