FOURIER TRANSFORMS MATH FALL 2016

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1 FOUIE TANSFOMS MATH. 413 FALL 16 T. PZEBINDA Contents 1. eview of Complex numbers 1.1. The field of complex numbers 1.. Multiplication in polar coordinates 1.3. The absolute value 1.4. Some geometry 3. eview of Linear Algebra Definition of a vector space 4.. Definition of a subspace of a vector space 5.3. Linear combination of vectors, subspace generated by a set of vectors, a basis of a vector space 6.4. Scalar product Finite Fourier Transform Finite cyclic group The space L (Z(N)) Definition and basic properties 1 4. Fourier series Trigonometric polynomials L 1 theory L theory Pointwise convergence of the Fourier series Some Applications of Fourier Series The isoperimetric inequality 5.. Heat equation on a circle 6. Fourier transform on The Fourier transform on the Schwartz space 6.. The Hermite functions 6.3. The Weyl-Heisenberg Uncertainty Principle The Fourier transform on L 1 () Some special functions, and the continuous and discrete data 8 7. Fourier transform on d The wave equation 3 Date: December 6, 16. 1

2 T. PZEBINDA 7.. The X-ray transform Three uncertainty principles. 34 eferences 34 Follow [ud64] and [em89] The field of complex numbers. 1. eview of Complex numbers C z = x + iy (x, y). As a vector space over, C =. Multiplication The inverse (x 1 + iy 1 )(x + iy ) = x 1 x y 1 y + i(x 1 y + x y 1 ). (x + iy) 1 = x x + y i y x + y. With these operations C is a field. Complex conjugation is an automprphism of C. The subset x + iy = x iy {z C; z = z} is a subfield isomorphic to the field of real numbers. Theorem 1. Let n be a positive integer. Then for any complex numbers a, a 1,..., a n with a n there is a z C such that Example. z = 1,... a n z n a =. 1.. Multiplication in polar coordinates. The nth roots of 1 z = x + iy = re iθ, x = r cos θ, y = r sin θ, r >, θ. Example. z = 5, z 5 = 3, The absolute value. preserves multiplication. Triangle inequality r 1 e iθ 1 r e iθ = r 1 r e i(θ 1+θ ). z n = e πik n (k =, 1,,..., n 1). z = x + iy = x + y = zz z 1 z = z 1 z. z 1 + z z 1 + z.

3 1.4. Some geometry. Sketch the sets desribed by FOUIE TANSFOMS FALL 16 3 z = 1, z 1, z 1, z = z, z = z, te it, e πi 4,... Problem 1. Define the square root function in terms of the polar coordinates on the complex plane with the negative real axis removed by e iθ = e iθ ( >, π < Θ < π). (1) Shows that in the Cartesian coordinates this function is given by x x + y + x + y x + iy = + i sign(y) x (x + iy C \ ). () Indeed, there are trigonometric half angle formulas: ( ) ( ) θ 1 + cos θ θ 1 cos θ cos =, sin = sign(θ) Hence, if then and x = cos(θ) and y = sin(θ), ( π < θ < π). ( ) θ cos = 1 + cos θ = 1 + x + x = ( ) θ sin = 1 cos θ + x sign(θ) = sign(y) and the formula follows, because = x + y. Problem. Check that the set of the two trigonomentric formulas are equaivalent to the single formula This is straightforward: cos(α + β) = cos α cos β sin α sin β (3) sin(α + β) = sin α cos β + cos α sin β e i(α+β) = e iα e iβ. (4) e i(α+β) = cos(α + β) cos(α + β) i sin(α + β) sin(α + β) and by explicit multiplication of complex numbers Hnece the claim follows. e iα e iβ = (cos α cos β sin α sin β) + i(sin α cos β + cos α sin β).

4 4 T. PZEBINDA Problem 3. Find all the solutions of the equation z 6 = 8 and draw them on the complex plane. The solutions are z k = e πik/6 (k =, 1,, 3, 4, 5). Problem 4. Sketch the set of points z C determined by the condition z 1 + i 1. This is the set of points outside the open disc or radius 1, centered at 1 i. Problem 5. Prove that for z C \ {1}, 1 + z + z + + z n = 1 zn+1 1 z (n = 1,, 3,... ). Notice that (1 + z + z + + z n )(1 z) = 1 + z + z + + z n (z + z + + z n + z n+1 ) = 1 z n+1. Problem 6. Deduce from Problem 5 that n= z n = 1 1 z (z C, z < 1). Since z < 1, we have lim n z n =. Therefore 1 + z + z + + z n +... = lim (1 + z + z + + z n 1 z n+1 ) = lim n n 1 z = 1 1 z. Follow [Lan87].. eview of Linear Algebra..1. Definition of a vector space. Fix a field F = or C. A set V is a vector space over F if it is equipped with two operations V V (u, v) u + v V, F V (a, v) av V

