Some Calderón–Zygmund kernels and their relations to Wolff capacities and rectifiability

Abstract

We consider the Calderón–Zygmund kernels $K_ {\alpha ,n}(x)=(x_i^{2n-1}/|x|^{2n-1+\alpha })_{i=1}^d$ in ${\mathbb R}^d$ for $0<\alpha \le 1$ and $n\in \mathbb {N}$. We show that, on the plane, for $0<\alpha <1$, the capacity associated to the kernels $K_{\alpha ,n}$ is comparable to the Riesz capacity $C_{\frac{2}{3}(2-\alpha ),\frac{3}{2}}$ of non-linear potential theory. As consequences we deduce the semiadditivity and bilipschitz invariance of this capacity. Furthermore we show that for any Borel set $E\subset {\mathbb R}^d$ with finite length the $L^2(\mathcal {H}^1 \lfloor E)$-boundedness of the singular integral associated to $K_{1,n}$ implies the rectifiability of the set E. We thus extend to any ambient dimension, results previously known only in the plane.

Appendix 1: Proofs of Propositions 3.1 and 3.3

For simplicity we let n odd. Then for $0<\alpha \le 1$

$$\begin{aligned} K^i_{\alpha ,n}(x)=\frac{x_i^{n}}{|x|^{n+\alpha }}, \quad x =(x_1,\ldots ,x_d) \in {\mathbb R}^d{\setminus }\{0\}. \end{aligned}$$

Proof of Proposition 3.1

Write $a=y-x$ and $b=z-y$; then $a+b=z-x$. Without loss of generality we can assume that $|a|\le |b|\le |a+b|$. A simple computation yields

$$\begin{aligned} \begin{aligned}&p^i_{\alpha ,n}(x,y,z)\\&\quad = K^i_{\alpha ,n}(x-y)\,K^i_{\alpha ,n}(x-z) + K^i_{\alpha ,n}(y-x)\,K^i_{\alpha ,n}(y-z) + K^i_{\alpha ,n}(z-x)\,K^i_{\alpha ,n}(z-y)\\&\quad = K^i_{\alpha ,n}(-a)\,K^i_{\alpha ,n}(-a-b) + K^i_{\alpha ,n}(a)\,K^i_{\alpha ,n}(-b) + K^i_{\alpha ,n}(a+b)\,K^1_{\alpha ,n}(b)\\&\quad =K^i_{\alpha ,n}(a+b)K^i_{\alpha ,n}(a)+K^i_{\alpha ,n}(a+b)K^i_{\alpha ,n}(b)-K^i_{\alpha ,n}(a)K^i_{\alpha ,n}(b)\\&\quad =\frac{(a_i+b_i)^na_i^n|b|^{n+\alpha }+(a_i+b_i)^nb_i^n|a|^{n+\alpha }-a_i^nb_i^n|a+b|^{n+\alpha }}{|a|^{n+\alpha }|b|^{n+\alpha }|a+b|^{n+\alpha }}. \end{aligned} \end{aligned}$$

(1.32)

If $a_ib_i=0$ the proof is immediate. Take for example $a_i=0$. Then we trivially obtain

$$\begin{aligned} p^i_{\alpha ,n}(x,y,z)=\frac{b_i^{2n}}{|b|^{n+\alpha }|a+b|^{n+\alpha }} \approx \frac{ \ M_i^{2n}}{L(x,y,z)^{2\alpha +2n}}. \end{aligned}$$

(1.33)

To prove the upper bound inequality in (3.8) we distinguish two cases.

Case $a_ib_i>0:$ Without loss of generality assume $a_i> 0$ and $b_i> 0$. In case $a_i<0$ and $b_i<0$,

$$\begin{aligned} \begin{aligned}&(a_i+b_i)^na_i^n|b|^{n+\alpha }+(a_i+b_i)^nb_i^n|a|^{n+\alpha }-a_i^nb_i^n|a+b|\\&\quad \quad =(|a_i|+|b_i|)^n|a_i|^n|b|^{n+\alpha }+(|a_i|+|b_i|)^n|b_i|^n|a|^{n+\alpha }-|a_i|^n|b_i|^n|a+b|^{n+\alpha } \end{aligned} \end{aligned}$$

and thus it can be reduced to the case where both coordinates are positive.

