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Charles Fefferman's Multiplier Problem for the Ball Charles Fefferman's Multiplier Problem for the Ball Dean Katsaros, [email protected] Decemeber, 2017 Prepared in partial completion of a reading course on Harmonic Analysis at the University of Mas- sachusetts, Amherst. Very little of the below is original work, the main contributions are making omitted details explicit. This was the exercise intended in the study. The material and exposition below was discussed extensively with Professor Andrea Nahmod. 1 Contents 1 Motivation 2 1.1 The Setting . .2 1.2 Basic Results for the Fourier Transform . .3 2 The Lemma of Y. Meyer. 4 2.1 Proof of the Lemma . .4 3 Proof of Theorem 1.1 5 3.1 The Contradiction with Lemma 3.1 . .5 3.2 Proof of Lemma 3.1 . .6 4 Explicit Counterexample to the Disk Conjecture. 8 Appendix 14 A Vector Valued Inequalities 14 A.1 `2-Valued extensions of Linear Operators. 14 B Lower Bound on χH (dχR)(x). 17 B.1 The calculation. 17 C Mikhlin Multiplier Theorem 20 C.1 Motivation. 20 C.2 Some Lp bounds for multipliers . 20 D Littlewood-Paley square-function estimate. 24 D.1 Probabilistic Preliminaries . 24 D.2 The Littlewood-Paley Theorem . 25 1 1 Motivation 1.1 The Setting 1 Consider the operator T defined by ^ Tc f(x) = χjx|≤1(x)f(x): (1.1.1) All functions have fourier transforms, denoted by the hat notation as above. It is a fundamental problem to recover a function from its fourier transform f^. In particular, define Z _ ^ 2πiξ·x ^ SRf(x) := χjx|≤R(x) f(ξ)e dξ = χjx|≤R(x) f(ξ) (x); (1.1.2) which corresponds to a partial sum in the fourier series of f, noting that this spatial indicator commutes with the inverse fourier transform. Worded differently, SRf(x) is the image, in space, of the set of frequen- cies associated to f restricted to the ball of radius R . p The question is then whether, for f 2 L , SRf ! f in jj · jjLp(Rd) as R ! 1. The choice of p turns out to be crucial. Suppose that p = 2. Then, since L2 is a Hilbert Space, we can utilize Plancheral's Theorem. ^ (SR\f − f)(ξ) = −χjxj>R(ξ)f(ξ); (1.1.3) so using Plancheral's, ^ lim jjSRf − fjj2 = lim jj(SR\f − f)(ξ)jj2 = lim jjχjxj>R(ξ)f(ξ)jj2 = 0 R!1 R!1 R!1 since χjxj>R ! 0 as R diverges. If f is smooth with compact support, the result is also affirmative. p d If f is only assumed to be in L (R ), then use the density of smooth functions of compact support on d 1 d p d 1 d 1 R C0 (R ) in L (R ) to find g 2 C0 (R ) such that jjf − gjjp < /2. g 2 C0 means that SRg ! g in Lp for any p, so lim sup jjSRf − fjjp ≤ lim sup (jjSRf − gjjp + jjf − gjjp) = lim sup (jjSR(f − g)jjp + jjf − gjjp) ≤ R!1 R!1 R!1 ≤ sup jjSRjjopjjf − gjjp + /2 < sup jjSRjjop(/2) + /2: R R This shows that if SR were bounded, then we would have convergence of SRf to f. If SR were not bounded, p then there is a function on L such that jjSRfjjp = 1 for each R. This means that jjSRf − fjjp ≥ jjSRfjjp + jjfjjp = 1, and so, Result 1.1. Boundedness of SR for all R, which is equivalent to boundedness of S1 =: T above, is equivalent p d to the convergence of SRf to f in L (R ). It is with this in mind that Charles Fefferman studied the operator (1.1). Although Some progress was made towards a proof that T is bounded, ultimately Fefferman was able to provide a proof that T is only bounded on L2 (d > 1). Theorem 1.2. T is bounded only on L2 (d > 1). 1Credit to https://lewko.wordpress.com/2009/08/03/lp-convergence-of-fourier-transforms/ for the example. 2 By assuming that T is bounded, a certain vector valued inequality on sequences of operators related to T , due to Y. Meyer, can be proven (Lemma 2.1). Fefferman is able to construct a counterexample to Lemma 2.1 via a geometrical result on Besicovitch sets (Lemma 3.1). p 2 The first reduction we need is that it suffices to disprove L , p > 2, boundedness on R . de Leeuw proved the following Theorem p j+i Theorem 1.3. Let m(x; y) be a Fourier multiplier for L (R ). Then for almost every x, mx(y) = m(x; y) p j is a Fourier multiplier for L (R ) and the multiplier norm of mx does not exceed that of m. In particular, the restriction is possible for each x such that (x; y) is a Lebesgue point of m for almost all y 2 Rj. proof. see [4]. p d d−1 Using this theorem, we know that L boundedness of T on R implies boundedness on R , and the case p > 2 yields p < 2 by duality2. It will be useful to record some basics of the Fourier Transform and Fourier Series. 