Fourier Analysis, Distributions, and Constant-Coefficient Linear PDE

Transcription

1 Fourier Analysis, Distributions, and Constant-Coefficient Linear PDE Michael E. Taylor Contents. Introduction 1. Fourier series 2. Harmonic functions and holomorphic functions in the plane 3. The Fourier transform 4. Distributions and tempered distributions 5. The classical evolution equations 6. Radial distributions, polar coordinates, and Bessel functions 7. The method of images and Poisson s summation formula 8. Homogeneous distributions and principal value distributions 9. Elliptic operators 1. Local solvability of constant-coefficient PDE 11. The discrete Fourier transform 12. The fast Fourier transform A. The mighty Gaussian and the sublime gamma function Introduction Fourier analysis is perhaps the most important single tool in the study of linear partial differential equations. It serves in several ways, the most basic and historically the first being to give specific formulas for solutions to various linear PDE with constant coefficients, particularly the three classics, the Laplace, wave, and heat equations: (.1) u = f, 2 u u u = f, t2 u = f, t with = 2 / x / x 2 n. The Fourier transform accomplishes this by transforming the operation of / x j to the algebraic operation of multiplication by iξ j. Thus the equations (.1) are transformed to algebraic equations and to ODE with parameters. Before introducing the Fourier transform of functions on Euclidean space R n, we discuss the Fourier series associated to functions on the torus T n in 1. Methods developed to establish the Fourier inversion formula for

2 2 Fourier series, in the special case of the circle S 1 = T 1, provide for free a development of the basic results on harmonic functions in the plane, and we give such results in 2, noting that these results specialize further to yield standard basic results in the theory of holomorphic functions of one complex variable, such as power-series expansions and Cauchy s integral formula. In 3 we define the Fourier transform of functions on R n and prove the Fourier inversion formula. The proof shares with the argument for Fourier series in 1 the property of simultaneously yielding explicit solutions to a PDE, this time the heat equation. It turns out that representations of solutions to such PDE as listed in (.1) are most naturally done in terms of objects more general than functions, called distributions. We develop the theory of distributions in 4; Fourier analysis works very naturally with the class of distributions known as tempered distributions. Section 5, in some sense the heart of this chapter, derives explicit solutions to the classical linear PDE (.1) via Fourier analysis. The use of Fourier analysis and distribution theory to represent solutions to these PDE gives rise to numerous interesting identities, involving both elementary functions and special functions, such as the gamma function and Bessel functions, and we present some of these identities here, only a smattering from a rich area of classical analysis. Further development of harmonic analysis in Chapter 8 will bring in additional studies of special functions. Fourier analysis and distribution theory are also useful tools for general investigations of linear PDE, in cases where explicit formulas might not be obtainable. We illustrate a couple of cases of this in the present chapter, discussing the existence and behavior of parametrices for elliptic PDE with constant coefficients, and applications to smoothness of solutions to such PDE, in 9, and proving local solvability of general linear PDE with constant coefficients in 1. Fourier analysis and distribution theory will acquire further power in the next chapter as tools for investigations of existence and qualitative properties of solutions to various classes of PDE, with the development of Sobolev spaces. Sections 11 and 12 deal with the discrete Fourier transform, particularly with Fourier analysis on finite cyclic groups. We study this both as an approximation to Fourier analysis on the torus and Euclidean space, sometimes useful for numerical work, and as a subject with its intrinsic interest, and with implications for number theory. In 12 we give a brief description of fast algorithms for computing discrete Fourier transforms. 1. Fourier series Let f be an integrable function on the torus T n, naturally isomorphic to R n /Z n and to the Cartesian product of n copies S 1 S 1 of the circle.

3 Its Fourier series is by definition a function on Z n given by 1 (1.1) ˆf(k) = (2π) n f(θ)e ik θ dθ, T n 1. Fourier series 3 where k = (k 1,..., k n ), k θ = k 1 θ k n θ n. We use the notation (1.2) Ff(k) = ˆf(k). Clearly, we have a continuous linear map (1.3) F : L 1 (T n ) l (Z n ), where l (Z n ) denotes the space of bounded functions on Z n, with the sup norm. If f C (T n ), then we can integrate by parts to get (1.4) k α 1 ˆf(k) = (2π) n (D α f)(θ) e ik θ dθ, T n where k α = k α 1 1 kαn n, and (1.5) D α = D α 1 1 Dα n n, D j = 1. i θ j It follows easily that (1.6) F : C (T n ) s(z n ), where s(z n ) consists of functions u on Z n which are rapidly decreasing, in the sense that, for each N, (1.7) p N (u) = sup k Z n k N u(k) <. Here, we use the notation k = ( 1 + k 2) 1/2, where k 2 = k kn. 2 If we use the inner product (1.8) (f, g) = (f, g) L 2 = 1 (2π) n f(θ)g(θ) dθ, for f, g C (T n ), or more generally for f, g L 2 (T n ), and if on s(z n ), or more generally on l 2 (Z n ), the space of square summable functions on Z n, we use the inner product (1.9) (u, v) = (u, v) l 2 = k Z n u(k)v(k), we have the formula (1.1) (Ff, u) l 2 = (f, F u) L 2, T n

4 4 valid for f C (T n ), u s(z n ), where (1.11) F : s(z n ) C (T n ) is given by (1.12) ( F u ) (θ) = k Z n u(k) e ik θ. Another identity that we will find useful is 1 (1.13) (2π) n e ik θ e il θ dθ = δ kl, T n where δ kl = 1 if k = l and δ kl = otherwise. Our main goal here is to establish the Fourier inversion formula (1.14) f(θ) = k Z n ˆf(k) e ik θ, the sum on the right in (1.14) converging in the appropriate function space, depending on the nature of f. Let us single out another space of functions on T n, the trigonometric polynomials: (1.15) T P = { a(k)e ik θ } : a(k) = except for finitely many k. k Z n Clearly, (1.16) F : T P c (Z n ), where c (Z n ) consists of functions on Z n which vanish except at a finite number of points; this follows from (1.13). The formula (1.12) gives (1.17) F : c (Z n ) T P, and the formula (1.13) easily yields (1.18) FF = I on c (Z n ), and even (1.19) FF = I on s(z n ). By comparison, the inversion formula (1.14) states (1.2) F F = I, on C (T n ), or some other space of functions on T n, as specified below. Before getting to this, let us note one other implication of (1.13), namely, if (1.21) f j (θ) = k ϕ j (k)e ik θ

