ON THE NORM OF THE OPERATOR ai + bh ON L p (R)

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1 ON THE NOM OF THE OPEATO ai + bh ON L p () YONG DING, LOUKAS GAFAKOS, AND KAI ZHU 1 Abstract We provide a direct proof of the following theorem of Kalton, Hollenbeck, and Verbitsky [7]: let H be the Hilbert transform and let a, b be real constants Then for 1 < p < the norm of the operator ai + bh from L p () to L p () is equal to ( max x ax b + (bx + a) tan 2p p + ax b (bx + a) tan 2p p x + tan 2p p + x tan 2p p ) 1 p Our proof avoids passing through the analogous result for the conjugate function on the circle, as in [7], and is given directly on the line We also provide new approximate extremals for ai + bh in the case p > 2 1 Introduction In this note we revisit the celebrated result of Kalton, Hollenbeck, and Verbitsky [7] concerning the value of the norm of the operator ai + bh from L p () to L p () for 1 < p < and a, b real constants We provide a self-contained direct proof of this result on the real line The original proof in [7] was given for the conjugate function on the circle in lieu of the Hilbert transform and the corresponding result for the line was obtained from the periodic case via a transference-type argument due to Zygmund [13, Ch XVI, Th 38] known as blowing up the circle Here we work directly with the Hilbert transform on the line, using an idea contained in [4] and [6], which is based on applying subharmonicity on the boundary of a suitable family of discs that fill up the upper half space as their radii tend to infinity The main estimates needed for our proof (Lemmas 31 and 32) are as in [7] but are included in this note for the sake of completeness (with a minor adjustment) The new contributions of this article are contained in Sections 4 and 5 In Section 4 we use a limiting argument and subharmonicity to prove the claimed bound for ai + bh We obtain the approximate extremals for the operators ai+bh in Section 5; these are based on these for the Hilbert transform which first appeared in Gohberg and Krupnik [5] for 1 < p < 2 and were also used by Pichorides [12] We find new approximate extremals for the Hilbert transform for 2 < p < in Section 5 and we use them to construct corresponding approximate extremals for ai + bh for this range of p s We note that the case a = 0, b = 1 of this result was proved by Pichorides [12] and B Cole (unpublished, see [2]), while the case a = 0, b = 1, p = 2 m, m = 1, 2,, was obtained four years earlier by Gohberg and Krupnik [5] For a short history on this topic we refer to Laeng [9] It is noteworthy that Mathematics 2010 Subject Classification: Primary 42A45 Secondary 42A50, 42A99 Keywords and phases: Best constants, Hilbert transform The first author is supported by NSFC( , , ),SFDP( ) and the Fundamental esearch Funds for the Central Universities(2014KJJCA10) The second author would like to acknowledge the support of Simons Foundation The third author would like to thank the China Scholarship Council for its support 1 Corresponding author 1

2 2 YONG DING, LOUKAS GAFAKOS, AND KAI ZHU 1 the operator norm of the Hilbert transform on L p is also the norm of other operators, for instance of the segment multiplier; on this see De Carli and Laeng [1] 2 The norm of ai + bh Denote the identity operator by I The Hilbert transform on the real line is defined by Hf(x) = pv 1 f(t) x t dt for a smooth function with compact support For a, b, define (1) B p = max x ax b + (bx + a) tan γ p + ax b (bx + a) tan γ p x + tan γ p + x tan γ p where γ = 2p B p can be defined equivalently by cos(θ + θ (2) B p = (a 2 + b 2 ) p/2 0 ) p + cos(θ + θ 0 + p max ) p 0 θ 2 cos θ p + cos(θ +, p ) p where tan θ 0 = b/a Our goal is to provide a proof of the following result in [7]: Theorem 1 [7] Let 1 < p < and for a, b Then for all smooth functions with compact support f on the line we have (ai + bh)f p L p () B p f p L p () where the constant B p is sharp In other words, ai + bh L p () L p () = B 1 p p As ai + bh maps real-valued functions to real-valued functions, in view of the Marcinkiewicz and Zygmund theorem [11] (see also [3, Theorem 551]), the norm of ai + bh on L p () and on L p (C) are equal 1 Thus we may work with a nice real-valued function f in the proof of Theorem 1 3 Some Lemmas In this section we provide two auxiliary results that are crucial in the proof of the main theorem Lemma 31 [7] Suppose p > 1/2, p 1, and F is a p-homogeneous continuous function on C Suppose there is a sector S so that F is subharmonic on S and superharmonic on the complementary sector S Suppose further there is no nontrivial sector on which F is harmonic Suppose that F (z) + F (e i/p z) 0 for all z, and there exists z 0 0 so that F (z 0 )+F (e i/p z 0 ) = 0 Then there is a continuous p-homogeneous subharmonic function G with G(z) F (z) for all z C Proof Lemma 31 is a restatement of Theorem 35 in [7] We only provide a sketch below making a minor modification in the proof in [7] (ie, definition of h in (3)) We can suppose there exists z 0 with z 0 = 1 so that F (z 0 ) + F (e i/p z 0 ) = 0 Let z 0 = e it 0, z 1 = e i(t 0+/p) Write F (re it ) = r p f(pt), where f is a 2p-periodic function 1 for operators that do not map real-valued functions to real-valued functions, these norms may not be equal; for instance this is the case for the iesz projections, see [8]

