1 Fourier transform as unitary equivalence

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1 Tel Aviv University, 009 Intro to functional analysis 1 1 Fourier transform as unitary equivalence 1a Introduction b Exponential map c Exponential map as an analytic vector-function 4 1d Fourier transform e Operators commuting with shifts f Inverse Fourier transform g Convolution operators h Fourier transform and convolution i Gaussian functions j Explicit formula k List of results l List of formulas Functional analysis treats functions as points in the function space. A useless neologism? Not at all. Rather, a way to use geometric intuition for the benefit of analysis. Surprisingly, simple geometric symmetries of a special, seemingly not notable two-dimensional surface in Hilbert space lead naturally to the famous Fourier transform (and some other useful things. 1a Introduction On the finite group Z/nZ, Fourier transform amounts to the basis of eigenvectors of the shift U : C n C n, U(f 0, f 1,...,f n, f n 1 = (f 1, f,...,f n 1, f 0. The eigenvalue e πil/n corresponds to the eigenvector ( 1, e πil/n, e πi l/n,..., e πi (n 1l/n. Defining unitary F : C n C n by F(f 0, f 1,..., f n 1 = = 1 ( f k, e πik/n f k, n k k k e πik /n f k,..., k e πik (n 1/n f k

2 Tel Aviv University, 009 Intro to functional analysis we get FU = V F where V is diagonal, V (f 0, f 1,...,f n 1 = ( f 0, e πi/n f 1,..., e πi(n 1/n f n 1. Thus, F diagonalizes U, FUF 1 = V. On the compact group T = R/Z we have Fourier series, f(x = k Z c k = 1 0 c k e πikx, f(xe πikx dx. The unitary operator F : L (T l (Z, Ff = c, diagonalizes shifts U(a : L (T L (T, U(af : x f(x + a; namely, FU(aF 1 = V (a, V (a : l (Z l (Z, (V (ac k = e πiak c k. On the noncompact group R the situation is similar in principle, but more complicated technically, since the shifts have continuous spectrum. 1 We ll see that the Fourier transform is a unitary operator F : L (R L (R that diagonalizes shifts U 1 (a : L (R L (R, U 1 (af : t f(t + a; namely, FU 1 (af 1 = V 1 (a, V 1 (a : L (R L (R, V 1 (af : t e ibt f(t. In fact, Ff : t 1 π + e ist f(s for f L 1 (R L (R. In order to overcome technical difficulties we ll go round. An orthonormal basis is a family of vectors with special (and extremely simple scalar products. Quite different family of vectors with special (and still simple scalar products, the so-called exponential map, will be instrumental. See also List of results, List of formulas, and Index. 1b Exponential map Two special geometric objects are introduced in this section: a curve ψ( in a Hilbert space, and the corresponding curve w( on the sphere. Placing appropriately the latter curve in the function space L (R we extend symmetries of the curve to shifts of the functions. 1 The function x e iax is not an eigenvector, since it does not belong to L (R.

3 Tel Aviv University, 009 Intro to functional analysis 3 Toy model Consider a map (vector-function, parametrized curve ψ : R E, where E is a Euclidean space, satisfying ψ(x, ψ(y = 1 + xy + 1 x y. Existence: ψ(x = ( 1, x, x / R 3. Uniqueness: c k ψ 1 (x k = c k ψ (x k. General form: ψ(x = e 0 + xe 1 + x e where e 0, e 1, e are orthonormal. By the way, e 0 = ψ(0, e 1 = ψ (0, e = ψ (0/. The subspace spanned by all ψ(x is 3-dimensional. Exponential map on R Consider a map (vector-function ψ 0 : R H, where H is a Hilbert space over R, satisfying (1b1 ψ 0 (x, ψ 0 (y = exp(xy/ ; ψ 0 (x, ψ 0 (y = exp(xy. Existence: ψ 0 (x = ( 1, x, x /!, x 3 / 3!,... l. As before: uniqueness; ψ 0 (x = k=0 x k k! e k ; e k are orthonormal; the power series converges (in norm for all x; e k = k/ k! ψ (k 0 (0; the subspace spanned by all ψ 0(x is infinite-dimensional. Assume it to be the whole H. Alternatively: (1b w 0 (x = ψ 0(x ψ 0 (x = e x /4 ψ 0 (x, w 0 (x, w 0 (y = exp( (x y /4 ; the latter is shift-invariant. Action of shifts: U 0 (a = U (w 0 (a Unitary(H for a R, U 0 (aw 0 (x = w 0 (x a ; U 0 (a + b = U 0 (au 0 (b. Observation: + dq = π/a, thus the vector-function w e aq 1 : R L (R, w 1 (x a : q 4 a x π e (q