5 such that for any u, v, w in V and any a, b, c in F FOUIE TANSFOMS FALL 16 5 (u + v) + w = u + (v + w) c(u + v) = cu + cv (a + b)u = au + bu (ab)u = a(bu) u + = u u + v = v + u 1u = u. and there is V such that u + = u and for every u V there is u V such that u + ( u) =. Examples: C n, complex valued functions defined on a set, matrices. Counterexamples: any finite set except {}, graph of a non-linear function,.... Problem 7. Show from the axioms of a vector space that the zero vector is unique. Suppose and are to zeros in the vector space. Then = + = + =. Problem 8. Show from the axioms of a vector space that the negative of a vector is unique. Suppose u has tow negatives u and u. Then u = + ( u ) = (u + ( u)) + ( u ) = (( u) + u) + ( u ) = ( u) + (u + ( u )) = u + = u. Problem 9. Show from the axioms of a vector space that the scalar zero times any vector is the zero vector. Indeed, v = ()v = (v) = v + v. Hence = v v = v (v + v) = (v v) + v = + v = v... Definition of a subspace of a vector space. A subset U V is called a (vector) subspace if for any u, v U, the sum u + v U and for any scalar c F, cu U. Examples of subspaces and subsets which are not subspaces. Problem 1. Give an example of a subset of which is not a vector space. Any single point other than the origin. Problem 11. Describe all vector subspaces of. Here is the list: the origin, the whole space and any straight line passing through the origin.

6 6 T. PZEBINDA Problem 1. Describe all vector subspaces of C 3. The origin the whole space C 3 ; any complex line Cv where v C 3 is not zero; and complex plane Cu + Cv, where u and v are linearly independent..3. Linear combination of vectors, subspace generated by a set of vectors, a basis of a vector space. Fact: if U and W are subspaces of V then so are U W and U + W... Theorem. Every vector space V has a basis. If V has a finite basis than every other basis is finite and has the same number of elements. The number of elements in a basis of V is called the dimension of V, denoted dim V. Notation dim V = means that V does not have any finite basis. Examples: functions on a set, matrices,... In applications some bases are better than other. One says that V is the direct sum of two subspaces U and W, denoted V = U W if V = U + W and U W = {}. Examples, bases, dimensions add. Problem 13. Find dim M m,n (F). dim M m,n (F) = mn. Problem 14. Let SM m,n (F) M m,n (F) be the subspace of the symmetric matrices. Find dim SM m,n (F). dim SM m,n (F) = n(n+1)..4. Scalar product. If F = then a function V V u, v (u, v) is called a scalar product if for any u, v, w V and any c, The scalar product is positive definite if (u, v) = (v, u) (u + w, v) = (u, v) + (w, v) (cu, v) = c(v, u). (v, v) (v V) and (v, v) = if and only if v =. Examples: n with different scalar products, L (S). If F = C then a function V V u, v (u, v) is called a scalar product or inner product or hermitian product if for any u, v, w V and any c, (u, v) = (v, u) (u + w, v) = (u, v) + (w, v) (cu, v) = c(v, u).

7 The scalar product is positive definite if FOUIE TANSFOMS FALL 16 7 (v, v) (v V) and (v, v) = if and only if v =. Examples: C n with different scalar products, L (S). If (, ) is a positive definite scalar product then we define the norm and we have the Cauchy inequality and the triangle inequality u = (u, u) (u, v) u v (5) u + v u + v. (6) Problem 15. Let F = or C. Let V be a vector space over F and let (, ) denote a positive scalar product on V. If S V is any subset define S = {u V; (u, v) = ; for all v S}. Show that S is a subspace of V and that { {} if S, S S = if / S. If u, w S, then for any v V and any c C = (u, v) + (w, v) = (u + w, v), = c(u, v) = (cu, v). Thus u + v S and cu S. Also, if u is orthogonal to itself, then Hence u =. u = (u, u) =. Problem 16. Let F = or C. Let V be a vector space over F and let (, ) denote a positive scalar product on V. Show that if U V is a subspace of V and dim V <, then V = U U. As we have just shown, U U = {}. Let u 1,..., u n be an orthonormal basis of U. Fix v V. Then for any j (u j, v (v, u 1 )u 1 (v, u 1 )u... (v, u 1 )u n ) = Hence v ((v, u 1 )u 1 + (v, u 1 )u (v, u 1 )u n ) U. But v = ((v, u 1 )u 1 + (v, u 1 )u (v, u 1 )u n ) + (v ((v, u 1 )u 1 + (v, u 1 )u (v, u 1 )u n )), so we are done. Follow [SS3, Chapter 7]. 3. Finite Fourier Transform.

8 8 T. PZEBINDA 3.1. Finite cyclic group. Fix N =, 3, 4,... and let Z(N) = {, 1,,..., N 1}. This is an abelian group with the addition modulo N. Also, there is the multiplication modulo N in Z(N). Let C N = {z C; z N = 1}. This is an abelian group with the multiplication of complex numbers as the group operation. The map Z(N) k e πik N CN is a group isomorphism. Examples, N = 3, 4, 7. Problem 17. Solve for x Z(5) 4x =. There is only one solution x = 1 Z(5). 3.. The space L (Z(N)). This space consists of all functions f : Z(N) C and is equipped with the inner product Cauchy inequality 1 N Equivalently Translations N 1 k= (f, g) = 1 N f(k)g(k) 1 N N 1 k= N 1 k= f(k)g(k). f(k) 1 N N 1 k= N 1 f(k)g(k) N 1 f(k) N 1 g(k). k= multiplication my a function m k= k= T t f(k) = f(k t) (t, k Z(N)), M m f(k) = m(k)f(k) (k Z(N)). Examples: the indicator function of a subset A Z(N) { 1 if k A, I A (k) = if k / A, g(k).

