Introduction to Harmonic Analysis

Transcription

1 } Introduction to Harmonic Analysis LECTURE NOTES } Chengchun Hao Institute of Mathematics, AMSS, CAS Updated: April 8, 26

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3 Contents The Fourier Transform and Tempered Distributions The L theory of the Fourier transform The L 2 theory and the Plancherel theorem Schwartz spaces The class of tempered distributions Characterization of operators commuting with translations Interpolation of Operators Riesz-Thorin s and Stein s interpolation theorems The distribution function and weak L p spaces The decreasing rearrangement and Lorentz spaces Marcinkiewicz interpolation theorem The Maximal Function and Calderón-Zygmund Decomposition Two covering lemmas Hardy-Littlewood maximal function Calderón-Zygmund decomposition Singular Integrals Harmonic functions and Poisson equation Poisson kernel and Hilbert transform The Calderón-Zygmund theorem Truncated integrals Singular integral operators commuted with dilations The maximal singular integral operator *Vector-valued analogues Riesz Transforms and Spherical Harmonics The Riesz transforms Spherical harmonics and higher Riesz transforms v

4 - vi - Contents 5.3 Equivalence between two classes of transforms The Littlewood-Paley g-function and Multipliers The Littlewood-Paley g-function Fourier multipliers on L p The partial sums operators The dyadic decomposition The Marcinkiewicz multiplier theorem Sobolev Spaces Riesz potentials and fractional integrals Bessel potentials Sobolev spaces References Index

5 LECTURE NOTES ON Introduction to Harmonic Analysis c 26 by Chengchun Hao hcc@amss.ac.cn Chapter The Fourier Transform and Tempered Distributions In this chapter, we introduce the Fourier transform and study its more elementary properties, and extend the definition to the space of tempered distributions. We also give some characterizations of operators commuting with translations.. The L theory of the Fourier transform We begin by introducing some notation that will be used throughout this work. R n denotes n-dimensional real Euclidean space. We consistently write x = (x, x 2,, x n ), ξ = (ξ, ξ 2,, ξ n ), for the elements of R n. The inner product of x, ξ R n is the number x ξ = n j= x jξ j, the norm of x R n is the nonnegative number x = x x. Furthermore, dx = dx dx 2 dx n denotes the element of ordinary Lebesgue measure. We will deal with various spaces of functions defined on R n. The simplest of these are the L p = L p (R n ) spaces, p <, of all measurable functions f such that f p = ( f(x) p dx ) /p R <. The number n f p is called the L p norm of f. The space L (R n ) consists of all essentially bounded functions on R n and, for f L (R n ), we let f be the essential supremum of f(x), x R n. Often, the space C (R n ) of all continuous functions vanishing at infinity, with the L norm just described, arises more naturally than L = L (R n ). Unless otherwise specified, all functions are assumed to be complex valued; it will be assumed, throughout the note, that all functions are (Borel) measurable. In addition to the vector-space operations, L (R n ) is endowed with a multiplication making this space a Banach algebra. This operation, called convolution, is defined in the following way: If both f and g belong to L (R n ), then their convolution h = f g is the function whose value at x R n is

6 The Fourier Transform and Tempered Distributions h(x) = f(x y)g(y)dy. R n One can show by an elementary argument that f(x y)g(y) is a measurable function of the two variables x and y. It then follows immediately from Fibini s theorem on the interchange of the order of integration that h L (R n ) and h f g. Furthermore, this operation is commutative and associative. More generally, we have, with the help of Minkowski s integral inequality F (x, y)dy L p x F (x, y) L p x dy, the following result: Theorem.. If f L p (R n ), p [, ], and g L (R n ) then h = f g is well defined and belongs to L p (R n ). Moreover, h p f p g. Now, we first consider the Fourier transform of L functions. Definition.2. Let ω R \ {} be a constant. If f L (R n ), then its Fourier transform is F f or ˆf : R n C defined by F f(ξ) = e ωix ξ f(x)dx (.) R n for all ξ R n. We now continue with some properties of the Fourier transform. Before doing this, we shall introduce some notations. For a measurable function f on R n, x R n and a we define the translation and dilation of f by τ y f(x) =f(x y), (.2) δ a f(x) =f(ax). (.3) Proposition.3. Given f, g L (R n ), x, y, ξ R n, α multiindex, a, b C, ε R and ε, we have (i) Linearity: F (af + bg) = af f + bf g. (ii) Translation: F τ y f(ξ) = e ωiy ξ ˆf(ξ). (iii) Modulation: F (e ωix y f(x))(ξ) = τ y ˆf(ξ). (iv) Scaling: F δ ε f(ξ) = ε n δ ε ˆf(ξ). (v) Differentiation: F α f(ξ) = (ωiξ) α ˆf(ξ), α ˆf(ξ) = F (( ωix) α f(x))(ξ). (vi) Convolution: F (f g)(ξ) = ˆf(ξ)ĝ(ξ). (vii) Transformation: F (f A)(ξ) = ˆf(Aξ), where A is an orthogonal matrix and ξ is a column vector. (viii) Conjugation: f(x) = ˆf( ξ). Jean Baptiste Joseph Fourier (2 March May 83) was a French mathematician and physicist best known for initiating the investigation of Fourier series and their applications to problems of heat transfer and vibrations. The Fourier transform and Fourier s Law are also named in his honor. Fourier is also generally credited with the discovery of the greenhouse effect.

7 .. The L theory of the Fourier transform Proof. These results are easy to be verified. We only prove (vii). In fact, F (f A)(ξ) = e ωix ξ f(ax)dx = e ωia y ξ f(y)dy R n R n = e ωia y ξ f(y)dy = e ωiy Aξ f(y)dy = ˆf(Aξ), R n R n where we used the change of variables y = Ax and the fact that A = A and det A =. Corollary.4. The Fourier transform of a radial function is radial. Proof. Let ξ, η R n with ξ = η. Then there exists some orthogonal matrix A such that Aξ = η. Since f is radial, we have f = f A. Then, it holds by (vii) in Proposition.3. F f(η) = F f(aξ) = F (f A)(ξ) = F f(ξ), It is easy to establish the following results: Theorem.5 (Uniform continuity). (i) The mapping F is a bounded linear transformation from L (R n ) into L (R n ). In fact, F f f. (ii) If f L (R n ), then F f is uniformly continuous. Proof. (i) is obvious. We now prove (ii). By ˆf(ξ + h) ˆf(ξ) = e ωix ξ [e ωix h ]f(x)dx, R n we have ˆf(ξ + h) ˆf(ξ) e ωix h f(x) dx R n e ωix h f(x) dx + 2 f(x) dx x r x >r ω r h f(x) dx + 2 f(x) dx x r x >r =:I + I 2, since for any θ e iθ = (cos θ ) 2 + sin 2 θ = 2 2 cos θ = 2 sin(θ/2) θ. Given any ε >, we can take r so large that I 2 < ε/2. Then, we fix this r and take h small enough such that I < ε/2. In other words, for given ε >, there exists a sufficiently small δ > such that ˆf(ξ + h) ˆf(ξ) < ε when h δ, where ε is independent of ξ. Ex..6. Suppose that a signal consists of a single rectangular pulse of width and height. Let s say that it gets turned on at x = and 2 turned off at x =. The standard name for this normalized rectangular pulse 2 is