4 Tel Aviv University, 009 Intro to functional analysis 4 satisfies (1b. Indeed, + = w 1 (x (qw 1 (y + (q dq = exp ( (q a x (q a y adq = π + = exp ( (q x (q y dq = π + ( ( exp q x + y (x + y (x + y dq = π = π ( π exp (x y. Traditionally one chooses a = 1/; the vector-functions w 1, ψ 1 : R L (R, w 1 (x : q π 1/4 e (q x / = π 1/4 exp ( q x + xq, (1b3 ψ 1 (x : q π 1/4 exp ( q + xq x 4 satisfy (1b, (1b1 respectively. It is not evident whether the vectors ψ 1 (x span L (R or not. 1 But anyway, our shifts U 1 (a = U (w 1 (a conform to the usual shifts on L (R; indeed, ( U 1 (aw 1 (x (q = w 1 (x a(q = w 1 (x(q+a, since w 1 (x(q is a function of q x only. Thus, for every f of the spanned subspace. U 1 (af : q f(q + a 1c Exponential map as an analytic vector-function Using analytic vector-functions on the complex plane we turn to special twodimensional surfaces in the Hilbert space, thus gaining additional symmetries. In the function space the shifts appear to be diagonalized. Also, some useful systems of functions generate L (R. Now we assume H to be a Hilbert space over C and note that the unitary operator e k i k e k leads to another vector-function x x k k! i k e k satisfying (1b1. Not exciting, unless we turn to ψ 1, try ψ 1 (ix : q π 1/4 exp ( q, + ixq + x 4 ψ 1 (ix ψ 1 (ix : q π 1/4 exp ( q + ixq = e ixq ψ 1 (0(q 1 We ll see that they do.

5 Tel Aviv University, 009 Intro to functional analysis 5 and observe that our shifts U 1 (a turn into multiplication operators. Really?! Consider a vector-function ψ 0 : C H satisfying (1c1 ψ 0 (z 1, ψ 0 (z = exp(z 1 z /. Existence: ψ 0 (z = ( 1, z, z /!, z 3 / 3!,... l. As before: uniqueness; ψ 0 (z = k=0 z k k! e k ; e k are orthonormal; the power series converges (in norm for all z, which means that ψ 0 is an entire vector-function; e k = k/ k! ψ (k 0 (0; the subspace spanned by all ψ 0 (z is infinite-dimensional, and we assume it to be the whole H. The vector-function R x ψ 0 (x H satisfies (1b1. Another vectorfunction R x ψ 0 (ix H also satisfies (1b1, since ixiy = xy. We introduce a vector-function ψ 1 : C L (R by (1c ψ 1 (z : q π 1/4 exp ( q z + zq 4 and note that ψ 1 (x, ψ 1 (y = exp(xy/ for x, y R. Does it mean that ψ 1 (z 1, ψ 1 (z = exp(z 1 z / for z 1, z C? This equality could be checked by a calculation, but it is more interesting to get it from an excursion into the theory of analytic vector-functions (which provides much more than just this equality. Entire vector-functions on C 1c3 Definition. An entire vector-function ϕ : C H is a vector-function of the form ϕ(z = z k h k in the sense that ϕ(z k=0 n z k h k 0 k=0 as n for every z C; here H is a Hilbert space, and h k H.

