TOPICS IN FOURIER ANALYSIS-III. Contents

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1 TOPICS IN FOUIE ANALYSIS-III M.T. NAI Contents 1. Fourier transform: Basic properties 2 2. On surjectivity 7 3. Inversion theorem 9 4. Proof of inversion theorem The Banach algebra L 1 ( Fourier-Plancheral Transform on L 2 ( Problems 22 eferences 24 NOTES PEPAED FO PAT OF AN ELECTIVE COUSE FOUIE ANALYSIS FO M.SC. STUDENTS OF IIT MADAS, JULY-NOVEMBE,

2 2 M.T. NAI 1. Fourier transform: Basic properties Definition 1.1. Let d N and f L 1 ( d. The Fourier transform of f is the function ˆf : d C defined by ˆf(ξ = f(xe ix.ξ dx, ξ d. d Here, x.ξ denotes the dot product of x, ξ d, i.e., if x = (x 1,..., x d, ξ = (ξ 1,..., ξ d are in d, then x.ξ := x 1 ξ 1 + x d ξ d. Note that the above integral exists for every ξ d, and hence the function ˆf is well-defined. Theorem 1.2. The map f ˆf is a bounded linear operator 1 from L 1 ( d to B( d. Proof. Clearly ˆf f 1 for all f L 1 ( d. Hence, ˆf B( d for all f L 1 ( d and ˆf f 1. Also f ˆf is a linear operator L 1 ( to B(. Example 1.3. Let d = 1 and f = χ (0,1. Then { ˆf(ξ = 1 1 e ixξ dx = Example 1.4. Let d = 1 and f(x = e x2 /2. Then ˆf(ξ = e (x2 /2+ixξ dx = e ξ2 /2 More generally, if f(x = e x 2 /2, x d, then ˆf(ξ = 2, ξ = 0, 2 sin ξ ξ ξ 0. e (x+iξ2 /2 dx = e ξ2 /2 e ( x 2 /2+ix ξ dx = ( d/2 e ξ 2 /2. e x2 /2 dx = e ξ2 /2. In the above examples we see that ˆf is uniformly continuous and ˆf(ξ 0 as ξ. This is, in fact, true for any f L 1 ( d. Theorem 1.5. Let f L 1 ( d. Then ˆf is uniformly continuous. 1 Here, B( d denotes the space of all bounded complex valued functions with sup-norm. This space is a Banach space.

3 TOPICS IN FOUIE ANALYSIS 3 Proof. For ξ, h d, we have ˆf(ξ + h ˆf(ξ = f(x[e ix (ξ+h e ix.ξ ]dx = d f(xe ix.ξ [e ix h 1]dx d Thus, Note that with f L 1 ( d. Hence, by DCT, ˆf(ξ + h ˆf(ξ f(x e ix h 1 dx. d f(x e ix h 1 0 as h 0, f(x e ix h 1 2 f(x d f(x e ix h 1 dx 0 h d as h 0. Hence, for every ε > 0, there exists δ > 0 such that ˆf(ξ + h ˆf(ξ f(x e ix h 1 dx < ε d for every h with h < δ. Thus, for every ε > 0, there exists δ > 0 such that ˆf(ξ + h ˆf(ξ < ε for every h with h < δ and for every ξ d. Thus, ˆf is uniformly continuous. Combining Theorems 1.2 and 1.5, we have Theorem 1.6. The map f ˆf is a bounded linear operator 2 from L 1 ( d to C b ( d with norm atmost 1. In fact, we have Theorem 1.7. Let f L 1 ( d. Then ˆf(ξ 0 as ξ. In particular 3, ˆf C 0 ( d for every f L 1 ( d. For its proof we shall make use of the following lemma. 2 C b ( d denotes the space of all bounded, continuous complex valued functions on with sup-norm. This space is a Banach space. 3 C 0 ( d denotes the space of all continuous functions h : d C such that h(x 0 as x. This space is a Banach space.

4 4 M.T. NAI Lemma 1.8. Let 1 p < and f L p ( d. Then lim y 0 d f(x y f(x p dx = 0. In particular, if we define f y (x = f(x y for x d, then the map y f y from d to L p ( d is continuous. Proof. For y d and f : d C, let f y (x := f(x y, x d. Thus, we have to prove that f f y p 0 as y 0 for f L p ( d. So, let f L p ( d and ε > 0. By denseness of C c ( d in L p ( d, there exists g C c ( d such that f g p < ε. Then, we also have f y g y p < ε. Hence, f f y p f g p + g g y p + g y f y p < 2ε + g g y p. Now, by the uniform continuity of g, there exists δ > 0 such that g(x g(x y < ε for all x d and y d with y < δ. Also, there exists a parallelepiped K d such that g(x = 0, g(x y = 0 for every x K and y < δ. Hence, g g y p p = g(x g(x y p dx ε p m(k for every y with y < δ. Thus, K f f y p < 2ε + [m(k] 1/p ε so that lim y 0 f f y p = 0. Also, for y, y 0 d, Hence, y f y is continuous on d. f y f y0 p = f f y y0 p 0 as y y 0. Proof of Theorem 1.7. We observe that for every ξ 0, ( f x πξ e ix.ξ dx = f(ye i(y+πξ/ ξ 2.ξ dx = f(ye iy.ξ e iπ dx = ξ ˆf(ξ. d 2 d d Thus, ( f x πξ e ix.ξ dx = ξ ˆf(ξ. d 2 Hence, ˆf(ξ = 1 ( [f(x f x πξ ]e ix.ξ dx 2 ξ d 2 so that ˆf(ξ 1 ( f(x f x πξ dx = f f 2 ξ d 2 πξ/ ξ 2 1. By Lemma 1.8, f f πξ/ ξ as ξ. Thus, ˆf(ξ 0 as ξ.