Notice that since $|a|\le |b|\le |a+b|$, $a_i\le |a|$ and $0<\alpha <1$,

$$\begin{aligned} p^i_{\alpha ,n}(x,y,z)= & {} \frac{(a_i+b_i)^nb_i^n}{|b|^{n+\alpha }|a+b|^{n+\alpha }}+\frac{a_i^n\left( (a_i+b_i)^n|b|^{n+\alpha }-b_i^n|a+b|^{n+\alpha }\right) }{|a|^{n+\alpha }|b|^{n+\alpha }|a+b|^{n+\alpha }}\\&\le \frac{1}{|b|^{\alpha }|a+b|^{\alpha }}+\frac{(a_i+b_i)^n-b_i^n}{|a|^\alpha |a+b|^{n-\alpha }}\\&\le \frac{1}{|b|^{\alpha }|a+b|^{\alpha }}+\frac{a_i^\alpha }{|a|^\alpha }\frac{\sum _{k=1}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) a_i^{k-\alpha }b_i^{n-k}}{|b|^{n+\alpha }}\\&\le \frac{1}{|b|^{\alpha }|a+b|^{\alpha }}+\frac{\sum _{k=1}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) |a|^{k-\alpha }|b|^{n-k}}{|b|^{n+\alpha }}\\&\le \frac{1}{|b|^{\alpha }|a+b|^{\alpha }}+\frac{B(n)|b|^{n-\alpha }}{|b|^{n+\alpha }} \le \frac{B(n,\alpha )}{|a+b|^{2\alpha }}, \end{aligned}$$

where the last inequality comes from $|a+b|\le 2|b|$, which follows from the triangle inequality and the fact that $|a|\le |b|$.

Case $a_i b_i <0:$ Without loss of generality we can assume that $a_i<0$, $b_i> 0$ and $b_i\le |a_i|$, the other cases follow analogously by interchanging the roles of $a_i$ and $b_i$.

$$\begin{aligned} \begin{aligned} p^i_{\alpha ,n}(x,y,z)&= \frac{(|a_i|-b_i)^n|a_i|^n|b|^{n+\alpha }-(|a_i|-b_i)^nb_i^n|a|^{n+\alpha }+|a_i|^nb_i^n|a+b|^{n+\alpha }}{|a|^{n+\alpha }|b|^{n+\alpha }|a+b|^{n+\alpha }}\\&\le \frac{(|a_i|-b_i)^n|a_i|^n|b|^{n+\alpha }+|a_i|^nb_i^n|a+b|^{n+\alpha }}{|a|^{n+\alpha }|b|^{n+\alpha }|a+b|^{n+\alpha }}\\&\le \frac{|a|^{2n}(|b|^{n+\alpha }+|a+b|^{n+\alpha })}{|a|^{n+\alpha }|b|^{n+\alpha }|a+b|^{n+\alpha }}=|a|^{n-\alpha }\left( \frac{1}{|a+b|^{n+\alpha }}+\frac{1}{|b|^{n+\alpha }}\right) \\&\le \frac{2^{2\alpha }+1}{|a+b|^{2\alpha }}. \end{aligned} \end{aligned}$$

since $b_1\le |a_1|$, $0<|a_1|-b_1<|a_1|<|a|$ and $|a|\le |b|\le |a+b|$.

We now prove the lower bound estimate in (3.8).

