Hardy-Littlewood-Sobolev for "componentwise product" of Riesz kernels

Question

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 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 $\|K_{(\alpha_1,\ldots,\alpha_d)}f_c\|_q=c^{-\sum_i\alpha_i-d/q}\|K_{(\alpha_1,\ldots,\alpha_d)}f\|_q$.

Answer 1

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.