Proof

1. (i) Write $y=y_{0}+b\,A_{\alpha }$ , where $y_{0}\in H_{\alpha }$ and $\langle x,A_{\alpha }\rangle =\alpha (x)$ for all $x\in \mathfrak {a}$ . Observe that $b=\alpha (y)/|\alpha |^{2}$ . Then, for $z\in H_{\alpha }$ , we have

   $$ \begin{align*} |y-z|^{2}=|y_{0}-z|^{2}+b^{2}|A_{\alpha}|^{2} \end{align*} $$
   which attends its minimum when $z=y_{0}$ . Hence, $\displaystyle d(y,H_{\alpha })=b\,|\alpha |= \frac {\alpha (y)}{|\alpha |}$ .

2. (ii) Follows by direct computation.▪

Proof Indeed,

$$ \begin{align*} |x-\sigma_{\alpha} y|^{2}\leq |x-y|^{2}+|y-\sigma_{\alpha} y|^{2}\asymp\phi_{\alpha}(x,y)^{2} \end{align*} $$

while $|x-\sigma _{\alpha } y|^{2}\geq |x-y|^{2}$ and

$$ \begin{align*} |x-\sigma_{\alpha} y|^{2}=|\sigma_{\alpha} x-y|^{2}\geq d(y,H_{\alpha})^{2} \end{align*} $$

(geometric argument: the straight line joining y with $\sigma _{\alpha } x$ crosses $H_{\alpha }$ ; if $\ell (t)=ty+(1-t)\,\sigma _{\alpha } x$ then $\alpha (\ell (0))=\alpha (y)\geq 0$ while $\alpha (\ell (1))=-\alpha (x)\leq 0$ ).▪

Proof We first prove by induction on the length of the minimal representation of w as a product of reflections by simple roots that (3.1) holds where $a_{i}^{w}(y)$ is a linear combination of simple roots with integer coefficients for each i.

If the length of w is 1 then $w=\sigma _{\alpha _{i}}$ for a simple root $\alpha _{i}$ and

$$ \begin{align*} y-\sigma_{\alpha_{i}}\,y=2\,\frac{\alpha_{i}(y)}{|\alpha_{i}|^{2}}\,A_{\alpha_{i}} \end{align*} $$

and the result holds in that case with $a_{i}^{w}(y)=\alpha _{i}(y)$ .

Let us assume that the result holds for $k\geq 1$ and let $w=\sigma _{\alpha _{j}}\,w_{0}$ , where $|w_{0}|=k$ .

$$ \begin{align*} y-\sigma_{\alpha_{j}}\,w_{0}y &=\sigma_{\alpha_{j}}\,(y-w_{0}y)+y-\sigma_{\alpha_{j}}y =\sigma_{\alpha_{j}}\,\Big(\sum_{i=1}^{r}\,a_{i}^{w_{0}}(y)\,\frac{2A_{\alpha_{i}}}{|\alpha_{i}|^{2}}\Big)+2\,\frac{\alpha_{j}(y)}{|\alpha_{j}|^{2}}\,A_{\alpha_{j}}\\ &=\sum_{i=1}^{r}\,a_{i}^{w_{0}}(y)\,\sigma_{\alpha_{j}}\Big(\frac{2A_{\alpha_{i}}}{|\alpha_{i}|^{2}}\Big)+2\,\frac{\alpha_{j}(y)}{|\alpha_{j}|^{2}}\,A_{\alpha_{j}}\\ &=\sum_{i=1}^{r}\,a_{i}^{w_{0}}(y)\,\frac{2A_{\alpha_{i}}}{|\alpha_{i}|^{2}} {-\sum_{i=1}^{r}\,a_{i}^{w_{0}}(y)\,2\,\frac{\langle\alpha_{j},\alpha_{i}\rangle}{|\alpha_{i}|^{2}}\,2\,\frac{A_{\alpha_{j}}}{|\alpha_{j}|^{2}} }+2\,\frac{\alpha_{j}(y)}{|\alpha_{j}|^{2}}\,A_{\alpha_{j}}. \end{align*} $$

Note that $2\,\frac {\langle \alpha _{j},\alpha _{i}\rangle }{|\alpha _{i}|^{2}}$ is an integer since $\Phi $ is crystallographic.

We now need to prove that $a_{i}^{w}(y)$ is a linear combination of simple roots with non-negative integer coefficients for each i.

Suppose that this is not the case. To fix matters, we can assume that $a_{1}$ contains a term, say $b_{2}\,\alpha _{2}$ , with $b_{2}<0$ . Choose x, $y\in \overline {\mathfrak {a}^+}$ such that $\alpha _{2}(y)=t>0$ , $\alpha _{k}(y)=0$ for $k\not =2$ and $\alpha _{1}(x)=t>0$ , $\alpha _{k}(x)=0$ for $k\not =1$ . Then using (3.1),

$$ \begin{align*} \langle x,y-wy\rangle =b_{2}\alpha_{2}(y)\alpha_{1}(x)=b_{2}t^{2}<0 \end{align*} $$

which cannot be true (refer to Remark 2.1). The result follows.▪

Proof Let the notation be as in the Proposition 3.5. By (3.1) and (4.3), we have

$$ \begin{align*} |x-wy|^{2}=|x-y|^{2}+2\,\langle x,y-wy\rangle =|x-y|^{2}+2\sum_{i=1}^{r}\frac{a_{i}^{w}(y)\alpha_{i}(x)}{|\alpha_{i}|^{2}}, \end{align*} $$

and the first inequality follows since not all $a_{i}^{w}(y)$ are nul.