Case $a_ib_i > 0:$ As explained in the proof of the upper bound inequality the proof can be reduced to the case when $a_i>0$ and $b_i>0$. Setting $t=|b|/|a|$ in (1.32) and noticing that $|a+b|/|a| \le 1+t$ we get

$$\begin{aligned} {p^i_{\alpha ,n}}(x,y,z) \ge \frac{a_i^n(a_i+b_i)^nt^{n+\alpha }-b_i^n a_i^n (1+t)^{n+\alpha }+b_i^n (a_i+b_i)^n }{|b|^{n+\alpha }|a+b|^{n+\alpha }}:=\frac{f_1(t)}{|b|^{n+\alpha }|a+b|^{n+\alpha }}. \end{aligned}$$

(1.34)

Then it readily follows that the unique zero of

$$\begin{aligned} f'_1(t)=a_i^n(n+\alpha )(t^{n+\alpha -1} (a_i+b_i)^n-b_i^n (1+t)^{n+\alpha -1}) \end{aligned}$$

is

$$\begin{aligned} t^*=\frac{1}{\left( \frac{a_i}{b_i}+1 \right) ^{\frac{n}{n+\alpha -1}}-1}>0. \end{aligned}$$

Moreover $f_1$ attains its minimum at $t^*$ because $f_1'(0)=-(n+\alpha )a_i^n b_i^n$ and $\lim _{t \rightarrow \infty } f_1'(t)=\lim _{t \rightarrow \infty } ((b_i+a_i)^n-b_i^n)t^{n+\alpha -1}=+\infty$.

We first consider the case when $t^*>1$. Then we deduce that

$$\begin{aligned} 0<\frac{a_i}{b_i}<2^{\frac{n + \alpha -1}{n}}-1<1, \end{aligned}$$

(1.35)

the last inequality coming from $\alpha <1$. Therefore $a_i<b_i$. Setting $s=a_i/b_i$ we obtain

$$\begin{aligned} f_1(t) =b_i^{2n} \left( s^n(1+s)^nt^{n+\alpha }-s^n (1+t)^{n+\alpha }+(s+1)^n\right) \end{aligned}$$

and it follows easily that

$$\begin{aligned} f_1(t^*) =b_i^{2n} (1+s)^n \left( 1-\frac{s^n}{((s+1)^{\frac{n}{n+\alpha -1}}-1)^{n+\alpha -1}} \right) . \end{aligned}$$

A direct computation shows that the function

$$\begin{aligned} g_1(s)=1-\frac{s^n}{\left( (s+1)^{\frac{n}{n+\alpha -1}}-1\right) ^{n+\alpha -1}} \end{aligned}$$

is decreasing. Then, by (1.35), $g_1$ attains its minimum at $s=2^{\frac{n + \alpha -1}{n}}-1$. Therefore

$$\begin{aligned} f_1(t) \ge f_1(t^*) \ge b_i^{2n} \left( 1- \left( 2^{\frac{n + \alpha -1}{n}}-1\right) ^n\right) :=b_i^{2n}A_1(n,\alpha ) \end{aligned}$$

(1.36)

and since $\alpha <1$, $A_1(n, \alpha ) >0.$

We now consider the case when $t^*\le 1$ and notice that as $t \ge 1$ we have $f_1(t) \ge f_1(1)$. As before for $s=\min \{a_i, b_i\} /\max \{a_i, b_i\}$

$$\begin{aligned} \begin{aligned} f_1(t) \ge f_1(1)&= a_i^n (a_i+b_i)^n-a_i^n b_i^n 2^{n+\alpha }+b_i^n (a_i+b_i)^n \\&=(a_i+b_i)^n (\max \{a_i,b_i\})^n \left( s^n-\left( \frac{s}{s+1} \right) ^n 2^{n+\alpha }+1 \right) \\&:=(a_i+b_i)^n (\max \{a_i,b_i\})^n g_2(s). \end{aligned} \end{aligned}$$

(1.37)