8 The Fourier Transform and Tempered Distributions Π(x) rect(x) := {, if 2 < x < 2,, otherwise. 2 It is also called, variously, the normalized boxcar function, the top hat function, the indicator function, or the characteristic function for the interval ( /2, /2). The Fourier transform of this signal is Π(ξ) = R e ωixξ Π(x)dx = /2 /2 e ωixξ dx = e ωixξ ωiξ /2 2 /2 x = 2 ωξ sin ωξ 2 when ξ. When ξ =, Π() = /2 dx =. By l Hôpital s rule, /2 lim ξ ωξ sin 2 Π(ξ) = lim 2 ξ ωξ ω ωξ cos 2 2 = lim 2 ξ ω = = Π(), so Π(ξ) is continuous at ξ =. There is a standard function called sinc 2 that is defined by sinc(ξ) = sin ξ. In this notation Π(ξ) = sinc ωξ. ξ 2 Here is the graph of Π(ξ). 2π ω 2π ω ξ Remark.7. The above definition of the Fourier transform in (.) extends immediately to finite Borel measures: if µ is such a measure on R n, we define F µ by letting F µ(ξ) = e ωix ξ dµ(x). R n Theorem.5 is valid for this Fourier transform if we replace the L norm by the total variation of µ. The following theorem plays a central role in Fourier Analysis. It takes its name from the fact that it holds even for functions that are integrable according to the definition of Lebesgue. We prove it for functions that are absolutely integrable in the Riemann sense. 3 2 The term sinc (English pronunciation:["sink]) is a contraction, first introduced by Phillip M. Woodward in 953, of the function s full Latin name, the sinus cardinalis (cardinal sine). 3 Let us very briefly recall what this means. A bounded function f on a finite interval [a, b] is integrable if it can be approximated by Riemann sums from above and below in such a way that the difference of the integrals of these sums can be made as small as we wish. This definition is then extended to unbounded functions and infinite intervals by taking limits; these cases are often called improper integrals. If I is any interval and f is a function on I such that the (possibly improper) integral f(x) dx has a finite value, then f is said to be absolutely integrable on I. I

9 .. The L theory of the Fourier transform Theorem.8 (Riemann-Lebesgue lemma). If f L (R n ) then F f as ξ ; thus, in view of the last result, we can conclude that F f C (R n ). The Riemann-Lebesgue lemma states that the integral of a function like the left is small. The integral will approach zero as the number of oscillations increases. Proof. First, for n =, suppose that f(x) = χ (a,b) (x), the characteristic function of an interval. Then b ˆf(ξ) = e ωixξ dx = e ωiaξ e ωibξ, as ξ. a ωiξ Similarly, the result holds when f is the characteristic function of the n- dimensional rectangle I = {x R n : a x b,, a n x n b n } since we can calculate F f explicitly as an iterated integral. The same is therefore true for a finite linear combination of such characteristic functions (i.e., simple functions). Since all such simple functions are dense in L, the result for a general f L (R n ) follows easily by approximating f in the L norm by such a simple function g, then f = g + (f g), where F f F g is uniformly small by Theorem.5, while F g(ξ) as ξ. Theorem.8 gives a necessary condition for a function to be a Fourier transform. However, that belonging to C is not a sufficient condition for being the Fourier transform of an integrable function. See the following example. Ex..9. Suppose, for simplicity, that n =. Let ln ξ, ξ > e, g(ξ) = ξ, ξ e, e g(ξ) = g( ξ), ξ <. It is clear that g(ξ) is uniformly continuous on R and g(ξ) as ξ. Assume that there exists an f L (R) such that ˆf(ξ) = g(ξ), i.e., g(ξ) = Since g(ξ) is an odd function, we have g(ξ) = e ωixξ f(x)dx = i e ωixξ f(x)dx. sin(ωxξ)f(x)dx = sin(ωxξ)f (x)dx, where F (x) = i[f( x) f(x)] L (R). Integrating g(ξ) over (, N) ξ yields N g(ξ) ( N ) ξ dξ = sin(ωxξ) ( ωxn ) sin t F (x) dξ dx = F (x) dt dx. ξ t Noticing that

10 The Fourier Transform and Tempered Distributions N sin t lim dt = π N t 2, and by Lebesgue dominated convergence theorem,we get that the integral of r.h.s. is convergent as N. That is, which yields e lim N g(ξ) ξ N g(ξ) ξ dξ = π 2 dξ < since e N g(ξ) ξ N F (x)dx <, dξ =. However, g(ξ) dξ lim dξ = lim N e ξ N e ξ ln ξ =. This contradiction indicates that the assumption was invalid. We now turn to the problem of inverting the Fourier transform. That is, we shall consider the question: Given the Fourier transform ˆf of an integrable function f, how do we obtain f back again from ˆf? The reader, who is familiar with the elementary theory of Fourier series and integrals, would expect f(x) to be equal to the integral C e ωix ξ ˆf(ξ)dξ. (.4) R n Unfortunately, ˆf need not be integrable (for example, let n = and f be the characteristic function of a finite interval). In order to get around this difficulty, we shall use certain summability methods for integrals. We first introduce the Abel method of summability, whose analog for series is very well-known. For each ε >, we define the Abel mean A ε = A ε (f) to be the integral A ε (f) = A ε = e ε x f(x)dx. (.5) R n It is clear that if f L (R n ) then lim A ε (f) = f(x)dx. On the other ε R n hand, these Abel means are well-defined even when f is not integrable (e.g., if we only assume that f is bounded, then A ε (f) is defined for all ε > ). Moreover, their limit lim A ε(f) = lim e ε x f(x)dx (.6) ε R n may exist even when f is not integrable. A classical example of such a case is obtained by letting f(x) = sinc(x) when n =. Whenever the limit in (.6) exists and is finite we say that fdx is Abel summable to this limit. R n A somewhat similar method of summability is Gauss summability. This method is defined by the Gauss (sometimes called Gauss-Weierstrass) means G ε (f) = f(x)dx. (.7) ε R n e ε x 2 We say that R n fdx is Gauss summable (to l) if

11 .. The L theory of the Fourier transform lim G ε(f) = lim e ε x 2 f(x)dx (.6 ) ε ε R n exists and equals the number l. We see that both (.6) and (.6 ) can be put in the form M ε,φ (f) = M ε (f) = Φ(εx)f(x)dx, (.8) R n where Φ C and Φ() =. Then f(x)dx is summable to l if R n lim ε M ε (f) = l. We shall call M ε (f) the Φ means of this integral. We shall need the Fourier transforms of the functions e ε x 2 and e ε x. The first one is easy to calculate. Theorem.. For all a >, we have ( ) n ω F e a ωx 2 (ξ) = (4πa) n/2 e ξ 2 4a. (.9) 2π Proof. The integral in question is e ωix ξ e a ωx 2 dx. R n Notice that this factors as a product of one variable integrals. Thus it is sufficient to prove the case n =. For this we use the formula for the integral of a Gaussian: R e πx2 dx =. It follows that e ωixξ e aω2 x 2 dx = = ω e ξ2 4a e a(ωx+iξ/(2a))2 e ξ2 4a dx +iξ/(2a) +iξ/(2a) = ω e ξ2 4a π/a e ax2 dx e πy2 dy ( ) ω = (4πa) /2 e ξ2 4a, 2π where we used contour integration at the next to last one. The second one is somewhat harder to obtain: Theorem.. For all a >, we have ( ) n ω F (e a ωx c n a ) = 2π (a 2 + ξ 2 ), c (n+)/2 n = Γ ((n + )/2) π (n+)/2. (.) Proof. By a change of variables, i.e., F (e a ωx ) = e ωix ξ e a ωx dx = (a ω ) n e ix ξ/a e x dx, R n R n we see that it suffices to show this result when a =. In order to show this, we need to express the decaying exponential as a superposition of Gaussians, i.e.,