5 Combining Theorems 1.2 and 1.7 we obtain: TOPICS IN FOUIE ANALYSIS 5 Theorem 1.9. The map f ˆf is a bounded linear operator from L 1 ( d to C 0 ( d. Theorem Suppose f L 1 ( is differentiable with f L 1 (. Then f (ξ = (iξ ˆf(ξ, ξ. For its proof, we shall make use of the following lemma. Lemma If f L 1 ( such that f L 1 (, then f(x 0 as x ±. Proof. Observe that for any a > 0, f(a = f(0 + a 0 f (xdx. Hence, lim f(a exists, since f L 1 (. Now, suppose that lim f(a = β 0. Then a a there exists α > 0 such that f(x > β /2 for all x α. Hence a α f(x (a αβ/2 for all a α. This contradicts the fact that f L 1 (. Thus, we have proved that f(x 0 as x. Similarly, we can prove that f(x 0 as x. Proof of Theorem We have f (ξ = f (xe ixξ dx. Since f L 1 (, a f (xe ixξ dx = lim f (xe ixξ dx. a a By integration by parts, for a > 0, a f (xe ixξ dx = [ e ixξ f(x ] a a + (iξ f(xe ixξ dx. a a [ By Lemma 1.11, lim a ± e ixξ f(x ] a = 0. Hence, a a f (xe ixξ dx = (iξ f(xe ixξ dx. lim a Thus, we have proved that f (ξ = (iξ ˆf(ξ. a a Theorem 1.10 has multi-variable analogue: For z = (z 1,... z d C d and α := (α 1,..., α d N d 0, define z α := z α 1 1 z α 2 2 z α d d, α := α α d and ( ( α α1 α2 ( αd ϕ :=. x 1 x 2 x d

6 6 M.T. NAI For k N, we write f C k ( d iff α f exists and continuous for every α N d 0 with α k. Theorem If f L 1 ( d such that α f exists a.e. and α f L 1 ( d for some multi-index α N d 0, then α f(ξ = (iξ α ˆf(ξ for every ξ d. Theorem Suppose f L 1 ( such that x g(x := xf(x belongs to L 1 (. Then ˆf is differentiable and ( ˆf (ξ = iĝ(ξ, ξ. Proof. For ξ, h, we have ˆf(ξ + h ˆf(ξ = f(x[e ix (ξ+h e ix.ξ ]dx = f(xe ix.ξ [e ix h 1]dx d Hence, ˆf(ξ + h ˆf(ξ ( e = xf(xe ix.ξ ix h 1 dx ih ix.h Note that e ix h 1 1 as h 0 ix.h and, since e ix h 1 1, ix.h ( e ix h 1 xf(xe ix.ξ xf(x. ix.h Hence, by DCT, ˆf(ξ + h ˆf(ξ ih as h 0. = ( e xf(xe ix.ξ ix h 1 dx ( xf(xe ix.ξ dx ix.h More generally, Theorem If f L 1 ( d such that d x k f(x dx < for some k N, then ˆf C k ( d and for all α N d 0 with α k. α ˆf(ξ = d ( ix α f(xe ix.ξ dx Let us introduce the following operators: For f L 1 ( d, (e h f(x = e ih x f(x, (τ h f(x = f(x h, (f(x = f( x, (D t f(x = f(tx.

