content_image/301325.jpg content_image/301326.jpg content_image/301327.jpg content_image/301328.jpg content_image/301329.jpg content_image/301330.jpg
where 

<img src='content_image/301325.jpg'>

Equalities (6.3) and (6.6) show that − 2 πa$_{ii}$ (0) = q$_{i}$ (0 $_{, ξ}$′ $_{i}$) , (2 ≤ i ≤ n ) . Then by Lemma 5, we have q$_{i}$ (0 $_{, ξ}$′ $_{i}$) → 0 as $_{ξ}$′ i → 0 and thus a$_{ii}$ (0) = 0 , (2 ≤ i ≤ n ) . 

In order to complete the proof, let $_{ξ}$′ $_{ij}$(1) = (0 , · · · , 0 ︸ ︷︷ ︸ i − 2 , 1 , 0 , · · · , 0 ︸ ︷︷ ︸ , j − i − 1 1 , 0 , · · · , 0 ︸ ︷︷ ︸ n − j ) ∈ R n − $^{1}$, i ̸ = j, and $_{ξ}$′ ij = $_{εξ}$′ $_{ij}$(1) ∈ G ′ . In this case, by a$_{ii}$ (0) = 0, a$_{ij}$ (0) = a$_{ji}$ (0) and (6.1), we have 

for $_{ξ}$′ = $_{ξ}$′ $_{ij}$, where 

<img src='content_image/301326.jpg'>

<img src='content_image/301327.jpg'>

and the indices i, j ∈ { 2 , ..., n } are fixed. Then, in the same way as above, equation (6.7) and Lemma 6 yield 

<img src='content_image/301328.jpg'>

where $^{U}$± $_{q}$(0 $_{, ξ}$′ $_{ij}$) = ∂$_{x}$ 1 q ( 0 , ± 0 $_{, ξ}$′ ij ) . From (6.7), by virtue of Lemmata 2, 3 we have 

<img src='content_image/301329.jpg'>

Hence, 

<img src='content_image/301330.jpg'>