5 1. Fourier series 5 are elements of T P, or more generally, if ϕ j s(z n ), then we have the Parseval identity (1.22) (f 1, f 2 ) L 2 = k Z n ϕ 1 (k)ϕ 2 (k); in particular, the Plancherel identity (1.23) f j 2 L 2 = k Z n ϕ j (k) 2, for f j T P, or more generally for any f j of the form (1.21) with ϕ j s(z n ). In particular, the map F given by (1.12), and satisfying (1.11) and (1.17), has a unique continuous extension to l 2 (Z n ), and (1.24) F : l 2 (Z n ) L 2 (T n ) is an isometry of l 2 (Z n ) onto its range. Part of the inversion formula will be that the map (1.24) is also surjective. Let us note that if f j T P, satisfying (1.21), then (1.13) implies ˆf j (k) = ϕ j (k), so we have directly in this case: (1.25) F F = I on T P. One approach to more general inversion formulas would be to establish that T P is dense in various function spaces, on which F F can be shown to act continuously. For more details on this approach, see the exercises at the end of 1 and 2 in the Functional Analysis appendix. Here, we will take a superficially different approach. We will make use of such basic results from real analysis as the denseness of C(T n ) in L p (T n ), for 1 p <. Our approach to (1.14) will be to establish the following Abel summability result. Consider (1.26) J r f(θ) = k Z n ˆf(k) r k e ik θ, where k = k k n, r [, 1). We will show that (1.27) J r f f, as r 1, in the appropriate spaces. The operator J r in (1.26) is defined for any f L 1 (T n ), if r < 1, and we have the formula (1.28) J r f(θ) = (2π) n f(θ ) r k e ik (θ θ ) dθ. T n k Z n The sum over Z n inside the integral can be written as r k e ik (θ θ ) = P n (r, θ θ ) (1.29) k Z n = p(r, θ 1 θ 1) p(r, θ n θ n),

6 6 where (1.3) p(r, θ) = k= = 1 + r k e ikθ ( r k e ikθ + r k e ikθ) k=1 1 r 2 = 1 2r cos θ + r 2. Then we have the explicit integral formula J r f(θ) = (2π) n (1.31) f(θ )P n (r, θ θ ) dθ T n = (2π) n f(θ θ )P n (r, θ ) dθ. T n Let us examine p(r, θ). It is clear that the numerator and denominator on the right side of (1.3) are positive, so p(r, θ) > for each r [, 1), θ S 1. Of course, as r 1, the numerator tends to ; as r 1, the denominator tends to a nonzero limit, except at θ =. Since it is clear that π (1.32) (2π) 1 p(r, θ) dθ = (2π) 1 r k e ikθ dθ = 1, S 1 we see that, for r close to 1, p(r, θ) as a function of θ is highly peaked near θ = and small elsewhere, as in Fig π Figure 1.1 We are now prepared to prove the following result giving Abel summability (1.27).

7 1. Fourier series 7 Proposition 1.1. If f C(T n ), then (1.33) J r f f uniformly on T n as r 1. Furthermore, for any p [1, ), if f L p (T n ), then (1.34) J r f f in L p (T n ) as r 1. The proof of (1.33) is an immediate consequence of (1.31) and the peaked nature of p(r, θ) near θ = discussed above, together with the observation that, if f is continuous at θ, then it does not vary very much near θ. The convergence in (1.34) is in the L p -norm, defined by ] 1/p. (1.35) g L p = [(2π) n g(θ) p dθ We have the well-known triangle inequality in such a norm: (1.36) g 1 + g 2 L p g 1 L p + g 2 L p, and this implies, via (1.31) and (1.32), J r f L p = (2π) n P n (r, θ )τ θ f dθ L p T n (1.37) (2π) n P n (r, θ ) τ θ f L p dθ T n where = f L p, (1.38) τ θ f(θ) = f(θ θ ), which implies τ θ f L p = f L p. In other words, (1.39) J r L(L p ) 1, 1 p <, where we are using the operator norm on L p : (1.4) T L(L p ) = sup { T f L p : f L p 1}. Using this, we can deduce (1.34) from (1.33), and the denseness of C(T n ) in each space L p (T n ), for 1 p <. Indeed, given f L p (T n ), and given ε >, find g C(T n ) such that f g L p < ε. Note that, generally, g L p g sup. Now we have (1.41) J r f f L p J r (f g) L p + J r g g L p + g f L p < ε + J r g g L + ε, making use of (1.39). By (1.33), the middle term is < ε if r is close enough to 1, so this proves (1.34).

8 8 Corollary 1.2. If f C (T n ), then the Fourier inversion formula (1.14) holds. Proof. In such a case, as noted, we have ˆf s(z n ), so certainly the right side of (1.14) is absolutely convergent to some f # C(T n ). In such a case, one a fortiori has (1.42) lim ˆf(k)r k e ik θ = f # (θ). r 1 k Z n But now Proposition 1.1 implies (1.42) is equal to f(θ) (i.e., f # = f), so the inversion formula is proved for f C (T n ). As a result, we see that (1.43) F : s(z n ) C (T n ) is surjective, as well as injective, with two-sided inverse F : C (T n ) s(z n ). This of course implies that the map (1.24) has dense range in L 2 (T n ); hence (1.44) F : l 2 (Z n ) L 2 (T n ) is unitary. Another way of stating this is (1.45) {e ik θ : k Z n } is an orthonormal basis of L 2 (T n ), with inner product given by (1.8). Also, the inversion formula (1.46) F F = I on C (T n ) implies (1.47) Ff l 2 = f L 2, so therefore F extends by continuity from C (T n ) to a map (1.48) F : L 2 (T n ) l 2 (Z n ), unitary, inverting (1.44). The denseness C (T n ) L 2 (T n ) L 1 (T n ) implies that this F coincides with the restriction to L 2 (T n ) of the map (1.3). Note that the fact that (1.44) and (1.48) are inverses of each other extends the inversion result of Corollary 1.2. We devote a little space to conditions implying that the Fourier series (1.14) is absolutely convergent, weaker than the hypothesis that f C (T n ). Note that since e ik θ = 1, the absolute convergence of (1.14) implies uniform convergence. By (1.4), we see that (1.49) f C l (T n ) = ˆf(k) C k l, which in turn clearly gives absolute convergence provided (1.5) l n + 1.

9 Exercises 9 Using Plancherel s identity and Cauchy s inequality, we can do somewhat better: Proposition 1.3. If f C l (T n ), then the Fourier series for f is absolutely convergent provided (1.51) l > n 2. Proof. We have ˆf(k) = k l k l ˆf(k) k k [ (1.52) k 2l] 1/2 [ k 2l ˆf(k) 2] 1/2 k [ C k 2l ˆf(k) 2] 1/2, k as long as (1.51) holds. The square of the right side is dominated by C k γ ˆf(k) 2 = C D γ f 2 L 2 (1.53) k γ l γ l so the proposition is proved. k C f 2 C l, Sharper results on absolute convergence of Fourier series will be given in Chapter 4. See also some of the exercises below for more on convergence when n = 1. Exercises 1. Given f, g L 1 (T n ), show that with u(θ) = (2π) n 2. Given f, g C(T n ), show that ˆf(k)ĝ(k) = û(k), T n (fg)(k) = m f(ϕ)g(θ ϕ) dϕ. ˆf(k m)ĝ(m). 3. Using the proof of Proposition 1.3, show that every f Lip(S 1 ) has an absolutely convergent Fourier series.

