4 Riesz Kernels. Since the functions k i (ξ) = ξ i. are bounded functions it is clear that R

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1 4 Riesz Kernels. A natural generalization of the Hilbert transform to higher dimension is mutiplication of the Fourier Transform by homogeneous functions of degree 0, the simplest ones being R i f(ξ) = ξ i ξ ˆf(ξ) (4.) Since the functions k i (ξ) = ξ i are bounded functions it is clear that R ξ i are bounded operators from L 2 (R d )intol 2 (R d ). On the other hand k i are not continuous at ξ = 0, and therefore the formal kernel K i with the representation R i f(x) = K i (x y)f(y)dy (4.2) R d can not be in L (R d ). Lemma. The kernels K i ( ) are given by where c d is a constant depending on the dimension. x i K i (x) =c d (4.3) x d+ Proof. We will begin with the following calculation. For any ɛ>0,d>δ >0 e i<x,ξ> e ɛ x 2dx = R x d d δ Γ( d δ ) e i<x,ξ> t d δ 2 e (t+ɛ) x 2 dxdt 2 R d 0 = c d Γ( d δ 2 ) 0 e ξ 2 d δ 4(t+ɛ) t 2 (t + ɛ) d 2 dt (4.4) If we let ɛ 0inequation(??) lim e i<x,ξ> e ɛ x 2 ɛ 0 R x d d δ dx = c d Γ( d δ 2 ) 0 e ξ 2 4t t δ 2 dt = c dγ( δ 2 ) Γ( d δ 2 ) ξ δ

2 If we let f ɛ,δ (x) = (d δ)x j e ɛ x 2 Γ(, lim x d+2 δ ɛ 0 fɛ,δ (x) =ic δ 2 ) d Γ( d δ )ξ j ξ δ. Finally we 2 let δ>. It is not difficult to see that for any smooth function f(x) withcompact support (R j f)(x) = y l y j y j [f(x + y) f(x)]dy + f(x + y)dy y d+ y l y d+ is independent of l because y j dy = 0 for any shell S = {l S y d+ y l 2 }. It is a smooth function of x. Forlargex, the first term is 0, and the second integral can be estimated by, [ x j y j R y x x j ] f(y) dy d d+ x d+ C x d+ if we use that f has compact support and satisfies R d f(y)dy =0. Itisnow easy to compute R j f(ξ) = c d ξ j ξ ˆf(ξ) The next step is to show that the kernels K i satisfy condition of equation (??). Let us take x, y R d and write y = rω where r = y and ω = y y Sd. x y C y x i y i x y x i dx = d+ x d+ + C x y C y x rω Cr + x rω Cr x rω Cr x rω Cr σ(x y) x y σ(x) d x dx d σ(x rω) σ(x rω) dx x rω d x d σ(x rω) σ(x) dx x d x rω d x dx d σ(x rω) σ(x) dx x d 2

3 where σ(x) = x i. If we make the substitution x = x rx we get x i y i x y C y x y x i d+ x dx C d+ x ω C x ω d x d dx σ(x ω) σ(x ) + dx x d x ω C The estimate is clearly uniform in r. If C is large enough 0 and ω are excluded from the domain of integrartion. For large x we get an extra cancellation in both the integrals to make them converge with a bound that is uniform in ω. For the second integral we need only that σ satisfies a Hölder condition on S d. We have therefore proved the following theorem. Theorem 4.. If the kernel K(x) is given by K(x) = σ( x ) x (4.5) x d and σ( ) satisfies a Hölder condition on S d and has mean 0 on S d, then convolution by K defines a bounded operator from L p (R d ) into L p (R d ) for all p in the range <p<. In particular the Riesz transforms 4. given by 4.2 with kernels 4.3 are bounded operators in every L p in the same range. 5 Sobolev Spaces. In dealing with differential equations we often come across solutions that do not have the smoothness necessary to be a solution in the ordinary sense. To illustrate it by an example, suppose we want to solve the equation u = i u xi x i = f (5.) on R d.ifd = the equation reduces to u xx = f which is easy to solve. We need only to integrate f twice, and if f has d continuous derivatives u will have d+2 continuous derivatives. On R d it is conceivable that each u xi x i may be singular, but somehow the singularities cancel miraculously to produce a much nicer f. Working formally with Fourier trnasforms ξ 2 û(ξ) = f(ξ) 3

4 and û xi x j (ξ) = ξ iξ j ξ 2 f(ξ) In other words u xi x j = R i R j f It says that for <p<, iff L p we can expect u to have two derivatives in L p, but if f is bounded and continuous one should not expect u to have two continuous derivatives. In fact on d = 2, one can construct a counter example, i.e. a function f which is continuous such that the soulution u of Poisson s equation exhibits a singularity of the individual second derivatives at 0, that of course cancel to produce a continuous f. The Sobolev spaces W p k (Rd ) are defined as the space of functions u on R d such that u and all its partial derivatives D n x D n d x d u of order n = n + + n d are in L p. We could start with C functions with compact support on R d and complete it in the norm u k,p = n,...n d n=n + +n d k D n x D n d x d u p (5.2) If u L p and D i u = D xi u L p u should be more regular than an L p function. Let us consider the operator Âu(ξ) = û(ξ) ( + ξ 2 ) 2 and consider its represenation by the kernel (Au)(x) = u(x + y)a(y)dy R d where e i<x,ξ> a(x) =c d dx = c d e i<x,ξ> e t(+ ξ 2 ) dt R d ( + ξ 2 ) 2 π R d 0 t e t = k d e x 2 k d 4t dt = e t x 2 e dt y d 4t 0 t d t d+ 2