6 Tel Aviv University, 009 Intro to functional analysis 6 Clearly, h k must satisfy k h k 0 as k. The one-dimensional case H = C conforms to the usual notion of (scalar entire function. 1c4 Exercise. If ϕ(z = k=0 zk h k is an entire vector-function then ξ(z = k=0 (k + 1zk h k+1 is an entire vector-function, and ϕ = ξ in the sense that for every z C, ϕ(z + z ϕ(z z ξ(z 0 as z 0+, z C. Prove it. Hint: (z 1 z ξ( (1 uz 0 + uz 1 du = ϕ(z1 ϕ(z 0. 1c5 Corollary. If ϕ(z = k=0 zk h k is an entire vector-function then ϕ is infinitely differentiable, and Thus, h n = 1 n! ϕ(n (0 for n = 0, 1,,... ϕ(z = k=0 z k k! ϕ(k (0. If entire vector-functions ϕ 1, ϕ : C H satisfy ϕ (n 1 (0 = ϕ(n (0 for all n, then ϕ 1 = ϕ. 1c6 Proposition. For arbitrary entire vector-functions ϕ 1, ϕ : C H, (a if ϕ 1 (x = ϕ (x for all x R then ϕ 1 = ϕ ; (b moreover, if the set {z C : ϕ 1 (z = ϕ (z} has at least one (finite accumulation point then ϕ 1 = ϕ. 1c7 Proposition. For every entire vector-function ϕ : C H, (a the closed subspace spanned by all ϕ(z contains all ϕ (z and moreover, all ϕ (n (z; (b the closed subspace spanned by all ϕ(z is equal to the closed subspace spanned by all ϕ (n (0; (c the closed subspace spanned by ϕ (n (z for a given z and all n does not depend on z; (d the closed subspace spanned by ϕ(x for all x R is equal to the closed subspace spanned by ϕ(ix for all x R; (e moreover, for an arbitrary set A C having at least one (finite accumulation point, the closed subspace spanned by ϕ(z for z A does not depend on the choice of A.

7 Tel Aviv University, 009 Intro to functional analysis 7 If ϕ : C H 1 is an entire vector-function and U : H 1 H is a linear isometric embedding (of the Hilbert space H 1 to the Hilbert space H then z Uϕ(z is an entire vector-function. 1c8 Theorem. Let H 1, H be Hilbert spaces and ϕ 1 : C H 1, ϕ : C H entire vector-functions. If then ϕ 1 (x, ϕ 1 (y = ϕ (x, ϕ (y for all x, y R ϕ 1 (z 1, ϕ 1 (z = ϕ (z 1, ϕ (z for all z 1, z C. Proof (sketch. We may assume that the closed spanned subspaces are the whole spaces. We take unitary U : H 1 H such that Uϕ 1 (x = ϕ (x for all x R (recall uniqueness. It follows that Uϕ 1 (z = ϕ (z for all z C. Back to the exponential map on C 1c9 Exercise. If f : C C is an entire function and ϕ : C H an entire vector-function then f ϕ : z f(zϕ(z is an entire vector-function. Prove it. 1c10 Exercise. The vector-function ϕ : C L (R defined by ϕ(z : q exp( q / + zq is entire. Prove it. Hint: e q / n (zq k k=0 k! exp( q / + zq, a majorant in L ; use the dominated convergence theorem. 1c11 Corollary. The vector-function ψ 1 : C L (R defined by (1c is entire. By the theorem, ψ 1 (z 1, ψ 1 (z = exp(z 1 z / for all z 1, z C ; especially, ψ 1 (ix, ψ 1 (iy = e xy/ = ψ 1 (x, ψ 1 (y for x, y R. Defining w 1 : C L (R by w 1 (z = ψ 1(z ψ 1 (z = /4 ψ e z 1 (z : q π 1/4 exp ( q z + zq 4 z 4 (not an analytic function we get ( 1 w 1 (z 1, w 1 (z = exp z 1z 1 4 z z ( = exp 1 4 z 1 z + i Im (z 1z ; especially, w 1 (ix, w 1 (iy = e x y /4 = w 1 (x, w 1 (y for x, y R.