12 The Fourier Transform and Tempered Distributions e γ = π e η η e γ2 /4η dη, γ >. (.) Then, using (.9) to establish the third equality, ( ) e ix t e x dx = e ix t e η π e x 2 /4η dη dx R n R n η = ) e η e π η (R ix t e x 2 /4η dx dη n = π =2 n π (n )/2 e η η ((4πη) n/2 e η t 2) dη e η(+ t 2) η n 2 dη =2 n π ( (n )/2 + t 2) n+ 2 e ζ ζ n+ 2 dζ ( ) n + =2 n π (n )/2 Γ 2 ( + t 2 ). (n+)/2 Thus, ( ) F (e a ωx ) = (a ω ) n (2π) n n c n ω ( + ξ/a 2 ) = c n a (n+)/2 2π (a 2 + ξ 2 ). (n+)/2 Consequently, the theorem will be established once we show (.). In fact, by changes of variables, we have e γ e η e γ2 /4η dη π η = 2 γ π e γ(σ 2σ )2 dσ (by η = γσ 2 ) = 2 γ e γ(σ 2σ )2 dσ (by σ π 2σ2 2σ ) γ ( = e γ(σ 2σ )2 + ) dσ (by averaging the last two formula) π 2σ 2 γ = e γu2 du (by u = σ π 2σ ) =, (by e πx2 dx = ) R which yields the desired identity (.). We shall denote the Fourier transform of a >, by W and P, respectively. That is, ( ) n ( ) n ω 2π e a ωx 2 and ω 2π e a ωx, W (ξ, a) = (4πa) n/2 e ξ 2 c n a 4a, P (ξ, a) =. (.2) (a 2 + ξ 2 )(n+)/2

13 .. The L theory of the Fourier transform The first of these two functions is called the Weierstrass (or Gauss- Weierstrass) kernel while the second is called the Poisson kernel. Theorem.2 (The multiplication formula). If f, g L (R n ), then ˆf(ξ)g(ξ)dξ = f(x)ĝ(x)dx. R n R n Proof. Using Fubini s theorem to interchange the order of the integration on R 2n, we obtain the identity. Theorem.3. If f and Φ belong to L (R n ), ϕ = ˆΦ and ϕ ε (x) = ε n ϕ(x/ε), then e ωix ξ Φ(εξ) ˆf(ξ)dξ = R n ϕ ε (y x)f(y)dy R n for all ε >. In particular, ( ) n ω e ωix ξ e ε ωξ ˆf(ξ)dξ = P (y x, ε)f(y)dy, 2π R n R n and ( ) n ω e ωix ξ e ε ωξ 2 ˆf(ξ)dξ = W (y x, ε)f(y)dy. 2π R n R n Proof. From (iii) and (iv) in Proposition.3, it implies (F e ωix ξ Φ(εξ))(y) = ϕ ε (y x). The first result holds immediately with the help of Theorem.2. The last two follow from (.9), (.) and (.2). Lemma.4. (i) R n W (x, ε)dx = for all ε >. (ii) R n P (x, ε)dx = for all ε >. Proof. By a change of variable, we first note that W (x, ε)dx = (4πε) n/2 e x 2 4ε dx = W (x, )dx, R n R R n n and c n ε P (x, ε)dx = dx = P (x, )dx. R n R (ε n 2 + x 2 )(n+)/2 R n Thus, it suffices to prove the lemma when ε =. For the first one, we use a change of variables and the formula for the integral of a Gaussian: R e πx2 dx = to get W (x, )dx = (4π) n/2 e x 2 4 dx = (4π) n/2 e π y 2 2 n π n/2 dy =. R n R n R n For the second one, we have P (x, )dx = c n dx. R n R ( + x n 2 )(n+)/2

14 - -. The Fourier Transform and Tempered Distributions Letting r = x, x = x/r (when x ), S n = {x R n : x = }, dx the element of surface area on S n whose surface area 4 is denoted by ω n and, finally, putting r = tan θ, we have dx = R ( + x n 2 )(n+)/2 =ω n S n ( + r 2 ) (n+)/2 dx r n dr r n dr ( + r 2 )(n+)/2 π/2 =ω n sin n θdθ. But ω n sin n θ is clearly the surface area of the sphere of radius sin θ obtained by intersecting S n with the hyperplane x = cos θ. Thus, the area of the upper half of S n is obtained by summing these (n ) dimensional areas as θ ranges from to π/2, that is, π/2 ω n sin n θdθ = ω n 2, which is the desired result by noting that /c n = ω n /2. S n x n+ O θ cos θ sin θ S n x Theorem.5. Suppose ϕ L (R n ) with R n ϕ(x)dx = and let ϕ ε (x) = ε n ϕ(x/ε) for ε >. If f L p (R n ), p <, or f C (R n ) L (R n ), then for p f ϕ ε f p, as ε. In particular, the Poisson integral of f: u(x, ε) = P (x y, ε)f(y)dy R n and the Gauss-Weierstrass integral of f: s(x, ε) = W (x y, ε)f(y)dy R n converge to f in the L p norm as ε. Proof. By a change of variables, we have ϕ ε (y)dy = ε n ϕ(y/ε)dy = ϕ(y)dy =. R n R n R n Hence, (f ϕ ε )(x) f(x) = [f(x y) f(x)]ϕ ε (y)dy. R n Therefore, by Minkowski s inequality for integrals and a change of variables, we get 4 ω n = 2π n/2 /Γ (n/2).

