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- ILMIY AXBOROTNOMA MATEMATIKA 2016-yil, 1-son
- 2. Proof of the main results Proof of Theorem 1.
Theorem 2. Let ε be an arbitrary real positive number and 1 + ⊂ n R S ε be a smooth hypersurface given as the graph of a smooth function ( ) n n x x x x ,..., , 1 2 1 1 εφ + = + and φ satisfies the conditions 10 ILMIY AXBOROTNOMA MATEMATIKA 2016-yil, 1-son ( ) 0 0 = φ , ( ) 0 0 = ∇ φ and ( ) 0 0 ≠ φ m d , where 2 ≥ m . Then there exists a neighborhood U of the origin, such that for any fixed function ( ) U С ∞ ∈ 0 ψ and for any p>m there exists a constant C p >0 such that the following estimate p p L p p L f C f M 1 − ≤ ε ε holds for all ) ( 1 0 + ∞ ∈ n R C f , where C p does not depend on ε . 2. Proof of the main results Proof of Theorem 1. Without loss of generality we assume that ( ) 0 0 2 1 2 ≠ ∂ ∂ x φ . Let us write the averaging operator ε t A in the form ( ) ( ) ( ) ( ) ( ) ∫ + − − − = + n R n n n n t dx dx dx x x x x x x t y tx y tx y f y f A ... ,..., , ,..., , 1 ,..., , 2 1 2 1 1 2 1 1 2 2 1 1 ψ εφ ε , (2) where t > 0, ) ( 0 U C ∞ ∈ ψ is a fixed function, ( ) ( ) n n x x x Ф x x x ,..., , 1 ,..., , 2 1 2 2 1 1 ψ ε ψ ⋅ ∇ + = and n R U ∈ denotes the sufficiently small neighborhood of the origin. For a small n θ let us consider the following equation ( ) ( ) 0 cos ,..., , 1 sin 2 1 = + + n n n n x x x x θ εφ θ (3) with respect to x n . By the implicit function theorem we have that the last equation has an unique smooth solution ) , , ,..., , ( 1 2 1 ε θ n n n x x x x − for 1 2 1 ,..., , − n x x x , n θ and 0 > ε sufficiently small, such that ( ) 0 0 ,..., 0 , 0 = n x and ( ) . 0 0 ,..., 0 , 0 ≠ ∂ ∂ n n x θ Thus, in the integral (2) we can use the following change of variables ( ) ( ) ( ) ε θ θ , , ,..., , , ,..., , , ,..., , 1 2 1 1 2 1 1 2 1 n n n n n n x x x x x x x x x x − − − and then obtain ( ) ( ) ( ) ( ) ( ) , ... , , ,..., , ) , , ,..., , ( ,..., , 1 ,..., , 1 2 1 1 2 1 1 2 1 2 1 1 2 2 1 1 n R n n n n n n n n t d dx dx dx x x x x x x x x x t y tx y tx y f y f A n θ ε θ ψ ε θ εφ ε ∫ − − − + + − − − = (4) where ( ) ( ) ( ) ε θ ε θ ψ ε θ ψ , , ,..., , ) , , ,..., , ( , ,..., , : , , ,..., , 1 2 1 1 2 1 1 2 1 1 1 2 1 n n n n n n n n n x x x J x x x x x x x x x x − − − − = and ( ) ε θ , , ,..., , 1 2 1 n n x x x J − denotes the Jacobian. Let us write the integral (4) as an iterated integral ( ) ( ) ∫ − = n n n b b n t t d y f A y f A θ θ ε , (5) where b n is a positive number and n t A θ denotes the following averaging operator ( ) ( ) ( ) ( ) ( ) ∫ − − − − + + − − − = 1 1 2 1 1 2 1 1 2 1 1 2 2 1 1 .... , , ,..., , , , ,..., , 1 ,..., , : n n R n n n n n n n n t dx dx dx x x x x x x t y tx y tx y f y f A ε θ ψ ε θ εφ θ , where ( ) ( ) ) , , ,..., , ( ,..., , , , ,..., , 1 2 1 2 1 1 2 1 ε θ φ ε θ φ n n n n n n x x x x x x x x x − − = . Now, we define the rotation operator in the form ( ) ) sin cos , cos sin , ,..., , ( : 1 1 1 2 1 n n n n n n n n n x x x x x x x f x f R n θ θ θ θ θ + + − + − = , which acts izometrically on ( ) 1 + n P R L . The operator n n R A t θ θ can be written in the form ( ) ( ) ( ) ( ) ( ) ( ( ) ( ) ( ) ( ) ) ( ) . ... , , ,..., , sin , , ,..., , 1 cos , cos , , ,..., , 1 sin , ,..., , 1 2 1 1 2 1 1 2 1 1 1 2 1 1 1 1 2 2 1 1 1 − − − + − + − − × + − + − + − − − − − − = ∫ − n n n n n n n n n n n n R n n n n n n n n n n t dx dx dx x x x x x x t y tx y x x x t y tx y tx y tx y tx y f y f R A n n n ε θ ψ θ ε θ εφ θ θ ε θ εφ θ θ θ 11 ILMIY AXBOROTNOMA MATEMATIKA 2016-yil, 1-son Using the equation in (3) we have the following formula ( ) ( ) ( ) ( ) ( ) ( ) ( ) . ... , , ,..., , sin cos , cos , , ,..., , 1 sin , ,..., , 1 2 1 1 2 1 1 1 2 1 1 1 1 2 2 1 1 1 − − + − + − − + + − − − − − − = ∫ − n n n n n n n n R n n n n n n n n n n t dx dx dx x x x y y x x x t y tx y tx y tx y tx y f y f R A n n n ε θ ψ θ θ θ ε θ εφ θ θ θ Then we obtain ( ) ( ) ( ) ( ) . ... , , ,..., , , cos , , ,..., , 1 , , ... , , 1 2 1 1 2 1 1 1 2 1 1 1 2 2 1 1 1 − − + − − − − × × + + − − − = ∫ − n n n n R n n n n n n n n t dx dx dx x x x y x x x t y tx y tx y tx y f y f R A R n n n n ε θ ψ θ ε θ εφ θ θ θ (6) Now, for a small 1 − n θ we consider the equation ( ) ( ) 0 cos cos , , ,..., , 1 sin 1 1 1 2 1 1 = + + − − − − − n n n n n n n x x x x θ θ ε θ εφ θ (7) with respect to x n-1 . By the implicit function theorem we have that the equation (7) has an unique smooth solution ) , , , ,..., , ( 1 2 2 1 1 ε θ θ n n n n x x x x − − − for 2 2 1 ,..., , − n x x x , 1 − n θ , n θ and 0 > ε sufficiently small, such that ( ) 0 0 ,..., 0 , 0 1 = − n x and ( ) . 0 0 ,..., 0 , 0 1 1 ≠ ∂ ∂ − − n n x θ Then to the integral in (6) we use the change of variables as ( ) ( ) n n n n n n n n n x x x x x x x x x x θ ε θ θ θ θ ), , , , ,..., , ( , ,..., , , , ,..., , 1 2 2 1 1 2 2 1 1 2 2 1 − − − − − − andobtain that ( ) ( ) ( ) ( ) , ... , , , ,..., , , cos , , , ,..., , 1 , , ... , , 1 2 2 1 1 2 2 1 1 1 1 2 2 1 1 1 1 2 2 1 1 1 − − − − − + − − − − − − × × − + + − − − − = ∫ − n n n n n n R n n n n n n n n n t d dx dx dx x x x y x x x t y tx y tx y tx y f y f R A R n n n n θ ε θ θ ψ θ ε θ θ εφ θ θ θ (8) where ( ) ( ) ( ) ε θ ε θ θ φ ε θ θ φ , , , , , ,..., , , ,..., , , , , ,..., , 1 2 2 1 1 2 2 1 1 2 2 1 1 n n n n n n n n n n n x x x x x x x x x x − − − − − − − = , ( ) ( ) ( ) ε θ θ ε θ ε θ θ ψ ε θ θ ψ , , , ,..., , ) , ), , , , ,..., , ( , ,..., , : , , , ,..., , 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 1 n n n n n n n n n n n n n n x x x J x x x x x x x x x x − − − − − − − − − = and ( ) ε θ θ , , , ,..., , 1 2 2 1 n n n x x x J − − denotes the Jacobian. Let us write the integral (8) as an iterated integral ( ) ( ) ∫ − − − − − − = 1 1 1 1 n n n n n n b b n t t d y f A y f R A R θ θ θ θ θ , (9) where b n-1 > 0 and 1 − n t A θ denotes the following averaging operator ( ) ( ) ( ) ( ) . ... , , , ,..., , , cos , , , ,..., , 1 , , ... , , : 2 2 1 1 2 2 1 1 1 1 2 2 1 1 1 1 2 2 1 1 2 1 − − − − + − − − − − ∫ − − + + − − − = n n n n n R n n n n n n n n n t dx dx dx x x x y x x x t y tx y tx y tx y f y f A n n ε θ θ ψ θ ε θ θ εφ θ Now, we define the rotation operator in the form ( ) ) , sin cos , cos sin , ,..., , ( : 1 1 1 1 1 1 1 2 2 1 1 + − − − − − − − + − = − n n n n n n n n n n x x x x x x x x f x f R n θ θ θ θ θ , which acts izometrically on ( ) 1 + n P R L . Using the equation in (7) we can write the operator 1 1 − − n n R A t θ θ in the following form ( ) ( ) ( ) ( ) ) ( ) . ... , , , ,..., , , sin cos , cos cos , , , ,..., , 1 sin , , ... , , 2 2 1 1 2 2 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 2 2 2 2 1 1 2 1 1 − − − − + − − − − − − − − − − − − + + + − − − − − = ∫ − − − n n n n n n n n n n R n n n n n n n n n n n n t dx dx dx x x x y y y x x x t y tx y tx y tx y tx y f y f R A n n n ε θ θ ψ θ θ θ θ ε θ θ εφ θ θ θ By using similar arguments we have ( ) ( ) ( ) ( ) . ... , , , ,..., , , , cos cos , , , ,..., , 1 , ,..., , 2 2 1 1 2 2 1 1 1 1 1 2 2 1 1 1 2 2 2 2 1 1 2 1 1 1 − − − − + − − − − − − − − × × + − − − − = ∫ − − − − n n n n n R n n n n n n n n n n n t dx dx dx x x x y y x x x t y tx y tx y tx y f y f R A R n n n n ε θ θ ψ θ θ ε θ θ φ ε θ θ θ Repeating the process n-1 times we get for a small 2 θ , the following equation 12 |
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