9 FOUIE TANSFOMS FALL 16 9 Define the Kronecker delta δ n = NI {n}. Then any f L (Z(N)) can be written uniquely as a linear combination of the Kronecker deltas: f = N 1 k= f(k) 1 N δ k. The set 1 N δ, 1 1 δ 1,... δ N 1 (7) N N is an orthonormal basis of f L (Z(N)). This basis is perfectly localized in time n Z(N). Define convolution of functions f g(n) = 1 N Problem 18. Show that f(n k)g(k) k Z(N) δ a δ b = δ a+b (a, b Z(N)). We compute δ a δ b (x) = 1 δ a (x y)δ b (y) N y = 1 NI a (x y)ni b (y) = 1 N N NI a(x b)n y = NI a+b (x) = δ a+b (x). Problem 19. Show that for any f L (Z(N)) δ n f = T n f. We compute δ n f(x) = 1 N = y δ n (x y)f(y) y I n (x y)f(y) = f(x n).

10 1 T. PZEBINDA 3.3. Definition and basic properties. Define the Finite Fourier Transform by Ff(n) = 1 N 1 f(k)e πi N nk (f L (Z(N), n, k Z(N)). (8) N k= F is a linear map which transforms translations into modulations: If we let then (9) may be written shortly as It follows from (5) that F(T l )f(n) = e πi N nl Ff(n). (9) e l (n) = e πi N nl FT l = M e l F. (1) N 1 1 { e πi N 1 if n =, nk = N if n. k= Hence we deduce Plancherel s formula Also, and (11) Ff = f (f L (Z(N)). (1) F(f g) = NF(f)F(g) (f, g L (Z(N)) (13) This last formula follows from the following equation Problem. Prove the formula (11). This follows from Problem 5 with z = e πi N. Problem 1. Deduce (1) from (11). F 4 = I. (14) F f(n) = f( n) (f, g L (Z(N)). (15) We compute Ff = 1 F(n) N n = 1 1 f(k)e πi N kn 1 N n N k N l 1 = f(k) 1 f(l) 1 N N N k = 1 f(k) = f. N n l n f(l)e πi N ln e πi N (l k)n

11 Problem. Deduce (15) from (11). We compute FFf(n) = FOUIE TANSFOMS FALL = = l Problem 3. Deduce (14) from (15). Let 1 Ff(k)e πi N kn N k 1 N k f(l) 1 N 1 N k f(n) = f( n). We have shown that F =. Since = I, we are done. l f(l)e πi N lk e πi N kn e πi N (n+l)k = f( n). Problem 4. Show that the only possible eigenvalues of the Fourier transform F are ±1 and ±i (Here i = 1.) Since F 4 = I, any eigenvalue λ C of F has to satisfy the equation λ 4 = 1. The solutions of this equation are 1, i, 1 and i. Problem 5. Give an example of a non-zero function f L (Z(4)) such that Ff = f. (This is an eigenvector corresponding to the eigenvalue 1.) The indicator function of the subset {, } {, 1,, 3} = Z(4) will do. Problem 6. Write down the matrix of F with respect to the ordered basis (7). The matrix is M(F) = 1 N ( e πi N mn ) m,n N 1. Follow [SS3, Chapters,3]. 4. Fourier series 4.1. Trigonometric polynomials. Let Then 1 e n (x) = e πix (x ). (16) e n (x)e m (x) dx = { 1 if m = n, if m = n. (17)

12 1 T. PZEBINDA Hence for any two finite sums In particular Notice that 1 a n e n (x) m n 1 a n e n (x) n b m e m (x) dx = a n b n. (18) n dx = a n. (19) cos(nx) = 1 e n(x) + 1 e n(x) (n Z, x ) sin(nx) = 1 i e n(x) 1 i e n(x) (n Z, x ). () Hence the name of the section. Also ( 1 a k = ) a n e n (x) e k (x) dx. (1) n Let us denote the space of all the trigonometric polynomials by T P (sorry for the coincidence). Thus { } T P = a n e n ; where only finitely many a n C are nonzero. n Z Problem 7. Let f = a n e n T P. n Z Show that f(x) = for all x if and only if a n = for all n Z. n Since the claim follows. Problem 8. Let Show that a n = 1 f(x)e n (x) dx f = n Z a n e n T P. a n 1 f(x) dx.

13 FOUIE TANSFOMS FALL This is easy: a n = Problem 9. Let 1 f(x)e n (x) dx 1 f(x)e n (x) dx = 1 f = n Z a n e n T P. f(x) dx. Show that 1 f(x) dx = n Z a n. 1 f(x) dx = 1 a m e m ( x) dx a n e n (x) n Z m Z = 1 a n a m e n (x)e m ( x) dx n Z m Z = n Z a n a n. Define convolution f g(x) = Problem 3. Let 1 f(x y)g(y) dy (f, g T P, x ). () f = a n e n, g = b n e n T P. n Z n Z Show that f g = n Z a n b n e n.

