LECTURE NOTES 4 FOR 247A The Hilbert transform - UCLA

1 LECTURE NOTES 4 FOR 247 ATERENCE Hilbert transformIn this set of NOTES we begin the theory ofsingular integral operators- operatorswhich are almost integral operators, except that their kernelK(x,y) just barelyfails to be integrable near the diagonalx=y. (This is in contrast to, say, fractionalintegral operators such asTf(y) := Rd1|x y|d sf(x)dx, whose kernels go to infinityat the diagonal but remain locally integrable.) These operators arisenaturally incomplex analysis, Fourier analysis, and also in PDE (in particular, theyincludethezeroth order pseudodifferential operatorsas a special case, about which morewill be said later). It is here for the first time that we must make essential use ofcancellation: cruder tools such as Schur s test which do not take advantage ofthesign of the kernel will not be effective we turn to the general theory, let us study the prototypical singular integraloperator1, theHilbert transformHf(x) := Rf(x t)tdt:= lim 01 |t|> f(x t)tdt.

2 (1)It is not immediately obvious thatHf(x) is well-defined even for nice functionsf. But iffisC1and compactly supported, then we can restrict the integral to acompact intervalt [ R,R] for some largeR(depending onxandf), and usesymmetry to write |t|> f(x t)tdt= <|t|<Rf(x t) f(x) mean-value theorem then shows thatf(x t) f(x)tis uniformly bounded onthe intervalt [ R,R] for fixedf,x, and so the limit actually exists from thedominated convergence theorem. A variant of this argument shows thatHfis alsowell-defined forfin the Schwartz class, though it does not map the Schwartz classto itself. Indeed it is not hard to see that we have the asymptoticlim|x| xHf(x) =1 Rffor such functions, and so iffhas non-zero mean thenHfonly decays like 1/|x|atinfinity. In particular, we already see thatHis not bounded onL1. However, wedo see thatHat least maps the Schwartz class toL2(R).

3 1 The other prototypical operator would be the identity operatorTf=f, but that is too trivialto be worth TAOLet us first make comment on some algebraic properties of the Hilbert commutes with translations and dilations (but not modulations):HTransx0= Transx0H; Dilp H=HDilp .(2)In fact, it is essentially the only such operator (see Q1). It is also formally skew-adjoint, in that forC1, compactly supportedf,g: RHf(x)g(x)dx= Rf(x)Hg(x)dx;we leave the verification as an Hilbert transform is connected to complex analysis (and in particular to Cauchyintegrals) by the following (Plemelj formulae).Letf C1(R)obey a qualitative decay boundf(x) =Of(hxi 1)(say - these conditions are needed just to makeHfwell-defined).Then for anyx R12 ilim 0 Rf(y)y (x i )dy= f(x) +iHf(x) translation invariance we can takex= 0. By taking complex conjugateswe may assume that the sign is +.

4 Our task is then to show thatlim 012 i Rf(y)y i dy 12f(0) i2 |y|> f(y) ydy= by 2 iand making the change of variablesy= w, we reduce toshowinglim 0 Rf( w)(1w i 1|w|>11w)dw if(0) = computation shows that the expression in parentheses is absolutely integrableand R(1w i 1|w|>11w) = we reduce to showing thatlim 0 R(f( w) f(0))(1w i 1|w|>11w)dw= claim then follows from dominated suppose thatfnot only obeys the hypotheses of the Plemelj formulae, butalso extends holomorphically to the upper half-plane{z C: Im(z) 0}andobeys the decay boundf(z) =Of(hzi 1) in this region. Then Cauchy s theoremgives12 ilim 0 Rf(y)y (x+i )dy=f(x+i )and12 ilim 0 Rf(y)y (x i )dy= 0and thus by either of the Plemelj formulae we see thatHf= ifin this case. Inparticular, comparing real and imaginary parts we conclude that Im(f) =HRe(f) LECTURE NOTES 43and Re(f) = HIm(f).

5 Thus for reasonably decaying holomorphic functions on theupper half-plane, the real and imaginary parts of the boundary value are connectedvia the Hilbert transform . In particular this shows that such functions are uniquelydetermined by just the real part of the boundary above discussion also strongly suggests the identityH2= 1. This can bemade more manifest by the following Fourier representation of the Hilbert S(R), then Hf( ) = isgn( ) f( )(3)for (almost every) R. (Recall thatHflies inL2and so its Fourier transformis defined via density by Plancherel s theorem.)ProofLet us first give a non-rigorous proof of this identity. Morally speaking wehaveHf=f 1 xand so since Fourier transforms ought to interchange convolutionand product Hf( ) = f( ) 1 x( ).On the other hand we have 1( ) = ( )so on integrating this (using the relationship between differentiationand multipli-cation) we get 1 2 ix( ) =12sgn( )and the claim we turn to the rigorous proof.