The second inequality follows from a simple geometric argument since x and $wy$ are not in the same Weyl chamber, the straight line between x and $wy$ must cross a wall of the Weyl chamber (similarly for y and $w^{-1}\,{x}$ ).▪

Proof Recall that by Proposition 3.5

$$ \begin{align*} y-w\,y=\sum_{i=1}^{r} \,2\,\frac{a_{i}^{w}(y)}{|\alpha_{i}|^{2}}\,A_{\alpha_{i}}. \end{align*} $$

We first show that there exists $i_{0}$ such that $\alpha _{i_{0}}(y)>0$ and $a_{i_{0}}^{w}(y)>0$ . Suppose by contradiction that this is not true, so

$$ \begin{align*} y-w\,y=\sum_{\alpha_{i}(y)=0} \,2\,\frac{a_{i}^{w}(y)}{|\alpha_{i}|^{2}}\,A_{\alpha_{i}}. \end{align*} $$

Then

$$ \begin{align*} \langle y, y-w\,y \rangle=\sum_{\alpha_{i}(y)=0} \,2\,\frac{a_{i}^{w}(y)}{|\alpha_{i}|^{2}}\,\alpha_{i}(y)=0, \end{align*} $$

so $\langle y, y \rangle =\langle y, w\,y \rangle $ . As $\|wy\|=\|y\|$ , this is only possible if $y=wy$ .

Since $\alpha _{i_{0}}\not \in \Phi _{y}\supseteq \Phi _{x}$ , it follows that

$$ \begin{align*} \qquad\qquad\quad\langle x, y-w\,y \rangle=\sum_{i=1}^{r} \,2\,\frac{a_{i}^{w}(y)}{|\alpha_{i}|^{2}}\,\alpha_{i}(x)\ge 2\,\frac{a_{i_{0}}^{w}(y)}{|\alpha_{i_{0}}|^{2}}\,\alpha_{i_{0}}(x)>0. \end{align*} $$

▪

Proof Suppose that $a_{k}^{w}=\sum _{i=1}^{r} n_{i} \alpha _{i}$ with $n_{k}=0$ . Consider $x_{0}$ , $y_{0}$ with $\Phi _{x_{0}}=\Phi _{y_{0}}$ generated by $\{\alpha _{i}\colon i\not = k\}$ . Then, we have

$$ \begin{align*} \langle x_{0},y_{0}-w\,y_{0}\rangle=\sum_{i=1}^{r} \,2\,\frac{a_{i}^{w}(y_{0})}{|\alpha_{i}|^{2}}\,\alpha_{i}(x_{0})= 2\,\frac{a_{k}^{w}(y_{0})}{|\alpha_{k}|^{2}}\,\alpha_{k}(x_{0})=0 \end{align*} $$

since $a_{k}^{w}(y_{0})=\sum _{i\not =k} n_{i} \alpha _{i}(y_{0})=0$ . By Corollary 3.9, we see that $w\in W_{x_{0}}$ with $W_{x_{0}}$ generated by $\sigma _{\alpha _{i}}$ , $i\not =k$ . This implies that $A_{\alpha _{k}}$ does not appear in the decomposition (3.1) of $y-wy$ , thus, equivalently, that $a_{k}^{w}=0$ .▪

Proof The map $z\mapsto |x-y|^{2}+2\langle x,y-z\rangle $ is linear in $z\in C(y)$ , the convex hull of $W\cdot y$ . Therefore, it attains its maximum and its minimum at points in $W\cdot y$ . When $z=wy$ , the denominator $|x-y|^{2}+2\langle x,y-z\rangle $ in (4.1) equals

(4.3) $$ \begin{align} |x-y|^{2}+2\langle x,y-w\,y\rangle=|x-wy|^{2}. \end{align} $$

The smallest value is $|x-y|^{2}$ and the largest is $|x-w_{max}y|^{2}$ .▪

Proof Note that by Lemma 3.1(i), $\alpha (x)=|\alpha |\,d(x,H_{\alpha })\leq |\alpha |(|x-y|+d(y,H_{\alpha })) =|\alpha ||x-y|+\alpha (y)$ so $\alpha (x)\leq |x-y|\,\max _{\alpha>0}|\alpha |+D\,|x-y|=D^{\prime }\,|x-y|$ . Using (4.3), Proposition 3.5 and Remark 3.8,

$$ \begin{align*} |x-wy|^{2}&=|x-y|^{2}+2\langle x,y-wy\rangle = |x-y|^{2}+2\sum_{i=1}^{r}2\,a_{i}^{w}(y)\alpha_{i}(x)/|\alpha_{i}|^{2}\\ &\leq |x-y|^{2}+4\,r\,M\,\max_{\alpha>0}\,\alpha(y)\,\max_{\alpha>0}\,\alpha(x) \leq (1+4\,r\,MD\,D^{\prime})|x-y|^{2}. \end{align*} $$