5 decays very rapidly at, is smooth for x 0 and has a singularity of x d near the origin for d 2 and a logarithmic singularity at 0 when d =. In particular a( ) L q for q< d. By Hölder s inequality, A will map L d p into L for p>d. If d = p>the result is false. Let us take d =2 and a nonnegative function f with compact support such that f L 2 but f(x) dx =. We saw that Af has a singularity at 0. Let us consider R d x u = D (Af). Clearly u 2 2 = û 2 2 = + ξ ˆf(ξ) 2 dξ ˆf = f 2 2 R 2 ξ 2 By Young s inequality any K L q maps L p L p provided =. p p q Therefore f W,p implies f L p so long as <. By induction p p d f W k,p implies that f W,p implies f L p so long as < k). p p d Therefore on R d, f W k,p implies the continuity of f if k> d. p Actually one can prove a stronger result to the effect that if =. p p d then W,p L p as long as <p <. This requires the following theorem. Theorem 5.. Let T a be the operator of convolution by the kernel x a d on R d. (T a f)(x) = y a d f(x + y)dy (5.3) R d Then T a is bounded from L p to L p provided <p< d a and p = p a d. Proof. First, we note that for a>0, T a is well defined on bounded functions with compact support. We start by proving a weak type inequlity of the form µ[x : (T a f)(x) l] C f q p l q For any choice of <p< d let f L a p. We can assume without loss of generality that f 0. We write (T a f)(x) = y a d f(x + y)dy + y a d f(x + y)dy y ρ u + u 2 y ρ 5

6 and estimate u,u 2 by u p C ρ a f p ( ) u 2 y p (a d) p dy f p = C 2 ρ a d+ d p f p y ρ We can now pick ρ =( 2C 2 f p ) p d ap l where q = pd d ap or q = p a d. and estimate sup u 2 (x) l x 2 µ[x : u (x) l 2 ] 2p C p ρ ap f p p l p = C 3 ( f p l = C 3 ( f p l ) ap 2 d ap +p Now, an application of Marcinkiewicz interpolation gives boundedness from L p to L q in the same range and with the same relation between p and q. We can also define the fractional derivative operarors ( D a f)(x) = ) q R d f(x + y) f(x) y d+a dy (5.4) for 0 <a<2. A calculation shows that in terms of Foirier transforms it is multiplication by e i<ξ, y> dy = c R y d d+a d,a ξ a Therefore D a and T a are essentially (upto a constant) inverses of each other. If r>0 is written as k + a, wherek is a nonnegative integer and 0 a<, then one defines the norm corresponding to r th derivative by u r,p = D P n D n d d u p + i n i k P D n D n d d u a,p (5.5) i n i=k This way the Sobolev spaces W r,p are defined for fractional derivatives as well. 6

7 Theorem 5.2. The inclusion map is well defined and bounded from W r,p into W s,q provided s<r, <p<q<, and r s. The extreme q p d value of q = is allowed if > r s. q p d Proof. We can assume without loss of generality that 0 <r s<. We can go from W r,p to W s,q in a finite number of steps, with 0 <r s<ateach step. We write I = c d,a T a D a where a = r s. By definition D a maps W r,p boundedly into W s,p. By the earlier theorem T a maps W s,p boundedly into W s,q. Although we proved it for s = 0, it is true for every s because T a commutes with D a.thecaeq = iscoveredaswellbythisargument. 6 Generalized Functions. Let us begin with the space W,2. This is a Hilbert Space with the inner product d <f,g> = [fḡ + f xi ḡ xi ]dx = f hdx R d R d where h = g d g x i x i.sinceg W,2, g xi L 2 and g xi x i is the derivative of an L 2 function. In fact since we can write fg xi dx as f xi gdx, Any derivative of an L 2 function can be thought of as a bounded linear functional on the space W,2. A simlar reasoning applies to all the spaces W r,p. The dual space of W r,p is W r,q where + =. p q For a function to be in L p its singularities as well as decay at must be controlled. We can get rid of the condition at by demaniding that f be in L p (K) for every bounded set K or equivalently by insisting that φf L p for every C function φ with compact support. This definition makes sense for W r,p as well. We say that f Wr,p loc if φf W r,p for every C function φ with compact support. One needs to check that on W r,p mutiplication by a smooth function is a bounded linear map. One can use Leibnitz s rule if r is an integer. For 0 <r< we need the following lemma. Lemma 2. If f W r,p and φ C r with r<r i.e. φ is a bounded function satisfying φ(x) φ(y) C x y r, for all x, y, then φf W r,p. Proof. We need to prove g(x) = R d φ(y)f(y) φ(x)f(x) y x d+r dy 7

8 is in L p.wecanwrite φ(y)f(y) φ(x)f(x) =φ(x)[f(y) f(x)] + [φ(y) φ(x)]f(y). The contribution of first term is easy to control. To control the second term it is sufficient to show that φ(y) φ(x) sup dy < x R y x d d+r which is not hard. We split the integral into two regions x y and x y >, use the Hölder property of φ to obtain an estimate on the integral over x y and the boundedness of φ to get an estimate over x y >, both of which are uniform in x. 8