8 Tel Aviv University, 009 Intro to functional analysis 8 1c1 Lemma. The closed subspace spanned by {ψ 1 (ix : x R} is the whole L (R. Proof (sketch. ψ 1 (ix(q = const(x e ixq e q / ; in L with the weight e q the closed subspace spanned by the functions q e ixq contains all periodic functions (recall Fourier series, therefore, all functions. 1c13 Corollary. The closed subspace spanned by {ψ 1 (x : x R} is the whole L (R. Recall the shift operators U 1 (a = U (w 1 (a defined by U 1 (aw 1 (x = w 1 (x a for a R and satisfying U 1 (af : q f(q + a for every f of the spanned subspace. Now we see that U 1 (a Unitary(L (R, U 1 (af : q f(q+a for all f L (R, a R. Replacing w 1 (x with w 1 (ix we get another one-parameter group of unitary operators V 1 (b defined by V 1 (bw 1 (iy = w 1 ( i(y + b for b R; they act on the spanned subspace, thus, on the whole L (R. Taking into account that we get w 1 (iy(q = π 1/4 exp ( q + iyq, ( w 1 i(y + b (q = e ibq w 1 (iy(q V 1 (b Unitary(L (R, V 1 (bf : q e ibq f(q for all f L (R, b R. 1d Fourier transform Now we are in position to define a unitary operator F : L (R L (R by Fψ 1 (x = ψ 1 ( ix for x R. We have FU 1 (aw 1 (x = Fw 1 (x a = w 1 ( i(x a = V1 (aw 1 ( ix = V 1 (afw 1 (x, which means that FU 1 (a = V 1 (af, that is, FU 1 (af 1 = V 1 (a for a R. We see that the two one-parameter unitary groups are unitarily equivalent. Shift operators are thus diagonalized!

9 Tel Aviv University, 009 Intro to functional analysis 9 1e Operators commuting with shifts Every operator commuting with shifts results from shifts. Also, the Fourier transform is the only operator diagonalizing the shifts. A linear electric circuit transforms a signal by an operator commuting with (time shifts, provided that the circuit does not change in time. Also a linear evolution equations with constant coefficients, such as the heat equation and the Schrödinger equation, lead to evolution operators commuting with (space shifts. 1e1 Theorem. Let a bounded linear operator A : L (R L (R satisfy U 1 (aa = AU 1 (a for all a R. Then A is a strong limit of a sequence of linear combinations of operators U 1 (a. That is, there exist c (n k f L (R n k=1 C and a (n k R such that c (n k U ( (n 1 a k f Af 0 as n. 1e Remark. (a The theorem fails for norm convergence of operators (as we ll see later. (b The theorem fails for L (R C (as we can see immediately. Proof (sketch. The proof consists of four steps. First step: we may replace U 1 ( with V 1 (. Proof: unitary equivalence. Second step: A is a multiplication operator, that is, ϕ L (R f L (R Af = ϕ f. Proof: define f 0 (t = e t /, g 0 = Af 0 and ϕ = g 0 /f 0, then AV 1 (bf 0 = V 1 (bg 0 = ϕ V 1 (bf 0 for all b. Linear combinations of functions V 1 (bf 0 are dense in L ; it follows that ϕ L (since A is bounded and Af = ϕ f for all f L. Third step: it is sufficient to find trigonometric polynomials p n such that p n ϕ almost everywhere and sup n,q p n (q <. Proof: then for every f L we have p n f ϕ f 0 by the dominated convergence theorem, since p n f ϕ f const f. Last step. We know that trigonometric polynomials are dense in L with weight e q. Moreover, having ϕ L (R we get (by the same Fourier series argument, via Feier sums trigonometric polynomials p n such that p n ϕ and p n (q ϕ(q e q dq 0 as n. Taking a subsequence we get n pn (q ϕ(q e q dq <, which implies n p n(q ϕ(q < almost everywhere.