The result then follows from (4.2) in Proposition 4.1, noting that for all $\alpha>0$ , we have $|x-y|\leq \phi _{\alpha }(x,y)\leq (1+D)\,|x-y|$ .▪

Proof Recall that, by Lemma 3.1(ii), $|x-\sigma _{\alpha } y|^{2}=|x-y|^{2}+C\,\alpha (x)\,\alpha (y)$ , where $C=4/|\alpha |^{2}$ . Using the formula $a^{d}-b^{d}=(a-b)\,\sum _{k=0}^{d-1}\,a^{k}\,b^{d-1-k}$ , and the fact that $|x-y|^{d}\leq |x-\sigma _{\alpha } y|^{d}$ on E, we have

$$ \begin{align*} T_{1}(x,y)&=\frac{(|x-\sigma_{\alpha} y|^{2})^{d}-(|x-y|^{2})^{d}}{\alpha(x)\,\alpha(y)\,|x-y|^{d}\,|x-\sigma_{\alpha} y|^{d}\,(|x-y|^{d}+|x-\sigma_{\alpha} y|^{d})}\\ &=C\,\frac{\sum_{k=0}^{d-1}\,|x-\sigma_{\alpha} y|^{2\,k}\,|x-y|^{2\,(d-1-k)}}{|x-y|^{d}\,|x-\sigma_{\alpha} y|^{d}\,(|x-y|^{d}+|x-\sigma_{\alpha} y|^{d})}\\ &\asymp \frac{|x-\sigma_{\alpha} y|^{2\,(d-1)}}{|x-y|^{d}\,|x-\sigma_{\alpha} y|^{2\,d}}=\frac{1}{|x-y|^{d}\,|x-\sigma_{\alpha} y|^{2}}. \end{align*} $$

▪

Proof Let $c_{0}=\frac {c}{N^{r}}$ and let $S_{0}$ be the set of simple roots $\alpha _{i}$ such that such that

(5.2) $$ \begin{align} c_{0}\,\alpha_{i}(y)< |x-y|. \end{align} $$

Let $\Phi ^{\prime }$ be generated by the simple roots such that (5.2) holds (there are at most r such simple roots). If $c_{0}\,\alpha (y)< |x-y|$ for every $\alpha \in \Phi ^{\prime }$ then $\Phi ^{\prime }$ satisfies the condition (5.1) with $c^{\prime }=c_{0}$ except perhaps $c_{0}\,\beta (y)=|x-y|$ for some $\beta \in \Phi _{0}^+$ . In that case, we choose $0<\delta <c-c_{0}$ such that the inequalities $ (c_{0}+\delta )\,\alpha (y)< |x-y|$ hold for every $\alpha \in \Phi ^{\prime }_+$ . We then take $c^{\prime }=c_{0}+\delta $ .

Suppose $c_{0}\,\alpha (y)\geq |x-y|$ for some $\alpha \in \Phi ^{\prime }$ . In that case, since $\alpha $ is not one of the $\alpha _{i}$ ’s from (5.2), it is not simple and we have $c_{0}\,\alpha _{j}>|x-y|/N$ for some $j\in S_{0}$ . We set $c_{1}=N\,c_{0}$ and repeat the process.

Let $S_{1}$ be the set of simple roots $\alpha _{i}$ such that such that

(5.3) $$ \begin{align} c_{1}\,\alpha_{i}(y)< |x-y|. \end{align} $$

Let $\Phi ^{\prime }$ be generated by the simple roots such that (5.3) holds (there are at most $r-1$ such simple roots). If $c_{1}\,\alpha (y)< |x-y|$ for every $\alpha \in \Phi ^{\prime }$ then $\Phi ^{\prime }$ satisfies the condition (5.1) with $c^{\prime }=c_{1}$ (if $c_{1}\,\beta (y)=|x-y|$ for some $\beta \in \Phi _{0}$ , we proceed as in the previous step).

Suppose $c_{1}\,\alpha (y)\geq |x-y|$ for some $\alpha \in \Phi ^{\prime }$ . In that case, we have $c_{1}\,\alpha _{j}(y)>|x-y|/N$ for some $j\in S_{1}$ . We set $c_{2}=N\,c_{1}$ and repeat the process.

Since we decrease the number of simple roots involved each time, the process must stop eventually either because we have achieved (5.1) or because we ran out of simple roots.▪

Proof We use the decomposition (3.1) of $y-wy$ . Let $\alpha _{i}$ , $i=1,\ldots , s=| \Phi ^{\prime }_+|$ be the simple positive roots from $\Phi ^{\prime }$ and $\alpha _{s+j}$ the simple positive roots from $\Phi _{0}^+$ . We have

(5.4) $$ \begin{align} y-w\,y=\sum_{i=1}^{s} \,2\,\frac{a_{i}^{w}(y)}{|\alpha_{i}|^{2}}\,A_{\alpha_{i}} + \sum_{j=1}^{r-s} \,2\,\frac{a_{s+j}^{w}(y)}{|\alpha_{s+j}|^{2}}\,A_{\alpha_{s+j}}. \end{align} $$

In order to prove the lemma it is sufficient to show that the second sum does not vanish since by Corollary 3.10, a nonzero function $a_{s+j}^{w}$ has a nonzero term $n\, \alpha _{s+j}$ .

