x λ ϕ(x)dx x λ ϕ(x)dx = xλ+1 λ + 1 ϕ(x) u = xλ+1 λ + 1 dv = ϕ (x)dx (x))dx = ϕ(x)

Transcription

1 1. Distributions A distribution is a linear functional L : D R (or C) where D = C c (R n ) which is linear and continuous Topology on D. Let K R n be a compact set. Define D K := {f D : f outside K} f K,m = sup D s f(x). x K, s m Then f n f means (1) suppf n K same K for all n (2) for any D s, D s f n D s f uniformly. T is continuous means for every K there is C K and m > so that 1.2. The distribution x λ +. T (ψ) C ψ K,m ψ D K. T λ (x) := x λ ϕ(x)dx is well defined for λ C such that Reλ > 1. T λ (x) = x λ ϕ(x)dx = xλ+1 λ + 1 ϕ(x) 1 λ + 1 x λ+1 ϕ (x)dx du = x λ dx v = ϕ(x) So u = xλ+1 λ + 1 dv = ϕ (x)dx T λ (x) = 1 λ + 1 T λ+1 The right hand side makes sense for Re(λ + 1) > 1 or Reλ > 2. In this way we can extend T λ to Reλ > 2, but there is a pole at λ = 1. lim (λ + 1)T λ = T = λ 1 So δ is the residue of T λ at λ = 1. (ϕ (x))dx = ϕ(x) 1 = ϕ()

2 Distributions with support at. Definition. We say a distribution T vanishes in an open set U, if T (ϕ) = for any function ϕ C c (R), with support in U. The support of a distribution is the smallest closed set F such that T vanishes in the open set U = R\F. Theorem. A distribution T such that suppt = {} is a derivative of δ. Proof. Let ψ be a function in D which is 1 on ( 1/2, 1/2), and vanishes outside ( 1, 1). Then write ψ ε (x) = ψ(x/ε). ψ ε (x) is identically 1 in ( ε, ε). Let f ε = f ψ ε. Because of the support property, T (f) = T (f ε ) Indeed this is because f f ε near, so T (f f ε ) =. Because of continuity, for any compact K, T (f) C sup D j f. j <k,x K We will apply this to the functions f ε, with ε, so we can take K = [ 1, 1]. By direct calculation, we see that D j f ε is of the form ε j D j ψ D i f, i + j k. Assume that D j f() = for all j k. By Taylor s formula, D s f ε Cε k+1 s So T (f) = T (f ε ), and T (f ε ) C ε. Now suppose f is arbitrary. ( f j () f k = x )ψ j Cc j! j k and has the same first k derivatives as f. So by the previous argument, T (f f k ) =. But then T (f) = T (f k ) = ( x f (j) j ) ()T j! ψ, ( ) x and the T j ψ are fixed constants independent of f. Thus we have j! proved that (1.3.1) T (f) = j k c j D j (f)().

3 3 2. Fourier Inversion 2.1. Recall that the Fourier transform (2.1.1) F x (f) := e iξx f(x) dx takes the Schwartz space S to itself. We would like to compute its inverse. The main theorem is the following. Theorem. The operator satisfies F ξ (φ) := 1 e iξx φ(ξ) dξ F ξ F x (f) = f, F x F ξ (φ) = φ. We will treat one of the equations, the other one is the mirror image. The natural thing to try is to write out the composition, and change the order of integration: (2.1.2) F ξ F x (f)(y) = 1 e e iξy e iξx iξ(x y) f(x) dx dξ = f(x) dξ dx. But the inner integral does not make sense. Formally, (2.1.3) e iξ(x y) dξ is the Fourier transform of the function φ(ξ) = 1, at x y : (2.1.4) φ(x y) = e i(x y)ξ 1 dξ. If we write ψ x (ξ) := 1 eixξ, then (2.1.3) is also ψ( y). To deal with the inability to just change the order of integration, and evaluate the inner integral, we resort to distributions. Definition. Let T be a tempered distribution. We define the Fourier transform T to be T (f) := T ( f). This is well defined, because f S. When T = L φ, this coincides with the usual Fourier transform. (2.1.5) L φ (f) = = φ(ξ) f(x) φ(x) dx. e iξx f(x) dx dξ = f(x) e iξx φ(x) dx dξ =

4 4 The change of order of integration is justified because both f and φ are in S From this point of view, Fourier inversion comes down to asking What distribution satisfies the property δ = f? The equation is the same as f() = T ( f). Given such a T, we can implement Fourier inversion as follows. Given x, define the translation by x operator L x as (2.2.1) L x (f)(y) := f(x + y). Then (2.2.2) f(x) = (L x (f))() = T ( L x (f)). On the other hand, L x (f)(ξ) = e iξy L x (f)(y) dy = e iξy f(x + y) dy = (2.2.3) e iξ(u x) f(u) du = e iξx f(ξ). So equation (2.2.2) becomes (2.2.4) f(x) = T ( e iξx f(ξ)). ( T acts on the RHS as a function of ξ) Let h t (x) := e x2 t 2. This function is in S. Then, (2.3.1) lim e x 2 t t 2 f(x) = f(x), e x2 t 2 f(x) f(x), so by the Lesbegue dominated convergence theorem, (2.3.2) lim L ht (f) = lim e x2 t 2 f(x) dx = t t On the other hand, (2.3.3) L ht ( f) = L pt (f) f() f(x) dx. by formula (4.1.6) and lemma 4.1 in L. Gross s notes. The conclusion is that T = 1. Putting all of this together, we get the formula for Forier inversion.