10 Tel Aviv University, 009 Intro to functional analysis 10 Now we can see that the theorem fails for norm convergence of operators; it happens because trigonometric polynomials are not dense in L (R. As a by-product we get the general form of an operator commuting with all V 1 (b, just the multiplication operator L (R f ϕ f, ϕ L (R, and, more important, the general form of an operator commuting with all U 1 (a, L (R f F 1 (ϕ Ff, ϕ L (R. 1e3 Corollary. If an operator commutes with all U 1 (a and all V 1 (b then it is scalar. Proof: it is multiplication by ϕ, and ϕ is shift-invariant. 1e4 Corollary. If an invertible operator F 1 satisfies F 1 U 1 (af 1 1 = V 1 (a for all a R then F 1 = cf for some c C. 1f Proof: F 1 1 F commutes with all U 1 (a and V 1 (b. The same holds for U(a and V (b in general. Inverse Fourier transform The relation Fψ 1 (z = ψ 1 ( iz holds for z R, therefore, for all z C. In particular, Fψ 1 (ix = ψ 1 (x for x R. We note that ψ 1 (z 1, ψ 1 (z = exp(z 1 z / = ψ 1 ( z 1, ψ 1 ( z, introduce a unitary operator J : L (R L (R by Jψ 1 (z = ψ 1 ( z and get FJψ 1 ( ix = ψ 1 (x, that is, FJFψ 1 (x = ψ 1 (x, which means that FJF = 1l and so, F 1 = FJ = JF. It remains to note that Jψ 1 (z(q = ψ 1 ( z(q = π 1/4 exp ( q z zq 4 = ψ 1 (z( q, thus, (Jf(q = f( q for f L (R. 1g Convolution operators Integral combinations of shift operators are convolution operators. We want to define an operator g(au 1 ( a da for every g L 1 (R. 1 One approach is, to integrate the vector-function a U 1 ( af (for a given 1 The minus sign is traditional.

11 Tel Aviv University, 009 Intro to functional analysis 11 f L (R with the weight g. Another approach is the classical convolution formula (f g(q = g(af(q a da. Integrating the vector-function 1g1 Lemma. The one-parameter unitary group U 1 ( is strongly continuous. That is, f U 1 (af 0 as a 0 for every f L (R. Continuity at other points follows easily from continuity at 0. First proof. Using the unitary equivalence we replace U 1 ( with V 1 ( ; now, f(q e ibq f(q dq 0 as b 0 by the dominated convergence theorem. 1g Remark. However, 1l U 1 (a = for all a 0, since ess sup t R 1 e ibt = for all b 0. Another proof. The claim holds evidently on a dense set of f, which is sufficient. Given f L (R, we have a bounded continuous vector-function R a U 1 ( af, and may consider the improper Riemann-Stieltjes integral + U 1 ( af dg(a = lim lim C + n n 1 k= n ( ( k + 1 ( k G n C G U n C 1 ( k n C f, where G(a = a g(q dq; norm convergence in L (R is proved in the same way as for scalar-valued functions (and of course, it holds for arbitrary partitions and points. This integral will be denoted by g(au 1 ( af da. Clearly, g(au 1 ( af da g 1 f, and we define a bounded linear operator g(au 1 ( a da as f g(au 1 ( af da. Thus, g(au 1 ( a da g 1 ; we have a bounded linear map from L 1 (R to the space of operators L (R L (R. Clearly, g(au 1 ( a da commutes with shifts. It is easy to guess that g(au 1 ( af da : q g(af(q a da, but not so easy to prove it, since f(t cannot be treated as a continuous linear functional of f L. Classical convolution formula

