MIT18_102s09_lec25

# MIT18_102s09_lec25 - MIT OpenCourseWare http/ocw.mit.edu...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 144 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 25. Tuesday, May 12: Fourier transform Last time I showed the completeness of the orthonormal sequence formed by the eigenfunctions of the harmonic oscillator. This allows us to prove some basic facts about the Fourier transform, which we already know is a linear operator (25.1) ( R ) , u ˆ( ξ ) = e ixξ u ( x ) dx. L 1 ( R ) −→ C ∞ Namely we have already shown the effect of the Fourier transform on the ‘ground state’: (25.2) F ( u )( ξ ) = √ 2 πe ( ξ ) . By a similar argument we can check that (25.3) F ( u j )( ξ ) = √ 2 πi j u j ( ξ ) ∀ j ∈ N . As usual we can proceed by induction using the fact that u j = Cu j − 1 . The integrals involved here are very rapidly convergent at infinity, so there is no problem with the integration by parts in (25.4) T d e − ixξ du j − 1 F ( dx u j − 1 ) = lim dx dx T →∞ − T T T = lim ( iξ ) e − ixξ u j − 1 dx + e − ixξ u j − 1 ( x ) = ( iξ ) F ( u j − 1 ) , T →∞ − T − T de − ixξ d F ( xu j − 1 ) = i dξ u j − 1 dx = i dξ F ( u j − 1 ) . Taken together these identities imply the validity of the inductive step: (25.5) F ( u j ) = F (( − d + x ) u j − 1 ) = ( i ( − d + ξ ) F ( u j − 1 ) = iC ( √ 2 πi j − 1 u j − 1 ) dx dξ so proving (25.3)....
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MIT18_102s09_lec25 - MIT OpenCourseWare http/ocw.mit.edu...

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