Let $d\in\mathbb N$ and $0<\alpha<d$. Define the Riesz kernel $K\_\alpha(x):=|x|^{\alpha-d}$, and the associated convolution operator
$$K\_\alpha f(x):=\int\frac{f(y)}{|x-y|^{d-\alpha}}~dy.$$
The classical [Hardy-Littlewood-Sobolev inequality](https://en.wikipedia.org/wiki/Sobolev_inequality#Hardy%E2%80%93Littlewood%E2%80%93Sobolev_lemma) states that we have the following boundedness property: for every $q>1$,
$$\|K\_\alpha f\|\_q\leq C\|f\|\_p$$
for some universal constant $C>0$ and $1/p=1/q+\alpha/d$.




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Now let $0<\alpha\_1,\ldots,\alpha\_d<1$, and suppose instead that we consider the convolution operator given by the kernel
$$K\_{(\alpha\_1,\ldots,\alpha\_d)}(x)=\prod\_{i=1}^d|x\_i|^{\alpha\_i-1},\qquad x=(x\_1,\ldots,x\_d),$$
which is what I call a "**componentwise product of Riesz Kernels**."



> 
> **Question.** Given $q>1$, does there exist some constant $C>0$ and a $p>1$ such that $$\|K\_{(\alpha\_1,\ldots,\alpha\_d)} f\|\_q\leq C\|f\|\_p?\tag{1}$$
> 
> 
> 


By a simple scaling argument, we can see that if (1) holds, then it must be for $1/p=1/q+\sum\_i\alpha\_i/d$. Indeed, letting $f\_c(x):=f(cx)$, then $\|f\_c\|\_p=c^{-d/p}\|f\|\_p$
and $\|K\_{(\alpha\_1,\ldots,\alpha\_d)}f\_c\|\_q=c^{-\sum\_i\alpha\_i-d/q}\|K\_{(\alpha\_1,\ldots,\alpha\_d)}f\|\_q$.



Hardy-Littlewood-Sobolev for "componentwise product" of Riesz kernels
I realise I was too optimistic in my comment: no such ineqaulity holds in general. Indeed, consider $f(x) = \prod\_j f\_j(x\_j)$. Then
$$ K\_{(\alpha\_1,\ldots,\alpha\_d)} f(x) = \prod\_j K\_{\alpha\_j} f\_j(x\_j) $$
and so
$$ \|f\|\_p = \prod\_j \|f\_j\|\_p , \qquad \|K\_{(\alpha\_1,\ldots,\alpha\_d)} f\|\_q = \prod\_j \|K\_{\alpha\_j} f\_j\|\_q $$
So by the usual Hardy–Littlewood—Sobolev inequality (the "only if" part), in order to get the bound we are interested in, we would need
$$ \frac{1}{p} = \frac{1}{q} + \frac{\alpha\_j}{1} , $$
that is,
$$ \alpha\_1 = \alpha\_2 = \ldots = \alpha\_n \qquad \text{and} \qquad \frac{1}{p} = \frac{1}{q} + \frac{\alpha}{d},$$
with $\alpha = \alpha\_1 + \alpha\_2 + \ldots + \alpha\_d$. In this case, the "if part" of the Hardy–Littlewood—Sobolev inequality implies that indeed the desired inequality holds.