Suppose by contradiction that the second sum vanishes. Choose x and y with $\Phi _{x}=\Phi _{y}=\Phi ^{\prime }$ . Formula (5.4) implies that

$$ \begin{align*} \langle x,y-w\,y\rangle=\sum_{i=1}^{s} \,2\,\frac{a_{i}^{w}(y)}{|\alpha_{i}|^{2}}\,{\alpha_{i}(x)} =0, \end{align*} $$

since $\alpha _{i}(x)=0$ . By Corollary 3.9, we get $w\in W_{y}=W^{\prime }$ .▪

Proof By Remark 3.11, if $\alpha \in \Phi _{0}^+$ then

$$ \begin{align*} \alpha(x)\geq \alpha(y)-C_{1}\,|x-y|\geq \alpha(y)-C_{1}\,c\,\alpha(y) =(1-C_{1}\,c)\,\alpha(y)\geq \alpha(y)/2. \end{align*} $$

On the other hand, $\alpha (y)\geq \alpha (x)-C_{1}\,|x-y|\geq \alpha (x)-C_{1}\,c\,\alpha (y) \geq \alpha (x)-2\,C_{1}\,c\,\alpha (x) =(1-2\,C_{1}\,c)\,\alpha (x)\geq \alpha (x)/2$ . The rest follows from Lemma 5.3.▪

Proof In that case, $\Phi ^{\prime }=\emptyset $ , $\Phi _{0}=\Phi $ , $W_{0}=W\setminus \{id\}$ , and by Corollary 5.5, for every $w\in W_{0}$ , equation (5.5) holds and

$$ \begin{align*} \frac{1-|x|^{2}}{w_{d}\,\pi(x)\,\pi(y)} \,\frac{1}{|x-y|^{d}}&\geq P^{W}(x,y)\\ &\geq\frac{1-|x|^{2}}{|W|\,w_{d}\,\pi(x)\,\pi(y)} \,\left[\frac{1}{|x-y|^{d}}-\sum_{w\not=id}\,\frac{1}{|x-w\,y|^{d}} \right]\\ &\geq\frac{1-|x|^{2}}{|W|\,w_{d}\,\pi(x)\,\pi(y)} \,\left[\frac{1}{|x-y|^{d}}-\frac{c^{d}}{(K_{0}/2)^{d/2}}\,\sum_{w\not=id}\,\frac{1}{|x-y|^{d}} \right]\\ &\geq(1-c^{d}\,|W|/(K_{0}/2)^{d/2})\,\frac{1-|x|^{2}}{|W|\,w_{d}\,\pi(x)\,\pi(y)} \,\frac{1}{|x-y|^{d}}\\ &\geq\frac{1}{2}\,\frac{1-|x|^{2}}{|W|\,w_{d}\,\pi(x)\,\pi(y)} \,\frac{1}{|x-y|^{d}}. \end{align*} $$

Hence, the conjecture holds in this case since $|x-\sigma _{\alpha }\,y|^{2}=|x-y|^{2}+C\,\alpha (x)\,\alpha (y)\asymp \alpha (x)\,\alpha (y)$ by the hypothesis $|x-y|\leq c\,\alpha (y)$ and Corollary 5.5.▪

Proof First note that

(5.7) $$ \begin{align} \partial^{x}_{\alpha}\,|x-y|^{P} =P\,\alpha(x-y)\,|x-y|^{P-2} \end{align} $$

and

(5.8) $$ \begin{align} \partial^{x}_{\alpha}\,\left[|x-y|^{d+2\,I}\,|x-w\,y|^{-d}\right] &=(d+2\,I)\,\alpha(x-y)\,|x-y|^{d+2\,I-2}\,|x-w\,y|^{-d}\nonumber \\ &\quad+(-d)\,\alpha(x-w\,y)|x-y|^{d+2\,I}|x-w\,y|^{-d-2}. \end{align} $$

In the computations below, it is helpful to remember that $|x-w\,y|=|w^{-1}\,x-y|$ . Using formulas (5.7) and (5.8), we obtain for each $w\in W_{0}$ ,