10 1 4. Show that for any f L 1 (T n ), ˆf(k) as k. (Hint: Given ε >, pick f ε C (T n ), f f ε L 1 < ε. Compare ˆf ε(k) and ˆf(k).) This result is known as the Riemann-Lebesgue lemma. 5. For f L 1 (S 1 ), set (1.54) S N f(θ) = N k= N ˆf(k)e ikθ. Show that S N f(θ) = (1/2π) π π f(θ ϕ)d N (ϕ) dϕ, where (1.55) D N (θ) = N k= N e ikθ = sin(n )θ sin 1 2 θ. (Hint: To evaluate the sum, recall how to sum a finite geometrical series.) D N (θ) is called the Dirichlet kernel. See Fig Figure Let f L 1 (S 1 ) have the following property of vanishing at θ = : f(θ) sin 1 2 θ = g(θ) L1 ( π, π). Show that S N f() as N. (Hint. Adapt the Riemann-Lebesgue lemma to show that g L 1 ( π, π) π π g(θ) sin(n + 1 )θ dθ as N.) 2 7. Deduce that if f L 1 (S 1 ) is Lipschitz continuous at θ, then S N f(θ ) f(θ ) as N. Furthermore, if f is Lipschitz on an open interval J S 1, then S N f f uniformly on compact subsets of J.

11 Exercises Let f L (S 1 ) be piecewise Lipschitz, with a finite number of simple jumps. Show that S N f(θ) f(θ) at points of continuity. If f has a jump at θ j, with limiting values f ± (θ j ), show that (1.56) S N f(θ j ) 1 [ ] f + (θ j ) + f (θ j ), 2 as N. (Hint: By Exercise 7, it remains only to establish (1.56). Show that this can be reduced to the case θ j = π, f(θ) = θ, for π θ < π. Verify that this function has Fourier series ( 1) k 2 sin kθ.) k k=1 Alternative: Reduce to the case θ j = and note that S N f() depends only on the even part of f, (1/2)[f(θ) + f( θ)]. 9. Work out the Fourier series of the function f Lip(S 1 ) given by f(θ) = θ, π θ π. Examining this at θ =, establish that 1 (1.57) k = π k=1 1. One can obtain Fourier coefficients of functions θ k and θ k on [ π, π] in terms of the Fourier coefficients of Show that, for n, k= q k (θ) = θ k 1 k! ˆq k(n) (is) k = 1 2πin on [, π], on [ π, ]. [ ]( ( 1) n e πis 1 1 s ) 1, n and use this to work out the Fourier series for these functions. Apply this to Exercise 9, and to the calculation at the end of Exercise Assume that g L 1 (S 1 ) has uniformly convergent Fourier series (S N g g) on compact subsets of an open interval J S 1. Show that whenever f L 1 (S 1 ) and f = g on J, then f also has uniformly convergent Fourier series on compact subsets of J. (Hint: Apply Exercise 7 to f g.) This result is called the localization principle for Fourier series. 12. Suppose f is Hölder continuous on S 1, that is, f C r (S 1 ), for some r (, 1), which means f(ϕ + θ) f(ϕ) C θ r. Show that f has uniformly convergent Fourier series on S 1. (Hint: We have f(θ + ϕ) f(ϕ) sin 1 θ sin C 1 θ (1 r) 2. 2

12 12 Apply Exercise 6.) 13. If ω : [, ) [, ) is continuous and increasing and ω() =, we say a function f on S 1 is continuous with modulus of continuity ω provided f(ϕ + θ) f(ϕ) Cω( θ ). Formulate the most general condition you can to establish uniform convergence of the Fourier series of a function with such a modulus of continuity. Note that Exercise 12 deals with the case ω(s) = s r, r (, 1). 14. Consider the Cesaro sum of the Fourier series of f: C N f(θ) = N k= N ( 1 k ) ˆf(k)e ikθ = 1 N 1 S l f(θ). N N l= Show that C N f(θ) = (1/2π) π f(θ ϕ)fn (ϕ)dϕ, where π (1.58) F N (θ) = 1 N N 1 l= N 1 1 D l (θ) = N sin 1 θ 2 l= ( sin l + 1 ) ( θ = 1 sin N 2 N 2 θ sin 1 2 θ ) 2. The function F N (θ) is called the Fejer kernel (see Fig. 1.3). Modify the proof of Proposition 1.1 to show that C N f F in B, for f B, where B is one of the Banach spaces C(S 1 ) or L p (S 1 ), 1 p <. (Hint: To evaluate the second sum in (1.58), use sin(l )θ = Im eiθ/2 e ilθ and sum a finite geometrical series. Also use the identity 2 sin 2 z = 1 cos 2z.) Figure 1.3

13 2. Harmonic functions and holomorphic functions in the plane Harmonic functions and holomorphic functions in the plane The method of proof of the Abel summability (1.26) (1.27) of Fourier series, specialized to T 1 = S 1, has important implications for the theory of harmonic functions on a domain Ω R 2, which we will discuss here. In the case of S 1, let us rewrite (1.26), (2.1) J r f(θ) = as (2.2) (PI f)(r, θ) = k= k= ˆf(k) r k e ikθ, ˆf(k) r k e ikθ. The function u(r, θ) = PI f(r, θ) is called the Poisson integral of f. If we use polar coordinates in the complex plane C = R 2, (2.3) z = r e iθ, then (2.2) becomes (2.4) (PI f)(z) = ˆf(k)z k + ˆf( k)z k k= k=1 = (PI + f)(z) + (PI f)(z), defined on the unit disk z < 1. Note also, from (1.3), that (2.5) PI f(z) = 1 z 2 f(w) 2π w z 2 ds(w), S 1 the integral being with respect to arclength on S 1. Recall that if f L 1 (S 1 ), the function ˆf(k) is bounded, so both power series in (2.4) have radius of convergence at least 1. Clearly, on the unit disk, v(z) = (PI + f)(z) is holomorphic and w(z) = (PI f)(z) is antiholomorphic. In other words, v and w belong to C (D), where D = {z C : z < 1}, and (2.6) where (2.7) Note that v z =, z = 1 ( 2 x + i ), y z w z z = z = on D, z = 1 ( 2 x i ). y z = 1 4,

14 14 where is the Laplace operator on R 2, a special case of the Laplace operator introduced in Chapter 2. Since v, w C (D), we have v = and w =, and hence (2.8) (PI f) =. In light of the results of 1, we have the following. Proposition 2.1. If f C(S 1 ), then (2.9) u = PI f(z) C (D) C(D) is harmonic, with boundary value f, that is, u solves the Dirichlet problem (2.1) u = in D, u D = f. One should expect that if f has extra smoothness on S 1, so does PI f on D. The following result is crude compared to results established in Chapters 4 and 5, but it will be of some interest. Proposition 2.2. For l = 1, 2, 3,..., we have (2.11) PI : C l+1 (S 1 ) C l (D). Proof. We begin with the case l = 1. Since we know from (2.4) that PI f C (D), we need merely check smoothness near D = S 1. Clearly, (2.12) f PI f = PI θ θ, so if f C 1 (S 1 ), then f/ θ C(S 1 ) and we have ( / θ)pi f continuous on D. Also, by (2.2), (2.13) r PI f = PI (Nf), r where N f is characterized by the Fourier series representation (2.14) Nf(θ) = ˆf(k) k e ikθ. Thus k= (2.15) Nf = i f Hf = ih θ θ, where H has the Fourier series representation (2.16) Hg(θ) = (sgn k)ĝ(k)e ikθ. k=