15 .. The L theory of the Fourier transform - - f ϕ ε f p f(x y) f(x) p ε n ϕ(y/ε) dy R n = f(x εy) f(x) p ϕ(y) dy. R n We point out that if f L p (R n ), p <, and denote f(x t) f(x) p = f (t), then f (t), as t. 5 In fact, if f D(R n ) := C (R n ) of all C functions with compact support, the assertion in that case is an immediate consequence of the uniform convergence f (x t) f (x), as t. In general, for any σ >, we can write f = f + f 2, such that f is as described and f 2 p σ, since D(R n ) is dense in L p (R n ) for p <. Then, f (t) f (t) + f2 (t), with f (t) as t, and f2 (t) 2σ. This shows that f (t) as t for general f L p (R n ), p <. For the case p = and f C (R n ), the same argument gives us the result since D(R n ) is dense in C (R n ) (cf. [Rud87, p.7, Proof of Theorem 3.7]). Thus, by the Lebesgue dominated convergence theorem (due to ϕ L and the fact f (εy) ϕ(y) 2 f p ϕ(y) ) and the fact f (εy) as ε, we have lim f ϕ ε f p lim f (εy) ϕ(y) dy = lim f (εy) ϕ(y) dy =. ε ε R n R n ε This completes the proof. With the same argument, we have Corollary.6. Let p. Suppose ϕ L (R n ) and ϕ(x)dx =, R n then f ϕ ε p as ε whenever f L p (R n ), p <, or f C (R n ) L (R n ). Proof. Once we observe that (f ϕ ε )(x) =(f ϕ ε )(x) f(x) = (f ϕ ε )(x) f(x) ϕ ε (y)dy R n = [f(x y) f(x)]ϕ ε (y)dy, R n the rest of the argument is precisely that used in the last proof. In particular, we also have Corollary.7. Suppose ϕ L (R n ) with ϕ(x)dx = and let ϕ R n ε (x) = ε n ϕ(x/ε) for ε >. Let f(x) L (R n ) be continuous at {}. Then, lim f(x)ϕ ε (x)dx = f(). ε R n Proof. Since R n f(x)ϕ ε (x)dx f() = R n (f(x) f())ϕ ε (x)dx, then we may assume without loss of generality that f() =. Since f is continuous at {}, then for any η >, there exists a δ > such that 5 This statement is the continuity of the mapping t f(x t) of R n to L p (R n ).

16 The Fourier Transform and Tempered Distributions f(x) < η, ϕ whenever x < δ. Noticing that ϕ(x)dx ϕ R n, we have f(x)ϕ ε (x)dx η ϕ ε (x) dx + f ϕ ε (x) dx R ϕ n x <δ x δ η ϕ + f ϕ(y) dy ϕ =η + f I ε. But I ε as ε. This proves the result. y δ/ε From Theorems.3 and.5, we obtain the following solution to the Fourier inversion problem: Theorem.8. If both Φ and its Fourier transform ϕ = ˆΦ are integrable and R n ϕ(x)dx =, then the Φ means of the integral ( ω /2π) n R n e ωix ξ ˆf(ξ)dξ converges to f(x) in the L norm. In particular, the Abel and Gauss means of this integral converge to f(x) in the L norm. We have singled out the Gauss-Weierstrass and the Abel methods of summability. The former is probably the simplest and is connected with the solution of the heat equation; the latter is intimately connected with harmonic functions and provides us with very powerful tools in Fourier analysis. ( ω 2π ) n Since s(x, ε) = e ωix ξ e ε ωξ 2 ˆf(ξ)dξ R converges in L to f(x) as n ε > tends to, we can find a sequence ε k such that s(x, ε k ) f(x) for a.e. x. If we further assume that ˆf L (R n ), the Lebesgue dominated convergence theorem gives us the following pointwise equality: Theorem.9 (Fourier inversion theorem). If both f and ˆf are integrable, then ( ) n ω f(x) = e ωix ξ ˆf(ξ)dξ, 2π R n for almost every x. Remark.2. We know from Theorem.5 that ˆf is continuous. If ˆf is integrable, the integral R n e ωix ξ ˆf(ξ)dξ also defines a continuous function (in fact, it equals ˆf( x)). Thus, by changing f on a set of measure, we can obtain equality in Theorem.9 for all x. It is clear from Theorem.8 that if ˆf(ξ) = for all ξ then f(x) = for almost every x. Applying this to f = f f 2, we obtain the following uniqueness result for the Fourier transform: Corollary.2 (Uniqueness). If f and f 2 belong to L (R n ) and ˆf (ξ) = ˆf 2 (ξ) for ξ R n, then f (x) = f 2 (x) for almost every x R n.

17 .. The L theory of the Fourier transform We will denote the inverse operation to the Fourier transform by F or ˇ. If f L, then we have ( ) n ω ˇf(x) = e ωix ξ f(ξ)dξ. (.3) 2π R n We give a very useful result. Theorem.22. Suppose f L (R n ) and ˆf. If f is continuous at, then ( ) n ω f() = ˆf(ξ)dξ. 2π Moreover, we have ˆf L (R n ) and ( ) n ω f(x) = e ωix ξ ˆf(ξ)dξ, 2π R n for almost every x. Proof. By Theorem.3, we have ( ) n ω e ε ωξ ˆf(ξ)dξ = P (y, ε)f(y)dy. 2π R n R n From Lemma.4, we get, for any δ >, P (y, ε)f(y)dy f() = P (y, ε)[f(y) f()]dy R R n n P (y, ε)[f(y) f()]dy + P (y, ε)[f(y) f()]dy y <δ =I + I 2. R n y δ Since f is continuous at, for any given σ >, we can choose δ small enough such that f(y) f() σ when y < δ. Thus, I σ by Lemma.4. For the second term, we have, by a change of variables, that I 2 f sup P (y, ε) + f() P (y, ε)dy y δ y δ c n ε = f + f() P (y, )dy, (ε 2 + δ 2 )(n+)/2 y δ/ε ( ) n as ε. Thus, ω e ε ωξ ˆf(ξ)dξ 2π R f() as ε. On the other n hand, by Lebesgue dominated convergence theorem, we obtain ( ) n ( ) n ω ω ˆf(ξ)dξ = lim e ε ωξ ˆf(ξ)dξ = f(), 2π R 2π n ε R n which implies ˆf L (R n ) due to ˆf. Therefore, from Theorem.9, it follows the desired result. An immediate consequence is Corollary.23. i) R n e ωix ξ W (ξ, ε)dξ = e ε ωx 2. ii) R n e ωix ξ P (ξ, ε)dξ = e ε ωx.

18 The Fourier Transform and Tempered Distributions Proof. Noticing that (( ω W (ξ, ε) = F 2π we have the desired results by Theorem.22. ) n e ε ωx 2 ), and P (ξ, ε) = F (( ) n ) ω e ε ωx, 2π We also have the semigroup properties of the Weierstrass and Poisson kernels. Corollary.24. If α and α 2 are positive real numbers, then i) W (ξ, α + α 2 ) = R n W (ξ η, α )W (η, α 2 )dη. ii) P (ξ, α + α 2 ) = R n P (ξ η, α )P (η, α 2 )dη. Proof. It follows, from Corollary.23, that ( ) n ω W (ξ, α + α 2 ) = (F e (α +α 2 ) ωx 2 )(ξ) 2π ( ) n ω = F (e α ωx 2 e α 2 ωx 2 )(ξ) 2π ( ) n ) ω = F (e α ωx 2 e ωix η W (η, α 2 )dη (ξ) 2π R ( ) n n ω = e ωix ξ α ωx 2 e e ωix η W (η, α 2 )dηdx 2π R n R ( ( ) n n ω = e ωix (ξ η) e α ωx dx) 2 W (η, α 2 )dη R n R 2π n = W (ξ η, α )W (η, α 2 )dη. R n A similar argument can give the other equality. Finally, we give an example of the semigroup about the heat equation. Ex..25. Consider the Cauchy problem to the heat equation u t u =, u() = u (x), t >, x R n. Taking the Fourier transform, we have û t + ωξ 2 û =, Thus, it follows, from Theorem., that û() = û (ξ). u =F e ωξ 2t F u = (F e ωξ 2t ) u = (4πt) n/2 e x 2 /4t u =W (x, t) u =: H(t)u. Then, we obtain H(t + t 2 )u =W (x, t + t 2 ) u = W (x, t ) W (x, t 2 ) u =W (x, t ) (W (x, t 2 ) u ) = W (x, t ) H(t 2 )u =H(t )H(t 2 )u, i.e., H(t + t 2 ) = H(t )H(t 2 ).