$$ \begin{align*} &{\partial^{y}_{\alpha}\,\partial^{x}_{\alpha}\,|x-y|^{d+2\,I}\,|x-w\,y|^{-d}}\\ &=-2\,(d+2\,I)\,|x-y|^{d+2\,I-2}\,|x-w\,y|^{-d}\\ &\quad+{(d+2\,I-2)}\,(d+2\,I)\,\alpha(x-y)\,\alpha(y-x)\,|x-y|^{d+2\,I-4}\,|x-w\,y|^{-d}\\ &\quad-d\,(d+2\,I)\,\alpha(x-y)\,\alpha(y-w^{-1}\,x)\,|x-y|^{d+2\,I-2}\,|x-w\,y|^{-d-2}\\ &\quad-{(d+2\,I-2)}\,(d+2\,I)\,\alpha(y-x)\,\alpha(x-w\,y)\,|x-y|^{d +2\,I-2}\,|x-w\,y|^{-d-2}\\ &\quad-{(d+2\,I-2)}\,(d+2\,I)\,\alpha(y-x)\,\alpha(x-w\,y)\,|x-y|^{d+2\,I-2}\,|x-w\,y|^{-d-2}\\ &\quad+d\,\partial_{\alpha}^{y}(\alpha(w\,y))\, |x-y|^{d+2\,I}\,|x-w\,y|^{-d-2}\\ &\quad+d\,(d+2)\,\alpha(x-w\,y)\,\alpha(y-w^{-1}\,x)\,|x-y|^{d+2\,I}\,|x-w\,y|^{-d-4}. \end{align*} $$

Observe that after each application of an operator $\partial ^{y}_{\alpha }\,\partial ^{x}_{\alpha }$ , where $\alpha $ is a root from $\Phi ^{\prime \prime }$ , we get a sum of terms which are products of constants times terms in $\alpha (x-y)$ , $\alpha (w^{-1}\,x-y)$ , $\alpha (x-w\,y)$ , $|x-y|^{d+2\,I-2\,{k}}$ , and $|x-w\,y|^{-d-{2l}}$ . We used the fact that a directional derivative of a linear term such as $\alpha (x-y)$ is a constant.

By applying for the first time $\partial ^{y}_{\alpha }\,\partial ^{x}_{\alpha }$ , the degree of homogeneity of the function $|x-y|^{d+2\,I}\,|x-w\,y|^{-d}$ went from $2\,I$ to $2\,I-2$ .

Each new application of an operator $\partial ^{y}_{\alpha }\,\partial ^{x}_{\alpha }$ decreases the degree by 2 until we get the degree of 0 once we have applied $\partial ^{x}({\pi ^{\prime \prime }})\,\partial ^{y}({\pi ^{\prime \prime }})$ .

We use the estimates (3.2) from Remark 3.11. From the fact that $|\alpha (x-y)|\leq C_{1}\,|x-y|$ , $|\alpha (x-w\,y)|\leq C_{1}\,|x-w\,y|$ , $|\alpha (w^{-1}\,x-y)|\leq C_{1}\,|w^{-1}\,x-y|=C_{1}\,|x-w\,y|$ , we note that $\left |\partial ^{x}({\pi ^{\prime \prime }})\,\partial ^{y}({\pi ^{\prime \prime }})\,R(x,y)\right |$ is bounded by a finite sum of terms of the form $C\, |x-y|^{d+2\,I-\,{j}}\,|x-w\,y|^{-d-{(2I-\,j)}}$ , or, equivalently,

$$ \begin{align*} C\, \frac{|x-y|^{d+i}}{|x-w\,y|^{d+i}}, \end{align*} $$

where $0\leq i\leq 2\,I$ . The rest follows.▪

Proof First note that $\partial _{\alpha _{k}}\,\langle \alpha ,x\rangle =\langle \alpha ,\alpha _{k}\rangle $ . The result then follows easily using induction on M and Leibnitz rule.▪

Proof Setting $\alpha _{i}(x)=0$ for $i=1$ , …, M in formula (5.9) gives

$$ \begin{align*} \partial_{\alpha_{M}}\,\cdots\,\partial_{\alpha_{1}}\,\pi^{\prime}(x) =\left[\sum_{\sigma\in S_{M}}\,\prod_{i=1}^{M}\,\langle\alpha_{\sigma(i)},\alpha_{i}\rangle\right]\,\prod_{\alpha\in\Phi^{\prime}_+\setminus\{\alpha_{1},\dots,\alpha_{M}\}}\,\langle\alpha,x\rangle. \end{align*} $$

The rest is straightforward: the last product cancels on S because it contains the term $ \langle \alpha _{M+1},x\rangle $ .▪

Proof It is clear that $\partial (\pi ^{\prime })\,\pi ^{\prime }(x)$ is a constant given that the degree of the operator $\partial (\pi ^{\prime })$ is equal to the degree of $\pi ^{\prime }(x)$ . Now, let

$$ \begin{align*} S_{\lambda}(x)= \frac{\sum_{w\in W}\,\epsilon(w)\,e^{\langle \lambda,w\,x\rangle}}{\pi^{\prime}(x)}=\frac{\sum_{w\in W}\,\epsilon(w)\,e^{\langle w^{-1} \lambda,x\rangle}}{\pi^{\prime}(x)}, \end{align*} $$

where W is generated by the reflections $\sigma _{\alpha }$ , $\alpha \in \Phi ^{\prime }$ . Now, $S_{\lambda }$ is an analytic function (the numerator is W skew-symmetric and analytic; in fact $S_{\lambda }$ is a constant multiple of $\pi (\lambda )$ times a spherical function). If we compute $\partial (\pi ^{\prime })\,(\pi ^{\prime }(x)\,S_{\lambda }(x))$ and set $x=0$ , we get

$$ \begin{align*} \left.\partial(\pi^{\prime})\,(\pi^{\prime}(x)\,S_{\lambda}(x))\right|{}_{x=0}=\sum_{w\in w}\,\epsilon(w)\,\pi^{\prime}(w^{-1}\,\lambda)=|W|. \end{align*} $$