7 TOPICS IN FOUIE ANALYSIS 7 Theorem The following results hold: For f L 1 ( d, h d, 0 t, (1 êhf = τ h ˆf, (2 τ h f = e h ˆf, (3 f = ˆf, (4 D t f = t d D 1/t ˆf, 0 t. Proof. (1 For ξ, h d, we have ê h f(ξ = e h f(xe ix.ξ dx = e ih.x f(xe ix.ξ dx = f(xe ix(ξ h dx. d d d Thus, êhf(ξ = ˆf(ξ h = (τ h ˆf(ξ. (2 For ξ, h d, we have τ h f(ξ = τ h f(xe ix.ξ dx = d f(x he ix.ξ dx = d f(ye i(y+h.ξ dx. d Thus, τ h f(ξ = f(ye i(y+h.ξ dx = e ih.ξ d f(ye iy.ξ dx = e ih.ξ ˆf(ξ = (e h ˆf(ξ. d (3 For ξ d, we have f(ξ = (f(xe ix.ξ dx = f( xe ix.ξ dx = f(ye iy.ξ dx = ˆf( ξ. d d d Hence, (f(ξ = ˆf( ξ = ( ˆf(ξ. (4 For ξ d and 0 t, we have ( D t f(ξ = (D t f(xe ix.ξ dx = f(txe ix.ξ dx = 1 f(ye iy.ξ/t. d t d d d Thus, ( D t f(ξ = 1 t d ˆf( ξ t = 1 t d (D 1/t ˆf(ξ. 2. On surjectivity We have seen (see Theorem 1.9 the map f ˆf is a bounded linear operator from L 1 ( d to C 0 ( d. A natural question is whether this operator is onto. We show that the answer is negative by using the following proposition. Proposition 2.1. If f L 1 ( is an odd function, then ˆf is an odd function and there exists M > 0 such that ˆf(ξ ξ dξ M r

8 8 M.T. NAI for all r, with 0 < r <. Proof. Let f L 1 ( is an odd function. It can be easily seen that 4, ˆf(ξ = 2i 0 f(x sin(ξxdx. Thus, ˆf is odd. Let r > 0. Then, we have ˆf(ξ ( 1 dξ = 2i r ξ r ξ 0 ( = 2i f(x = 2i 0 0 f(x We know that there exists M 0 > 0 such that Thus, r ˆf(ξ ξ dξ 2 0 f(x x rx f(x sin(ξxdx dξ r ( x rx b a sin(s s sin(ξx dξ dx ξ sin(s ds s sin x x dx. dx M 0 for all (a, b. ds dx 2M 0 f 1. Theorem 2.2. The bounded linear operator f ˆf is a bounded linear operator from L 1 ( d to C 0 ( d is not onto. Proof. By Proposition 2.1, it is enough to construct an odd function g C 0 ( such that g(t dt r t as. A candidate 5 for such a function is the odd extension of g defined by { t/e, 0 t e, g(t := 1/ ln(t, t > e. Note that e g(t dt = ln(ln( as. t 4 If f is a real valued even function then ˆf(ξ = 2 0 f(x cos(ξxdx. 5 This example is taken from the book Classical Fourier Transforms, Springer Verlag, 1989 by K. Chandrasekharan.

9 TOPICS IN FOUIE ANALYSIS 9 3. Inversion theorem Another question one may ask is whether f can be recovered from ˆf. Here is an answer for this. Theorem 3.1. (Inversion theorem Suppose f L 1 (. If ˆf L 1 ( and if g(x := 1 ˆf(te itx dt, x, then g C 0 ( and f = g a.e., i.e., f(x = 1 ˆf(te itx dt, a.a. x. In particular if f C 0 ( L 1 ( such that ˆf L 1 (, then f(x = 1 ˆf(te itx dt x. An immediate corollary to the above theorem is the following. Corollary 3.2. The bounded linear operator f ˆf from L 1 ( into C 0 ( is injective. Proof. By the above theorem, we can infer that if f L 1 ( such that ˆf = 0, then f satisfies the assumption that ˆf L 1 ( so that f = 0 a.e. Now we give two examples dealing with the cases in which the conditions of inverse theorem are satisfied/not satisfied. { 1 x, x 1, Example 3.3. Let f(x = 0, x > 1. Then we see that ˆf(t = 1 1 (1 x e itx dx = 2 Thus both f and ˆf belong to L 1 (. 1 0 (1 x cos(txdx = sin2 (t/2 (t/2 2. { 1, x 1, Example 3.4. Let f(x = 0, x > 1. ˆf(t = 1 1 Thus f L 1 ( but ˆf L 1 (. e itx dx = Then we see that ( sin t cos(txdx = 2. t

10 10 M.T. NAI 4. Proof of inversion theorem For the proof of the inverse theorem, we shall make use of the following fact: If (f n is a Cauchy sequence in L 1 (, then there exists an f L 1 ( and a sequence (k n in N such that f kn f a.e. We are also going to make use of an integrable function ϕ : with the following properties: (i 0 < ϕ(x 1 for all x ; (ii For λ > 0, ϕ λ (x := ϕ(λx 1 as λ 0; (iii The function ψ defined by is non-negative and satisfies ψ(x = 1 ϕ(te itx dt, ψ(xdx = 1. x For example, one may take ϕ(x = e x. Clearly, this function satisfies properties (i and (ii above. To see (iii, we note that 0 ψ(x = e t e itx dt = e t e itx dt + e t e itx dt = x, 2 and hence Thus, (iii is also satisfied. ψ(x dx = 1 π 0 dx 1 + x 2 = 1. Proof of inversion theorem. Let ϕ : be a function satisfying the conditions (i-(iii above. For λ > 0, let f λ (x := 1 ϕ λ (t ˆf(te itx dt, x. We show that f λ f 1 0 as λ 0. (1 Then 6 there exists a sequence (λ n of positive reals such that λ n 0 and f λn f a.e.. (2 6 ecall from measure theory (cf. [1] that every Cauchy sequence in L p ( d with 1 p < has a subsequence which converges almost every where to some function in L p ( d.

