fourier

# fourier - A NALYSIS TOOLS W ITH APPLICATIONS 379 20 Fourier...

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ANALYSIS TOOLS WITH APPLICATIONS 379 20. Fourier Transform The underlying space in this section is R n with Lebesgue measure. The Fourier inversion formula is going to state that (20.1) f ( x )= μ 1 2 π n Z R n dξe iξx Z R n dyf ( y ) e iyξ . If we let ξ =2 πη, th ismaybewr ittenas f ( x )= Z R n dηe i 2 πηx Z R n dyf ( y ) e iy 2 πη andwehaveremovedthemultiplicativefactorof ¡ 1 2 π ¢ n in Eq. (20.1) at the expense of placing factors of 2 π in the arguments of the exponential. Another way to avoid writing the 2 π ’s altogether is to rede f ne dx and andthisiswhatwewilldohere. Notation 20.1. Let m be Lebesgue measure on R n and de f ne: d x = μ 1 2 π n dm ( x ) and d ξ μ 1 2 π n dm ( ξ ) . To be consistent with this new normalization of Lebesgue measure we will rede f ne k f k p and h f,g i as k f k p = μZ R n | f ( x ) | p d x 1 /p = Ã μ 1 2 π n/ 2 Z R n | f ( x ) | p dm ( x ) ! 1 /p and h f,g i := Z R n f ( x ) g ( x ) d x when fg L 1 . Similarly we will de f ne the convolution relative to these normalizations by f F g := ¡ 1 2 π ¢ n/ 2 f g, i.e. f F g ( x )= Z R n f ( x y ) g ( y ) d y = Z R n f ( x y ) g ( y ) μ 1 2 π n/ 2 dm ( y ) . The following notation will also be convenient; given a multi-index α Z n + , let | α | = α 1 + ··· + α n , x α := n Y j =1 x α j j ,∂ α x = μ ∂x α := n Y j =1 μ ∂x j α j and D α x = μ 1 i | α | μ ∂x α = μ 1 i ∂x α . Also let h x i := (1 + | x | 2 ) 1 / 2 and for s R let ν s ( x )=(1+ | x | )

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380 BRUCE K. DRIVER 20.1. Fourier Transform. De f nition 20.2 (Fourier Transform) . For f L 1 , let ˆ f ( ξ )= F f ( ξ ):= Z R n e ix · ξ f ( x ) d x (20.2) g ( x )= F 1 g ( x )= Z R n e ix · ξ g ( ξ ) d ξ = F g ( x ) (20.3) The next theorem summarizes some more basic properties of the Fourier trans- form. Theorem 20.3. Suppose that f,g L 1 . Then (1) ˆ f C 0 ( R n ) and ° ° ° ˆ f ° ° ° u k f k 1 . (2) For y R n , ( τ y f )ˆ( ξ )= e iy · ξ ˆ f ( ξ ) where, as usual, τ y f ( x ):= f ( x y ) . (3) The Fourier transform takes convolution to products, i.e. ( f F g ) ˆ = ˆ f ˆ g. (4) For f,g L 1 , h ˆ f,g i = h f, ˆ g i . (5) If T : R n R n is an invertible linear transformation, then ( f T ) ( ξ )= | det T | 1 ˆ f ( ¡ T 1 ¢ ξ ) and ( f T ) ( ξ )= | det T | 1 f ( ¡ T 1 ¢ ξ ) (6) If (1+ | x | ) k f ( x ) L 1 , then ˆ f C k and α ˆ f C 0 for all | α | k. Moreover, (20.4) α ξ ˆ f ( ξ )= F [( ix ) α f ( x )] ( ξ ) for all | α | k.
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fourier - A NALYSIS TOOLS W ITH APPLICATIONS 379 20 Fourier...

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