1.2 Basic Results for the Fourier Transform We define the Fourier transform of f via f^(ξ) = R f(x)e−2πix·ξdx with inverse formally given as f(x) = Rd R f^(ξ)e2πi·ξdξ. This is motivated by the expansion of f in the orthogonal basis e2πikx: Rd 1 X 2πikx f(x) = cke : (1.2.1) −∞ −2πimx Assume uniform convergence of 1.2.1. Then, term by term integration against e returns cm = R −2πimx ^ P1 ^ 2πikx3 f(x)e dx. Hence, with our definition of f, the fourier series of f is given by −∞ f(k)e . We now record some basic facts for the fourier transform that will be useful in manipulations. f[∗ g = f^· g:^ (1.2.2) f(\x + h)(ξ) = f^(ξ)e2πih·ξ (1.2.3) fe\2πih·x(ξ) = f^(ξ − h) (1.2.4) λ[−df(λ−1ξ) = f^(λξ) (1.2.5) d Define the Schwartz class S(R ) as the set of functions f that are smooth and such that for each n n α @ β d multi-Index α 2 N and β 2 N , the function x @x f is bounded on R . Via the Lemma, d 2 d Lemma 1.4. S(R ) is dense in L (R ). it can be shown that d 2 d Theorem 1.5. The Fourier transform on S(R ) has a unique extension to a unitary mapping of L (R ) to itself such that ^ jjfjjL2 = jjfjjL2 : (1.2.6) Equation 1.2.6 is konwn as Plancheral's Theorem, a very useful and important fact when passing between a function and its fourier transform. 2This is a typical application of the TT ∗ duality lemma, an appropriate version of which can be found in [5] 3It is a fundamental question of fourier analysis to ask when this sum is equal to f itself. 3 2 The Lemma of Y. Meyer. 2.1 Proof of the Lemma We know that the ball multiplier (1.1) is bounded on L2. Charles Fefferman proved that T is bounded only on L2 for spatial dimension n > 1. The first step of the proof of this result is to show the following lemma, which can proven if we assume that jjT fjjp ≤ Cjjfjjp for p > 2. 2 2 Lemma 2.1. Let v1; v2; ··· be a sequence of unit vectors in R , and let Hj = fx 2 R x · vj ≥ 0g p 2 be the upper half-plane defined by vj. Define a sequence of operators T1;T2; ··· on L (R ) by setting ^ Tdjf(x) = χHj (x)f(x). Then, for any sequence of functions f1; f2; ··· , 0 11=2 0 11=2 X 2 X 2 @ jTjfjj A ≤ C @ jfjj A : (2.1.1) j j p p r r r proof. Replace T via the following. Define T by T f(x) = χ r (x)f^(x). With D the disk of radius j j dj Dj j r 2 1 2 r and center rvj. limr!1 Tj f(x) = Tjf(x) for x 2 R . For f 2 C0 (R ), this holds via estimates on r Tdjf − Tdj f . To see this, fix j and write Z Z r −2πix·ξ −2πi·xξ T f − T f = χ r (x) f(x)e dx − χ (x) f(x)e dx ≤ dj dj Dj Hj Z −2πix·ξ ≤ χDr (x) − χH (x) f(x)e dx: j j 1 Now, we have assumed that f 2 C , and so eventually as r ! 1, χ r (x) − χ (x) 6= 0 in a set A that 0 Dj Hj r r has A \ suppf = ;. I.e., Dj has moving center a scalar multiplie of vj, so in particular the center of Dj lies r in H , r eventually is large enough that D does not intersect suppf. Hence, either χ r (x) − χ (x) = 0, j j Dj Hj or, when r is large enough that the difference of these indicators is nonzero for a significant set but far from the support of f, f = 0 for large enough r to move the support of χDr (x) − χH (x) sufficiently far j j into Hj. Now, we will use this to show that Z Z r ^ 2πix·ξ ^ 2πix·ξ Tjf − Tj f = f(ξ)e dξ − f(ξ)e dξ ! 0: r Hj Dj Consider Z Z Z Z r −2πix·ξ 2πiξ·x −2πix·ξ 2πiξ·x Tif − T f = χH (ξ) f(x)e dx e dξ − χDr (ξ) f(x)e dx e dξ = j j j Z Z Z 2πiξ·x r 2πiξ·x r = Tdjfe dξ − Tdfe dξ ≤ Tdjf − Tdf dξ: j j PN 2 Then, Fatou's Lemma applied to the nonnegative partial sums SN (T f) = j jTjfjj now shows that 0 11=2 0 11=2 X 2 X r 2 jTjfjj ≤ lim inf T fj : (2.1.2) @ A r!1 @ j A j j p p 4 r ^ ^ 1 2 Now, Tjf − T f ≤ χDr (ξ) − χH (ξ) jfj, and χDr (ξ) − χH (ξ) jfj 2 L ( ) for r large enough since d dj i i i i R r f was assumed to be smooth.
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