Proofs of Lemmas 1.3.5 and 1.3.8

1.3 Interpolation

1.3.4 Proofs of Lemmas 1.3.5 and 1.3.8

Proof of Lemma 1.3.5.Define analytic functions

G(z) =F(z)(B^1−z₀ B^z₁)⁻¹ and G_n(z) =G(z)e^(z2−1)/n.

Since F is bounded on the closed unit strip andB^1−z₀ B^z₁ is bounded from below, we conclude thatGis bounded by some constantM on the closed strip. Also,Gis bounded by one on its boundary. Since

|G_n(x+iy)| ≤Me^−y2/ne^(x2−1)/n≤Me^−y2/n,

we deduce thatG_n(x+iy)converges to zero uniformly in 0≤x≤1 as|y| →∞. Se- lecty(n)>0 such that for|y| ≥y(n),|G_n(x+iy)| ≤1 uniformly inx∈[0,1]. By the maximum principle we obtain that|G_n(z)| ≤1 in the rectangle[0,1]×[−y(n),y(n)];

hence|G_n(z)| ≤1 everywhere in the closed strip. Lettingn→∞, we conclude that

|G(z)| ≤1 in the closed strip.

Having disposed of the proof of Lemma 1.3.5, we end this section with a proof of Lemma 1.3.8.

Proof of Lemma 1.3.8.Recall the Poisson integral formula U(z) = 1

2π Z +π

−π

U(Re^iϕ) R²−ρ²

|Re^iϕ−ρe^iθ|²dϕ, z=ρe^iθ, (1.3.34) which is valid for a harmonic functionUdefined on the unit diskD={z: |z|<1}

when|z|<R<1. See Rudin [229, p. 258].

Consider now a subharmonic functionu onD that is continuous on the circle

|ζ|=R<1. WhenU=u, the right side of (1.3.34) defines a harmonic function on the set{z∈C: |z|<R}that coincides withuon the circle|ζ|=R. The maximum principle for subharmonic functions (Rudin [229, p. 362]) implies that for|z|<R<

1 we have u(z)≤ 1

2π Z +π

−π u(Re^iϕ) R²−ρ²

|Re^iϕ−ρe^iθ|²dϕ, z=ρe^iθ. (1.3.35) This is valid for all subharmonic functionsuonDthat are continuous on the circle

|ζ|=Rwhenρ<R<1.

It is not difficult to verify that h(ζ) = 1

πilog

i1+ζ 1−ζ

is a conformal map fromDonto the stripS= (0,1)×R. Indeed,i(1+ζ)/(1−ζ) lies in the upper half-plane and the preceding complex logarithm is a well defined holomorphic function that takes the upper half-plane onto the stripR×(0,π). Since F◦his a holomorphic function onD, log|F◦h|is a subharmonic function onD.

Applying (1.3.35) to the functionz7→log|F(h(z))|, we obtain log|F(h(z))| ≤ 1

2π Z _+π

−π

log|F(h(Re^iϕ))| R²−ρ²

R²−2ρRcos(θ−ϕ) +ρ²dϕ (1.3.36) whenz=ρe^iϕand|z|=ρ<R. Observe that when|ζ|=1 andζ 6=±1,h(ζ)has real part zero or one. It follows from the hypothesis that

log|F(h(ζ))| ≤Ae^a|Imh(ζ)|=Ae^a

Im_πi¹ log

i^1+ζ_1−ζ

≤Ae

a π

log|^1+ζ_1−ζ|

.

Therefore, log|F(h(ζ))| is bounded by a multiple of |1+ζ|^−a/π+|1−ζ|^−a/π, which is integrable over the set|ζ|=1, sincea<π. Fix nowz=ρe^iθ withρ<R and letR→1 in (1.3.36). The Lebesgue dominated convergence theorem gives that

log|F(h(ρe^iθ))| ≤ 1 2π

Z _+π

−π log|F(h(e^iϕ))| 1−ρ²

1−2ρcos(θ−ϕ) +ρ²dϕ. (1.3.37) Settingx=h(ρe^iθ), we obtain that

ρe^iθ=h⁻¹(x) =e^πix−i

e^πix+i=−i cos(πx) 1+sin(πx)=

cos(πx) 1+sin(πx)

e^−i(π/2), from which it follows thatρ = (cos(πx))/(1+sin(πx))and θ=−(π/2), when 0<x≤¹₂, whileρ=−(cos(πx))/(1+sin(πx))andθ=π/2, when¹₂≤x<1. In either case we easily deduce that

1−ρ²

1−2ρcos(θ−ϕ) +ρ²= sin(πx) 1+cos(πx)sin(ϕ). Using this we write (1.3.37) as

log|F(x)| ≤ 1 2π

Z π

−π

sin(πx)

1+cos(πx)sin(ϕ)log|F(h(e^iϕ))|dϕ. (1.3.38) We now change variables. On the interval[−π,0)we use the change of variables iy=h(e^iϕ)or, equivalently,e^iϕ=−tanh(πy)−isech(πy). Observe that asϕranges from−π to 0,yranges from+∞to−∞. Furthermore,dϕ=−πsech(πy)dy. We have

1 2π

Z 0

−π

sin(πx)

1+cos(πx)sin(ϕ)log|F(h(e^iϕ))|dϕ

=1 2

Z ∞

−∞

sin(πx)

cosh(πy)−cos(πx)log|F(iy)|dy.