19 .2. The L 2 theory and the Plancherel theorem The L 2 theory and the Plancherel theorem The integral defining the Fourier transform is not defined in the Lebesgue sense for the general function in L 2 (R n ); nevertheless, the Fourier transform has a natural definition on this space and a particularly elegant theory. If, in addition to being integrable, we assume f to be square-integrable then ˆf will also be square-integrable. In fact, we have the following basic result: Theorem.26 (Plancherel theorem). If f L (R n ) L 2 (R n ), then ˆf 2 = ( ω 2π ) n/2 f 2. Proof. Let g(x) = f( x). Then, by Theorem., h = f g L (R n ) and, by Proposition.3, ĥ = ˆfĝ. But ĝ = ˆf, ( thus) ĥ = ˆf 2. Applying Theorem n.22, we have ĥ L (R n ) and h() = ω 2π Rn ĥ(ξ)dξ. Thus, we get ( ) n ˆf(ξ) ω 2 dξ = ĥ(ξ)dξ = h() R n R 2π ( n ) n ω = f(x)g( x)dx 2π R ( ) n n ( ) n ω ω = f(x)f(x)dx = f(x) 2 dx, 2π R 2π n R n which completes the proof. Since L L 2 is dense in L 2, there exists a unique bounded extension, F, of this operator to all of L 2. F will be called the Fourier transform on L 2 ; we shall also use the notation ˆf = F f whenever f L 2 (R n ). A linear operator on L 2 (R n ) that is an isometry and maps onto L 2 (R n ) is called a unitary operator. It is an immediate consequence of Theorem.26 ( ω 2π ) n/2 F is an isometry. Moreover, we have the additional property that that ( ) n/2 ω 2π F is onto: Theorem.27. ( ω 2π ) n/2 F is a unitary operator on L 2 (R n ). ( ) n/2 Proof. Since ω 2π F is an isometry, its range is a closed subspace of L 2 (R n ). If this subspace were not all of L 2 (R n ), we could find a function g such that ˆfgdx = for all f L 2 and g R n 2. Theorem.2 obviously extends to L 2 ; consequently, fĝdx = ˆfgdx = for all f L 2. R n R n But this implies that ĝ(x) = for almost every x, contradicting the fact that ( ) n/2 ĝ 2 = ω g 2. 2π

20 The Fourier Transform and Tempered Distributions Theorem.27 is a major part of the basic theorem in the L 2 theory of the Fourier transform: Theorem.28. The inverse of the Fourier transform, F, can be obtained by letting ( ) n ω (F f)(x) = (F f)( x) 2π for all f L 2 (R n ). We can also extend the definition of the Fourier transform to other spaces, such as Schwartz space, tempered distributions and so on..3 Schwartz spaces Distributions (generalized functions) aroused mostly due to Paul Dirac and his delta function δ. The Dirac delta gives a description of a point of unit mass (placed at the origin). The mass density function is such that if its integrated on a set not containing the origin it vanishes, but if the set does contain the origin it is. No function (in the traditional sense) can have this property because we know that the value of a function at a particular point does not change the value of the integral. In mathematical analysis, distributions are objects which generalize functions and probability distributions. They extend the concept of derivative to all integrable functions and beyond, and are used to formulate generalized solutions of partial differential equations. They are important in physics and engineering where many non-continuous problems naturally lead to differential equations whose solutions are distributions, such as the Dirac delta distribution. Generalized functions were introduced by Sergei Sobolev in 935. They were independently introduced in late 94s by Laurent Schwartz, who developed a comprehensive theory of distributions. The basic idea in the theory of distributions is to consider them as linear functionals on some space of regular functions the so-called testing functions. The space of testing functions is assumed to be well-behaved with respect to the operations (differentiation, Fourier transform, convolution, translation, etc.) we have been studying, and this is then reflected in the properties of distributions. We are naturally led to the definition of such a space of testing functions by the following considerations. Suppose we want these operations to be defined on a function space, S, and to preserve it. Then, it would certainly have to consist of functions that are indefinitely differentiable; this, in view of part (v)

21 .3. Schwartz spaces in Proposition.3, indicates that each function in S, after being multiplied by a polynomial, must still be in S. We therefore make the following definition: Definition.29. The Schwartz space S (R n ) of rapidly decaying functions is defined { as } S (R n ) = ϕ C (R n ) : ϕ α,β := sup x α ( β ϕ)(x) <, α, β N n, x R n where N = N {}. (.4) If ϕ S, then ϕ(x) C m ( + x ) m for any m N. The second part of next example shows that the converse is not true. Ex..3. ϕ(x) = e ε x 2, ε >, belongs to S ; on the other hand, ϕ(x) = e ε x fails to be differential at the origin and, therefore, does not belong to S. Ex..3. ϕ(x) = e ε(+ x 2 ) γ belongs to S for any ε, γ >. Ex..32. S contains the space D(R n ). But it is not immediately clear that D is nonempty. To find a function in D, consider the function { e f(t) = /t, t >,, t. Then, f C, is bounded and so are all its derivatives. Let ϕ(t) = f( + t)f( t), then ϕ(t) = e 2/( t2) if t <, is zero otherwise. It clearly belongs to D = D(R ). We can easily obtain n-dimensional variants from ϕ. For examples, (i) For x R n, define ψ(x) = ϕ(x )ϕ(x 2 ) ϕ(x n ), then ψ D(R n ); (ii) For x R n, define ψ(x) = e 2/( x 2) for x < and otherwise, then ψ D(R n ); (iii) If η C and ψ is the function in (ii), then ψ(εx)η(x) defines a function in D(R n ); moreover, e 2 ψ(εx)η(x) η(x) as ε. Ex..33. We observe that the order of multiplication by powers of x,, x n and differentiation, in (.4), could have been reversed. That is, ϕ S if and only if ϕ C and sup x R n β (x α ϕ(x)) < for all multi-indices α and β of nonnegative integers. This shows that if P is a polynomial in n variables and ϕ S then P (x)ϕ(x) and P ( )ϕ(x) are again in S, where P ( ) is the associated differential operator (i.e., we replace x α by α in P (x)). Ex..34. Sometimes S (R n ) is called the space of rapidly decaying functions. But observe that the function ϕ(x) = e x2 e iex is not in S (R). Hence, rapid decay of the value of the function alone does not assure the membership in S (R).

