f (x)e inx dx 2π π for 2π period functions. Now take we can take an arbitrary interval, then our dense exponentials are 1 2π e(inπx)/a.

Transcription

1 7 Fourier Transforms (Lecture with S. Helgason) Fourier series are defined as f () a n e in, a n = π f ()e in d π π for π period functions. Now take we can take an arbitrary interval, then our dense eponentials are π e(inπ)/. Then for L (, ) f () e inπ inπ Let ( ) f (y)e inπy nπ = g Then f () g(nπ/)e(inπ)/ (π/), then think of nπ/ as a new variable u = π π nπ/, du = Δ, and let define g(u) = f ()e iu d and we hope that f (y) π e f () g(u)e iu du π 4

2 Definition. f L () define the fourier transform fˆ by fˆ(y) = f ()e iy d, y fˆ is bounded and uniformly continuous because fˆ(y) f () e iy d f () d < f L () and to show uniform continuity i(y+h) fˆ(y + h) fˆ(y) = f ()(e e iy )d = f ()e iy (e ih )d f () e ih d f () so by Lebesgue dominated convergence theorem, we can take limits f () e ih ih lim d = lim f () e d = 0 h 0 h 0 so f is continuous, and in fact uniformly continuous, because of the bound f () (the y dropped out). Definition. Schwarz Space function f is a Schwarz function if for each m 0, l 0 m dl f d l, bounded on Note 3 facts C () S, e / S and S L (). lso, it is not too hard to see by c the definition that if f S() then for some C. Lemma. f S, g() = f (), then fˆ(y) = f ()e iy d, f () C( + ) N d fˆ(y) = f ()( i)e iy d In fact f ()( i)e iy d is uniformly convergent, and now ( dfˆ ) (y) = i f (y), f (y) = i dfˆ 43

3 Lemma. f S, h = df /d Then by integration by parts ( ) [ ] ˆ df df h(y) = = e iy d = f ()e iy d d t k f (t)(ξ) = ( i) k fˆ(ξ) dξ k ( ) df (y) = iyfˆ(y). d f ()( iy)(e iy )d and as, f () 0, so then the epression above winds up being iyfˆ(y). Thus So sum up what we have done above in the following theorem Theorem. d k d n f (t) (ξ) = (iξ) n fˆ(ξ) dt n This helps in differential equations, the fourier transforms interchanges differentiation and multiplication, and vice versa. But the function we deal with has to be small at infinity, however the Schwarz space allows for more general fourier transforms, like polynomials. However we don t know that functions are fourier transforms of L, that is L () =?. However, for Schwarz functions... Theorem. f fˆ is a bijection of S onto S and iy f () = fˆ(y)e π This is the fourier inversion formula. Before the proof we present an eample Eample. f the characteristic function of [, ]. Then fˆ(y) = e iy d = sin y y Then we would like to show that f = sin y e iy π y But there is a snag at y = 0. Consider sin y sin y (cos y + i sin y) = cos y y y 44

4 (since i sin y (sin y/y) is odd). So then to take care of y = 0 sin y sin( + )y sin( )y cos y = + 0 y 0 y y But sin αy = sgn α 0 y so as we get sgn( + ) sgn( ) = χ [,]. We continue with the proof of the theorem Proof. If f S, then fˆ S. nd the following are both fourier transforms of Schwarz functions: ( ) ( ) d df (y) fˆ = ±i yf (y), = iyfˆ(y) By iterations of these we can show that y m dl fˆ is bounded, in fact l d n fˆ m y n is the fourier transform of a Schwarz function, so it is bounded and thus fˆ is a Schwarz function. ssume the inversion formula is proved. Then f fˆ is one to one, also f fˆ is surjective because g S, gˇ = f, fˆ = g, since fˆ f is one to one. Now we need to prove the inversion formula. First we find the fourier transform of e = f (). This is the solution to the differential equation df f + = 0 d Now if h is a solution then ( ) d dh (e h) = e h + = 0 d d So e h is constant, then h = Ce, now ( ) df dfˆ f + = i + iyfˆ(y) = 0 d So fˆ satisfies the same differential equation as f, so fˆ(y) = Ce y. Calculate this at the origin to find the constant fˆ(0) = π, so We need a few lemmas to finish the proof fˆ(y) = πe y 45

5 Lemma. Symmetry Lemma g, f S then f ()ĝ()d = fˆ(y)g(y) Proof. We can just use Fubini s theorem with ( ) f ()e iy d g(y) Lemma. f S, a, f a () = f ( + a) then fˆ(y + a) = fˆa (y) = e iay fˆ(y) Proof. Just by definition of the transform and change of variables. Lemma. f S, f a () = f (/a), then Proof. Same as above fˆ(y/a) = fˆa() = afˆ(ay) Now a special case of the symmetry law gives a fˆ(y)e y a = πa f ()e d Now if we do a change of variables and let s = a, then the above becomes ( s ) s fˆ(y)e y a = f e ds a Let a. Then the above becomes s fˆ(y) = πf (0) e ds = πf (0) so the inversion holds at point 0. Use lemma about f a (), so f a (0) = f (a) = e iay fˆ(y) Theorem. Let f S, then π f () L = π f () d = fˆ(y) = fˆ(y) L Proof. Observe take g = ( fˆ). Then by symmetry ĝ = f (), and π f ()f()d = fˆ(y)fˆ(y) f () = fˆ(y)e iy = (fˆ)(). π π 46