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By H. Jacquet, R. P. Langlands

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Extra resources for Automorphic Forms on GL(2) (part 2)

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22)  (2, m)  f m (2)  g m (2)  245m  481m2  2415m3 .  9 29 30  m1 ( 60 48  48 m  48 m2  48 m3 )  then we have and  245m  481m2  2415m3  m 1 p 35 24( m  m 1 ) 1 ]m 2 amp . 26) p 1  m 1    p  [1  m 1 p 35 1 24 ( m  m ) 1 ]m 2 amp . 28). The above methods and results are also mentioned in (Yang AM 1997)[6]-(Yang HJ 1997)[9]. 20) { [ sin(  )  2 m1/ s 1 m1/ r ]m s amp } p { n r bnq }q , m 1  n 1 m 1  r  1 p equivalent inequalities: the following equivalent inequalities:  q   sin( / r ) .

21)  ( k (m, n)bn ) q n 1    [k (r )  O(n  2 )]n r bnq . 8). 8) are all equivalent.  n 1  )p (2) For 0  p  1 , we have the following equivalent inequalities: 1 1 m 1  m 1   J q {  (m)m s amp } p . 8) is valid. By the  lder’s inequality, we have reverse H o 1 p 1 s 1  k p (r ) [1  k (1r ) O (m 1 )]m s amp . 6), both the above inequalities still keep the strict-sign inequalities. 8). 24) p m 1   m 1 mq / r 1 [1 k (1r ) O ( m 3 )]q1   k q (r ) n r bnq .

S, n) :  k (m, n)( mn ) , m, n  N. If am , bn m 1  is a positive number, we define the weight coefficients  (r , m) and  ( s, n) as follows  (r , m) :  k (m, n)( ) p 1 s n 1 m 1 1 1 r m 1    (m)m  0 m n n 1  I   k (m, n)ambn k (r ) :  k (1, u )u r du  p / s 1 (2) for 0  p  1 , we have the following reverse equivalent inequalities: 1 p   J :  np1 ( n ) ( k (m, n)am ) p  I   k (m, n)[ mn1/( rp ) am ][ mn 1/( sq ) bn ] n 1 m 1 n 1 Bicheng Yang All rights reserved - © 2011 Bentham Science Publishers Ltd.

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