Let $K:\mathbb{R}^3\backslash\{0\}\times\mathbb{R}^3\backslash\{0\}\rightarrow\mathbb{C}$, such that $K(x,y)=K(y,x)$ and $K(x,y)=|x|^{-1}|y|^{-1}H(x,y)$, with $H$ locally bounded. Let $T$ be the (singular) integral operator with kernel $K$, i.e. $$T(f)(x)=\int_{\mathbb{R}^3}K(x,y)f(y)dy$$ Suppose that $T$ is unitary on $L^2(\mathbb{R}^3)$ and bounded from $w^{-1}L^1(\mathbb{R}^3)$ to $wL^{\infty}(\mathbb{R}^3)$, where $w(x)=1+|x|^{-1}$.

Can i deduce, by means of some interpolation tecnique, that $T$ is bounded from ${z_p}^{-1}L^p(\mathbb{R}^3)$ to ${z_p}L^q(\mathbb{R}^3)$? (with $1<p<2$, $1/p+1/q=1$and $z_p$ an appropriate weight function depending on $p$)

EDIT: As Christian pointed out, one cand find $z$ such that the operator with integral kernel $K(x,y)/(z(x)z(y))$ is bounded from $L^2$ to $L^2$ and from $L^1$ to $L^{\infty}$. Hence, a typical interpolation argument allows us to take $z_p=z$ for every $p$.

However, i'm interested about the dependence of an "optimal" $z_p$ on $p$. Hopefully, something like $z_p=1+|x|^{-\alpha(p)}$, with $\alpha(p)>0$ and decreasing to $0$ as $p\rightarrow 2$.

Thank you for any suggestion