The left hand side is equal to $\partial (\pi ^{\prime })\,(\pi ^{\prime }(x))\,S(0)$ since all terms involving derivatives of $S_{\lambda }$ are eliminated by one of the roots (a consequence of Corollary 5.11). We can therefore conclude that $\partial (\pi ^{\prime })\,(\pi ^{\prime }(x))\not =0$ .▪

Proof We use the Cauchy Mean Value Theorem several times as in the proof of the multivariate de l’Hospital rule proposed by Lawlor [Reference Lawlor23]. More precisely, if x, $y\not \in H_{\alpha }$ , then, denoting by $x_{\alpha }$ and $y_{\alpha }$ the projections of x and y on $H_{\alpha }$ ,

$$ \begin{align*} \frac{R(x,y)}{\alpha(x)\,\alpha(y)}=\frac1{\alpha(y)}\frac{R(x,y)-R(x_{\alpha},y)}{\alpha(x)-\alpha(x_{\alpha})}= \frac1{\alpha(y)}\frac{\partial_{\alpha}^{x}R(x^{\prime},y)}{\partial_{\alpha}^{x}\alpha(x^{\prime})} \end{align*} $$

for an intermediate point $x^{\prime }\in \mathfrak {a}^+\setminus H_{\alpha }$ . The same reasoning allows us to repeat the argument with $\partial _{\alpha }^{y}$ :

$$ \begin{align*} \frac{\partial_{\alpha}^{x}\,R(x^{\prime},y)}{\alpha(y)} =\frac{\partial_{\alpha}^{x}\,R(x^{\prime},y)-\partial_{\alpha}^{x}\,R(x^{\prime},y_{\alpha})}{\alpha(y)-\alpha(y_{\alpha})} =\frac{\partial_{\alpha}^{y}\,\partial_{\alpha}^{x}\,R(x^{\prime},y^{\prime})}{\partial_{\alpha}^{y}\alpha(y^{\prime})}. \end{align*} $$

This was done for a single $\alpha $ but we can repeat the process for all the roots $\alpha $ , $\alpha \in {\Phi ^{\prime \prime }}$ . This is made possible by Corollary 5.11. Note that $\partial ^{x}(\pi ^{\prime \prime })\,\partial ^{y}(\pi ^{\prime \prime })\,(\pi ^{\prime \prime }(x)\,\pi ^{\prime \prime }(y))\not =0$ by Lemma 5.12.▪

Proof This result follows directly from Corollary 5.5, Lemma 5.9, and Proposition 5.14.

Proof We use induction on the rank r of the root system. The result is true in rank 1 by Lemma 4.4.

Assume that it is true for ranks 1,…, $r-1$ , $r\geq 2$ . Let $\Phi $ be a rank r complex root system. The induction hypothesis means that there exist constants $K_{1}$ and $K_{2}$ such that for $x\in B^+,y\in S^+$ , we have

(5.11) $$ \begin{align} K_{1}\leq |x-y|^{d}\,\prod_{\alpha\in \Phi^{\prime}_+}\phi_{\alpha}(x,y)^{2}\,\frac{\sum_{w\in W^{\prime}}\,\frac{\epsilon(w)}{|x-w\,y|^{d}}}{\pi^{\prime}(x)\,\pi^{\prime}(y)}\leq K_{2} \end{align} $$

for every basic root subsystem $\Phi ^{\prime }$ of $\Phi $ which is of lower rank. Since the number of such subsystems is finite, we may assume that $K_{1}$ and $K_{2}$ are independent from the choice of $\Phi ^{\prime }$ (depend only on $\Phi $ ).

Fix c such that

$$ \begin{align*} 0<c\leq \min\left\{1,\frac{1}{4\,C_{1}},\frac{\sqrt{K_{0}/2}}{\sqrt[d]{2\,|W|}}\right\}, \end{align*} $$

where $C_{1}$ is as in Remark 3.11, $K_{0}$ is as in Lemma 5.3 and such that

(5.12) $$ \begin{align} (1+2\,N\,C_{1})^{2M}\,(2+C_{1}^{-1})^{2M}\,K\,c^{d}\leq K_{1}/2, \end{align} $$

where K is as in Proposition 5.14 (we consider all the basic root subsystems $\Phi ^{\prime \prime }\subseteq \Phi ^{\prime }$ for all possible basic root subsystems $\Phi ^{\prime } $ of $\Phi $ and take the largest corresponding value of K). Recall that N denotes the maximal length of positive roots in $\Phi $ and M is the number of all positive roots in $\Phi $ . By Lemma 5.1, if $S_{\Phi ^{\prime },c^{\prime }}$ is as in (5.10), we have

$$ \begin{align*} B^+\times S^+=\bigcup_{\Phi^{\prime},c^{\prime}}\,S_{\Phi^{\prime},c^{\prime}}, \end{align*} $$

where the union is taken over all basic root subsystems $\Phi ^{\prime }$ of $\Phi $ and $c^{\prime }$ with $c/N^{r}\leq c^{\prime }\leq c$ . Since the number of such basic root subsystems is finite, we only have to prove the Conjecture 1 for a specific choice of $\Phi ^{\prime }$ where x, y satisfy

