On the dimension-free control of higher order truncated Riesz transforms by higher order Riesz transforms

Fix a positive integer $k$. Let $R_k$ be a higher order Riesz transform of order $k$ on ${\mathbb{R}}^d$ and let $R_k^t,$ $t>0,$ be the corresponding truncated Riesz transform. We study the relation between $\mathrm{\parallel }{R}_{k}f{\mathrm{\parallel }}_{L^p({\mathbb{R}}^d)}$ and $\mathrm{\parallel }{R}_k^{t}f{\mathrm{\parallel }}_{L^p({\mathbb{R}}^d)}$ for $p=1$, $p=\mathrm{\infty },$ and $p=2.$ We do this by analyzing the factorization operator $M_k^t$ defined by the relation $R_k^t=M_k^t{R}_k\mathrm{.}$ The operator $M_k^t$ is a convolution operator associated with an $L^1$ radial kernel $b_{k,d}^t(x)=t^{-d}{b}_{k,d}(x\mathrm{/}t),$ where $b_{k,d}(x):=b_{k,d}^1(x)\mathrm{.}$ We prove that $b_{k,d}\ge 0$ only for $k=1,2.$ We also show that for fixed $k\ge 3$, $$\underset{d\to \mathrm{\infty }}{\lim }\mathrm{\parallel }{b}_{k,d}{\mathrm{\parallel }}_{L^1({\mathbb{R}}^d)}=\mathrm{\infty }\mathrm{.}$$ This contrasts with the cases $k=1,2$, where it is known that $\mathrm{\parallel }{b}_{k,d}{\mathrm{\parallel }}_{L^1({\mathbb{R}}^d)}=1$. Finally, we show that for any positive integer $k$, the Fourier transform of $b_{k,d}$ is bounded in absolute value by $1.$ This implies the contractive estimate $$\mathrm{\parallel }{R}_k^{t}f{\mathrm{\parallel }}_{L^2({\mathbb{R}}^d)}\le \mathrm{\parallel }{R}_{k}f{\mathrm{\parallel }}_{L^2({\mathbb{R}}^d)}$$ and an analogous estimate for general singular integrals with smooth kernels for radial input functions $f\mathrm{.}$