3 ON THE NOM OF THE OPEATO ai + bh 3 on By Proposition 33 in [7], if I is any interval so that e ix/p S for x I, then f is trigonometrically convex on I, and if e ix/p S for x I, then f is trigonometrically concave on I At least one of z 0, z 1 is contained in S; let us suppose that z 0 S The function f(x) + f(x + ) has minimum at pt 0, hence f (pt 0 ) + f (pt 0 + ) 0, f +(pt 0 ) + f +(pt 0 + ) 0 This implies that there exist a and b such that a + b = 0 and f (pt 0 ) a f +(pt 0 ) and f (pt 0 + ) b f +(pt 0 + ) Now define (3) h(x) = f(pt 0 ) cos(x pt 0 ) + a sin(x pt 0 ) Then by Lemma 31 in [7], h f on a neighborhood of pt 0 Lemma 32 in [7] implies that h(x) f(x) for pα + 2p x < pt and for pt 0 x pβ By the Phragmén-Lindelöf theorem ([10]) we obtain that h f in a neighborhood of [pt 0, pt 0 + ] Let T = re iθ : r > 0, t 0 < θ < t 0 + }, define p H(reiθ ) = r p h(pθ) for t 0 < θ < t 0 + p and H(z) if z T, G(z) = F (z) if z / T Then G(z) F (z) for all z C and G is subharmonic on both T and its complementary sector T It is easy to see G is then subharmonic on C\0} since h f in a neighborhood of pt 0 and pt 0 + Finally h(x) + h(x + ) = 0 implies that G(z) + G(e i/p z) 0 for all z Integrating over a circle around 0 yields the subharmonicity of G at 0 Next we have a version of Lemma 42 in [7] in which we provide an explicit formula for the subharmonic function G Lemma 32 Let B p be given by (1), T = re it : r > 0, t 0 < t < t 0 + }, where t p 0 is the value that makes right part of (2) attain its maximum Let z = re it, z 0 = re it 0, G(z) = G(re it ) be -periodic of t and when 0 < t < : B p ez 0 p 1 sgn(ez 0 )e[( z z 0 ) p z 0 ] aez 0 + bimz 0 p 1 G(z) = sgn(aez 0 + bimz 0 )(ae[( z z 0 ) p z 0 ] + bim[( z z 0 ) p z 0 ]), if z T or e i z T B p ez p aez + bimz p, if both z, e i z / T Then G(z) is subharmonic and satisfies (4) aez + bimz p B p ez p G(z) Proof The case b = 0 is trivial, so we assume b 0, and we may further assume that a 2 + b 2 = 1 Let F (z) = B p ez p aez + bimz p Then F (re it ) = r p f(t), where f(t) = B p cos t p a cos t + b sin t p is -periodic and continuously differentiable The definition in (2) implies that We observe that F 0 is equivalent to min [f(t) + f(t + /p)] = 0 0 t 2 B p ez p 2 aez + bimz p 2 In order for F (z) to be subharmonic, the following must be true: a + b tan t p 2 B p We can see that for p 2 there will be two separate double sectors where F (z) is subharmonic, and superharmonic in their complement So let p = p/2, t 0 = 2t 0, define