It follows easily that the only non-zero root of

$$\begin{aligned} g_2'(s)=n s^{n-1}\left( 1-\frac{2^{n+\alpha }}{(1+s)^{n+1}}\right) \end{aligned}$$

is

$$\begin{aligned} s^*= 2^{\frac{n+\alpha }{n+1}}-1. \end{aligned}$$

Since $\alpha \in (0,1)$, then $2^{\frac{n+\alpha -1}{n}}-1<s^*<1.$ Furthermore notice that $g_2'(2^{\frac{n+\alpha -1}{n}}-1)<0$ and $g_2'(1)>0$. Hence $g_2$ attains its minimum at $s^*$. Therefore

$$\begin{aligned} \begin{aligned} g_2(s)\ge g_2(s^*)&=\left( 2^{\frac{n+\alpha }{n+1}}-1\right) ^n \left( 1-\frac{2^{n+\alpha }}{2^{\frac{(n+\alpha )n}{n+1}}} \right) +1 \\&=1-\left( 2^{\frac{n+\alpha }{n+1}}-1\right) ^{n+1}:=A_2(n,\alpha )>0, \end{aligned} \end{aligned}$$

(1.38)

the positivity of the constant $A_2(n,\alpha )$ coming from inequality $\alpha <1$. Therefore (1.34) together with (1.36), (1.37) and (1.38) imply that

$$\begin{aligned} {p^i_{\alpha ,n}}(x,y,z) \ge A(n,\alpha ) \frac{M_i^{2n}}{L(x,y,z)^{2n+2a}}, \end{aligned}$$

(1.39)

for some positive constant $A(n,\alpha )$. Hence we have finished the proof when $a_i \, b_i > 0$.

Case $a_i \, b_i < 0:$ Setting $t=|b| /|a|$ and using (1.34) we get that

$$\begin{aligned} \begin{aligned} {p^i_{\alpha ,n}}(x,y,z)&=\frac{1}{|b|^{n+\alpha }|a+b|^{n+\alpha }}(a_i^n(a_i+b_i)^nt^{n+\alpha }-b_i^n a_i^n (1+t)^{n+\alpha }+b_i^n (a_i+b_i)^n ) \\&\ge \frac{1}{|b|^{n+\alpha }|a+b|^{n+\alpha }}(a_i^n(a_i+b_i)^nt^{n+\alpha }-b_i^n a_i^n t^{n+\alpha }+b_i^n (a_i+b_i)^n )\\&:= \frac{f_2(t)}{|b|^{n+\alpha }|a+b|^{n+\alpha }}. \end{aligned} \end{aligned}$$

(1.40)

Notice that $f_2$ is an increasing function because $a_i^2+a_ib_i>a_ib_i$ and n is odd:

$$\begin{aligned} \begin{aligned} f'_2(t)&=(n+\alpha )t^{n+\alpha -1}(a_i^n(a_i+b_i)^n-a_i^nb_i^n)=(n+\alpha )t^{n+\alpha -1}((a_i^2+a_ib_i)^n-a_i^nb_i^n) > 0. \end{aligned} \end{aligned}$$

Therefore since $t \ge 1$ we have that

$$\begin{aligned} f_2(t) \ge f_2(1)=(a_i^n+b_i^n)(a_i+b_i)^n-b_i^n a_i^n. \end{aligned}$$

We assume that $|b_i| \ge |a_i|$, the case where $|b_i| <|a_i|$ can be treated in the exact same manner. We first consider the case where $a_i>0$ and $b_i<0$. Let

$$\begin{aligned} h(r)=(r-|b_i|)^n(r^n-|b_i|^n)+r^n|b_i|^n. \end{aligned}$$

Then

$$\begin{aligned} h'(r)=n\left( (r-|b_i|)^{n-1}(r^n-|b_i|^n)+r^{n-1}((r-|b_i|)^n+|b_i|^n)\right) . \end{aligned}$$

Notice that

$$\begin{aligned} h'(|b_i|/2)=0. \end{aligned}$$

Furthermore

$$\begin{aligned} h'(r)>0 \text{ for } |b_i|/2 <r\le |b_i|. \end{aligned}$$

To see this notice that since $0<|b_i|-r<r$, then $(|b_i|-r)^{n-1} <r^{n-1}$.Therefore since $r^n-|b_i|^n<0$

$$\begin{aligned} h'(r)>n(r^{n-1}(r^n-|b_i|^n)+r^{n-1}((r-|b_i|)^n+|b_i|^n))= nr^{n-1}(r^n-(|b_i|-r)^n)>0. \end{aligned}$$