14 14 T. PZEBINDA f g(x) = = 1 1 f(x y)g(y) dy b k e k (y) dy a n e n (x y) n Z k Z = 1 a n b k e n (x y)e k (y) dy n Z k Z = 1 a n b k e n (x) e n ( y)e k (y) dy n Z k Z = n Z a n b n e n (x). 4.. L 1 theory. Let L 1 denote the space of all functions f : C such that f 1 := 1 f(x) dx <, (3) f(x 1) = f(x) (x ). (4) Define convolution in L 1 by (). Check its properties including the norm inequality f g 1 f 1 g 1 (f, g L 1 ). (5) Define Fourier transform Ff(n) = Check that 1 f(x)e n (x) dx. (6) F(f g)(n) = Ff(n)Fg(n) (n Z). (7) Examples of discontinuous functions.... Show that Ff(n) f 1 (f L 1, n Z). (8) Problem 31. Let T > be a positive number. Suppose f : is a T -periodic function integrable on any bounded interval. Show that for any a T f(x) dx = a T +a f(x) dx. There is an integer n and a real number b, b < T, such that a = nt + b. Therefore, by a change of variables, T +a f(x) dx = T f(x + a) dx = T f(x + nt + b) dx = a T f(x + b) dx,

15 FOUIE TANSFOMS FALL where the last equality follows from the fact that f is T -periodic. Again, by a change of variables T f(x + b) dx = T +b and T +b f(x) dx = T b f(x) dx = b T f(t + x) dx = b f(x) dx + b T +b T f(x) dx, f(x) dx where the last equality follows from the fact that f is T -periodic. Altogether T +a a f(x) dx = b f(x) dx + T b f(x) dx = T f(x) dx. Problem 3. Let T y f(x) = f(x y). Thus T y is the translation operator that translates any function by y to the right. Check that FT y f(n) = e πiyn Ff(n) We compute: FT y f(n) = = 1 1 T y f(x)e πixn dx = f(x)e πi(x+y)n dx = = e πiyn Ff(n). Problem 33. Let < p < 1 and let f p (x) = Compute the Fourier coefficients 1 Ff p (n) 1 f(x y)e πixn dx 1 f(x)e πixn e πiyn dx = e πiyn f(x)e πixn dx { 1 p if x p, if p < x 1. (n Z). This is straightforward and for n Ff p (n) = p p = 1 p Ff() = 1 p 1 p e πixn dx = 1 e ixn dx p p 1 ( e πinp e πinp) = sin(πnp) πin πnp.

16 16 T. PZEBINDA Problem 34. Show that for a continuous function g lim g f p () = g(). p Notice that Hence, g f p () g() = 1 p g(x) dx g() = 1 p p p g f p () g() 1 p Since g is continuous which proves the claim. = max x p p p p (g(x) g()) dx. g(x) g() dx 1 p ( ) max g(x) g() p p p x p g(x) g(). lim p ( ) max g(x) g() = x p Problem 35. Show that for a continuous function h and any y lim h f p (y) = h(y). p dx Fix y and let g(x) = h(y x). Then, by Problem 3, 4.3. L theory. Define L = {f : C; h f p (y) = g f p () p g() = h(y). 1 f(x) dx <, f(x 1) = f(x), x }. The Cauchy inequality implies that the folowing integral defining an inner product on L is absolutely convergent As usual set (f, g) = 1 f(x)g(x) dx (f, g L ). 1 f = f(x) dx (f L ). This is a norm and the functions e n, (16), form an orthonormal set in L. Define Ff(n) = (f, e n ) = 1 f(x)e n (x) dx = 1 f(x)e πixn dx (f L, n Z).

17 FOUIE TANSFOMS FALL This extends our definition of the Fourier transform from the trigonometric polynomials to L. Lemma 3. Fix an integer N =, 1,,... Then for any complex numbers c n f F(n)e n f c n e n (f L ). n N n N n N n N Proof. see page 78 in the text. The point is that f c n e n = f F(n)e n + b n e n, where b n = F(n) c n and (f n N n N F(n)e n, n N b n e n ) =. n N n N By taking c n = in the above proof we see that f = f F(n)e n + F(n)e n, and hence Theorem 4. For any f L, f n N F(n) (f L ). (9) f = n Z F(n)e n in the sense of the mean square convergence: lim f F(n)e n =. N In particular n N f = n N F(n) Pointwise convergence of the Fourier series. Follow [SS3, sect..]. For N =, 1,,... we define Dirchlet kernel N D N (x) = e πinx and the Fejer kernel F N (x) = n= N N 1 n= D n (x)

18 18 T. PZEBINDA and show that F N (x) = 1 N ( ) sin(nπx). sin(πx) This last formula is used to show that for any < δ < 1 there is a positive constant c δ such that F N (x) 1 Nc δ (δ x 1 ). Furtheremore, strainght from the defnintion we see that 1 F N (x) dx = 1. These last two properties imply that for any continuous 1-periodic function f : C lim max f(x) f F N(x) =. N x Notice that f F N is a trigonometric polynomial. Hence we have the folowing theorem. Theorem 5. Any complex-valued continuous 1-periodic function may be approximated uniformly by trigonometric polynomials. Since the continuous functions are dense in L, Theorem 5 implies Theorem 4. However one should not be too optymistic because there is the following theorem, whose proof may be found in the text. Theorem 6. There is a continuous 1-periodic function f : C and a point x such that f(x) n Z F(n)e n (x). Theorem 7. For any differentiable 1-periodic function f : C f(x) = lim F(n)e n (x) (x ). N n N Let g : C be the π-periodic function defined by Then, Fg() = and for n, g(x) = x if x < 1. Problem 36. Show that Fg(n) = ( 1)n+1 πin. n=1 1 n = π 6.