4 4 YONG DING, LOUKAS GAFAKOS, AND KAI ZHU 1 F (z) = F (z 1/2 ), then F is p-homogeneous and satisfies the hypotheses of Lemma 31 with p and t 0 Write F (re it ) = r p f( pt), where We can get f(t) = B p cos(t/p) p a cos(t/p) + b sin(t/p) p (5) f( p t 0 ) = B p cos t 0 p cos(t 0 θ) p, (6) f ( p t 0 ) = f +( p t 0 ) = B p cos t 0 p cos t 0 where tan θ = b/a By the proof of Lemma 31, let sin t 0 + cos(t 0 θ) p cos(t 0 θ) h(x) = f( p t 0 ) cos(x p t 0 ) + f +( p t 0 ) sin(x p t 0 ), sin(t 0 θ), then h(x) f(x) for all x in a neighborhood of [ p t 0, p t 0 + ] Let T = re it : r > 0, t 0 < t < t 0 + p }, and H(reit ) = r p h( pt) for t 0 < t < t 0 +, let p H(z) = H(re G(z) it ) = r p [ = f( p t 0 ) cos( pt p t 0 ) + f +( p t 0 ) sin( pt p t 0 )] if z T, F (z) = r p (B p cos t 2 p a cos t + b sin t 2 2 p ) if z / T So G is subharmonic and G(z) F (z) Now let G(z) = G(z 2 ) Clearly G is p- homogeneous and satisfies G(z) F (z) for all z C, so we get (4) Since z 2 is holomorphic, G(z) is also subharmonic The function G(z) = G(re it ) is -periodic, and by (5), (6) we have for 0 < t < : r p cos t 0 p [B p cos(p(t t 0 ) + t 0 ) cos(t 0 θ) p cos(p(t t 0 ) + t 0 θ)], cos t 0 cos(t 0 θ) G(z) = if z T or e i z T, r p (B p cos t p cos(t θ) p ), if both z and e i z / T, where tan θ = b/a Since z 0 = re it 0, it is equivalent to B p ez 0 p 1 sgn(ez 0 )e[( z z 0 ) p z 0 ] aez 0 + bimz 0 p 1 G(z) = sgn(aez 0 + bimz 0 )(ae[( z z 0 ) p z 0 ] + bim[( z z 0 ) p z 0 ]), if z T or e i z T, B p ez p aez + bimz p, if both z and e i z / T This completes the proof of the lemma If p = 2, then obviously 4 Proof of Theorem 1 ai + bh 2 L 2 () L 2 () = a2 + b 2 = B 2, so we can assume p 2 Consider the holomorphic extension of f(x) + ih(f)(x) on the upper half space given by u(z) + iv(z) = i + f(t) z t dt, u, v real-valued

5 ON THE NOM OF THE OPEATO ai + bh 5 Let G(z) be given by Lemma 32, our next step is to use Lemma 32 and replace z with h(z) = u(z) + iv(z) Since h(z) is holomorphic and G is subharmonic, it follows that G(h(z)) is subharmonic on the upper half space We note that ([4]) u(x + iy) + v(x + iy) By Lemma 32, we have that G(z) C z p, hence So (7) G(h(z)) C f 1 + x + y G(h(z)) C h(z) p C( u(z) + v(z) ) p C p f (1 + x + y ) p where z = x + iy The boundary values of G(h(z)) are G(h(x + i0)) The following part of the argument is based on [6] For > 100, consider the circle with center (0, ) and radius = 1, denote by and C U = i + e iφ : /4 φ 5/4} C L = i + e iφ : 5/4 φ 7/4} It follows from the subharmonicity of G(h(z)) that (8) G(h(z))ds + G(h(z))ds 2 G(h(i)) C U Clearly (7) implies that (9) G(h(i)) C 0 as, (1 + ) p and that (10) C U C L G(h(z))ds C 0 as (1 + ) p Letting in (8), and using (9), (10), we obtain (11) G(h(x))dx 0 provided (12) C L G(h(z))ds G(h(x))dx as To show (12), using parametric equations, the integral G(h(z))ds is equal to C L 2/2 ( )) (13) G h (x + i i 1 x2 dx 2/2 2 1 x2 2 In view of (7), for all > 100, the integrand in (13) is bounded by the integrable function C f (1 + x ) p since 1 x2 is bounded from below by 1/2 in the range 2