With an identical argument one sees that $h'(r)<0$ for $0<r \le |b_i|/2$. Hence it follows that, for $0<r\le |b_i|$,

$$\begin{aligned} h(r) \ge h(|b_i|/2) \ge \frac{|b_i|^{2n}}{2^n}. \end{aligned}$$

Since $a_i \in (0, |b_i|]$ we get that $f_2(1) \ge \frac{|b_i|^{2n}}{2^n}$ and by (1.40)

$$\begin{aligned} {p^i_{\alpha ,n}}(x,y,z) \ge 2^{-n} \frac{b_i^{2n}}{|b|^{n+\alpha }|a+b|^{n+\alpha }} \ge A_3(n) \frac{M_i^{2n}}{L(x,y,z)^{2n+2\alpha }}. \end{aligned}$$

(1.41)

The case where $a_i <0$ and $b_i>0$ is very similar. In this case instead of the function h we consider the function $l(r)=(r+b_i)^n(r^n+b_i^n)-r^nb_i^n$ for $-|b_i|/2\le x <0$ and we show that in that range,

$$\begin{aligned} l(r) \ge l(-|b_i|/2) \ge b_i^{2n}/2^n. \end{aligned}$$

Therefore as $a_i \in [-|b_i|,0)$, $f_2(1) \ge \frac{|b_i|^{2n}}{2^n}$ and we obtain from (1.40)

$$\begin{aligned} {p^i_{\alpha ,n}}(x,y,z) \ge 2^{-n} \frac{b_i^{2n}}{|b|^{n+\alpha }|a+b|^{n+\alpha }} \ge A_3(n) \frac{M_i^{2n}}{L(x,y,z)^{2n+2\alpha }}. \end{aligned}$$

(1.42)

Therefore the proof of the lower bound follows by (1.39), (1.41) and (1.42). $\square$

Remark 1

Notice that in the proof of the lower bound inequality when $a_ib_i <0$, we do not make use of the fact that $\alpha <1$. Therefore (1.41) and (1.42) remain valid in the case where $\alpha =1$.

Proof of Proposition 3.3

For simplicity we let $p^i_{1,n}:=p^i_n$ for $i=1,\ldots ,d$. Let $a=y-x, b=z-y$ then $a+b=z-x$ and without loss of generality we can assume that $|a|\le |b|\le |a+b|=1$. In case $x_i=y_i=z_i$, then trivially by (1.32), $p^i_n(x,y,z)=0$. Hence we can assume that $a_i \ne 0$ or $b_i \ne 0$ and, by (1.32), for $\alpha =1$ , assuming without loss of generality that $b_i \ne 0$, we get

$$\begin{aligned} p^i_{n}(x,y,z)= \frac{(a_i+b_i)^n b_i^n \left( \left( \frac{a_i}{b_i} \right) ^n |b|^{n+1}+|a|^{n+1}-\frac{a_i^n}{(a_i+b_i)^n}\right) }{|a|^{n+1}|b|^{n+1}}. \end{aligned}$$

(1.43)

If the points x, y, z are collinear then the initial assumption $|a|\le |b|\le |a+b|$ implies that $|a|+|b|=|a+b|$. Furthermore $b=\lambda a$ for some ${\lambda }\ne 0$. We provide the details in the case when $\lambda >0$ as the remaining case is identical. We have by (1.43)

$$\begin{aligned} \begin{aligned} p^i_{n}(x,y,z)&=\frac{(a_i+\lambda a_i)^n \lambda ^n a_i^n \left( \left( \frac{1}{\lambda } \right) ^n \lambda ^{n+1} |a|^{n+1}+|a|^{n+1}-\left( \frac{1}{1+{\lambda }}\right) ^n\right) }{|a|^{n+1}|b|^{n+1}} \\&=\frac{a_i^{2n} \lambda ^n \left( \left( (1+{\lambda })|a| \right) ^{n+1}-1\right) }{|a|^{n+1}|b|^{n+1}}\\&=\frac{a_i^{2n} \lambda ^n }{|a|^{n+1}|b|^{n+1}}\left( (1+{\lambda })|a|-1\right) \sum _{j=0}^{n}((1+{\lambda })|a|)^j=0 \end{aligned} \end{aligned}$$

because $(1+{\lambda })|a|-1=|a|+|b|-1=0$.