19 FOUIE TANSFOMS FALL Notice that We know from Theorem 4 that n=1 1 n = 1 Fg(n). n Z Fg(n) = n Z 1 1 x dx. The last integral is easy to compute and the formula follows. Problem 37. Let V be a vector space over the complex numbers with a positive definite hermitian product (, ) and the corresponding norm v = (v, v). Verify the following identity 4(u, v) = u + v u v +i u + iv i u iv. This is completely straightforward. Problem 38. Let V and V be two vector spaces over the field of complex numbers. Let (, ) be a positive definite hermitian product on V and let (, ) be a positive definite hermitian product on V. Denote by v = (v, v) and v = (v, v ) the corresponding norms. Suppose T : V V is a linear map such that Show that T v = v (v V). (T u, T v) = (u, v) (u, v V). We use the identity of the previous problem: 4(T u, T v) = T u + T v T u T v + i T u + it v i T u it v = T (u + v) T (u v) + i T (u + iv) i T (u iv) = u + v u v +i u + iv i u iv = 4(u, v). Problem 39. Deduce from Problem and Parseval s identity (Theorem 1.3 (ii) in the tex, or Theorem 4 above) that for any π-periodic, square integrable functions f, g : C 1 π π π f(x)g(x) dx = n Z Ff(n)Fg(n).

20 T. PZEBINDA Take V to be the space of all the π-periodic, square integrable functions f : C with the hermitian product 1 π f(x)g(x) dx π π and V to be the space of all the sequences a n C such that n Z a n <, with the hermitian product a n b n. n Z Since the linear map F : V V preserves the norm, the formula follows from Problem Some Applications of Fourier Series The isoperimetric inequality. Follow [SS3, Chapter 4, section 1]. Problem 4. Test the isoperimteric inequality when the curve is the boundary of a regular N-gon. As you can find in Wikipedia, the area of a regular N-gon of diameter is equal to A = 1 N sin( π N ) and the length of its boundary (perimeter) l = N sin( π N ). Hence 1 l A 4π = tan( π ) N π > 1. N Notice that the limit if N of the left hand side is Heat equation on a circle. Follow [SS3, Chapter 4, section 4]. 6. Fourier transform on The Fourier transform on the Schwartz space. Follow [SS3, Chapter 5, sections ]. 6.. The Hermite functions. Follow [HT9, Chapter 3, section.1]. Notice that the Fourier transform as a map from S() to itself satisfies the following identities FxF 1 = 1 πi, Here = d dx and x is the multiplication by x. Hence F F 1 = πix. (3) F(πx )F 1 = i(πx ). (31)

21 ecall the Gaussian g(x) = e πx FOUIE TANSFOMS FALL 16 1 and let h n (x) = (πx ) n g(x) (n =, 1,,...). (3) We see from (31) and from the fact that Fg = g that Fh n F 1 = ( i) n h n (n =, 1,,...). (33) ecall the positive definite inner product (f 1, f ) = f 1 (x)f (x) dx (f 1, f S()). For a linear map T : S() S() define T : S() S() by Then x = x and =. Hence (f 1, T f ) = (T f 1, f ) (f 1, f S()). (πx ) = (πx + ). Set a = (πx + ). Then h n = (a ) n h, with h = g. Notice that h = (h, h ) = e πx dx = e π(x ) d(x ) 1 = 1. (34) Also, by a short computation, Hence for k = 1,, 3,... ah =. (35) (h, h k ) = (h, (a )h k 1 ) = (ah, h k 1 ) = (, h k 1 ) =. (36) ecall the commutator [X, Y ] = XY Y X. An inductive argument shows that Since [x, ] = 1 we see that Hence, n 1 [X, Y n ] = Y j [X, Y ]Y n j 1 (n = 1,, 3,...). (37) j= [a, (a ) n ] = 4πn(a ) n 1 (n = 1,, 3,...). ah n = ([a, (a ) n ] + (a ) n a)h = [a, (a ) n ]h = 4πn(a ) n 1 h = 4πnh n 1 (n = 1,, 3,...). (38) Let 1 n m. We see from the above that (h n, h m ) = ((a )h n 1, h m ) = (h n 1, ah m ) = (h n 1, 4πmh m 1 ) = 4πm(h n 1, h m 1 ). (39) By combining (35), (36) and (39) we see that (h n, h m ) = (4π) n n! 1 δ m,n (n, m =, 1,,...). (4)

22 T. PZEBINDA Set v n = (4π) n n! h n (n, m =, 1,,...). (41) Then the functions v, v 1, v... form an orthonormal system of eigenvectors in S() for F: Fv n = ( i) n v n (n =, 1,,...). (4) One can prove, though this is non-trivial that this system is complete in L () in the sense that any f L () may be written uniquely as f = (f, v n )v n. (43) Problem 41. ecall the Hermite functions v n, (41). Compute the integral (v 1 (x) + v (x) + 3v 3 (x)) dx. n= We recognize that the above integral is equal to (v 1 + v + 3v 3, v 1 + v + 3v 3 ) = (v 1, v 1 ) + (v 1, v ) + (v 1, 3v 3 ) + (v, v 1 ) + (v, v ) + (v, 3v 3 ) + (3v 3, v 1 ) + (3v 3, v ) + (3v 3, 3v 3 ) = (v 1, v 1 ) + (v, v ) + (3v 3, 3v 3 ) = (v 1, v 1 ) + 4(v, v ) + 9(v 3, v 3 ) = = 14. Problem 4. Compute the Fourier transform F(v 1 + v + 3v 3 ). Since Fv j = ( i)v j We see that F(v 1 + v + 3v 3 ) = iv 1 v + i3v 3. Problem 43. Let = d. Let us identify a smooth functio p(x) with the multiplication dx operator f pf. Show that [, p] = p