6 6 YONG DING, LOUKAS GAFAKOS, AND KAI ZHU 1 of integration Then the Lebesgue dominated convergence theorem gives that (13) converges to (14) G(h(x))dx as Then replace z with h(x) = f(x) + ih(f)(x) in (4) and integrate (4) with respect to x, we get (15) af(x) + bh(f)(x) p dx B p f(x) p dx G(h(x))dx So by (11) we obtain (16) (ai + bh)f p L p () B p f p L p () 5 The sharpness of the constant B p To deduce that the constant B p is sharp, we need to show (17) ai + bh p L p () L p () B p The proof of (17) relies on finding suitable analytic functions in H p of the upper half space that will serve as approximate extremals Unlike the case of the circle, where the functions ( (1 + z)/(1 z) ) 1/p ɛ in H p of the unit disc serve this purpose for all 1 < p < (see [7]) as ɛ 0, we need to consider the cases p < 2 and p > 2 separately Case 1: 1 < p < 2 ecall the analytic function used in [5] (also used in [12]), ( F (z) = (z + 1) 1 i z + 1 ) 2γ/ z 1 on the upper half plane If 1 < p < 2 and /2p < γ < /2p, where p = p/(p 1), then F (z) belongs to H p (the Hardy Spaces) in the upper half plane Let f γ (x) = 1 ( ) 2γ/ x + 1 cos γ, x + 1 x 1 then we have F (x + i0) = f γ (x) + i 1 ( x+1 x+1 1 x+1 2γ/ x 1 ) sin γ when x > 1, ( x+1 2γ/ x 1 ) sin γ when x < 1 and since this is equal to the boundary values of a holomorphic function on the upper half plane, it follows that (tan γ)f γ (x) when x > 1, H(f γ )(x) = (tan γ)f γ (x) when x < 1, So consider a function of the form g γ = αf γ + βh(f γ ), where α, β Notice that H(g γ ) = αh(f γ ) βf γ, and the function ( x 1 2γ x + 1 2γ 1 ) p is integrable over the

7 ON THE NOM OF THE OPEATO ai + bh 7 entire line since /2p < γ < /2p, so for fixed α, β we have (ai + bh)g γ p L p () g γ p = (aα bβ)f γ + (aβ + bα)h(f γ ) p dx L p () αf γ + βh(f γ ) p dx = (aα bβ) + (aβ + bα) tan γ p A γ + (aα bβ) (aβ + bα) tan γ p B γ α + β tan γ p A γ + α β tan γ p B γ where A γ = x >1 f γ(x) p dx, B γ = x <1 f γ(x) p dx It is easy to get A γ B γ, so (ai + bh)g γ p L p () (18) and g γ p L p B () γ (aα bβ) + (aβ + bα) tan γ p + (aα bβ) (aβ + bα) tan γ p, A γ α + β tan γ p + α β tan γ p (ai + bh)g γ p L p () (19) Now we argue that g γ p L p A () γ (aα bβ) + (aβ + bα) tan γ p + (aα bβ) (aβ + bα) tan γ p B γ α + β tan γ p + α β tan γ p A γ (20) lim = 1 γ 2p B γ In fact, by the second mean value theorem for definite integrals, there exists (, 1) where 0 < < 1 so that Since x 2γp x + 1 2γp p 2 x + 1 p p dx dx = 1 2 p p p dx + dx + p x 1 dx as γ, we get 2p lim γ 2p x p 2 p x 1 dx = 1 p x 1 dx Clearly this implies (20) Combining (18) with (19) we obtain p p ai + bh p L p () L p () ( (aα bβ) + (aβ + bα) tan γ p + (aα bβ) (aβ + bα) tan γ p max α,β α + β tan γ p + α β tan γ p where γ = 2p Letting x = α/β in (1), we see that (17) holds, therefore the constant B p is sharp for 1 < p < 2 dx dx ) 1 p,