11 TOPICS IN FOUIE ANALYSIS 11 Since 0 ϕ λn (x 1 and ϕ λn (x 1 for every x, we have ϕ λn (t ˆf(te itx ˆf(te itx and so that by DCT, By (2 and (3, ϕ λn (t ˆf(te itx ˆf(t for all n N f λn (x := 1 ϕ λn (t ˆf(te itx dt 1 f(x = 1 ˆf(te itx dt for almost all x. ˆf(te itx dt. (3 Now, for proving (1, we first observe, applying Fubini s theorem, that f λ (x = 1 ϕ λ (t ˆf(te itx dt = 1 ( ϕ(λte itx f(ye iyt dy dt = 1 ( 1 λ ϕ(ueiux/λ f(ye iyu/λ dy du ( 1 1 = f(y λ ϕ(ueiu(x y/λ du dy = f(y 1 ( x y λ ψ dy λ = f(x τ 1 τ λ ψ( dτ λ = f(x λsψ(sds. Since ψ(sds = 1, we have f λ (x f(x = [f(x λs f(x]ψ(sds. Hence, appealing again to Fubini, f λ (x f(x dx ( f(x λs f(x ψ(sds dx ( f(x λs f(x dx ψ(sds

12 12 M.T. NAI Let ( g λ (s := f(x λs f(x dx ψ(s. By Lemma 1.8, g λ (s 0 as λ 0. Also, g λ (s 2 f 1 ψ(s with 2 f 1 ψ L 1 (. Hence, by DCT, f λ f 1 := f λ (x f(x dx g λ (sds 0 as λ The Banach algebra L 1 ( Now, we define a multiplicative structure on L 1 ( Definition 5.1. For f, g in L 1 ( d, we define the convolution of f and g, denoted by f g, is defined by (f g(x = f(x yg(y dy, d for almost all x d. Note that the above definition is meaningful only if we ensure the existence of the integral involved. For this purpose we shall use the following facts. If Y is a topological space and X is a measurable space, then a function f : X Y is measurable if and only if for every Borel set B in Y, f 1 (B is measurable. This follows by showing that S := {B Y : f 1 (B measrable } is a σ-algebra on Y, and Borel sets belong to S. If X, Y, Z are topological spaces and ϕ : X Y and ψ : Y Z are Borel measurable functions, then ψ ϕ : X Y is Borel measurable. If f is Lebesgue measurable, then there exists a Borel measurable function f 0 such that f = f 0 a.e. Theorem 5.2. Let f and g belong to L 1 ( d. Then (f g(x := f(x yg(y dy d is well-defined for almost all x d, f g L 1 ( d and f g 1 f 1 g 1.

13 TOPICS IN FOUIE ANALYSIS 13 Proof. We prove the result for d = 1. The same arguments work for any d N. First let us assume that f and g are Borel measurable. Note that the function (x, y x y is continuous from 2 to. Therefore, it can be seen that the functions (x, y f(x y and (x, y g(y are Borel measurable whenever f : and g : are Borel measurable. Hence, (x, y f(x yg(y is Borel measurable. Now, using the fact that f and g belong to L 1 ( and Fubini s theorem, we have ( ( Thus, f(x yg(y dy dx = f(x yg(y dx dy ( = g(y f(x y dx dy f 1 g 1. f(x yg(y dy < for all most all x, f g is well defined for almost all x and f g L 1 (. We also have f g 1 f 1 g 1. Now, let us assume that f and g are Lebesgue measurable. Then we know that there exist Borel measurable functions f 0 and g 0 such that f = g 0 and g = g 0 a.e. Hence, we obtain the conclusion of the theorem by applying the first part to f 0 and g 0. emark 5.3. In fact, f g can be defined for any f L p ( d and g L q ( d, where 1 p <. Theorem 5.4. The following hold. (1 f g = g f for all f, g L 1 ( d. (2 f (g h = (f g h for all f, g, h L 1 ( d. Theorem 5.5. If f, g L 1 ( d, then f g = ˆfĝ. Proof. Note that f g(ξ = (f g(xe ix.ξ dx d ( = f(x ye i(x y.ξ dx g(ydy d d = ˆf(ξĝ(ξ.