15 2. Harmonic functions and holomorphic functions in the plane 15 We claim that, for l, (2.17) H : C l+1 (S 1 ) C l (S 1 ), and hence N : C l+2 (S 1 ) C l (S 1 ). Given this, for f C 2 (S 1 ), the quantity (2.13) is seen to belong to C(D), and this finishes the l = 1 case of (2.11). In turn, since H commutes with / θ, it suffices to establish the l = case of (2.17). Now, by Proposition 1.3, the Fourier series for g C 1 (S 1 ) is absolutely convergent, giving (2.17). To prove the general case of (2.11), a short calculation yields ( (2.18) r ) j ( ) kpi (( ) kn f = PI f) j. r θ θ Note that N j = ( i) j ( / θ) j H o(j), where o(j) is zero if j is even and one if j is odd, so, for l, N j : C l+j+1 (S 1 ) C l (S 1 ), the left side being improved to C l+j if j is even. Therefore, if f C l+1 (S 1 ) and j + k = l, the right side of (2.18) is P I f jk with f jk C(S 1 ), which proves (2.11) in general. The implication f C l (S 1 ) PI f C l (D) does not quite work, as we will see later, essentially because H does not map C l (S 1 ) to itself. It is true that f C l,α (S 1 ) PI f C l,α (D), for α (, 1). This is a special case of Hölder estimates that will be established in 7 of Chapter 13. Similarly there are sharp results on regularity of PI f in Sobolev spaces, discussed in Chapter 4, and in much greater generality, in Chapter 5. It is important to know that PI f provides the unique solution to the Dirichlet problem (2.1). We will establish several general uniqueness results, starting with the following. Proposition 2.3. Let Ω R n be a bounded region with smooth boundary, say, Ω = D. Suppose u, v C 2 (Ω), with u = v = f on Ω, and u = v = in Ω. Then u = v on all of Ω. Proof. Set w = u v C 2 (Ω); w = on Ω. We can apply the Green identity (3.15) of Chapter 2, to write (2.19) (dw, dw) = ( w, w) + w w ν ds. By hypothesis the right side of (2.19) is. Thus w is constant on each component of Ω, and the boundary condition forces w =. In view of Proposition 2.2, we could apply this to u = PI f if f C 3 (S 1 ), but this is not a satisfactory result, and we will do much better below. Ω

16 16 A result related to our uniqueness question is the mean-value property, a special case of which is the following. Proposition 2.4. If f C(S 1 ), u = PI f, then (2.2) u() = 1 2π π π f(θ) dθ. Proof. It follows from the series (2.2) that u() = ˆf(), which gives (2.2). A more general result is the following. Proposition 2.5. If B R R n is the open ball of radius R, centered at the origin, with B R = S R, of area A(R), then for u C 2 (B R ) C(B R ), u =, we have (2.21) u() = 1 u(x) ds. A(R) S R Proof. We apply Green s identity [ ] (2.22) u v v u dx = Ω Ω [ u v ν v u ] ds, ν to Ω = B r, < r < R, v(x) = x 2, with v = 2n, to get, when u =, (2.23) n u(x) dx = r u(x) ds, B r S r noting that substituting v = 1 in (2.22) gives ( u/ ν) ds =. If we let Ω ϕ(r) = B r u(x) dx, this implies ϕ (r) = (n/r)ϕ(r), and hence ϕ(r) = Kr n, i.e., V (r) 1 B r u(x) dx is constant. Passing to the limit r gives (2.21). Second Proof. Define v C 2 (B R ) C(B R ) by v(x) = u(gx) dg, SO(n) where dg is Haar measure on the rotation group SO(n), defined in 5 of Appendix B. The Laplace operator is invariant under rotations, so v = on B R. The finction v is radial; v(x) = ṽ( x ). The formula = 2 r 2 + n 1 r r + 1 r 2 S,

17 2. Harmonic functions and holomorphic functions in the plane 17 (cf. (4.17) of Chapter 2), for in polar coordinates, where S Laplace operator on the unit sphere S n 1, gives ( d 2 dr 2 + n 1 d ) ṽ(r) =. r dr This is an Euler equation, whose solutions are is the A + Br 2 n, n 3, A + B log r, n = 2. Since v does not blow up at, we have B =, so v is constant. Clearly v(x) equals the right side of (2.21) for x = R, and v() = u(), so we again have (2.21). Corollary 2.6. For any Ω R n open, any u C 2 (Ω) harmonic, any ball B p, centered at p and contained in Ω, we have (2.24) u(p) = Avg Bp u(z). We can now prove the following important maximum principle for harmonic functions. Much more general versions of this will be given in Chapter 5. Proposition 2.7. Let Ω R n be connected and open, and let u C 2 (Ω) be harmonic and real-valued. Then u has no interior maximum, or minimum, unless u is constant. In particular, if Ω is bounded and u C(Ω), then (2.25) sup u(p) = sup u(q). p Ω q Ω Also, even for u complex-valued, (2.26) sup u(p) = sup u(q). p Ω q Ω Proof. That a non-constant, real harmonic function has no interior extremum is an obvious consequence of (2.24), and the other consequences, (2.25) and (2.26), follow immediately. Corollary 2.8. The uniqueness result of Proposition 2.3 holds for any bounded open Ω R n, with no smoothness on Ω, and for any harmonic (2.27) u, v C 2 (Ω) C(Ω). Proof. Apply the maximum principle to u v.

18 18 Here, we have used the mean-value property to prove the maximum principle. The more general maximum principle established in Chapter 5 will not use the mean-value property; indeed, together with a symmetry argument, it can be made a basis for a proof of the mean-value property. We will leave these considerations until Chapter 5. With our uniqueness result in hand, we can easily establish the following interior regularity result for harmonic functions. Proposition 2.9. Let Ω R 2 be open, and let u C 2 (Ω) be harmonic. Then in fact, u C (Ω); u is even real analytic on Ω. Proof. By translations and dilations, we can reduce to the case Ω = D, u C 2 (D). The uniqueness result of Corollary 2.8 implies (2.28) u = PI f, where f = u S 1. Then the conclusion that u is real analytic on D follows directly from the power-series expansion (2.4). Parenthetically, we remark that, by Corollary 2.8, the identity (2.28) holds for any u C 2 (D) C(D) harmonic in D. Using the results we have developed, via Fourier series, about harmonic functions, we can quickly draw some basic conclusions about holomorphic functions. If Ω C is open, f : Ω C is by definition holomorphic if and only if f C 1 (Ω) and f/ z =, where / z is given by (2.7). Clearly, if f C 2 (Ω) is holomorphic, then it is also harmonic, and so are its real and imaginary parts. Suppose u C 2 (D) C(D) is holomorphic in D. Then the series representation (2.4) is valid, since (2.28) holds. This series is a sum of two terms: u(z) = ˆf(k)z k + ˆf( k)z k (2.29) k= = u 1 (z) + u 2 (z), where u 1 / z = and u 2 / z =. But if we are given that u/ z =, then also u 2 / z =, so u 2 / x = u 2 / y =. Thus u 2 is constant, and since u 2 () =, this forces u 2 =. In other words, the holomorphic function u(z) has the power series (2.3) u(z) = k=1 a k z k, z D, k= where a k = ˆf(k), k. Note that differentiation of (2.3) gives (2.31) a k = 1 ( ) ku(). k! z