5 3. Oscillator Representation 3.1. Recall V = S(R), rapidly decreasing functions. The Lie algebra { ( ) ( ) ( ) } 1 1 sl(2, C) = e =, h =, f =. 1 1 acts on V by the formulas: 5 ϖ(h) = x x (3.1.1) ϖ(e) = i 2 x2 ϖ(f) = i 2 2 x For a linear space V, denote by EndV, the space of linear maps L : V V. This is again vector space, with the usual adition and scalar multiplication (3.1.2) (L 1 + L 2 )(v) := L 1 (v) + L 2 (v), (αl)(v) := αl(v). To say that a Lie algebra g acts on a space V, means that there is a linear map (3.1.3) ϖ : g End(V ) which also satisfies (3.1.4) ϖ([x, y]) = ϖ(x)ϖ(y) ϖ(y)ϖ(x). We can also exponentiate the operators in (3.1.1): (1) ω(e th )F (x) = e t/2 F (e t x) (2) ω(e te )F (x) = e itx2 /2 F (x) (3) ω(e tf )F (x) = convolution with ( 1+i 2 )(πt) 1/2 e ix2 /2t (1) and (2) are easy. (3) needs Fourier transform. The reason for the i is that (1) and (2) are unitary operators w.r. 2 to L 2 (R). We are interested in the operator 2iω(e f) = x 2 2 x k := i(e f). This is called the Hermite operator. We want eigenfunctions. If we write a = x + x, a + = x x,

6 6 we get [a, a + 1 ] = 2, 4 (a+ a + a a + ) = ϖ(k). This is also a Lie algebra representation. The Heisenberg algebra z acts by multiplication by 2. [a, (a + ) j ] = 2j(a + ) j 1 {p, q, z}, [p, q] = z, [p, z] =, [q, z] =. [a, a +2 ] = a a +2 a +2 a = a a +2 a + (2 + a a + ) = a a +2 2a + a + a a + = a a +2 (2 + a a + )a + 2a + = 4a +. Let v = e x2 /2. Then a v =. Set v j v j s(r). So a + v j = v j+1. Furthermore := (a + ) j v, j N. Then a v j = a (a + ) j v = 2ja +j 1 v = 2 j v j 1 (v j, v l ) = 2 l l!δ jl (v, v ) = 2 l l! πδ jl e x2 dx = π Conclusion. v j = P j (x)e x2 /2, P j a polynomial. P j called Hermite Polynomial. Theorem. Hermite functions form an orthogonal basis for L 2 (R). ϖ(k)v j = ( j + 1 2) vj 4. Review of some Linear algebra 4.1. Let A be an n n matrix with complex entries. The minimal polynomial is defined as the lowest degree polynomial m = m A m(t) such that m(a) =. The chracteristic polynomial, p = p A, is defined as (4.1.1) p(t) = det(ti A). Then (4.1.2) m(t) = (t λ i ) m i, p(t) = (t λ i ) n i. The λ i are called the (generalized) eigenvalues. The Cayley-Hamilton theorem implies that 1 m i n i.

7 4.2. The results in this section hold over any field. Assume that m = p q, with p, q polynomials, such that (p, q) = 1. The Euclidean algorithm implies that there are polynomials a, b such that (4.2.1) ap + bq = 1. Then (4.2.2) a(a)p(a) + b(a)q(a) = I. Let (4.2.3) V p := q(a)v, V q = p(a)v. The first conclusion is that (4.2.4) V = V p + V q, p(a)v p = (), q(a)v q = (). The second part is clear; for example p(a)v p = p(a)q(a)v = m(a)v = () by the definition of m. For the first part, (4.2.5) v = Iv = p(a)a(a)v + q(a)b(a)v. In addition V p V q = (); if v = p(a)x = q(a)y, then (4.2.6) v =a(a)p(a)v + b(a)q(a)v = a(a)p(a)q(a)y + b(a)q(a)p(a)x = =a(a)y + b(a)m(a)x = + =. So if we change bases in such a way that say the first r vectors form a basis of V p, and the last s vectors a basis of V q, then the matrix A becomes block diagonal ( ) Ap (4.2.7) A = A q The minimal polynomial of A p ois p, and the minimal polynomial of A q is q. The similar result holds for when m decomposes into more than two factors, mutually prime to each other We return to the setting of section 4.1. There is a basis such that the matrix A is block diagonal, A λ1... A (4.3.1) A = λ A λk Each A λi has minimal polynomial (t λ i ) m i, and each block has size n i. The spaces V λi are called generalized λ i eigenspaces. 7