23 FOUIE TANSFOMS FALL 16 3 where p is the derivative of p. For any test function f [, p]f = (pf) p( f) = p f + pf pf = p f. Hence [, p] coincides with the multiplication by p. Problem 44. Is there a non-zero function f S() such that Ff = 3f? No, because the only eigenvalues of the Fourier transform are ±1 and ±i. Problem 45. Are there a non-zero function f, g S() such that f(x)g(x) dx = 4 and Ff(x)Fg(x) dx = 6? No, because the Fourier transform preserves the scalar product given by the above integral. Problem 46. Is there a function f S() such that f(x) > for all x and Ff(x) > for all x? Yes, the Gaussian g(x) = e πx. Problem 47. Is there a function f S() such that f(x) > for all x and Ff() =? No, because Ff() = f(x) dx >. The energy oprator of a quantum oscillator may be defined as Notice that Hence (38) shows that H = (πx). H = 1 (aa + a a). (44) Hv n = 4π(n + 1 )v n (n =, 1,,...). (45) Thus the Hertmite functions are the states of the quantum oscillator at the energies 4π(n + 1 ).

24 4 T. PZEBINDA 6.3. The Weyl-Heisenberg Uncertainty Principle. Follow [Wic94]. Let f S() be a real valued function. Then, by the Cauchy inequality, ( xf(x) dx f (x) dx xf(x)f (x) dx). Moreover, xf(x)f (x) dx = x 1 ( ) f(x) 1 dx = f(x) dx. Hence xf(x) dx f (x) dx 1 4 f 4. If the equality holds, then there is a such that f (x) = af(x). Thus there is a constant c such that f(x) = ±e ax c. Since f is square integrable, a <. Let us write f as f(x) = e πixξ Ff(ξ) dξ, where Ff is the Fourier transform of f. Then f (x) = πiξe πixξ Ff(ξ) dξ. Therefore, Thus for f = 1, f (x) dx = 4π xf(x) dx A few more easy steps lead to the following theorem. ξ Ff(ξ) dξ. ξ Ff(ξ) dξ 1 16π. Theorem 8 (H. Weyl, 1931). Let f S() with f = 1. Set µ = x f(x) dx, σ = (x µ) f(x) dx, µ = ξ Ff(ξ) dξ, σ = (ξ µ) Ff(ξ) dx. Then σ σ 1 4π and the equality occurs if and only if f(x) = e ax bx c, where a > and b, c C. In other words, f is a constant multiple of a translation f(x) f(x + x ) and a modulation f(x) e πiy x f(x) of a Gaussian e ax.

25 FOUIE TANSFOMS FALL 16 5 Problem 48. Let n =, 1,,... be a polynomial and let { x f(x) = n e x 1 if x >, if x. Show that f is continuous. We need to show that limf(x) =. x This follows by a repetitive application of Hospital rule. Problem 49. Let P be a polynomial and let { P (x f(x) = 1 )e x 1 if x >, if x. Show that f is continuous. This follows from the previous problem because k P (x 1 ) = a n x n. n= Problem 5. Using the ideas of the previous two problems one may show that the function { e x 1 if x >, f(x) = if x is infinitely many times differentiable. Use this fact to crate a non-zero smooth compactly function g :. Let f be as in the previous problem. Set This is such an example. g(x) = f(1 x ) (x ) The Fourier transform on L 1 (). We define Ff(y) = e πixy f(x) dx (f L 1 ()). This integral is absolutely convergent and Ff f 1 (f L 1 ()), (46) where Ff = sup Ff(y) and f 1 = f(x) dx. y Let C () be the space of the continuous functions which have limit zero at infinity: C () = {f : C; f is continous and lim x f(x) = }.

26 6 T. PZEBINDA As we checked in class Ff(L 1 ()) C (). (47) This fact is known as the iemann - Lebesgue Lemma. Also, we checked that the Fourier transform of the function e π x (x ) (48) is equal to 1 1 (y ). (49) π y + 1 However at this point we don t know that in this new setting the Fourier transform is invertible, so we can t say that e π x = e πixy 1 1 dy (x ), (5) π y + 1 though the integral is absolutely convergent and the equation is expected. Computing the right hand side of (5) directly is not easy. Instead we check the following fact. Lemma 9. For any f L 1 () and any φ S(), Ff(y)φ(y) dy = f(x)fφ(x) dx. Proof. The left hand side is equal to e πixy f(x)φ(y) dx dy = = f(x)e πixy φ(y) dy dx e πixy f(x)φ(y) dy dx which coincides with the right hand side. The change of the order of the integration is justified by the absolute convergance of the double integral and the Fubini s Theorem. Corollary 1. The formula (5) holds. Proof. Let P (y) = 1 1 π y + 1. Let L(x) denote the left hand side and (x) the right hand side of (5). Fix a function φ S(). Lemma 9 and the formula (49) show that (x)φ(x) dx = e πixy P (y) dyφ(x) dx = P (y) e πixy φ(x) dx dy = P (y)fφ( y) dy = P ( y)fφ(y) dy = P (y)fφ(y) dy = FP (x)φ(x) dx = L(x)φ(x) dx.