19 2. Harmonic functions and holomorphic functions in the plane 19 By the usual method of translating and dilating coordinates, we deduce the following. Proposition 2.1. If Ω C is open, u C 2 (Ω) holomorphic, p Ω, and D p a disk centered at p, D p Ω, then on D p, u(z) is given by a convergent power-series expansion (2.32) u(z) = b k (z p) k. k= We can relax the C 2 -hypothesis to u C 1 (Ω). As much stronger and more general results are given in Chapter 5, we omit the details here. For further use, we record the following result, whose proof is trivial. Lemma If u C (Ω) is holomorphic, and P = a jk D j xd k y is any constant-coefficient differential operator, then P u is holomorphic in Ω. Proof. ( / z)p u = P ( u/ z). We can use the power-series representation (2.3) (2.32) to prove the fundamental result on uniqueness of analytic continuation, which we give below. Here is the first result, of a very general nature. Proposition Let Ω R n be open and connected, and let u be a real-analytic function on Ω. If p Ω and all derivatives D α u(p) =, then u = on all of Ω. Proof. Let K = {x Ω : D α u(x) = for all α }. Since u C (Ω), K is closed in Ω. However, for each p K, since u is given in a neighborhood of p by a power series u(q) = α u (α) (p) α! (q p) α, we also see that K is open in Ω. This proves the proposition. Our basic corollary for holomorphic functions is the following. Corollary Let Ω C be open and connected, u holomorphic on Ω. Let γ be a line segment contained in Ω. If u γ =, then u = on Ω.

20 2 Proof. Translating and rotating, we can assume γ is a segment in the real axis, with γ. Near, u(z) has a power-series expansion of the form (2.3), with a k given by (2.31). Using Lemma 2.11, we see that (2.33) ( z ) ku() ( ) ku(), = x which vanishes for all k. Thus u = on a nonempty open set in Ω, and the rest follows by Proposition Actually, a much stronger result is true. If Ω C is connected, p j Ω are distinct, p j p Ω, u is holomorphic in Ω, and u(p j ) = for each j, then u must vanish identically. In other words, u can have only isolated zeros if it does not vanish identically. Indeed, say u(p) =. If u is not identically zero, some coefficient in the series (2.32) is nonzero; let b m be the first such coefficient: (2.34) u(z) = (z p) m k= = (z p) m v(z), b m+k (z p) k where v(z) is holomorphic on D p and v() = b m. Thus, by continuity, v(ζ) for ζ p < ε if ε is sufficiently small, which implies u(ζ) if ζ p but ζ p < ε. A typical use of Corollary 2.13 is in computations of integrals. We will see an example of this in the next section. We end this section by recalling the classical Cauchy integral theorem and integral formula. Throughout, Ω will be a bounded open domain in C with smooth boundary. Stokes formula, proved in Chapter 1, 13, states (2.35) Ω dα = α, Ω for a 1-form α with coefficients in C 1 (Ω). If α = p dx + q dy, this gives the classical Green s formula (2.36) p dx + q dy = Ω Ω ( q x p ) dx dy. y

21 2. Harmonic functions and holomorphic functions in the plane 21 If u(x, y) C 1 (Ω) is a complex-valued function, we consequently have u dz = (u dx + iu dy) (2.37) Ω Ω = Ω = 2i Ω ( i u x u ) dx dy y u z dx dy. In the special case when u is holomorphic in Ω, we have the Cauchy integral theorem: Theorem If Ω C is bounded with smooth boundary and u C 1 (Ω) is holomorphic, then (2.38) u(z) dz =. Ω Using various limiting arguments, one can relax the hypotheses on smoothness of Ω and of u near Ω; we won t go into this here. Next we prove Cauchy s integral formula. Proposition With Ω as above, u C 2 (Ω) holomorphic in Ω, we have (2.39) u(ζ) = 1 u(z) dz, for ζ Ω. 2πi z ζ Proof. Write (2.4) Ω u(z)(ζ z) 1 = (ζ z) 1[ u(z) u(ζ) ] + u(ζ)(ζ z) 1 = v(z) + u(ζ)(ζ z) 1. By the series expansion for u(z) about ζ, we see that v(z) is holomorphic near ζ; clearly, it is holomorphic on the rest of Ω, and it belongs to C 1 (Ω), so v(z) dz =. Thus, to prove (2.39), it suffices to show that Ω (2.41) (z ζ) 1 dz = 2πi, for ζ Ω. Ω Indeed, if ε is small enough that B(ζ, ε) = {z C : z ζ ε} is contained in Ω, then Cauchy s theorem implies that the left side of (2.41) is equal to (2.42) (z ζ) 1 dz B(ζ,ε)

22 22 since (ζ z) 1 is holomorphic in z for z ζ. Making a change of variable, we see that (2.42) is equal to so the proof is complete. 2π ( ε e iθ ) 1 iε e iθ dθ = 2πi, As stated before, the C 2 -hypothesis can be relaxed to C 1. A function u(z) of the form u(z) = v(z)/(z a) k, k Z +, where v is holomorphic on a neighborhood O of a, is said to have a pole of order k at z = a if v(a). In such a case, a variant of the preceding calculations yields 1 2πi γ u(z) dz = 1 (k 1)! v(k 1) (a), the coefficient of (z a) k 1 in the power series of v(z) about z = a, if γ is a smooth, simple, closed curve about a such that v is holomorphic on a neighborhood of the closed region bounded by γ. This quantity is called the residue of u(z) at z = a. One can often evaluate integrals by evaluating residues. We give a simple illustration here; others are given in (2.48), (3.32), (A.14), and (A.15). Here we evaluate (2.43) dx 1 + x 2 = lim R γ R dz 1 + z 2, where γ R is the closed curve, going from R to R along the real axis, then from R to R counterclockwise on the circle of radius R centered at, that is, γ R = O R, where O R = {z : Re z >, z < R}. There is just one pole of (1 + z 2 ) 1 in O R, located at z = i. Since (1 + z 2 ) 1 = (z + i) 1 (z i) 1, we see that the residue of (1 + z 2 ) 1 at z = i is 1/2i, so dx 1 + x 2 = π. Exercises 1. Suppose u satisfies the following Neumann boundary problem in the disk D: (2.44) u = in D, u r = g on S1. If u = PI(f), show that f and g must be related by ĝ(k) = k ˆf(k), for all k Z, that is, (2.45) g = Nf,

