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ELLIPTIC COMPLEXES and INDEX THEORY 1. Sobolev Theory on R

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ELLIPTIC COMPLEXES and INDEX THEORY 1. Sobolev Theory on R ELLIPTIC COMPLEXES AND INDEX THEORY VISHWAMBHAR PATI 1. Sobolev Theory on Rn As general references for this section, see the books [Nar], Ch.3, [Gil], Ch.1, [Hor], Ch I, II and [Rud], Ch. 6,7. 1.1. Test functions and distributions. We introduce some standard notation. For a multi-index α = (α1; ::; αn) of length n, the symbol α := i αi, and α! := α1!; ::; αn!. For derivatives, we denote: j j P α1 αn α @ @ @ 1 dx = :::::: ; @j = ; Dj = @j @x1 @xn @xj p 1 − α1 α2 αn α Finally x1 x2 :::xn will be denoted simply by x . We can define some standard function spaces on Rn. For us, functions will always be complex valued. Definition 1.1.1 (Standard function spaces on Rn). The function spaces defined below are all complex vector spaces, and to define a topology on them, it is enough to define convergence to zero. (i): n n C1(R ) = smooth functions on R f g α Define fn 0 if dx fn 0 uniformly on compact sets for all α 0. This space is also sometimes denoted !by analysts. ! j |≥ E (ii): Rn Rn α C01( ) = f C1( ) : lim dx f = 0 for all α 0 x f 2 !1 j |≥ g This is the space of all smooth functions whose derivatives of all orders vanish at infinity. The topology n in this space is the subspace topology from C1(R ). It is often denoted . E0 (iii): n n C1(R ) = f C1(R ) : support f is compact c f 2 g Its topology is defined by fn 0 if there exists a compact set K such that suppfn K for all n and α ! ⊂ n d f 0 uniformly on K for all α 0. Note that this is not the subspace topology from C1(R ), for if x n ! j |≥ we define a take a non-zero function on R with compact support [ 1; 1] say, and let fn(x) = (x n), R − R − (which has support [n 1; n + 1] ), then fn 0 in C1( ), but not in Cc1( ). It is, in fact easily seen to be strictly finer than−the subspace topology!. This space is denoted by analysts, and also called the space of test functions. D (iv): Rn Rn α β ( ) = f C1( ) : sup x dxf(x) Cαβ for all α ; β 0 S f 2 x Rn j |≤ j j j |≥ g 2 This is called the Schwartz space of rapidly decreasing functions. Define the topology by declaring fn 0 if ! α β sup x dxfn(x) 0 x Rn j j! 2 for each α ; β 0. j j j |≥ 1 2 VISHWAMBHAR PATI It is an easy exercise to see that there are natural inclusions: n n n n C1(R ) (R ) C1(R ) C1(R ) c ⊂ S ⊂ 0 ⊂ all of which are continuous, and all of which are strict. The reader may also check that the inclusion C1 C1 c ⊂ is dense, (by using multiplication with cutoff functions φn which are identically 1 on a ball of radius n and identically zero outside a ball of radius say 2n), and hence all the inclusions above are dense. On or C1, we may introduce the L -norm defined by: S c p 1 p f := f p dx Lp k k Rn j j Z for 1 p < . Upon completing either of these two spaces with respect to this norm, one gets the Banach ≤ n 1 n space Lp(R ). For p = , this is false, as can be seen by looking at the non-zero constant functions. L (R ) is got by taking all measurable1 functions on Rn which are essentially bounded. 1 1.2. The Fourier Transform and Plancherel Theorem. In the sequel we will simply write C1 for n C1(R ), and so on, if no confusion is likely. Also, to eliminate annoying powers of 2π, we introduce the measure (volume element) dx on Rn by the formula: n=2 dx := (2π)− dx1:::dxn Definition 1.2.1. For f , define the Fourier Transform of f by the formula 2 S iξ:x f(ξ) := f(x)e− dx Rn Z which makes sense for any f L1, and in particularb for f . Here ξ:x = i ξixi is the usual Euclidean inner product of vectors in Rn. Similarly2 , for f , define the 2InvSerse Fourier Transform of f by the formula: 2 S P iξ:x f _(ξ) := f( ξ) = f(x)e dx − Rn Z b Finally, for f; g , define the convolution product 2 S f g(x) := f(x y)g(y)dy = f( z)g(z + x)dz ∗ Rn − Rn − Z Z It is easy to verify (taking limits inside the integral sign after appealing to Lebesgue's Dominated Convergence Theorem) that f g is also in , and that f g = g f. ∗ S ∗ ∗ Before proving the main proposition of this section, we need a couple of useful lemmas. Note that the Gaussian function jxj2 (x) = e− 2 is in . Also its integral (x)dx = 1. S Rn R Lemma 1.2.2. For the Gaussian above, we have = . Proof: We have: b 2 iξ:x jxj (ξ) = e− e− 2 dx Z jξj2 (x+iξ):(x+iξ) b = e− 2 e 2 dx Z Let x = (x1; ::; xn) and ξ = (ξ1; ::; ξn). By choosing a rectangular contour in C with vertices a; a; a+iξ1; a+ z2=2 − − iξ1, noting that the integral of the holomorphic function e− 1 around this contour is zero, and also that the contributions along the vertical edges ( a + it) and (a + it) for 0 t ξ1 (if ξ1 0) (resp. ξ1 t 0 if ξ 0) converge to zero as a , we see that:− ≤ ≤ ≥ ≤ ≤ ≤ 1 ! 1 2 2 (x1+iξ1) =2 x =2 e− dx1 = e− 1 dx1 = p2π R R Z Z ELLIPTIC COMPLEXES AND INDEX THEORY 3 apply the argument variable by variable to conclude that: 2 (x+iξ):(x+iξ)=2 n (xj +iξj ) =2 e− dx = (2π)− 2 ::: e− j dx1:::dxn = 1 Rn R R R Z Z Z Z P which proves our assertion. 2 Lemma 1.2.3 (Approximate identities). Let φ , such that φ(x) 0 for all x and φ(x)dx = 1. For > 0, define the approximate identity or mollifier: 2 S ≥ R n φ(x) = − φ(x/) Then for any f , we have f φ converges uniformly to f as 0. 2 S ∗ ! Proof: Since Rn φdx = 1 for each > 0, we have: R (f φ )(x) f(x) = φ (y) (f(x y) f(x)) dy j ∗ − j − − Z φ (y) (f(x y) f(x)) dy ≤ j − − j Z Let C > 0 be such that f(x) dx C. Now let η > 0 be any positive number. Choose a δ > 0 (by uniform continuity of f) such thatj f(x j y) ≤ f(x) η for all y δ, and all x. Then R j − − |≤ j |≤ φ(y) (f(x y) f(x)) dy η φ(y)dy η φ(y)dy = η y δ j − − j ≤ y δ ≤ Rn Zj |≤ Zj |≤ Z for all > 0. Now choose an 0 > 0 small enough so that y >δ φ(y)dy < η=2C for 0. Then, we have: j j ≤ R φ(y) (f(x y) f(x)) dy φ(y)(2C)dy η y >δ j − − j ≤ y >δ ≤ Zj j Zj j for < . Combining the integrals for y δ and y > δ, we get: 0 j j ≤ j j f φ (x) f(x) 2η for all < j ∗ − |≤ 0 independent of x. That is, f φ f uniformly as 0. 2 ∗ ! ! Remark 1.2.4. It follows from the above that one can take any non-negative compactly supported function φ and define the approximate identities φ. Similarly, by starting with the Gaussian defined above, we get that the functions: n x 2=22 (x) = − e−| j are approximate identities. Proposition 1.2.5. We have the following facts about the Fourier transform on the Schwartz class . S (i): The map f f is an isomorphism of with itself, of order 4. In fact, 7! S f ^ (x) = f( x); f _ (x) = f(x) for all f b − 2 S (The second formula is called the Fourier Inversion Formula.) b b (ii): For all multi-indices α, α α α α α (D f)^ (ξ) = ξ f; D f(ξ) = ( 1)j j (x f)^ x ξ − In particular, by the first formula, if P is an n-variable polynomial with complex coefficients, then for the constant coefficient differential operatorb P (D) wbe have: (P (D)f)^ = P (ξ)f b 4 VISHWAMBHAR PATI (iii): f g = (f g)^ ; f g = (fg)^ ∗ ∗ (iv): (Plancherel Theorem) The map fbb f is a unitaryb isomorphismb of to itself with respect to L2-norm. 7! S n Thus (in view of (i) above), it extends to a unitary isomorphism of L2(R ) to itself. b (v): (Riemann-Lebesgue Lemma) There is an inclusion: n n (L (R ))^ C (R ) 1 ⊂ 0 where the space on the right is the space of all continuous functions vanishing at , with the topology of uniform convergence on compact sets. 1 Proof: We first prove (ii). Since all derivatives of f are also in , and hence Lebesgue integrable, one can use the Lebesgue Dominated Convergence Theorem,2andS differentiateS under the integral sign to get: α α iξ:x Dξ f(ξ) = Dξ e− f(x) dx Rn Z α α iξ:x α α b = ( 1)j jx e− f(x)dx = ( 1)j j (x f)^ Rn − − Z In particular, we have that f is a smooth function.
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