14 14 M.T. NAI We already know that L 1 ( d is a Banach space. With respect to the multiplication (f, g f g, L 1 ( d a commutative Banach algebra. Theorem 5.6. The Banach algebra L 1 ( d does not have a multiplicative identity with respect to convolution. Proof. Suppose there exists ϕ L 1 ( d such that f ϕ = f for all f L 1 ( d. Then we have ˆf(ξ ˆϕ(ξ = ˆf(ξ for all ξ d. Taking f(x = e x 2 /2, we have ˆf(ξ = ( d/2 e ξ 2 /2, so that ( d/2 e ξ 2 /2 ˆϕ(ξ = ( d/2 e ξ 2 /2 for all ξ d. Hence, ˆϕ(ξ = 1 for all ξ d. This is a contradiction to the fact that ˆf C 0 ( d. However, it has an approximate identity.. Definition 5.7. By an approximate identity we mean a family {e λ : λ > 0} in L 1 ( d such that f e λ f 1 0 as λ 0. Now, we specify an approximate identity for L 1 (. The same can be extended to the case of L 1 ( d for appropriate changes. Here onwards, we consider a non-negative measurable function ψ which satisfies ψ(xdx = 1, and let ψλ(x := 1 ( x λ ψ, x, λ > 0. λ Then we have (f ψ λ (x = f(x yψ λ (ydy = f(x λsψ(sds. Indeed, (f ψ λ (x = = = = f(x yψ λ (ydy f(yψ λ (x ydy f(y 1 ( x y λ ψ dy λ f(x λsψ(sds. Theorem 5.8. The set {ψ λ : λ > 0} is an approximate identity for L 1 (. (

15 Proof. ecall from ( that (f ψ λ (x = Now, g λ (s := TOPICS IN FOUIE ANALYSIS 15 f(x yψ λ (ydy = f(x λsψ(sds. f(x λs f(s dx ψ(s 0 as λ 0, by Lemma 1.8, and g λ (s 2 f 1 ψ(s. Hence, by DCT, (f ψ λ (x f(x dx 0 as λ 0. Thus, f ψ λ f 1 0 as λ 0. In fact, something more is true. Lemma 5.9. If f L p ( for 1 p <, then f ψ λ L p ( and lim f ψ λ f p = 0. λ 0 Proof. Let f L p ( for 1 p <. We have already proved for p = 1. So, assume that 1 < p <. By (, (f ψ λ (x f(x = [f(x λs f(x]ψ(sds. Applying Hölder s inequality with respect to the probability measure dµ(s = ψ(sds, we have (f ψ λ (x f(x f(x λs f(x ψ(sds Hence, applying Fubini s theorem, (f ψ λ (x f(x p dx ( 1/p ( f(x λs f(x p ψ(sds ψ(sds ( 1/p f(x λs f(x ψ(sds p. = 1/q ( f(x λs f(x p ψ(sds dx ( ψ(s f(x λs f(x p dx ds. Since ψ(sds = 1, we have ( f(x λs f(x p dx ψ(sds (2 f p p (

16 16 M.T. NAI so that f ψ λ L p. Further, ψ(s f(x λs f(x p dx (2 f p p ψ(s and by Lemma 1.8, ψ(s f(x λs f(x p dx 0 as λ 0. Hence, by DCT, f ψ λ f p = (f ψ λ (x f(x p dx ψ(s f(x λs f(x p dxds 0 as λ 0. We observe that for each t, the linear functional ϕ t : L 1 ( defined by ϕ t (f = ˆf(t, f L 1 ( is a linear functional which also satisfies the multiplicative property: ϕ t (f g = ϕ t (fϕ t (g, f, g L 1 (. In other words, ϕ t is a complex homomorphism on Banach algebra L 1 (. This linear functional is continuous: For each t, ϕ t (f = ˆf(t f 1 for all f L 1 (. It also shows that ϕ t 1 for all t. In fact, this is true for every complex homomorphism on any Banach algebra as the following theorem shows. Theorem Let A be a complex Banach algebra and let ϕ : A C be a algebra homomorphism. Then ϕ is continuous and ϕ 1. Proof. We prove that f(a a for every a A: Suppose this is not true. then there exists a 0 A such that ϕ(a 0 > a 0. Let a = a 0 / ϕ(a 0. Then a < 1 and ϕ(a = 1. Since a n 0, the sequence (b n defined by b n = (a + a a n is convergent, say b n b. Note that Taking limits, ab = a + b. Hence, ab n = (a 2 + a a n+1 = a + b n a n+1. ϕ(a + f(b = ϕ(ab = ϕ(aϕ(b. Since ϕ(a = 1, we obtain 1 + ϕ(b = ϕ(b. This is impossible. The following theorem shows that Fourier transform is the only complex linear functional on L 1 ( having the multiplicative property. Theorem If ϕ : L 1 ( C is a multiplicative linear functional, that is, ϕ is a linear functional satisfying ϕ(f g = ϕ(fϕ(g f, g L 1 (, then there exists t such that ϕ(f = ˆf(t for all f L 1 (.

