Size without cancellation does not give a principal value

Since $k$ obeys the size bound with $A_1=1$ and $f={\mathbf{1}}_{B(0,1)}\in L^1\cap L^2$, [F2] applies, and polar coordinates [F1] give, for every $0<\epsilon <1$, $T_{\epsilon }f(0)={\int }_{\epsilon <\mathrm{\mid }y\mathrm{\mid }<1}\mathrm{\mid }y{\mathrm{\mid }}^{-n}dy={\int }_{\epsilon }^1{r}^{-n}\mathrm{\mid }{S}^{n-1}\mathrm{\mid }{r}^{n-1}dr=\mathrm{\mid }{S}^{n-1}\mathrm{\mid }{\int }_{\epsilon }^1\frac{dr}{r}=\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log \frac{1}{\epsilon },$ with $\mathrm{\mid }{S}^{n-1}\mathrm{\mid }>0$; the doubly truncated integral likewise equals $\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log (\min \{1,N\}\mathrm{/}\epsilon )$ for $0<\epsilon <\min \{1,N\}$, and it vanishes if $\epsilon \ge 1$; fixing $N=2$ and sending $\epsilon \downarrow 0$ proves $T^{\ast \ast }f(0)=+\mathrm{\infty }$.

Consequently ${\lim }_{\epsilon \downarrow 0}{T}_{\epsilon }f(0)=+\mathrm{\infty }$: the symmetric truncations do not converge at the origin, and the maximal truncated operator is infinite there, although every individual truncation is finite.

No principal-value distribution for $k$ exists along any sequence ${\delta }_j\downarrow 0$. Indeed, take the nonnegative Schwartz test $\phi (x)=e^{-\mathrm{\mid }x{\mathrm{\mid }}^2}$, with $\phi (0)=1$; by continuity of $\phi$ there is $\rho >0$ with $\phi \ge \phi (0)\mathrm{/}2$ on $B(0,\rho )$, so for every $\delta \in (0,\rho )$ polar coordinates give ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge \delta }k(x)\phi (x)dx\ge {\int }_{\delta <\mathrm{\mid }x\mathrm{\mid }<\rho }\mathrm{\mid }x{\mathrm{\mid }}^{-n}\frac{\phi (0)}{2}dx=\frac{\phi (0)}{2}\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log \frac{\rho }{\delta },$ which tends to $+\mathrm{\infty }$ as $\delta \downarrow 0$; hence the limit in [F3] fails for this $\phi$ and every sequence ${\delta }_j\downarrow 0$. Together with step 2.1 and [F2] this shows that the pointwise size condition by itself yields neither a principal-value distribution nor a finite maximal truncated operator, so principal-value cancellation is an additional requirement. In fact $k$ also satisfies Hörmander's condition: $\mathrm{\mid }\mathrm{\nabla }k(z)\mathrm{\mid }=n\mathrm{\mid }z{\mathrm{\mid }}^{-n-1}$ gives $\mathrm{\mid }k(x-y)-k(x)\mathrm{\mid }\le n{2}^{n+1}\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }x{\mathrm{\mid }}^{-n-1}$ on $\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }$, whose integral is bounded uniformly in $y$ by polar coordinates. Thus these kernel bounds alone do not assert a principal value or an associated $L^2$-bounded operator.