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 $1/p=1/q+\alpha/d$.

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