6 As the Hilbert transform is odd,a symmetryargument allows one to reduce to the case >0. It then suffices to show that f+iHf2has vanishing Fourier transform in this half-line. Define the Cauchy integraloperatorC f(x) =12 i Rf(y)y (x i ) Plemelj formulae show that these converge pointwise to f+iHf2as 0;sincefis Schwartz, it is also not hard to show via dominated convergence thatthey also converge inL2. Thus by theL2boundedness of the Fourier transform itsuffices to show that each of theC falso have vanishing Fourier transform on thehalf-line2. Fix >0 (we could rescale to be, say, 1, but we will not have need of2 This is part of a more general phenomenon, that functions which extend holomorphically tothe upper half-space in a controlled manner tend to have vanishing Fourier transform onR , whilethose which extend to the lower half-space have vanishing Fourier transform onR+.)

7 Thus thereis a close relationship between holomorphic extension and distribution of the Fourier on this TAOthis normalisation). We can truncate and defineC ,Rf(x) =12 i Rf(y)y (x i )1|y x|<Rdy;more dominated convergence shows thatC ,Rfconverges toC finL2asR + and so it will suffice to show that the Fourier coefficients ofC ,Rfconverge pointwiseto zero asR + on the half-line >0. From Fubini s theorem we easily compute C ,Rf( ) =12 i f( ) Re 2 iy (y i )1|y|<Rdy;but then by shifting the contour to the lower semicircle of radiusRand then lettingR we obtain the this proposition and Plancherel s theorem we conclude thatHis an isometry:kHfkL2(R)=kfkL2(R)for allf S(R).Because of this,Hhas a unique dense extension toL2(R), and (by another appli-cation of Plancherel s theorem to justify taking the limit) the formula(3) is validfor allf L2(R).

8 There is still the question of whether the original definition isvalid forf L2(R), or in other words whether the sequence of functions1 |t|> f(x t)tdt(note that Cauchy-Schwarz ensures that the integrand is absolutely integrable)converges as 0 toHfforf L2(R), and in what sense this convergence turns out that the convergence is true both pointwise almost everywhere andin theL2(R) metric sense, and in fact it follows from variants of the Calder on-Zygmund theory presented in this NOTES , but we shall not do so here (as it may becovered in the participating student seminar).From (3) we also see that we do indeed have the identityH2= 1 onL2(R), andthatHis indeed skew-adjoint onL2(R) (H = H); in particular,His us briefly mention why the Hilbert transform is also connected tothe theory ofpartial Fourier L2(R)andN+,N R.

9 Then N+ N f( )e2 ix d =i2(ModN+HMod N+f Mod N HModN f).ProofBy limiting arguments (and because everything is continuous onL2(R))it suffices to verify this identity forfin the Schwartz class. But then the claimfollows by taking the Fourier transform of both sides and using (3).It is thus clear that in order to address the question of the extentto which thepartial Fourier integrals N+ N f( )e2 ix d actually converges back tof, we willneed to understand the properties of the Hilbert transform , andin particular itsboundedness NOTES 45 The formula (3) suggests thatHbehaves like multiplication by iin some sense;informally,Hbehaves like ifor positive frequency functions and +ifor negativefrequency functions. Now multiplication by iis a fairly harmless operation - itis bounded on every normed vector space - so one might expectHto similarlybe bounded on more normed spaces than justL2(R); in particular, it might bebounded onLp(R).

10 This is indeed the case for 1< p < , and we now turn to thetheory which will generate such by example that the Hilbert transform is not boundedonL1(R) orL (R). on-Zygmund theoryIn the Hilbert transform , we have an operator which is bounded onL2(R), and wewish to extend this boundedness to otherLp(R) spaces as well. This is of course notautomatic - the enemy is that the operator might map a broad shallow functioninto a tall narrow function of comparableL2norm (which will increase theLpnorms markedly for allp >2), or conversely map a tall narrow function down to abroad shallow function of comparableL2norm (which will increase theLpnormsmarkedly forp <2). It turns out that these two phenomena are more or less dualto each other, and so when working with an operator which is comparable toits adjoint (which is for instance the case with the skew-adjoint Hilbert transform ,H = H) eliminating one of these enemies will automatically remove the operators such as the Hilbert transform , there are basically two approaches3toexclude these scenarios.