We will now turn our attention to the case when the points x, y, z are not collinear. We will consider several cases.

Case $a_ib_i>0.$ As in the proof of Proposition 3.8 we only have to consider the case when $a_i,b_i>0$. We first consider the subcase $0<|a|\le |b|<|a+b|=1$.

Setting $w=a_i/b_i$ in (1.43) we get

$$\begin{aligned} p^i_{n}(x,y,z)=\frac{(a_i+b_i)^n b_i^n}{|a|^{n+1}|b|^{n+1}} f(w) \end{aligned}$$

with

$$\begin{aligned} f(w)=w^n|b|^{n+1}+|a|^{n+1}-\left( 1+\frac{1}{w}\right) ^{-n}. \end{aligned}$$

Notice that the only non-vanishing admissible root of the equation

$$\begin{aligned} f'(w)=nw^{n-1}\left( |b|^{n+1}-\left( \frac{1}{w+1}\right) ^{n+1} \right) =0 \end{aligned}$$

is $w=|b|^{-1}-1$. Furthermore it follows easily that

$$\begin{aligned} \lim _{w\rightarrow 0^+}f(w)=|a|^{n+1}>0 \quad \text {and}\quad \lim _{w \rightarrow +\infty }f(w)=+\infty \end{aligned}$$

hence $f:(0,\infty ) \rightarrow {\mathbb R}$ attains its minimum at $|b|^{-1}-1$. After a direct computation we get that

$$\begin{aligned} f(|b|^{-1}-1)=|a|^{n+1}-(1-|b|)^{n+1}. \end{aligned}$$

We can now write,

$$\begin{aligned} \begin{aligned} |a|^{n+1}&-(1-|b|)^{n+1}=|a|^{n+1}\left( 1-\left( \frac{1-|b|}{|a|}\right) ^{n+1} \right) \\&=|a|^{n+1}\left( 1- \frac{1-|b|}{|a|}\right) \sum _{j=0}^n \left( \frac{1-|b|}{|a|} \right) ^j=|a|^n(|a|-1+|b|) \sum _{j=0}^n \left( \frac{1-|b|}{|a|} \right) ^j. \\ \end{aligned} \end{aligned}$$

Therefore

$$\begin{aligned} p^i_{n}(x,y,z)\ge \frac{(a_i+b_i)^n b_i^n}{|a|^{n+1}|b|^{n+1}}|a|^n (|a|-1+|b|)\sum _{j=0}^n \left( \frac{1-|b|}{|a|} \right) ^j. \end{aligned}$$

(1.44)

Recall that, by Heron’s formula, the area of the triangle determined by $x,y,z \in {\mathbb R}^d$ is given by

$$\begin{aligned} \text {area}(T_{x,y,z})=\frac{1}{2} \sqrt{|a+b|^2|a|^2-\left( \frac{|a+b|^2+|a|^2-|b|^2}{2}\right) ^2}, \end{aligned}$$

where $a=y-x, b=z-y$ and $a+b=z-x$. Hence

$$\begin{aligned}&16 \, \text {area}(T_{x,y,z})^2=(2|a+b||a|-(|a+b|^2+|a|^2\\&\quad -|b|^2))(2|a+b||a|+|a+b|^2+|a|^2-|b|^2). \end{aligned}$$