(5.13) $$ \begin{align} c^{\prime}\,\alpha(y)< |x-y|< c^{\prime}\,\beta(y) \end{align} $$

for $\alpha \in \Phi ^{\prime }_+$ , $\beta \in \Phi _{0}^+$ and any $c^{\prime }$ such that $c/N^{r}\leq c^{\prime }\leq c$ .

Recall that c is fixed. We ensure that the constants in the estimates of the Conjecture 1 depend only on c, when $\Phi ^{\prime }$ is fixed and $c^{\prime }$ varies in the segment $[c/N^{r},c]$ .

We examine three possible cases:

$\underline {\Phi ^{\prime }=\emptyset }$ : in that case, Lemma 5.6 allows us to conclude.

$\underline {\Phi ^{\prime }=\Phi }$ : in that case, the result follows from (5.13) and from Proposition 4.3 (if we refer to the notation of Proposition 4.3, $D=1/c^{\prime }$ and the constants depend on D; this is where $c/N^{r}\leq c^{\prime }\leq c $ comes into play).

$\underline {1\leq \mbox {rank}(\Phi ^{\prime })< \mbox {rank}(\Phi )}$ : We have $\phi _{\beta }(x,y)=\beta (y)$ and $\beta (x)\,\beta (y)/2\leq \phi _{\beta }^{2}(x,y)\leq 2\,\beta (x)\,\beta (y)$ by Corollary 5.5. Hence,

(5.14) $$ \begin{align} 2^{-M}\,\prod_{\beta\in\Phi_{0}^+}\,\beta(x)\,\beta(y)\leq \prod_{\beta\in\Phi_{0}^+}\,\phi_{\beta}(x,y)^{2} \leq 2^{M}\,\prod_{\beta\in\Phi_{0}^+}\,\beta(x)\,\beta(y). \end{align} $$

We used the fact that $|\Phi _{0}^+|\leq |\Phi _+|\leq M$ .

For a basic subsystem $\Phi ^{\prime \prime }\subseteq \Phi ^{\prime }$ , let

$$ \begin{align*} X_{\Phi^{\prime\prime}}&=\{(x,y)\in S_{\Phi^{\prime},c^{\prime}}\colon \alpha(y)\leq 2 N\,C_{1}\,|x-y| \mbox{ for } \alpha\in \Phi^{\prime\prime}_+ \\&\qquad\qquad\qquad\qquad\mbox{ and }\alpha(y)>2\,C_{1}\,|x-y| \mbox{ for } \alpha\in \Phi^{\prime}_+\setminus\Phi^{\prime\prime}\} \end{align*} $$

and note that the union of all $X_{\Phi ^{\prime \prime }}$ for $\Phi ^{\prime \prime }\subseteq \Phi ^{\prime }_+$ gives $S_{\Phi ^{\prime },c^{\prime }}$ . Indeed, for given x and y, $\Phi ^{\prime \prime }$ is generated by the simple roots $\alpha _{i}$ such that $\alpha _{i}(y)\leq 2 \,C_{1}\,|x-y|$ (it may happen that $\Phi ^{\prime \prime }=\emptyset $ in which case $X_{\Phi ^{\prime \prime }}=\{(x,y)\in S_{\Phi ^{\prime },c^{\prime }}\colon \alpha (y)>2\,C_{1}\,|x-y|\ \mbox {for }\alpha \in \Phi ^{\prime }_+\}$ ).

If $(x,y)\in X_{\Phi ^{\prime \prime }}$ and $\alpha \in \Phi ^{\prime \prime }$ then

(5.15) $$ \begin{align} \phi_{\alpha}(x,y)^{2}\leq (1+2\,C_{1})^{2}\,|x-y|^{2}. \end{align} $$

If $(x,y)\in X_{\Phi ^{\prime \prime }}$ and $\alpha \in \Phi ^{\prime }\setminus \Phi ^{\prime \prime }$ then $\alpha (x)\geq \alpha (y)-C_{1}\,|x-y|> \alpha (y)/2$ . Hence, $|x-y| < (2\,C_{1})^{-1}\,\alpha (y)\leq 2\,(2\,C_{1})^{-1}\,\alpha (x)$ . Therefore,

(5.16) $$ \begin{align} \phi_{\alpha}(x,y)^{2}\leq (2+C_{1}^{-1})\,(1+(2\,C_{1})^{-1})\,\alpha(x)\,\alpha(y)\leq (2+C_{1}^{-1})^{2}\,\alpha(x)\,\alpha(y). \end{align} $$