8 8 YONG DING, LOUKAS GAFAKOS, AND KAI ZHU 1 Case 2: 2 < p < In this case, the function ( x 1 2γ x + 1 2γ 1 ) p used in Case 1 fails to be integrable over the entire line So we consider the following analytic function: F (z) = ( i(z 2 1) ) 2γ, which belongs to H p in the upper half plane when 2 < p < and /4p < γ < /2p Let f γ (x) = x + 1 2γ x 1 2γ cos γ, then we have It follows that F (x + i0) = f γ (x) + i H(f γ )(x) = x + 1 2γ x 1 2γ sin γ when x > 1, x + 1 2γ x 1 2γ sin γ when x < 1 (tan γ)f γ (x) when x < 1, (tan γ)f γ (x) when x > 1, Consider the function g γ = αf γ + βh(f γ ), where α, β Notice that the function ( x 1 2γ x + 1 2γ ) p is integrable over the entire line since /4p < γ < /2p, so for fixed α, β we have (ai + bh)g γ p L p () g γ p L p () = (aα bβ) + (aβ + bα) tan γ p A γ + (aα bβ) (aβ + bα) tan γ p B γ α + β tan γ p A γ + α β tan γ p B γ where A γ = x <1 f γ(x) p dx, B γ = x >1 f γ(x) p dx It is easy to see A γ B γ, so (ai + bh)g γ p L p () (21) and g γ p L p B () γ (aα bβ) + (aβ + bα) tan γ p + (aα bβ) (aβ + bα) tan γ p, A γ α + β tan γ p + α β tan γ p (ai + bh)g γ p L p () (22) g γ p L p A () γ (aα bβ) + (aβ + bα) tan γ p + (aα bβ) (aβ + bα) tan γ p B γ α + β tan γ p + α β tan γ p By the second mean value theorem for definite integrals, there exists (, 1) where 0 < < 1 so that = 4γp x 4γp 2 2 x + 1 x + 1 x 1 dx x 1 dx x + 1 x 1 x + 1 x 1 dx + dx + x + 1 x 1 dx dx x + 1 x 1

9 Since This implies 2γp x + 1 ON THE NOM OF THE OPEATO ai + bh 9 x 1 dx as γ, we have 2p lim γ 2p x 4γp Combining (21) and (22) we obtain 2 x + 1 x 1 dx = 1 x + 1 x 1 dx B γ lim = 1 γ 2p A γ ai + bh p L p () L p () ( (aα bβ) + (aβ + bα) tan max α,β 2p p + (aα bβ) (aβ + bα) tan 2p p α + β tan 2p p + α β tan 2p p ) 1 p Letting x = α/β in (1), so (17) holds, therefore the constant B p is sharp for 2 < p < eferences [1] L De Carli, E Laeng, Sharp L p estimates for the segment multiplier, Collect Math 51 (2000), [2] T W Gamelin, Uniform Algebras and Jensen Measures, London Math Soc Lecture Note Series, Vol 32, Cambridge Univ Press, Cambridge New York, 1978 [3] L Grafakos, Classical Fourier Analysis, 3r ed, GTM 249, Springer New York, 2014 [4] L Grafakos, Best bounds for the Hilbert transform on L p ( 1 ), Math es Let 4 (1997), [5] T Gokhberg, N Y Krupnik, Norm of the Hilbert transformation in the L p space, Funktsional Analiz i Ego Prilozhen 2 (1968), [6] L Grafakos, T Savage, Best bounds for the Hilbert transform on L p ( 1 ); A corrigendum, Math ess Let, 22 (2015), [7] B Hollenbeck, N J Kalton, I E Verbitsky, Best constants for some operators associated with the Fourier and Hilbert transforms, Studia Math 157 (2003), [8] B Hollenbeck, I E Verbitsky, Best constants for the iesz projection, J Funct Anal 175 (2000), [9] E Laeng, emarks on the Hilbert transform and on some families of multiplier operators related to it, Collect Math 58 (2007), [10] B Ya Levin, Lectures on entire functions, Transl Math Monogr 150, Amer Math Soc, Providence, I, 1996 [11] J Marcinkiewicz, A Zygmund, Quelques inégalités pour les opérations linéaires, Fund Math 32 (1939), [12] S K Pichorides, On the best values of the constants in the theorems of M iesz, Zygmund and Kolmogorov, Studia Math, 44 (1972), [13] A Zygmund, Trigonometric Series, Vol 2, Cambridge Univ Press, London, UK, 1968

10 10 YONG DING, LOUKAS GAFAKOS, AND KAI ZHU 1 Yong Ding, Laboratory of Mathematics and Complex Systems, School of Mathematical Sciences, Beijing Normal University, Ministry of Education of China, Beijing , China address: dingy@bnueducn Loukas Grafakos, Department of Mathematics, University of Missouri, Columbia MO 65211, USA address: grafakosl@missouriedu Kai Zhu, School of Mathematical Sciences, Beijing Normal University, Beijing , China address: kaizhu0116@126com