27 Thus for all φ S(). Let ψ C c FOUIE TANSFOMS FALL 16 7 (L(x) (x))φ(x) dx = () be non-negative, compactly supported near zero, with ψ(x) dx = 1. By taking φ(x) = t 1 ψ(t 1 x), and letting t > approach, we conclude that L(x) (x) = (x ). Problem 51. Spend 5 minutes trying to compute the right hand side of (5) directly. Problem 5. Compute the Fourier transform of the function { 1 if x 1, f(x) = if x > 1. Ff(y) = sin(πy) πy Problem 53. Compute the Fourier transform of the function { 1 x if x 1, f(x) = if x > 1.. Problem 54. ecall the gamma function Γ(s) = Show that ( ) sin(πy) Ff(y) =. πy e x x s 1 dx (s > ). Γ( 1 ) = π. We compute e x x 1 dx = e y dy = e y dy = π.

28 8 T. PZEBINDA Problem 55. Suppose both f : C is supported in the interval [a, b]. Let t >. Show that the function f t (x) = t 1 f(t 1 x) is supported in [ta, tb]. If f t (x) then a t 1 x b. Hence the conclusion follows. Problem 56. Suppose both f : C is supported in the interval [a, b]. Let t >. Show that the function T x f(x) = f(x x ) is supported in [a + x, b + x ]. If T x f(x) then a x x b. Hence the conclusion follows. Problem 57. Let f L 1 (). Show that for any x, supp Ff = supp FT x f. This is because FT x f is a multiple of Ff: FT x f(y) = e πix y Ff(y). Problem 58. Let f L 1 (). For t > define f (t) (x) = t 1 f(t 1 x). Check that Ff (t) = (Ff) (t 1 ). This is because t 1 f(t 1 x)e πixy dx = t 1 f(x)e πixty dx. Problem 59. Suppose both f S() and Ff are compactly supported. Show that f =. (Use the hint to exercise 1, page 167 in the text.) Using translations f(x) f(x x ) and dilations f(x) t 1 f(t 1 x) we may assume that supp f (, 1 ). (See Problems (55) - (58).) Then 1 f(x)e πinx dx = Ff(n) (n Z) is zero for n large enough. From the theory of Fourier series f(x) = Ff(n)e πinx (x [, 1]), n where the sum is finite. Hence f is a trigonometric polynomial. But trigonometric polynomials have only finitely many zeros, so f can t vanish on the whole interval [ 1, 1] unless f = Some special functions, and the continuous and discrete data. The following functions H t (x) = 1 e x 4t and P t (x) = 1 t (x, t > ) (51) 4πt π x + t are known as the Gaussian probability distribution and Poisson probability distribution, respectively. They are very important in the Theory of Probability and in Statistics.

29 FOUIE TANSFOMS FALL 16 9 Problem 6. Check that FH t (x) = e 4π tx. ecall the Gaussian g(x) = e πx and the fact that g = Fg. Since H t (x) = s 1 e π(s 1 x) = s 1 g(s 1 x), the usual computation shows that FH t (x) = Fg(sx) = e π(sx). Problem 61. Show that that FP t (x) = e πt x. Since P t (x) = t 1 P 1 (t 1 x), the usual computation shows that FP t (x) = FP 1 (tx). Then the formula follows from (5). ecall the convolution of two integrable function, (). Problem 6. Show that H t H s = H t+s (s, t > ). By applying the Fourier transform we see that the equation is equivalent to the e 4π tx e 4π sx = e 4π (t+s)x, which is obviously true. Problem 63. Show that P t P s = P t+s (s, t > ). By applying the Fourier transform we see that the equation is equivalent to the which is obviously true. e πt x e πs x = e π(t+s) x,

30 3 T. PZEBINDA Problem 64. Suppose f S() is such that for some fixed t > Show that f =. P t f =. By applying the Fourier transform we see that the equation is equivalent to the e πt x Ff(x) = (x ), Since the exponential is never zero we see that Ff =. Hence f =. The equation (53) below is known as the Poisson Summation Formula. Theorem 11. Suppose f S(). Then Equivalently, n Z f(x + n) = n Z n Z f(n) = n Z Ff(n)e πinx (x ). (5) Proof. For a proof of (53) and (5) see [SS3, pages 154 and 155]. Ff(n). (53) The next theorem shows thata function whose Fourier transform is compactly supported is determined by discrete data. Theorem 1. Suppose f S() is such that supp Ff [ 1, 1 ]. Then f(x) dx = Ff(n). (54) n Z and f(x) = n Z sin(π(x n)) f(n). (55) π(x n) Proof. The left hand side of (54) is equal to 1 Ff(x) dx = Ff(x) dx = c n, 1 n Z where c n is the Fourier coefficient of the function Ff. By definition, 1 c n = Ff(x)e πinx dx = Ff(x)e πinx dx = f( n). 1 This implies the equality (54).

31 FOUIE TANSFOMS FALL Furtheremore, Ff(y) = Ff(y)χ(y) = c n e πiny χ(y) = f( n)e πiny χ(y) n Z n Z = f(n)e πiny χ(y), n Z where Hence, χ(x) = { 1 if x 1, if x > 1. f(x) = f(n) e πiny χ(y)e πiyx dy n Z = 1 f(n) e πi(x n)y dy, n Z 1 which equals to the right hand side of (55). Problem 65. Suppose g S() is such that supp Fg [ a, a]. Show that g(x) = n Z g( 1 n)) n)sin(π(ax. a π(ax n) This follows from (55) by writting g(x) = t 1 f(t 1 x), where a = 1 t. (Done in class.) Problem 66. Let Show that H t (x) = n Z e 4π tn e πinx (x ). H t (x) >. This follows from (5) because the function H t (x) defined in (51) is in S() and has the Fourier transform computed in Problem 6. (Done in class.)