17 TOPICS IN FOUIE ANALYSIS 17 Proof. Let ϕ be a nonzero continuous linear functional on L 1 (. representation theorem, there exists a unique h L ( such that ϕ(f = f(xh(xdx for every f L 1 (. Using the fact that ϕ(f g = ϕ(fϕ(g, we obtain the relation g(yϕ(f y dy = ϕ(f g(yh(ydy Then by iesz where f y (x := f(x y. From this, ϕ(f y = ϕ(fh(y. Since y f y is continuous, we may assume, without loss of generality, that h is continuous. We also have ϕ(fh(x + y = ϕ(f x+y = ϕ((f x y = ϕ(f x h(y = ϕ(fh(xh(y. Hence, h(x + y = h(xh(y and hence, h(0 = 1, as h 0. Continuity of h implies the existence of δ > 0 such that δ h(y 0. Now, from h(x + y = h(xh(y we have 0 δ δ x+δ h(x h(ydy = h(x + ydy = h(tdt. 0 0 x Hence, h is differentiable so that from h(x + y = h(xh(y, h (x = h(xh (0, i.e., h (x = αh(x, α := h (0. Thus, h(x = e αx. Since h is bounded, α := iτ for some τ. Thus, h(x = e iτx for some τ. emark The above proof is adopted from udin[3]. 6. Fourier-Plancheral Transform on L 2 ( We have already observed that f ˆf is an injective bounded linear operator from L 1 ( into C 0 (. We have also given an example to show that this map is not surjective. Next theorem introduce the definition of Fourier transform of functions in L 2 ( by extending the known definition from L 1 ( L 2 ( to all of L 2 (. Lemma 6.1. Let f C b (. Then where ψ λ is as in Theorem 5.8. lim (f ψ λ(x = f(x for every x, λ 0

18 18 M.T. NAI Proof. Note that for every x, (f ψ λ (x f(x = [f(x λs f(x]ψ(sds. Note that f(x λs f(x ψ(s 2 f ψ(s with ψ L 1 ( and [f(x λs f(x]ψ(s 0 as λ 0 as f is continuous at x. Hence, the result follows by DCT. Theorem 6.2. The following hold. (i If f L 1 ( L 2 (, then ˆf L 2 ( and ˆf 2 = f 2. (ii The set Y := { ˆf : f L 1 ( L 2 (} is dense in L 2 (. Proof. (i Let f L 1 ( L 2 (. To show that ˆf 2 L 1 (. Note that ˆf 2 = ˆf ˆf, ˆf(ξ = f(xe ixξ dx = f(xe ixξ dx = f( xe xξ dx = f(xe itx dt, where f(x = f( x for x. Thus, ˆf(ξ = ˆf(ξ and hence, where g := f f. Note that g(x = f(x y f(ydy = where f τ (x := f(x τ, and ˆf 2 = ˆf ˆf = ˆf ˆ f = f f = ĝ, f(x yf( ydy = g(x = f x, f f 2 2. f(x + uf(udu = f x, f, Thus, g is bounded. By Lemma 1.8, τ f τ from to L 2 ( is continuous so that g is continuous. Hence, by Lemma 6.1, for every x. In particular, (g ψ λ (x g(x as λ 0 + (g ψ λ (0 g(0 = f 2 2 as λ 0 +.

19 TOPICS IN FOUIE ANALYSIS 19 But, taking ϕ is as in the proof of inversion theorem with corresponding ψ, (g ψ λ (x = g(y 1 ( x y λ ψ dy λ ( 1 1 = g(y λ ϕ(ueiu(x y/λ du dy = 1 ( 1 λ ϕ(ueiux/λ g(ye iyu/λ dy du = 1 ( ϕ(λte itx g(ye iyt dy dt = 1 ϕ(λtĝ(te itx dt. In particular, (g ψ λ (0 = 1 ϕ(λtĝ(tdt. ecall that ĝ(t = ˆf 2 0. Hence, assuming that ϕ(λt increases to 1 as λ 0 + (e.g., ϕ(x = e x, we obtain ϕ(λtĝ(t increases to ĝ(t as λ 0 +. Hence, by MCT (g ψ λ (0 = 1 ϕ(λtĝ(tdt 1 ĝ(tdt = 1 ˆf(t 2 dt as λ 0 +, We already have (g ψ λ (0 f 2 as λ 0 +. Thus, 1 ˆf(t 2 dt = f 2 2 so that ˆf L 2 ( and ˆf 2 2 = f 2 2. (ii It is enough to show that Y = {0}, i.e., to prove that w Y implies w = 0. So, let w Y. By Lemma 5.9, ψ λ w L 2 ( and ψ λ w w 2 0 as λ 0 +. Hence, it is enough to prove that ψ λ w = 0 for all λ > 0. Note that (ψ λ w(x := ψ λ (x yw(ydy, x. Taking ϕ and ψ as in the proof of inversion theorem, we have ψ λ (x y = 1 ϕ λ (te i(x yt dt = 1 [ϕ λ (te ix ]e iyt dt. Assuming ϕ L 1 ( L 2 ( (e.g. ϕ(t = e t, by the above relation, we have y ψ λ (x y is a Fourier transform of a function in L 1 ( L 2 ( so that it belongs to Y. Thus, ψ λ w = 0 for all λ > 0. This completes the proof. In view of the above theorem, we have the following.