22 The Fourier Transform and Tempered Distributions Theorem.35. The spaces C (R n ) and L p (R n ), p, contain S (R n ). Moreover, both S and D are dense in C (R n ) and L p (R n ) for p <. Proof. S C L is obvious by (.4). The L p norm of ϕ S is bounded by a finite linear combination of L norms of terms of the form x α ϕ(x). In fact, by (.4), we have ( ) /p ϕ(x) p dx R ( n /p ( ) /p ϕ(x) dx) p + ϕ(x) p dx x x > ( ) /p ( ) /p ϕ dx + x 2n ϕ(x) x 2np dx x x > ( ωn ) ( /p /p ωn = ϕ + x n (2p )n) 2n ϕ <. For the proof of the density, we only need to prove the case of D since D S. We will use the fact that the set of finite linear combinations of characteristic functions of bounded measurable sets in R n is dense in L p (R n ), p <. This is a well-known fact from functional analysis. Now, let E R n be a bounded measurable set and let ε >. Then, there exists a closed set F and an open set Q such that F E Q and m(q \ F ) < ε p (or only m(q) < ε p if there is no closed set F E). Here m is the Lebesgue measure in R n. Next, let ϕ be a function from D such that supp ϕ Q, ϕ F and ϕ. Then, ϕ χ E p p = ϕ(x) χ E (x) p dx dx = m(q \ F ) < ε p R n Q\F or ϕ χ E p < ε, where χ E denotes the characteristic function of E. Thus, we may conclude that D(R n ) = L p (R n ) with respect to L p measure for p <. For the case of C, we leave it to the interested reader. Remark.36. The density is not valid for p =. Indeed, for a nonzero constant function f c and for any function ϕ D(R n ), we have f ϕ c >. Hence we cannot approximate any function from L (R n ) by functions from D(R n ). This example also indicates that S is not dense in L since lim ϕ(x) = for all ϕ S. x From part (v) in Proposition.3, we immediately have

23 .3. Schwartz spaces Theorem.37. If ϕ S, then ˆϕ S. If ϕ, ψ S, then Theorem.37 implies that ˆϕ, ˆψ S. Therefore, ˆϕ ˆψ S. By part (vi) in Proposition.3, i.e., F (ϕ ψ) = ˆϕ ˆψ, an application of the inverse Fourier transform shows that Theorem.38. If ϕ, ψ S, then ϕ ψ S. The space S (R n ) is not a normed space because ϕ α,β is only a semi-norm for multi-indices α and β, i.e., the condition ϕ α,β = if and only if ϕ = fails to hold, for example, for constant function ϕ. But the space (S, ρ) is a metric space if the metric ρ is defined by ρ(ϕ, ψ) = 2 α β ϕ ψ α,β. + ϕ ψ α,β α,β N n Theorem.39 (Completeness). The space (S, ρ) is a complete metric space, i.e., every Cauchy sequence converges. Proof. Let {ϕ k } k= S be a Cauchy sequence. For any σ > and any γ N n, let ε = 2 γ σ, then there exists an N +2σ (ε) N such that ρ(ϕ k, ϕ m ) < ε when k, m N (ε) since {ϕ k } k= is a Cauchy sequence. Thus, we have ϕ k ϕ m,γ < σ + ϕ k ϕ m,γ + σ, and then sup γ (ϕ k ϕ m ) < σ x K for any compact set K R n. It means that {ϕ k } k= is a Cauchy sequence in the Banach space C γ (K). Hence, there exists a function ϕ C γ (K) such that lim ϕ k = ϕ, in C γ (K). k Thus, we can conclude that ϕ C (R n ). It only remains to prove that ϕ S. It is clear that for any α, β N n sup x K x α β ϕ sup x K Taking k, we obtain sup x K x α β (ϕ k ϕ) + sup x α β ϕ k x K C α (K) sup x K β (ϕ k ϕ) + sup x α β ϕ k. x K x α β ϕ lim sup ϕ k α,β <. k The last inequality is valid since {ϕ k } k= is a Cauchy sequence, so that ϕ k α,β is bounded. The last inequality doesn t depend on K either. Thus, ϕ α,β < and then ϕ S.

24 The Fourier Transform and Tempered Distributions Moreover, some easily established properties of S and its topology, are the following: Proposition.4. i) The mapping ϕ(x) x α β ϕ(x) is continuous. ii) If ϕ S, then lim h τ h ϕ = ϕ. iii) Suppose ϕ S and h = (,, h i,, ) lies on the i-th coordinate axis of R n, then the difference quotient [ϕ τ h ϕ]/h i tends to ϕ/ x i as h. iv) The Fourier transform is a homeomorphism of S onto itself. v) S is separable. Finally, we describe and prove a fundamental result of Fourier analysis that is known as the uncertainty principle. In fact this theorem was "discovered" by W. Heisenberg in the context of quantum mechanics. Expressed colloquially, the uncertainty principle says that it is not possible to know both the position and the momentum of a particle at the same time. Expressed more precisely, the uncertainty principle says that the position and the momentum cannot be simultaneously localized. In the context of harmonic analysis, the uncertainty principle implies that one cannot at the same time localize the value of a function and its Fourier transform. The exact statement is as follows. Theorem.4 (The Heisenberg uncertainty principle). Suppose ψ is a function in S (R). Then ( ) /2 xψ 2 ξ ˆψ ω ψ π 2 ω, and equality holds if and only if ψ(x) = Ae Bx2 where B > and A R. Moreover, we have ( ) /2 (x x )ψ 2 (ξ ξ ) ˆψ ω ψ π 2 ω for every x, ξ R. Proof. The last inequality actually follows from the first by replacing ψ(x) by e ωixξ ψ(x + x ) (whose Fourier transform is e ωix (ξ+ξ ) ˆψ(ξ + ξ ) by parts (ii) and (iii) in Proposition.3) and changing variables. To prove the first inequality, we argue as follows. Since ψ S, we know that ψ and ψ are rapidly decreasing. Thus, an integration by parts gives ψ 2 2 = ψ(x) 2 dx = x d = dx ψ(x) 2 dx ) dx. ( xψ (x)ψ(x) + xψ (x)ψ(x)

25 .4. The class of tempered distributions The last identity follows because ψ 2 = ψψ. Therefore, ψ x ψ(x) ψ (x) dx 2 xψ 2 ψ 2, where we have used the Cauchy-Schwarz inequality. By part (v) in Proposition.3, we have F (ψ )(ξ) = ωiξ ˆψ(ξ). It follows, from the Plancherel theorem, that ( ) /2 ( ) /2 ω ω ψ 2 = F (ψ ) 2 = ω ξ 2π 2π ˆψ 2. Thus, we conclude the proof of the inequality in the theorem. If equality holds, then we must also have equality where we applied the Cauchy-Schwarz inequality, and as a result, we find that ψ (x) = βxψ(x) for some constant β. The solutions to this equation are ψ(x) = Ae βx2 /2, where A is a constant. Since we want ψ to be a Schwartz function, we must take β = 2B <..4 The class of tempered distributions The collection S of all continuous linear functionals on S is called the space of tempered distributions. That is Definition.42. The functional T : S C is a tempered distribution if i) T is linear, i.e., T, αϕ + βψ = α T, ϕ + β T, ψ for all α, β C and ϕ, ψ S. ii) T is continuous on S, i.e., there exist n N and a constant c > such that T, ϕ c for any ϕ S. α, β n ϕ α,β In addition, for T k, T S, the convergence T k T in S means that T k, ϕ T, ϕ in C for all ϕ S. Remark.43. Since D S, the space of tempered distributions S is more narrow than the space of distributions D, i.e., S D. Another more narrow distribution space E which consists of continuous linear functionals on the (widest test function) space E := C (R n ). In short, D S E implies that E S D. Ex..44. Let f L p (R n ), p, and define T = T f by letting T, ϕ = T f, ϕ = f(x)ϕ(x)dx R n