23 Exercises 23 with N defined by (2.13). 2. Define the function k N by k N (θ) = k k 1 e ikθ. Show that k N L 2 (S 1 ) L 1 (S 1 ). Also show that, provided g L 2 (S 1 ) and (2.46) g(θ) dθ =, S 1 a solution to (2.45) is given by (2.47) f(θ) = (2π) 1 S 1 k N (θ ϕ)g(ϕ) dϕ = T g(θ). 3. If T is defined by (2.47), show that T : L p (S 1 ) L p (S 1 ) for p [1, ) and T : C l (S 1 ) C l (S 1 ) for l =, 1, 2, Given g C 1 (S 1 ), show that (2.44) has a solution u C 1 (D) if and only if (2.46) holds. If g C l (S 1 ), show that (2.44) has a solution u C l (D). Note: Regularity results of a more precise nature are given in Chapter 4, 4, in Exercise 1. See also Chapter 5, 7, for more general results. 5. Let Ω R 2 be a smooth, bounded, connected region. Show that if w C 2 (Ω), w = on Ω, and w/ ν = on Ω, then w is constant. (Hint: Use (2.19).) Note: One can weaken the C 2 -hypothesis to w C 1 (Ω); compare Proposition 2.2 of Chapter 5. For another type of relaxation, see Chapter 4, 4, Exercise Show that a C 1 -function f : Ω C is holomorphic if and only if, at each z Ω, Df(z), a priori a real linear map on R 2, is in fact complex linear on C. Note: This exercise has already been given in Chapter 1, Let f be a holomorphic function on Ω C, with f : Ω O, and let u be harmonic on O. Show that v = u f is harmonic on Ω. (Hint: For a short proof, write u locally as a sum of a holomorphic and an anti-holomorphic function.) 8. Let g(z) = 1 a kz k, and form the harmonic function u = 2 Re g = g + g. Show that, under appropriate hypotheses on (a k ), g S 1 = P +(u S 1), where P + is given by P +f(θ) = ˆf(k)e ikθ. k= 9. Find a holomorphic function on D that is unbounded but whose real part has a continuous extension to D. Reconsider this problem after reading 6 of Chapter Hence show that P + does not map C(S 1 ) to itself, nor does it map any C l (S 1 ) to itself, for any integer l. 11. Hence find f C 1 (S 1 ) such that PI f / C 1 (D). 12. Use the method of residues to calculate (2.48) e iξx dx, ξ. 1 + x2

24 24 (Hint: Write this as lim R γ R e iξz /(1 + z 2 ) dz, with γ R as in (2.43), given ξ. Then find the residue of e iξz /(1 + z 2 ) at z = i.) 13. Use the Poisson integral formula (2.5) to prove the following. Let u C 2 (D) C(D) be harmonic in the disk D = {z C : z < 1}. Assume u on D and u() = 1. Then (2.49) z = a [, 1) = u(z) 1 a 1 + a. This result is known as a Harnack inequality. Hint. Use the inequality w = 1, z = a [, 1) = 1 z 2 w z 2 1 a 1 + a. Note. By translating and scaling, if u is harmonic and on D R(p) = {z C : z p < R}, then z p = a [, R) = u(z) R a R + a u(p). 14. Using Exercise 13, show that if u is harmonic in the entire plane C and u on C, then u is constant. More generally, if there exists a constant K such that u K on C, then u is constant. This is a version of Liouville s theorem. See Proposition 4.6 for another version. 15. Using Exercise 13, show that there exists A (, ) with the following property. Let u be harmonic on D R(). Assume Then u() =, u(z) M on D R (). u(z) AM on D R/2 (). (Hint. Set v(z) = M u(z), so v(z) on D R (), v() = M. Say p D R/2 (), u(p) = inf DR/2 () u. Deduce that and from there that v(z) 1 3 (M u(p)) on D R/4(p), Avg DR () v (M u(p)), 3 while this average is equal to v() = M.) 16. Assume u is harmonic on C and u(z) C + C 1 z k, z C, with C, C 1 (, ). Take A from Exercise 15. Show that there exists C 2 such that u(z) C 2 AC 1 z k, z C. Note. In conjunction with Proposition 4.6, one gets that u(z) must be a polynomial of degree k in x and y. 17. Let Ω be the strip Ω = {z C : < Re z < 1}, with closure Ω. Let f be continuous on Ω and holomorphic on Ω. Show that if f is bounded, f(z) A on Ω = f(z) A on Ω.

25 3. The Fourier transform 25 (Hint. First prove the implication under the additional hypothesis that f(z) as z. Then consider f ε(z) = e εz2 f(z).) 18. In the setting of Exercise 17, show that if θ (, 1), f(iy) A, f(1 + iy) B, y R = f(θ + iy) A 1 θ B θ, y R. This result is known as the Hadamard three-lines lemma. g(z) = A z 1 B z f(z).) (Hint. Consider 3. The Fourier transform The Fourier transform is defined by (3.1) Ff(ξ) = ˆf(ξ) = (2π) n/2 f(x)e ix ξ dx when f L 1 (R n ). It is clear that (3.2) F : L 1 (R n ) L (R n ). This is analogous to (1.3). The analogue for C (T n ), and simultaneously for s(z n ), of 1, in this case is the Schwartz space of rapidly decreasing functions: (3.3) S(R n ) = { u C (R n ) : x β D α u L (R n ) for all α, β }, where x β = x β 1 1 xβn n, D α = D α 1 1 Dαn n, with D j = i / x j. It is easy to verify that (3.4) F : S(R n ) S(R n ) and (3.5) ξ α D β ξ Ff(ξ) = ( 1) β F(D α x β f)(ξ). We define F by (3.6) F f(ξ) = f(ξ) = (2π) n/2 f(x)e ix ξ dx, which differs from (3.1) only in the sign of the exponent. It is clear that F satisfies the mapping properties (3.2), (3.4), and (3.7) (Fu, v) = (u, F v), for u, v S(R n ), where (u, v) denotes the usual L 2 -inner product, (u, v) = u(x) v(x) dx. As in the theory of Fourier series, the first major result is the Fourier inversion formula. The following is our first version. Proposition 3.1. We have the inversion formula (3.8) F F = FF = I on S(R n ).

26 26 As in the proof of the inversion formula for Fourier series, via Proposition 1.1, in the present proof we will sneak up on the inversion formula by throwing in a convergence factor that will allow interchange of orders of integration (in the proof of Proposition 1.1, the orders of an integral and an infinite series were interchanged). Also, as we will see in 5, this method will have serendipitous applications to PDE. So, let us write, for f S(R n ), [ ] F Ff(x) = (2π) n f(y)e iy ξ dy e ix ξ dξ (3.9) = (2π) n lim f(y) e ε ξ 2 e i(x y) ξ dy dξ. ε We can interchange the order of integration on the right for any ε >, to obtain (3.1) F Ff(x) = lim f(y)p(ε, x y) dy, ε where (3.11) p(ε, x) = (2π) n e ε ξ 2 +ix ξ dξ. Note that (3.12) p(ε, x) = ε n/2 q(ε 1/2 x), where q(x) = p(1, x). In a moment we will show that (3.13) p(ε, x) = (4πε) n/2 e x 2 /4ε. The derivation of this identity will also show that (3.14) q(x) dx = 1. R n From this, it follows as in the proof of Proposition 1.1 that (3.15) lim f(y)p(ε, x y) dy = f(x), ε for any f S(R n ), even for f bounded and continuous, so we have proved F F = I on S(R n ); the proof that FF = I on S(R n ) is identical. It remains to verify (3.13). We observe that p(ε, x), defined by (3.11), is an entire holomorphic function of x C n, for any ε >. It is convenient to verify that (3.16) p(ε, ix) = (4πε) n/2 e x 2 /4ε, x R n,

