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220-HW5 - Problem 5 Let R n = R m × R k and write R n ∋...

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MATH 220: PROBLEM SET 5 DUE TUESDAY, OCTOBER 27, 2009 Problem 1. Suppose that f is (piecewise) continuous on R n with | x | N f ( x ) bounded for some N >n (or indeed simply that f L 1 ( R n )). Throughout this problem, a R n . (i) Let f a ( x ) = f ( x a ). Show that ( F f a )( ξ ) = e - ia · ξ ( F f )( ξ ). (ii) Let g a ( x ) = e ix · a f ( x ). Show that ( F g a )( ξ ) = ( F f )( ξ a ). (iii) Show that ( F - 1 f a )( x ) = e ia · x ( F - 1 f )( x ). (iv) Show that ( F - 1 g a )( x ) = ( F - 1 f )( x + a ). Problem 2. Use part (i) of Problem 1 to show that ( F ( j f ))( ξ ) = j ( F f )( ξ ) if f is C 1 and | x | N f , | x | N j f are bounded for some N >n . Problem 3. Find the Fourier transform on R of the following functions: (i) f ( x ) = H ( a − | x | ), where a> 0 and H is the Heaviside step function, so H ( t ) = 1 if t> 0, H ( t ) = 0 if t< 0. (ii) f ( x ) = H ( x ) e - ax , where a> 0. (iii) f ( x ) = | x | n e - a | x | , where a> 0, and n 0 integer. (iv) f ( x ) = (1 + x 2 ) - 1 . (Hint: use that if f = F - 1 g then g = F f by the Fourier inversion formula. Rewrite (1 + x 2 ) - 1 as partial fractions (factor the denominator).) Problem 4. Find the Fourier transform on R 3 of the function f ( x ) = | x | n e - a | x | , where a > 0 and n ≥ − 1 integer. (Hint: express the integral in the Fourier transform in polar coordinates.)
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Unformatted text preview: Problem 5. Let R n = R m × R k , and write R n ∋ x = ( y, z ) ∈ R m × R k . Suppose that f ∈ C 1 ( R m × R k ) and | z | K f , | z | K ∂ x j f are bounded for all j = 1 , . . ., n , and K > k . De±ne the partial Fourier transform of f by ( F z f )( y, ζ ) = i R k e-iz · ζ f ( y, z ) dz, y ∈ R m , ζ ∈ R k . Show that (i) ( F z D z j f )( y, ζ ) = ζ j ( F z f )( y, ζ ). (ii) ( F z D y j f )( y, ζ ) = ( D y j ( F z f ))( y, ζ ). Note : Under appropriate additional assumptions, as for the full Fourier transform, the formulae F z ( z j f ) = − D ζ j F z f , F z ( y j f ) = y j F z f , and the analogous formulae for ( F-1 ζ ψ )( y, z ) = (2 π )-k i R k e iz · ζ ψ ( y, ζ ) dz also hold, but you do not need to prove this (but you should know these!). 1...
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