26 The Fourier Transform and Tempered Distributions for ϕ S. It is clear that T f is a linear functional on S. To show that it is continuous, therefore, it suffices to show that it is continuous at the origin. Then, suppose ϕ k in S as k. From the proof of Theorem.35, we have seen that for any q, ϕ k q is dominated by a finite linear combination of L norms of terms of the form x α ϕ k (x). That is, ϕ k q is dominated by a finite linear combination of semi-norms ϕ k α,. Thus, ϕ k q as k. Choosing q = p, i.e., /p+/q =, Hölder s inequality shows that T, ϕ k f p ϕ k p as k. Thus, T S. Ex..45. We consider the case n =. Let f(x) = m k= a kx k be a polynomial, then f S since m T f, ϕ = a k x k ϕ(x)dx R k= m a k ( + x ) ε ( + x ) +ε x k ϕ(x) dx C k= k= R m a k ϕ k++ε, ( + x ) ε dx, so that the condition ii) of the definition is satisfied for ε = and n = m + 2. Ex..46. Fix x R n and a multi-index β N n. By the continuity of the semi-norm α,β in S, we have that T, ϕ = β ϕ(x ), for ϕ S, defines a tempered distribution. A special case is the Dirac δ-function: T δ, ϕ = ϕ(). The tempered distributions of Examples are called functions or measures. We shall write, in these cases, f and δ instead of T f and T δ. These functions and measures may be considered as embedded in S. If we put on S the weakest topology such that the linear functionals T T, ϕ (ϕ S ) are continuous, it is easy to see that the spaces L p (R n ), p, are continuously embedded in S. The same is true for the space of all finite Borel measures on R n, i.e., B(R n ). There exists a simple and important characterization of tempered distributions: Theorem.47. A linear functional T on S is a tempered distribution if and only if there exists a constant C > and integers l and m such that T, ϕ C for all ϕ S. R α l, β m ϕ α,β

27 .4. The class of tempered distributions Proof. It is clear that the existence of C, l, m implies the continuity of T. Suppose T is continuous. It follows from the definition of the metric that a basis for the neighborhoods of the origin in S is the collection of sets N ε,l,m = {ϕ : α l, β m ϕ α,β < ε}, where ε > and l and m are integers, because ϕ k ϕ as k if and only if ϕ k ϕ α,β for all (α, β) in the topology induced by this system of neighborhoods and their translates. Thus, there exists such a set N ε,l,m satisfying T, ϕ whenever ϕ N ε,l,m. Let ϕ = α l, β m ϕ α,β for all ϕ S. If σ (, ε), then ψ = σϕ/ ϕ N ε,l,m if ϕ. From the linearity of T, we obtain σ T, ϕ = T, ψ. ϕ But this is the desired inequality with C = /σ. Ex..48. Let T S and ϕ D(R n ) with ϕ() =. Then the product ϕ(x/k)t is well-defined in S by ϕ(x/k)t, ψ := T, ϕ(x/k)ψ, for all ψ S. If we consider the sequence T k := ϕ(x/k)t, then T k, ψ T, ϕ(x/k)ψ T, ψ as k since ϕ(x/k)ψ ψ in S. Thus, T k T in S as k. Moreover, T k has compact support as a tempered distribution in view of the compactness of ϕ k = ϕ(x/k). Now we are ready to prove more serious and more useful fact. Theorem.49. Let T S, then there exists a sequence {T k } k= S such that T k, ϕ = T k (x)ϕ(x)dx T, ϕ, as k, R n where ϕ S. In short, S is dense in S with respect to the topology on S. Proof. If h and g are integrable functions and ϕ S, then it follows, from Fubini s theorem, that h g, ϕ = ϕ(x) h(x y)g(y)dydx = g(y) h(x y)ϕ(x)dxdy R n R n R n R n = g(y) Rh(y x)ϕ(x)dxdy = g, Rh ϕ, R n R n where Rh(x) := h( x) is the reflection of h. Let now ψ D(R n ) with ψ(x)dx = and ψ( x) = ψ(x). Let ζ R n D(R n ) with ζ() =. Denote ψ k (x) := k n ψ(kx). For any T S, denote T k := ψ k T k, where T k = ζ(x/k)t. From above considerations, we know that ψ k T k, ϕ = T k, Rψ k ϕ. Let us prove that these T k meet the requirements of the theorem. In fact, we have

28 The Fourier Transform and Tempered Distributions T k, ϕ ψ k T k, ϕ = T k, Rψ k ϕ = ζ(x/k)t, ψ k ϕ = T, ζ(x/k)(ψ k ϕ) T, ϕ, as k, by the fact ψ k ϕ ϕ in S as k in view of Theorem.5, and the fact ζ(x/k) pointwise as k since ζ() = and ζ(x/k)ϕ ϕ in S as k. Finally, since ψ k, ζ D(R n ), it follows that T k D(R n ) S (R n ). Definition.5. Let L : S S be a linear continuous mapping. Then, the dual/conjugate mapping L : S S is defined by L T, ϕ := T, Lϕ, T S, ϕ S. Clearly, L is also a linear continuous mapping. Corollary.5. Any linear continuous mapping (or operator) L : S S admits a linear continuous extension L : S S. Proof. If T S, then by Theorem.49, there exists a sequence {T k } k= S such that T k T in S as k. Hence, for any ϕ S. LT k, ϕ = T k, L ϕ T, L ϕ := LT, ϕ, as k, Now, we can list the properties of tempered distributions about the multiplication, differentiation, translation, dilation and Fourier transform. Theorem.52. The following linear continuous operators from S into S admit unique linear continuous extensions as maps from S into S : For T S and ϕ S, i) ψt, ϕ := T, ψϕ, ψ S. ii) α T, ϕ := T, ( ) α α ϕ, α N n. iii) τ h T, ϕ := T, τ h ϕ, h R n. iv) δ λ T, ϕ := T, λ n δ /λ ϕ, λ R. v) F T, ϕ := T, F ϕ. Proof. See the previous definition, Theorem.49 and its corollary. Remark.53. Since F F T, ϕ = F T, F ϕ = T, F F ϕ = T, ϕ, we get F F = F F = I in S. Ex..54. Since for any ϕ S, F, ϕ =, F ϕ = (F ϕ)(ξ)dξ R ( ) n n ( ) n ω ω = e ωi ξ (F ϕ)(ξ)dξ 2π 2π R ( ) n n ( ) n ω ω = F F ϕ() = ϕ() 2π 2π