27 3. The Fourier transform 27 from which (3.13) follows by analytic continuation. Now p(ε, ix) = (2π) n e x ξ ε ξ 2 dξ (3.17) = (2π) n e x 2 /4ε e x/2 ε+ εξ 2 dξ = (2π) n ε n/2 e x 2 /4ε e ξ 2 dξ. To prove (3.16), it remains to show that (3.18) e ξ 2 dξ = π n/2. Indeed, if R n (3.19) A = e ξ2 dξ, then the left side of (3.18) is equal to A n. But for n = 2 we can use polar coordinates: 2π (3.2) A 2 = e ξ 2 dξ = e r2 r dr dθ = π. R 2 This completes the proof of the identity (3.16) and hence of (3.13). In light of (3.7) and the Fourier inversion formula (3.8), we see that, for u, v S(R n ), (3.21) (Fu, Fv) = (u, v) = (F u, F v). Thus F and F extend uniquely from S(R n ) to isometries on L 2 (R n ) and are inverses to each other. Thus we have the Plancherel theorem: R n Proposition 3.2. The Fourier transform (3.22) F : L 2 (R n ) L 2 (R n ) is unitary, with inverse F. The inversion formulas of Propositions 3.1 and 3.2 do not provide for the inversion of F in (3.2). We will obtain this as a byproduct of the Fourier inversion formula for tempered distributions, in the next section. We make a remark about the computation of the Fourier integral (3.11), done above via analytic continuation. The following derivation does not make any direct use of complex analysis. It suffices to handle the case ε = 1/2, that is, to show (3.23) Ĝ(ξ) = e ξ 2 /2 if G(x) = e x 2 /2, on R n.

28 28 We have interchanged the roles of x and ξ compared to those in (3.11) and (3.13). It suffices to get (3.23) in the case n = 1, by the obvious multiplicativity. Now the Gaussian function G(x) = e x2 /2 satisfies the differential equation ( d ) (3.24) dx + x G(x) =. By the intertwining property (3.5), it follows that (d/dξ + ξ)ĝ(ξ) =, and uniqueness of solutions to this ODE yields Ĝ(ξ) = /2 Ce ξ2. The constant C is evaluated via the identity (3.2); C = 1; and we are done. As for the necessity of computing the Fourier integral (3.11) to prove the Fourier inversion formula, let us note the following. For any g S(R n ) with g() = 1, (g(ξ) = e ξ 2 being an example), we have (replacing ε by δ 2 ), just as in (3.9), F Ff(x) = (2π) n lim f(y)g(δξ)e i(x y) ξ dy dξ δ (3.25) = lim f(y)h δ (x y) dy, δ where (3.26) h δ (x) = (2π) n g(δξ)e ix ξ dξ = (2π) n/2 δ n g(δ 1 x). By the peaked nature of h δ as δ, we see that the limit in (3.25) is equal to (3.27) C f(x), where (3.28) C = h 1 (x) dx = (2π) n/2 g(x) dx. The argument (3.25) (3.27) shows that C is independent of the choice of g S(R n ), and we need only find a single example g such that g(x) can be evaluated explicitly and then the integral on the right in (3.28) can be evaluated explicitly. In most natural examples one picks g to be even, so g = ĝ. We remark that one does not need to have g S(R n ) in the argument above; it suffices to have g L 1 (R n ), bounded and continuous, and such that ĝ L 1 (R n ). An example, in the case n = 1, is (3.29) g(ξ) = e ξ. In this case, elementary calculations give ( 2 ) 1/2 1 (3.3) ĝ(x) = π x ;

29 3. The Fourier transform 29 compare (5.21). In this case, (3.28) can be evaluated in terms of the arctangent. Another example, in the case n = 1, is (3.31) In this case, g(ξ) = 1 ξ if ξ 1, if ξ 1. (3.32) ĝ(x) = (2π) 1/2( sin 1 2 x ) 2, 1 2 x and (3.28) can be evaluated by the method of residues. The calculation of (3.32) can be achieved by evaluating 1 (1 ξ) cos xξ dξ via an integration by parts, though there is a more painless way, mentioned below. We now make some comments on the relation between the Fourier transform and convolutions. The convolution u v of two functions on R n is defined by u v(x) = u(y)v(x y) dy (3.33) = u(x y)v(y) dy. Note that u v = v u. If u, v S(R n ), so is u v. Also u v L p (R n ) u L 1 v L p, so the convolution has a unique, continuous extension to a bilinear map (3.34) L 1 (R n ) L p (R n ) L p (R n ), for 1 p < ; one can directly perceive that this also works for p =. Note that the right side of (3.1), for any ε >, is an example of a convolution. Computing the Fourier transform of (3.33) leads immediately to the formula (3.35) F(u v)(ξ) = (2π) n/2 û(ξ)ˆv(ξ). We also note that if (3.36) P = α k a α D α is a constant-coefficient differential operator, we have (3.37) P (u v) = (P u) v = u (P v)

30 3 if u, v S(R n ). This also generalizes; if u S(R n ), v L p (R n ), the first identity continues to hold; as we will see in the next section, so does the second identity, once we are able to interpret what it means. We mention the following simple application of (3.35), to a short calculation of (3.32). With g given by (3.31), we have g = g 1 g 1, where (3.38) g 1 (ξ) = 1 for ξ Thus (3.39) ĝ 1 (x) = (2π) 1/2 1/2 1/2 [ 1 2, 1 2 and then (3.32) follows immediately from (3.35). ], otherwise. e ixξ dξ = (2π) 1/2 sin 1 2 x 1 2 x, Exercises 1. Show that F : L 1 (R n ) C (R n ), where C (R n ) denotes the space of functions v, continuous on R n, such that v(ξ) as ξ. (Hint: Use the denseness of S(R n ) in L 1 (R n ).) This result is the Riemann-Lebesgue lemma for the Fourier transform. 2. Show that the Fourier transforms (3.1) and (3.22) coincide on L 1 (R n ) L 2 (R n ). 3. For f L 1 (R), set S R f(x) = (2π) 1/2 R ˆf(ξ)e ixξ dξ. Show that R where S Rf(x) = D R f(x) = R D R (x) = (2π) 1 e ixξ dξ = R D R(x y)f(y) dy, sin Rx πx. Compare Exercise 5 of Show that f L 2 (R) S R f f in L 2 -norm as R. 5. Show that there exist f L 1 (R) such that S Rf / L 1 (R) for any R (, ). (Hint: Note that D R / L 1 (R).) 6. For f L 1 (R), set R ( C R f(x) = (2π) 1/2 1 ξ ) ˆf(ξ)e ixξ dξ. R R R Show that C R f(x) = E R f(x), where R ( E R (x) = (2π) 1 1 ξ ) e ixξ dξ = 2 R πr Note that E R L 1 (R). Show that, for 1 p <, [ sin 1 2 Rx x f L p (R) = C Rf f in L p -norm, as R. We say the Fourier transform of f is Cesaro-summable if C Rf f as R. ] 2.