32 3 T. PZEBINDA Follow [SS3, Chapter 6]. 7. Fourier transform on d. Problem 67. ead [SS3, Chapter 6, sections 1,, 3] The wave equation. Problem 68. Let u(x, t) be a solution of the wave equation such that lim u(x, t) = lim x Show that the energy E(t) = is a constant function of t. u(x, t) = t u(x, t) (x d, t > ) x j u(x, t) = (j = 1,,.., d) x d t u(x, t) dx + d j=1 d xj u(x, t) dx It will suffice to check that the derivative of E(t) is zero: d E (t) = t u(x, t) t u(x, t) dx + xj u(x, t) t xj u(x, t) dx d d j=1 = d d t u(x, t) x j u(x, t) dx + xj u(x, t) t xj u(x, t) dx d j=1 j=1 d d ( ) = j=1 t u(x, t) x j u(x, t) dx + d xj u(x, t) t xj u(x, t) dx d d ( ) = xj t u(x, t) xj u(x, t) dx + d xj u(x, t) t xj u(x, t) dx d =. j=1 Problem 69. Check that the solution of the wave equation u(x, t) = t u(x, t) u(x, ) = f(x), t u(x, ) = g(x) (x 3, t > ) may be written as u(x, t) = 1 f(x ty) dσ(y) t grad(f)(x ty) y dσ(y) 4π S 4π S + t g(x ty) dσ(y). 4π S

33 FOUIE TANSFOMS FALL This follows from the fact that t f(x ty) dσ(y) = grad(f)(x ty) y dσ(y). S S Problem 7. Let u be as in Problem 69. Does u(, 1) depend on f() or g()? No because is not on the sphere of radius 1 centered at. Problem 71. Let u be as in Problem 69. Suppose f = and g. Show that u. This follows from the formula 7.. The X-ray transform. u(x, t) = t 4π S g(x ty) dσ(y). Problem 7. Draw a picture showing that the formula l t,θ = {t(cos θ, sin θ) + u(sin θ, cos θ); u } (t, θ [, π)) describes a line in passing through the point t(cos θ, sin θ) and perpendicular to the line (cos θ, sin θ). Done in class. Problem 73. Define the X-ray transform (adon transform) on by X(f)(t, θ) = f(t(cos θ, sin θ) + u(sin θ, cos θ)) du (56) (t, θ [, π), f S()). and the one dimensional Fourier transform of the X-ray transform by ˆX(f)(τ, θ) = e πiτt X(f)(t, θ) dt (τ, θ [, π), f S()). (57) Show that ˆX(f)(τ, θ) = Ff(τ cos θ, τ sin θ). (58) Notice that t = (cos θ, sin θ) (t(cos θ, sin θ) + u(sin θ, cos θ)). Hence, the left hand side of (58) is equal to e πiτ(cos θ,sin θ) (t(cos θ,sin θ)+u(sin θ, cos θ)) (59) f(t(cos θ, sin θ) + u(sin θ, cos θ)) du dt.

34 34 T. PZEBINDA Notice that (x, y) = t(cos θ, sin θ) + u(sin θ, cos θ) is a change of variables with Jacobian equal to 1. Hence (59) is equal too e πiτ(cos θ,sin θ) (x,y) f(x, y) dx dy = Ff(τ cos θ, τ sin θ). Problem 74. Show that π e πiτ(cos θ,sin θ) (x,y) ˆX(f)(τ, θ) dθ τ dτ = f(x, y). (6) By Problem 73 and integration in polar coordinates and Fourier inversion formula, the left hand side of (6) is equal to = π e πiτ(cos θ,sin θ) (x,y) Ff(τ cos θ, τ sin θ)τ dτ dθ e πi(x,y ) (x,y) Ff(x, y ) dx dy = f(x, y). Problem 75. Show that for any two functions f, g S(), X(f) = X(g) if and only if f = g. If X(f) = X(g) then ˆX(f) = ˆX(g). Hence the claim follows from Problem 74. Follow [OP4b] and [ÖP4a]. 8. Three uncertainty principles. eferences [HT9]. Howe and E. C. Tan. Non-Abelian Harmonic Analysis, Applications of SL (). Springer Verlag, 199. [Lan87] S. Lang. Linear Algebra. Springer Verlag, California, [ÖP4a] Özaydın and T. Przebinda. An Entropy Based Uncertainty Principle for a Locally Compact Abelian Group. J. Funct. Anal., 15:41 5, 4. [OP4b] E. Matusiak M. Ozaydin and T. Przebida. Three Uncertainty Principles. epresentation Theory of eal and p-adic Groups, edited by Eng-Chye Tan and Chen-Bo Zhu Singapore University Press, pages 1 18, 4. [em89] emmert,. Theory of Complex Functions. Springer-Verlag, [ud64] udin, W. Principles of Mathematical Analysis. McGraw-Hill, Inc, [SS3] E.M. Stein and. Shakarchi. Fourier Analysis: An Introduction (Princetone Lectures in Analysis). Princeton University Press, Princeton, NJ, 3. [Wic94] M. V. Wickerhauser. Adapted Wavelet Analysis from Theory to Software. A K Peters, 1994.