12 Tel Aviv University, 009 Intro to functional analysis 1 1g3 Lemma. Let f L (R and g L 1 (R, then for almost all q R, (a f(q ag(a da <, (b g(au 1 ( af da : q g(af(q a da. Proof. (a dq da f(q a g(a = da g(a dq f(q a = g 1 f <, therefore f(q a g(a da < for almost all q, which implies f(q ag(a da ( f(q a g(a da 1/ ( g(a da 1/ <. (b The function q f(q ag(a da belongs to L, since dq f(q ag(a da g 1 dq da f(q a g(a g 1 f. It is sufficient to prove the equality g(au 1 ( af da, h = q f(q ag(a da, h for all h L (R. The linear functional, h applies to the Riemann-Stieltjes integral, giving g(au 1 ( af da, h = g(a U 1 ( af, h da = da g(a dq f(q ah(t. The needed equality da g(a dq h(qf(q a = dq h(q da g(af(q a follows from Fubini s theorem, since dadq g(ah(qf(q a = = dq h(q h 1/ da g(af(q a q da g(af(q a 1/ <. Thus we have two equivalent definitions of the convolution f g for f L (R, g L 1 (R, f g = g(au 1 ( af da ; f g : t f(t sg(s ds. The latter formula shows that f g = g f for f, g L 1 (R L (R.

13 Tel Aviv University, 009 Intro to functional analysis 13 1h Fourier transform and convolution The convolution operator g(au 1 ( a da : f f g commutes with all U 1 (a, therefore it must be f F 1 (ϕ Ff for some ϕ L (R. We have F k c k U 1 ( a k F 1 = k c k V 1 ( a k ; the limiting procedure (strong convergence... gives ( F g(au 1 ( a da F 1 = g(av 1 ( a da, which means that ϕ(q = g(ae iaq da ; f g = F 1 (ϕ Ff, that is, F(f g = ϕ Ff for f L (R and g L 1 (R. 1i Gaussian functions We know that Fψ 1 (0 = ψ 1 (0 and ψ 1 (0 : q π 1/4 e q /. Now we rescale that function: f σ L (R, f σ (q = π 1/4 σ 1/ exp ( q σ for 0 < σ < ; note that f 1 = ψ 1 (0, Ff 1 = f 1. The functions f σ are normalized in L : f σ = 1. Also Gaussian functions normalized in L 1 are useful, g σ = f σ = 1/ π 1/4 σ 1/ f σ : q (π 1/ σ 1 exp ( q. f σ 1 σ 1i1 Lemma. Ff σ = f 1/σ for 0 < σ <. Proof. It is sufficient to prove it for σ 1, since it implies f σ = F 1 f 1/σ = FJf 1/σ = Ff 1/σ. Given σ > 1 we introduce δ = σ 1 and note that f 1 g δ = σ 1/ f σ (check it. On the other hand, F(f 1 g δ = ϕ δ Ff 1 where ϕ δ (q = gδ (ae iaq da = exp ( 1 δ q (check it. Thus, Ff σ = σ 1/ ϕ δ f 1 = f 1/σ (check it. 1i Corollary. Fg σ = 1 σ g 1/σ.

14 Tel Aviv University, 009 Intro to functional analysis 14 1j Explicit formula 1j1 Theorem. Ff : t 1 + e ist f(s ds for f L 1 (R L (R. π We use the Gaussian functions f σ, g σ. 1j Lemma. f g σ f in L as σ 0+, for every f L. Proof. f g σ f = g σ (a(u 1 ( af f da g σ (a U 1 ( af f da = ( ε,+ε + R\( ε,+ε 0. Proof of the theorem. F(f g σ = F(g σ f = ϕ Fg σ, where ϕ : t f(se ist ds. We have Fg σ = 1 σ g 1/σ : t (π 1/ e σ t / (π 1/ as σ 0+ pointwise (moreover, locally uniformly. On the other hand, ϕ 1 g σ 1/σ = ϕ Fg σ = F(f g σ F(f in L. It follows that ϕ L and F(f = (π 1/ ϕ. 1j3 Corollary. F(f g = (π 1/ (Ff (Fg for f L, g L 1 L. 1k 1k1 List of results Plancherel s theorem There exists a bounded linear operator (called Fourier transform F : L (R L (R such that Ff : t 1 + e ist f(s ds for all f L 1 (R L (R. π The operator F is isometric, which means the following. Plancherel s formula: Parseval s formula: 1k Inversion formula Ff = f for all f L (R. Ff, Fg = f, g for all f, g L (R. The isometric operator F is unitary, which means that F ( L (R = L (R. The inverse operator F 1 is given by F 1 = FJ = JF where J is defined by Jf : t f( t.