31 Exercises 31 In Exercises 7 13, suppose f S(R) has the following properties: f, f(x) dx = 1, and xf(x) dx =. Set F (ξ) = (2π)1/2 ˆf(ξ). The point of the exercises is to obtain a version of the central limit theorem. 7. Show that F () = 1, F () =, F () = 2a <. Also, ξ F (ξ) < Set F n (ξ) = F (ξ/ n) n. Relate (2π) 1/2 Fn (x) to the convolution of n copies of f. 9. Show that there exist A > and G C ([ A, A]) such that f(ξ) = e aξ2 G(ξ) for ξ A, and G() = 1, G () = G () =. Hence F n (ξ) = e aξ2 G(n 1/2 ξ) n, for ξ A n. 1. Show that G(ξ/ n) n 1 Cn α if ξ n (1/2 α)/3, for n large. Fix α (, 1 ), and set γ = (1/2 α)/3 (, 1 2 ) Show that, for ξ n γ, F (ξ/ n) an (1 2γ), for n large, so F n(ξ) e an2γ /4 = δ n. Deduce that F n (ξ) dξ C δ ξ n γ (n 1)/n n n, as n. 12. From Exercises 9 11, deduce that F n e aξ2 in L 1 (R) as n. 13. Deduce now that (2π) 1/2 Fn (4πa) 1/2 e x2 /4a in both C (R) and L 1 (R), as n. Relate this to the central limit theorem of probability theory. Weaken the hypotheses on f as much as you can. (Hint: In passing from the C -result to the L 1 -result, positivity of F n will be useful.) 14. With p ε(x) = (4πε) 1/2 e x2 /4ε, as in (3.13) for n = 1, show that, for any u(x), continuous and compactly supported on R, p ε u u uniformly as ε. Show that for each ε >, p ε u(x) is the restriction to R of an entire holomorphic function of x C. 15. Using Exercise 14, prove the Weierstrass approximation theorem: Any f C([a, b]) is a uniform limit of polynomials. (Hint: Extend f to u as above, approximate u by p ε u, and expand this in a power series.) 16. Suppose f S(R n ) is supported in B R = {x R n : x R}. Show that ˆf(ξ) is holomorphic in ξ C n and satisfies (3.4) ˆf(ξ + iη) C N ξ N e R η, ξ, η R n. 17. Conversely, suppose g(ξ) = ˆf(ξ) S(R n ) has a holomorphic extension to C n satisfying (3.4). Show that f is supported in x R. (Hint: With ω = x/ x, r, write (3.41) f(x) = (2π) n/2 ˆf(ξ + irω)e ix ξ rx ω dξ, with R n ˆf(ξ + irω) e ix ξ rx ω C N ξ N e r(r x ), and let r +.) This is a basic case of the Paley-Wiener theorem.

32 Given f L 1 (R n ), show that f is supported in B R if and only if ˆf(ξ) is holomorphic in C n, satisfying ˆf(ξ + iη) C e R η, ξ, η R n. Reconsider this problem after reading Show that f(x) = (Hint: See (A.13) (A.15).) 1 cosh x = ˆf(ξ) = π/2 cosh π 2 ξ. 4. Distributions and tempered distributions L. Schwartz s theory of distributions has proved to be not only a wonderful tool in partial differential equations, but also a device that lends clarity to many aspects of Fourier analysis. We sketch the basic concepts of distribution theory here, making use of such basic concepts as Fréchet spaces and weak topologies, which are treated in Appendix A, Functional Analysis. We begin with the concept of a tempered distribution. This is a continuous linear functional (4.1) w : S(R n ) C, where S(R n ) is the Schwartz space defined in 3. The space S(R n ) has a topology, determined by the seminorms (4.2) p k (u) = sup x k D α u(x). x R n The distance function (4.3) d(u, v) = α k k= 2 k p k (u v) 1 + p k (u v) makes S(R n ) a complete metric space; with such a topology it is a Fréchet space. For a linear map w as in (4.1) to be continuous, it is necessary and sufficient that, for some k, C, (4.4) w(u) C p k (u), for all u S(R n ). The action of w is often written as follows: (4.5) w(u) = u, w. The set of all continuous linear functionals on S(R n ) is denoted (4.6) S (R n ) and is called the space of tempered distributions. The space S (R n ) has a topology, called the weak topology, or sometimes simply the weak topology, in terms of which a directed family w γ converges

33 4. Distributions and tempered distributions 33 to w weakly in S (R n ) if and only if, for each u S(R n ), u, w γ u, w. One can also consider the strong topology on S (R n ), the topology of uniform convergence on bounded subsets of S(R n ), but we will not consider this explicitly. For more on the topology of S and S, see [H], [Sch], and [Yo]. We now consider examples of tempered distributions. There is a natural injection (4.7) L p (R n ) S (R n ), for any p [1, ], given by (4.8) u, f = u(x)f(x) dx, u S(R n ), f L p (R n ). Similarly any finite measure on R n gives an element of S (R n ). The basic example is the Dirac delta function δ, defined by (4.9) u, δ = u(). Also, each differential operator D j = i / x j acts on S (R n ), by the definition (4.1) u, D j w = D j u, w, u S, w S. Iterating, we see that each D α = D α 1 1 Dα n n acts on S : (4.11) D α : S (R n ) S (R n ), and we have (4.12) u, D α w = ( 1) α D α u, w for u S, w S. Similarly, u, fw = fu, w defines fw for w S, provided that f and each of its derivatives is polynomially bounded. To illustrate, consider on R the Heaviside function (4.13) H(x) = 1 if x, if x <. Then H L (R) S (R), and the definition (4.1) gives (4.14) d dx H = δ as a consequence of the fundamental theorem of calculus. The derivative of δ is characterized by (4.15) u, δ = u () in this case.

34 34 The Fourier transform F : S(R n ) S(R n ), studied in 3, extends to S by the formula (4.16) u, Fw = Fu, w ; we can also set (4.17) u, F w = F u, w to get (4.18) F, F : S (R n ) S (R n ). The maps (4.18) are continuous when S (R n ) is given the weak topology, as follows easily from the definitions. The Fourier inversion formula of Proposition 3.1 yields: Proposition 4.1. We have (4.19) F F = FF = I on S (R n ). Proof. Using (4.16) and (4.17), if u S, w S, u, F Fw = F u, Fw = FF u, w = u, w, and a similar analysis works for FF w. As an example of a Fourier transform of a tempered distribution, the definition gives directly (4.2) Fδ = (2π) n/2 ; the Fourier transform of the delta function is a constant function. One has the same result for F δ. By the Fourier inversion formula, (4.21) F1 = (2π) n/2 δ. Next, let us consider on the line R, for any ε >, (4.22) We have, by elementary calculation, H ε (x) = e εx for x, for x <. (4.23) Ĥ ε (ξ) = (2π) 1/2 e εx iξx dx = (2π) 1/2 (ε + iξ) 1, for each ε >. Now it is clear that (4.24) H ε H as ε, in S (R), in the weak topology. It follows that (4.25) Ĥ(ξ) = (2π) 1/2 lim ε (ε + iξ) 1 in S (R).