By H. Jacquet, R. P. Langlands

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

**Sample text**

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 m1/ 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 )]q1 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 : np1 ( 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.