15 Tel Aviv University, 009 Intro to functional analysis 15 1k3 First, Diagonalization of shifts FU(aF 1 = V (a for all a R, where operators U(a, V (a : L (R L (R are defined by U(af : t f(t + a, V (af : t e iat f(t. Second, this property characterizes F uniquely up to a coefficient. 1k4 Diagonalization of the Fourier transform There exists an orthonormal basis (e 0, e 1, e,... of L (R such that Namely, where ψ : C L (R is defined by 1k5 Fe k = ( i k e k for k = 0, 1,,... e k = k/ d k k! dz k ψ(z, z=0 ψ(z : q π 1/4 exp ( q Some complete systems in L (R z + zq. 4 The vectors e k mentioned above are a complete system in the sense that their linear combinations are dense in L (R. Also functions q exp ( (q x for all x R are a complete system. 1k6 here Fourier transform and convolution F(f g = (π 1/ (Ff (Fg for all f L (R, g L 1 (R L (R ; f g : t f(t sg(s ds. 1k7 Operators commuting with shifts A bounded linear operator on L (R commutes with all U(a if and only if it is of the form f F 1 (ϕ Ff for some ϕ L (R.

16 Tel Aviv University, 009 Intro to functional analysis 16 1k8 Vector-functions In addition we have several statements about vector-functions (both specific and general valued in a (general Hilbert space, their derivatives and integrals. See especially Sections 1c and 1g. 1l List of formulas x, y R; z 1, z C; ψ 0, w 0 : C H; (1l1 (1l (1l3 ( 1 ψ 0 (z 1, ψ 0 (z = exp z 1z ; w 0 (z = ψ 0(z ψ 0 (z = /4 ψ e z 0 (z ; ( w 0 (z 1, w 0 (z = exp 1 4 z 1 z + i Im (z 1z ; (1l4 w 0 (x, w 0 (y = e x y /4 = w 0 (ix, w 0 (iy ; U 0, V 0 : R Unitary(H; (1l5 (1l6 U 0 (aw 0 (x = w 0 (x a ; V 0 (bw 0 (iy = w 0 ( i(y + b ; ψ 1, w 1 : C L (R; (1l7 (1l8 (1l9 (1l10 ψ 1 (z : q π 1/4 exp ( q w 1 (z : q π 1/4 exp ( q z + zq ; 4 z + zq 4 z 4 w 1 (x : q π 1/4 exp ( 1 (q x ; w 1 (iy : q π 1/4 exp ( q + iyq ; U 1, V 1 : R Unitary(L (R, a, b R; ; (1l11 (1l1 U 1 (af : q f(q + a ; V 1 (bf : q e ibq f(q ;

17 Tel Aviv University, 009 Intro to functional analysis 17 F Unitary(L (R; (1l13 (1l14 (1l15 (1l16 (1l17 (1l18 (1l19 Fψ 1 (z = ψ 1 ( iz ; FU 1 (af 1 = V 1 (a ; F 1 = FJ = JF, Jf : q f( q ; Ff : t 1 + e ist f(s ds for f L 1 (R L (R ; π f g = g(au 1 ( af da : t f(t sg(s ds ; f g = g f for f, g L 1 (R L (R ; F(f g = (π 1/ (Ff (Fg for f L (R, g L 1 (R L (R. Index convolution, 1 entire, 5 shift, 3, 4, 8 f g, 1 F, 8 f σ, 13 g σ, 13 J, 10 ψ 0, 3, 5 ψ 1, 4, 5 U 0, 3 U 1, 4, 8 V 1, 8 w 0, 3 w 1, 4, 7