A scale integral estimate for mean-zero kernels

The difference form and the pointwise ball estimate. For $t>0$ one has $\int {\psi }_t=0$ by the change of variables $y\mapsto ty$, so for a.e. $x$ the convolution equals $g\ast {\psi }_t(x)={\int }_{{\mathbb{R}}^d}(g(x-y)-g(x)){\psi }_t(y)dy$, the subtracted term $\int g(x){\psi }_t(y)dy$ vanishing; this is well defined for a.e. $x$ by [F1] and the a.e. finiteness of $g$. Since $\mathrm{\parallel }{\psi }_t{\mathrm{\parallel }}_{\mathrm{\infty }}=t^{-d}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{supp}{\psi }_t\subseteq B(0,c_{\psi }t)$, the absolute value is at most $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}{t}^{-d}{\int }_{B(0,c_{\psi }t)}\mathrm{\mid }g(x-y)-g(x)\mathrm{\mid }dy$, and Holder [F2] over that ball gives $\mathrm{\mid }g\ast {\psi }_t(x){\mathrm{\mid }}^p\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}^p{t}^{-dp}({\omega }_d{c}_{\psi }^d{t}^d{)}^{p-1}{\int }_{B(0,c_{\psi }t)}\mathrm{\mid }g(x-y)-g(x){\mathrm{\mid }}^{p}dy=C(\psi )t^{-d}{\int }_{B(0,c_{\psi }t)}\mathrm{\mid }g(x-y)-g(x){\mathrm{\mid }}^{p}dy$.

Integrating in $x$ and $t$. Integrating the bound of step 1.1 over $x$ and using Tonelli [F3] to exchange the $x$- and $y$-integrals gives $\mathrm{\parallel }g\ast {\psi }_t{\mathrm{\parallel }}_p^p\le C(\psi )t^{-d}{\int }_{\mathrm{\mid }y\mathrm{\mid }\le c_{\psi }t}{\int }_{{\mathbb{R}}^d}\mathrm{\mid }g(x-y)-g(x){\mathrm{\mid }}^{p}dxdy=C(\psi )t^{-d}{\int }_{\mathrm{\mid }y\mathrm{\mid }\le c_{\psi }t}{D}_g(\mathrm{\mid }y\mathrm{\mid },\widehat{y})dy$, where $D_g(r,\omega ):={\int }_{{\mathbb{R}}^d}\mathrm{\mid }g(x+r\omega )-g(x){\mathrm{\mid }}^{p}dx$ and $\widehat{y}:=y\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$. Multiplying by $t^{-p}$ and integrating over $t\in (0,T)$, another application of Tonelli [F3] gives ${\int }_0^T{t}^{-p}\mathrm{\parallel }g\ast {\psi }_t{\mathrm{\parallel }}_p^{p}dt\le C(\psi ){\int }_{{\mathbb{R}}^d}{D}_g(\mathrm{\mid }y\mathrm{\mid },\widehat{y})\left({\int }_0^T{\mathbf{1}}_{\{\mathrm{\mid }y\mathrm{\mid }\le c_{\psi }t\}}{t}^{-p-d}dt\right)dy$; the inner integral vanishes for $\mathrm{\mid }y\mathrm{\mid }>c_{\psi }T$ and is at most ${\int }_{\mathrm{\mid }y\mathrm{\mid }\mathrm{/}{c}_{\psi }}^{\mathrm{\infty }}{t}^{-p-d}dt=\frac{c_{\psi }^{p+d-1}}{p+d-1}\mathrm{\mid }y{\mathrm{\mid }}^{-(p+d-1)}$, so the left-hand side is at most $C^{\mathrm{\prime }}(d,p,\psi ){\int }_{{\mathbb{R}}^d}{D}_g(\mathrm{\mid }y\mathrm{\mid },\widehat{y})\mathrm{\mid }y{\mathrm{\mid }}^{-(p+d-1)}dy$, a bound independent of $T$.

Identifying the weight and invoking the coordinate-direction form. The substitution $y=x+h$ together with translation invariance of Lebesgue measure and Tonelli [F3] gives ${\int }_{{\mathbb{R}}^d}{D}_g(\mathrm{\mid }y\mathrm{\mid },\widehat{y})\mathrm{\mid }y{\mathrm{\mid }}^{-(p+d-1)}dy={\int }_{{\mathbb{R}}^d}\mathrm{\parallel }g(\cdot +h)-g(\cdot ){\mathrm{\parallel }}_p^p\mathrm{\mid }h{\mathrm{\mid }}^{-d-p\theta }dh=[g{]}_{\theta ,p}^p$, because $p+d-1=d+p\theta$ at $\theta =1-1\mathrm{/}p$. By the comparability clause of [F4], $[g{]}_{\theta ,p}^p\le C_1{\sum }_{i=1}^d{\int }_0^{\mathrm{\infty }}{h}^{-p}{\int }_{{\mathbb{R}}^d}\mathrm{\mid }g(\xi +h{e}_i)-g(\xi ){\mathrm{\mid }}^{p}d\xi dh$ (again $1+p\theta =p$). Step 2.1 first gives the bound by the seminorm. Combining with both directions of [F4] gives both displayed inequalities, with the second constant independent of $T$ and of $g$.