Let

$$ \begin{align*} A&=\frac{|W|w_{d} P^{W}(x,y)}{\Omega(x,y)}=|x-y|^{d}\,\prod_{\gamma\in \Phi_+}\,\phi_{\gamma}(x,y)^{2}\,\frac{\sum_{w\in W}\,\frac{\epsilon(w)}{|x-w\,y|^{d}}}{\pi(x)\,\pi(y)}\\ &=\frac{\prod_{\beta\in \Phi_{0}^+}\,\phi_{\beta}(x,y)^{2}}{\prod_{\beta\in \Phi_{0}^+}\,\beta(x)\,\beta(y)} \,\left[\vphantom{\left. +\underbrace{|x-y|^{d}\,\prod_{\alpha\in \Phi^{\prime}_+}\,\phi_{\alpha}(x,y)^{2}\,\frac{ \sum_{w\in W_{0}}\,\frac{\epsilon(w)}{|x-w\,y|^{d}} } {\pi^{\prime}(x)\,\pi^{\prime}(y)}}_{\widetilde A}\right]}|x-y|^{d}\,\prod_{\alpha\in \Phi^{\prime}_+}\,\phi_{\alpha}(x,y)^{2}\,\frac{\sum_{w\in W^{\prime}}\,\frac{\epsilon(w)}{|x-w\,y|^{d}}}{\pi^{\prime}(x)\,\pi^{\prime}(y)} \right.\\&\qquad\left. +\underbrace{|x-y|^{d}\,\prod_{\alpha\in \Phi^{\prime}_+}\,\phi_{\alpha}(x,y)^{2}\,\frac{ \sum_{w\in W_{0}}\,\frac{\epsilon(w)}{|x-w\,y|^{d}} } {\pi^{\prime}(x)\,\pi^{\prime}(y)}}_{\widetilde A}\right]. \end{align*} $$

Now, using (5.11) and (5.14), we get

(5.17) $$ \begin{align} 2^{-M}\,[K_{1}-|\widetilde A|]\leq A\leq 2^{M}\,[K_{2}+|\widetilde A|]. \end{align} $$

Let us fix $\Phi ^{\prime \prime }\subset \Phi ^{\prime }$ . For $(x,y)\in X_{\Phi ^{\prime \prime }}$ , using inequalities (5.15) and (5.16) and next the condition (5.12) on c and Proposition 5.15, we obtain

$$ \begin{align*} |\widetilde A|&\leq (1+2\,NC_{1})^{2\,|\Phi^{\prime\prime}_+|}\,(2+C_{1}^{-1})^{2\,|\Phi^{\prime}_+\setminus\Phi^{\prime\prime}_+|}\, \left|\frac{R(x,y)}{\pi^{\prime\prime}(x)\,\pi^{\prime\prime}(y)}\right|\\ &\leq (1+2\,NC_{1})^{M}\,(2+C_{1}^{-1})^{M}\,K\,c^{d}\leq K_{1}/2. \end{align*} $$

This is enough to conclude from (5.17) that $A\asymp 1$ .▪

Proof In the rank 1 case with $d=2$ , we have

$$ \begin{align*} N^{W}(x,y)&=\frac{1}{2\,\pi}\,\frac{\ln|x-y|-\ln|x-\sigma\,y|} {\alpha(x)\,\alpha(y)} =-\frac{1}{4\,\pi}\,\frac{\ln\frac{|x-\sigma\,y|^{2}}{|x-y|^{2}}}{\alpha(x)\,\alpha(y)}\\ &=-\frac{1}{2\,\pi} \,\frac{1}{|x-y|^{2}} \,\frac{\ln(1+\psi_{\alpha,\alpha}(x,y))}{\psi_{\alpha,\alpha}(x,y)}. \end{align*} $$

In the $A_{2}$ case,

$$ \begin{align*} N^{W}(x,y)&=\frac{1}{2\,\pi}\,\sum_{w\in W}\,\epsilon(w)\,\frac{\ln|x-w\,y|} {\pi(x)\,\pi(y)} =\frac{1}{2\,\pi}\,\frac{\ln\frac{|x-y|\,|x-\sigma_{\alpha}\,\sigma_{\beta}\,y|\,|x-\sigma_{\beta}\,\sigma_{\alpha}\,y|}{|x-\sigma_{\alpha}\,y|\,|x-\sigma_{\beta}\,y|\,|x-\sigma_{\alpha+\beta}|}}{\pi(x)\,\pi(y)}\\ &=-\frac{1}{4\,\pi}\,\frac{\ln\frac{|x-\sigma_{\alpha}\,y|^{2}\,|x-\sigma_{\beta}\,y|^{2}\,|x-\sigma_{\alpha+\beta}|^{2}}{|x-y|^{2}\,|x-\sigma_{\alpha}\,\sigma_{\beta}\,y|^{2}\,|x-\sigma_{\beta}\,\sigma_{\alpha}\,y|^{2}}}{\pi(x)\,\pi(y)}\\ &=-\frac{C^{3}}{4\,\pi}\,\frac{1}{|x-y|^{6}} \frac{L(x,y)}{\prod_{\gamma>0}\,\psi_{\gamma,\gamma}(x,y)} \end{align*} $$

(using Lemma 3.1 repeatedly).▪