Plugging this identity into Menger’s curvature formula we get

$$\begin{aligned} \begin{aligned} c^2(x,y,z)&=\frac{16 \, \text {area}(T_{x,y,z})^2}{|a|^2|b|^2|a+b|^2}\\&=\frac{(2|a+b||b|-|a+b|^2-|a|^2+|b|^2)(2|a+b||b|+|a+b|^2+|a|^2-|b|^2)}{|a|^2|b|^2|a+b|^2}\\&=\frac{(|b|^2-(|a+b|-|a|)^2)((|a|+|a+b|)^2-|b|^2)}{|a|^2|b|^2|a+b|^2}\\&=\frac{(|b|-|a+b|+|a|)(|b|+|a+b|-|a|)(|a|+|a+b|-|b|)(|b|+|a+b|+|a|)}{|a|^2|b|^2|a+b|^2} \end{aligned} \end{aligned}$$

and since we are assuming $|a+b|=1$,

$$\begin{aligned} c^2(x,y,z)=\frac{(|b|+|a|-1)(|b|+1-|a|)(|a|+1-|b|)(|b|+1+|a|)}{|a|^2|b|^2}. \end{aligned}$$

(1.45)

By (1.44) and (1.45) we get that

$$\begin{aligned} \begin{aligned}&p^i_{n}(x,y,z)\ge \frac{(a_i+b_i)^n b_i^n}{|a|^{n+1}|b|^{n+1}}\frac{|a|^n|a|^2|b|^2}{(|b|+1-|a|)(|a|+1-|b|)(|b|+1+|a|)}\\ {}&\quad \sum _{j=0}^n \left( \frac{1-|b|}{|a|} \right) ^j c^2(x,y,z)\ge \frac{b_i^{2n}|a||b|}{|b|^n(|b|+1-|a|)(|a|+1-|b|)(|b|+1+|a|)}c^2(x,y,z), \end{aligned} \end{aligned}$$

the last inequality coming from $a_i+b_i\ge b_i$ and the fact that the sum above is $>$1. Using the triangle inequality, $1=|a+b|\le |a|+|b|$, and the fact that $1=|a+b|\le 2|b|$, we obtain

$$\begin{aligned} p^i_{n}(x,y,z)\ge \frac{b_i^{2n}}{12|b|^n}c^2(x,y,z)\ge c(n)\frac{ b_i^{2n}}{|b|^{2n}}c^2(x,y,z). \end{aligned}$$

To complete the proof in case $a_ib_i>0$, we are left with the situation $|b|=|a+b|=1$. By (1.43)

$$\begin{aligned} p^i_{n}(x,y,z)=\frac{(a_i+b_i)^n b_i^n \left( \left( \frac{a_i}{b_i} \right) ^n +|a|^{n+\alpha }-\left( \frac{a_i}{a_i+b_i}\right) ^n\right) }{|a|^{n+\alpha }}\ge (a_i+b_i)^n b_i^n \ge b_i^{2n}, \end{aligned}$$

because $\frac{a_i}{b_i}>\frac{a_i}{a_i+b_i}$ and thus $\left( \frac{a_i}{b_i} \right) ^n>\left( \frac{a_i}{a_i+b_i}\right) ^n.$ Hence

$$\begin{aligned} p^i_{n}(x,y,z) \ge \frac{b_i^{2n}}{|b|^{2n}}|b|^{-2} \gtrsim \frac{b_i^{2n}}{|b|^{2n}} c^2(x,y,z). \end{aligned}$$

Case $a_i b_i<0$. It follows from Remark 1 [see (1.42) with $\alpha =1$].

Case $a_i \,b_i=0$. Since we have assumed that $b\ne 0$ we have that $a_i=0$ and by (1.33), with $\alpha =1$, we are done.

Therefore we have shown that whenever x, y, z are not collinear and they do not lie in the hyperplane $x_i=y_i=z_i$, then $p^i_{n}(x,y,z)>0$. This finishes the proof of (i). Furthermore, we have shown that if this is the case, then

$$\begin{aligned} p^i_{n}(x,y,z) \ge C(n) \frac{ b_i^{2n}}{|b|^{2n}}c^2(x,y,z). \end{aligned}$$

(1.46)

For the proof of (ii) notice that since ${\measuredangle }(V_j, L_{y,z}) \ge \theta _0$ there exists some coordinate $i_0 \ne j$ such that

$$\begin{aligned} |b_{i_0}|=|y_{i_0}-z_{i_0}|\ge C(\theta _0)|y-z|=C(\theta _0)|b|, \end{aligned}$$

hence (ii) follows by (1.46). $\square$