20 20 M.T. NAI Theorem 6.3. (Fourier-Plancheral Theorem There exists a unique surjective continuous linear operator Φ from L 2 ( onto itself such that Φ(f = ˆf for every f L 1 ( L 2 ( and Φ(f 2 = f 2 for every f L 2 (. Proof. By Theorem 6.2, the map Define Φ 0 : L 1 ( L 2 ( L 2 ( defined by Φ 0 (f = ˆf, f L 1 ( L 2 (, is a continuous linear operator with its domain and range dense in L 2 (. ecall also that Φ 0 (f = ˆf 2 = f 2. Hence, it has a unique continuous linear extension Φ : L 2 ( L 2 (. Also, for f L 2 (, if (f n in L 1 ( L 2 ( is such that f f n 2 0, then we have Φ(f 2 = Φ( lim n f n 2 = lim n Φ(f n 2 = lim n ˆf n 2 = lim n fn 2 = f 2. This also shows that range of Φ is closed. Since range of Φ 0 is dense in L 2 (, Φ is onto. Definition 6.4. The operator Φ : L 2 ( L 2 ( in Theorem 6.3 is called the Fourier-Plancheral transform. Notation 6.5. Abusing the notation, for f L 2 (, Φ(f is also denoted by ˆf. This is somewhat justified by Theorem 6.9 below. Now, let us observe the following. Theorem 6.6. Let f L 1 ( d and for n > 0, let E n := {x d : x n}. Then ˆf(ξ = lim f(xe ix.ξ dx n E n for every ξ d. Proof. Let ξ d. Since n=1 E n = d and E n E n+1 for every n N, we have (χ En f(xe ix.ξ f(xe ix.ξ as n. Also, (χ En f(xe ix.ξ f(x, f L 1 ( d. Hence, by DCT, f(xe ix.ξ dx E n f(xe ix.ξ dx = ˆf(ξ d as n for every ξ d. What about for f L 2 ( d? Here is the answer.

21 TOPICS IN FOUIE ANALYSIS 21 Theorem 6.7. Let f L 2 ( and for n > 0, let E n := {x d : x n}. The following hold. (1 f n := χ En f L 1 ( L 2 ( for every n N. (2 f f n 2 0 as n. (3 ( ˆf n is a Cauchy sequence in L 2 (. (4 ˆf n Φ(f 2 0 as n, where Φ is the Fourier-Plancheral transform. (5 There exists a subsequence ( ˆf kn for ( ˆf n such that ˆf kn Φ(f a.e. on. Proof. (1 We have f n (x dx = χ En (x f(x dx f 2 [µ(e n ] 1/2, f n (x 2 dx f 2 2. Hence, f n := χ En f L 1 ( L 2 ( for every n N. (2 We have f f n (x 2 dx = [1 χ En (x]f(x 2 dx = χ E c n (x f(x 2 dx, χ E c n (x f(x 2 0 as n and χ E c n (x f(x 2 f(x 2 with f 2 L 1 ( so that by DCT, f f n 2 2 = f f n(x 2 dx 0 as n. (3 We have ˆf n ˆf m 2 = f n f m 2 = f n f m 2. Hence, by (2, ˆf n ˆf m 2 = f n f m 2 0 as n, m. (4 By (2 and Fourier-Plancheral theorem, we have ˆf n Φ(f 2 = Φ(f n Φ(f 2 = f n f 2 0 as n. (5 This follows from (4. emark 6.8. For f L 2 (, let P r f = fχ [ r,r], r > 0. Then P r f L 1 ( L 2 ( and P r satisfying : L 2 ( L 2 ( is an orthogonal projection P r f f 2 0 and P r f Φ(f 2 0 as r.