29 .4. The class of tempered distributions ( ) n ω = δ, ϕ, 2π we have ( ) n ω ˆ = δ, in S. 2π Moreover, ˇδ ( ) n = ω 2π. Ex..55. For ϕ S, we have ˆδ, ϕ = δ, F ϕ = ˆϕ() = e ωix ϕ(x)dx =, ϕ. R n Thus, ˆδ = in S. Ex..56. Since α δ, ϕ = α δ, ˆϕ = ( ) α δ, α ˆϕ = δ, F [(ωiξ) α ϕ] we have α δ = (ωiξ) α. = ˆδ, (ωiξ) α ϕ = (ωiξ) α, ϕ, Now, we shall show that the convolution can be defined on the class S. We first recall a notation we have used: If g is any function on R n, we define its reflection, Rg, by letting Rg(x) = g( x). A direct application of Fubini s theorem shows that if u, ϕ and ψ are all in S, then (u ϕ)(x)ψ(x)dx = u(x)(rϕ ψ)(x)dx. R n R n The mappings ψ (u ϕ)(x)ψ(x)dx and θ u(x)θ(x)dx are linear R n R n functionals on S. If we denote these functionals by u ϕ and u, the last equality can be written in the form: u ϕ, ψ = u, Rϕ ψ. (.5) If u S and ϕ, ψ S, the right side of (.5) is well-defined since Rϕ ψ S. Furthermore, the mapping ψ u, Rϕ ψ, being the composition of two continuous functions, is continuous. Thus, we can define the convolution of the distribution u with the testing function ϕ, u ϕ, by means of equality (.5). It is easy to show that this convolution is associative in the sense that (u ϕ) ψ = u (ϕ ψ) whenever u S and ϕ, ψ S. The following result is a characterization of the convolution we have just described. Theorem.57. If u S and ϕ S, then the convolution u ϕ is the function f, whose value at x R n is f(x) = u, τ x Rϕ, where τ x denotes the translation by x operator. Moreover, f belongs to the class C and it, as well as all its derivatives, are slowly increasing. Proof. We first show that f is C slowly increasing. Let h = (,, h j,, ), then by part iii) in Proposition.4,

30 The Fourier Transform and Tempered Distributions τ x+h Rϕ τ x Rϕ Rϕ τ x, h j y j as h, in the topology of S. Thus, since u is continuous, we have f(x + h) f(x) = u, τ x+hrϕ τ x Rϕ Rϕ u, τ x h j h j y j as h j. This, together with ii) in Proposition.4, shows that f has continuous first-order partial derivatives. Since Rϕ/ y j S, we can iterate this argument and show that β f exists and is continuous for all multi-index β N n. We observe that β f(x) = u, ( ) β τ x β Rϕ. Consequently, since β Rϕ S, if f were slowly increasing, then the same would hold for all the derivatives of f. In fact, that f is slowly increasing is an easy consequence of Theorem.47: There exist C > and integers l and m such that f(x) = u, τ x Rϕ C τ x Rϕ α,β. α l, β m But τ x Rϕ α,β = sup y R n y α β Rϕ(y x) = sup y R n (y +x) α β Rϕ(y) and the latter is clearly bounded by a polynomial in x. In order to show that u ϕ is the function f, we must show that u ϕ, ψ = f(x)ψ(x)dx. But, R n u ϕ, ψ = u, Rϕ ψ = u, Rϕ( x)ψ(x)dx R n = u, τ x Rϕ( )ψ(x)dx R n = u, τ x Rϕ ψ(x)dx = R n f(x)ψ(x)dx, R n since u is continuous and linear and the fact that the integral τ R n x Rϕ(y)ψ(x)dx converges in S, which is the desired equality..5 Characterization of operators commuting with translations Having set down these facts of distribution theory, we shall now apply them to the study of the basic class of linear operators that occur in Fourier analysis: the class of operators that commute with translations. Definition.58. A vector space X of measurable functions on R n is called closed under translations if for f X we have τ y f X for all y R n. Let X and Y be vector spaces of measurable functions on R n that are closed under translations. Let also T be an operator from X to Y. We say that T commutes with translations or is translation invariant if T (τ y f) = τ y (T f)

31 .5. Characterization of operators commuting with translations for all f X and all y R n. It is automatic to see that convolution operators commute with translations. One of the main goals of this section is to prove the converse, i.e., every bounded linear operator that commutes with translations is of convolution type. We have the following: Theorem.59. Let p, q. Suppose T is a bounded linear operator from L p (R n ) into L q (R n ) that commutes with translations. Then there exists a unique tempered distribution u such that T f = u f, f S. The theorem will be a consequence of the following lemma. Lemma.6. Let p. If f L p (R n ) has derivatives in the L p norm of all orders n +, then f equals almost everywhere a continuous function g satisfying g() C α f p, α n+ where C depends only on the dimension n and the exponent p. Proof. Let ξ R n. Then there exists a C n such that ( + ξ 2 ) (n+)/2 ( + ξ + + ξ n ) n+ C n α n+ ξ α. Let us first suppose p =, we shall show ˆf L. By part (v) in Proposition.3 and part (i) in Theorem.5, we have ˆf(ξ) C n( + ξ 2 ) (n+)/2 ξ α ˆf(ξ) α n+ =C n( + ξ 2 ) (n+)/2 C ( + ξ 2 ) (n+)/2 α n+ α n+ ω α F ( α f)(ξ) α f. Since ( + ξ 2 ) (n+)/2 defines an integrable function on R n, it follows that ˆf L (R n ) and, letting C = C ( + ξ 2 ) (n+)/2 dξ, we get R n ˆf C α f. α n+ Thus, by Theorem.9, f equals almost everywhere a continuous function g and by Theorem.5, ( ) n ω g() f 2π ˆf C α f. α n+

32 The Fourier Transform and Tempered Distributions Suppose now that p >. Choose ϕ D(R n ) such that ϕ(x) = if x and ϕ(x) = if x > 2. Then, it is clear that fϕ L (R n ). Thus, fϕ equals almost everywhere a continuous function h such that h() C α (fϕ). α n+ By Leibniz rule for differentiation, we have α (fϕ) = α! µ+ν=α µ!ν! µ f ν ϕ, and then α α! (fϕ) x 2 µ!ν! µ f ν ϕ dx µ+ν=α C sup ν ϕ(x) µ f(x) dx µ+ν=α A µ α x 2 x 2 x 2 µ f(x) dx AB µ α µ f p, where A ν ϕ, ν α, and B depends only on p and n. Thus, we can find a constant K such that h() K α f p. α n+ Since ϕ(x) = if x, we see that f is equal almost everywhere to a continuous function g in the sphere of radius centered at, moreover, g() = h() K α f p. α n+ But, by choosing ϕ appropriately, the argument clearly shows that f equals almost everywhere a continuous function on any sphere centered at. This proves the lemma. Now, we turn to the proof of the previous theorem. Proof of Theorem.59. We first prove that β T f = T β f, f S (R n ). (.6) In fact, if h = (,, h j,, ) lies on the j-th coordinate axis, we have τ h (T f) T f = T (τ ( ) hf) T f τh f f = T, h j h j since T is linear and commuting with translations. By part iii) in Proposition.4, τ hf f h j f x j in S as h and also in L p norm due to the density of S in L p. Since T is bounded operator from L p to L q, it follows that τ h (T f) T f h j T f x j in L q as h. By induction, we get (.6). By Lemma.6, T f equals almost everywhere a continuous function g f satisfying g f () C β (T f) q = C T ( β f) q β n+ β n+ h j