22 22 M.T. NAI Theorem 6.9. For f L 2 ( and r > 0, let g r (t := r f(xe itx dx, h r (x := r r r Φ(f(te itx dt for all x, t. Then g r L 2 (, h r L 2 (, and g r Φ(f 2 0, h r f 2 0 as r. Proof. Exercise. 7. Problems (1 Let f L 1 ( d. Prove the following: (a ˆf(ξ is well-defined for every ξ d. (b sup ˆf(ξ f 1. ξ d (c ξ ˆf(ξ is uniformly continuous on d. (d The map f ˆf is a linear operator from L 1 ( d to C b ( d with norm at most 1. (e ˆf(ξ 0 as ξ 0. (2 Let 1 p <. For f L p ( d and y d, let (τ y f(x := f(x y, x d. For each f L p ( d, prove the following: (a τ y f L p ( d for every y d. (b The map y τ y f from d to L p ( d is continuous. (3 Prove that, if f L 1 ( is differentiable with f L 1 (, then f (ξ = (iξ ˆf(ξ, ξ. (4 Suppose f L 1 ( such that x g(x := xf(x belongs to L 1 (. Prove that ˆf is differentiable and (5 Let ( ˆf (ξ = iĝ(ξ, ξ. (e h f(x = e ih x f(x, (τ h f(x = f(x h, (f(x = f( x, (D t f(x = f(tx. For f L 1 ( d, h d, 0 t, prove the following. (a êhf = τ h ˆf, (b τ h f = e h ˆf, (c f = ˆf, (d D t f = t d D 1/t ˆf, 0 t.

23 TOPICS IN FOUIE ANALYSIS 23 (6 Prove that the operator f ˆf from L 1 ( d to C 0 ( d is not onto. (7 Suppose f L 1 (. If ˆf L 1 ( and if g(x := 1 ˆf(te itx dt, x, then prove that f = g a.e. Also deduce the following. (a If f C 0 ( L 1 ( such that ˆf L 1 (, then f(x = 1 ˆf(te itx dt x. (b f ˆf is an injective operator. (8 Let ϕ : be an integrable function such that 0 < ϕ(x 1 for all x and ϕ(λx := ϕ(λx 1 as λ 0+. Prove that, for every g L 1 (, ϕ(λtg(te itx dt g(te itx dt. (9 Let ϕ be as in Problem(8 and let ψ be defined by ψ(x = 1 ϕ(te itx dt for x. If f L 1 ( is such that ˆf L 1 ( and if ψ(xdx = 1, then prove that 1 ϕ(λt ˆf(te itx dt = f(x λsψ(sds. (10 Let ψ L 1 ( be a non-negative function such that ψ(xdx = 1 and for f L 1 ( and λ > 0, let h λ (x := f(x λsψ(sds. Prove that h λ f 1 0 as λ 0. Deduce the following. (a There exists a sequence (λ n of positive real numbers such that h λn f a.e. (b If ϕ is as in Problem(8, f L 1 ( such that ˆf L 1 (, and f λ (x := ϕ(λt ˆf(te itx dt, then f λ f 1 0 as λ 0. (11 Let ϕ(x = e x. Verify the following. (a ϕ satisfies the conditions in Problem(8. (b ψ(x := 1 ϕ(teitx dt, x satisfies ψ(xdx = 1. (12 Let ψ be as in Problem(9. If f L p ( for 1 p <, then prove that f ψ λ L p ( and lim f ψ λ f p = 0. λ 0 (13 Let ψ be as in Problem(9 and f L (. Prove that if f is continuous at a point x, then lim(f ψ λ (x = f(x, where ψ λ is as in Problem(9. λ 0

24 24 M.T. NAI (14 Prove the following. (a If f L 1 ( L 2 (, then ˆf L 2 ( and ˆf 2 = f 2. (b The set Y := { ˆf : f L 1 ( L 2 (} is dense in L 2 (. (15 Prove that there exists a unique surjective continuous linear operator Φ from L 2 ( onto itself such that Φ(f = ˆf for every f L 1 ( L 2 (. Show also that Φ(f 2 = f 2 for every f L 2 (. (16 Let f L 1 ( d and for n > 0, let E n := {x d : x n}. Prove that f(xe ix.ξ dx f(xe ix.ξ dx = ˆf(ξ as n E n d for every ξ d. (17 Let f L 2 ( and for n > 0, let E n := {x d : x n}. Prove the following. (a f n := χ En f L 1 ( L 2 ( for every n N. (b f f n 2 0 as n. (c ( ˆf n is a Cauchy sequence in L 2 (. (d ˆf n Φ(f 2 0 as n, where Φ is the Fourier-Plancheral transform. (e There exists a subsequence ( ˆf kn for ( ˆf n such that ˆf kn Φ(f a.e. on. (18 For f L 2 ( and r > 0, let g r (ξ := r r f(xe ixξ dx, h r (x := r for all x, ξ. Then g r L 2 (, h r L 2 (, and r Φ(f(ξe ixξ dt g r Φ(f 2 0, h r f 2 0 as r. eferences [1] M.T. Nair, Measure and Integration, Notes for the MSc. Course, Jan-may, [2]. adha & S. Thangavelu, Fourier Series, Web-Course, NPTEL, IIT Madras, [3] W. udin, eal and Complex Analysis. [4] B. O. Turesson, Fourier Analysis, Distribution Thoery, and Wavelets, Lecture Notes, March, Department of Mathematics, I.I.T. Madras, Chennai , INDIA address: mtnair@iitm.ac.in