(1.3.39)

On the interval(0,π]we use the change of variables 1+iy=h(e^iϕ)or, equivalently, e^iϕ=−tanh(πy) +isech(πy). Observe that asϕranges from 0 toπ,yranges from

−∞to+∞. Furthermore,dϕ=πsech(πy)dy. Similarly, we obtain 1

2π Z _π

0

sin(πx)

1+cos(πx)sin(ϕ)log|F(h(e^iϕ))|dϕ

= 1 2

Z +∞

−∞

sin(πx)

cosh(πy) +cos(πx)log|F(1+iy)|dy.

(1.3.40)

Now add (1.3.39) and (1.3.40) and use (1.3.38) to conclude the proof.

Exercises

1.3.1.Generalize Theorem 1.3.2 to the situation in whichT is quasilinear, that is, it satisfies for someK>0,

|T(λf)|=|λ| |T(f)| and |T(f+g)| ≤K(|T(f)|+|T(g)|), for allλ∈C, and all f,gin the domain ofT. Prove that in this case, the constantA in (1.3.7) can be taken to beKtimes the constant in (1.3.8).

1.3.2.Let 1<p<r≤∞and suppose thatT is a linear operator that mapsL¹to L^1,∞with normA₀andL^rtoL^rwith normA₁. Prove thatTmapsL^ptoL^pwith norm at most

8(p−1)^−1pA

1p−1 r 1−1 r

0 A

1−1 p 1−1 r

1 .

Hint:First interpolate betweenL¹andL^rusing Theorem 1.3.2 and then interpolate betweenL^p+12 andL^rusing Theorem 1.3.4.

1.3.3.Let 0<p₀<p<p₁≤∞and letT be an operator as in Theorem 1.3.2 that also satisfies

|T(f)| ≤T(|f|), for all f ∈L^p0+L^p1.

(a) Ifp0=1 andp1=∞, prove thatT mapsL^ptoL^pwith norm at most p

p−1A

1 p

0A¹⁻

1 p

1 .

(b) More generally, ifp₀<p<p₁=∞, prove that the norm ofT fromL^ptoL^pis at most

p^1+1p

B(p₀+1,p−p0) p₀^p0(p−p₀)^p−p0

¹_p A

p0 p

0 A¹⁻

p0 p

1 ,

whereB(s,t) =^R₀¹x^s−1(1−x)^t−1dxis the usual Beta function.

(c) When 0<p₀<p₁<∞, then the norm ofT fromL^ptoL^pis at most

min

0<λ<1p^1p

B(p−p₀,p₀+1) (1−λ)^p0 +

p1−p+1 p1−p

λ^{p1 1}_p

A

1p−1 p1 p10−1 p1

0 A

p10−1 p p10−1 p1

1 .

Hint:Parts (a), (b): The hypothesis |T(f)| ≤T(|f|)reduces matters to nonneg- ative functions. For f ≥0 and for fixed α >0 write f = f₀+f₁, where f₀= f−λ α/A₁when f ≥λ α/A₁and zero otherwise for some 0<λ <1. Then we have that|{|T(f)|>α}| ≤ |{|T(f₀)|>(1−λ)α}|. Whenp₁<∞writef=f₀+f₁, where f₀= f−δ α when f ≥δ αand zero otherwise. Use that|{|T(f)|>α}| ≤

|{|T(f₀)|>(1−λ)α}|+|{|T(f₁)|>λ α}|and optimize overδ >0.

1.3.4.Let 0<α,β <π. LetT_zbe a family of linear operators defined on the strip S_a,b={z∈C: a≤Rez≤b}that is analytic on the interior ofS_a,b, in the sense of (1.3.22), continuous on its closure, and satisfies for allz∈S_a,b,

e−α|Imz|/(b−a)

log Z

Y

T_z(f)g dν

≤C_f,g<∞.

Let 1≤p0,q₀,p1,q1≤∞. Suppose that Ta+iy mapsL^p0(X)toL^q0(Y)with bound M0(y)andTb+iymapsL^p1(X)toL^q1(Y)with boundM1(y), where

sup

−∞<y<∞

e−β|y|/(b−a)logM_j(y)<∞, j=0,1.

Then fora<t<b,T_t mapsL^p(X)toL^q(Y), where 1

p=

b−t b−a

p0

+

t−a b−a

p1

and 1 q=

b−t b−a

q0

+

t−a b−a

q1

.

1.3.5.(Stein [251]) OnRⁿletK_λ(x₁, . . . ,x_n)be the function π

n−1

2 Γ(λ+1)

Γ(λ+ⁿ⁺¹₂ ) Z +1

−1 e^{2πis(x21+···+x2n)1/2}(1−s²)^λ+n−12 ds,

whereλis a complex number. LetT_λbe the operator given by convolution withK_λ. Show thatT_λ mapsL^p(Rⁿ)to itself for Reλ >(n−1)|¹₂−¹_p|.

Hint:Using the result in Appendix B.5, show that when Reλ=0,T_λmapsL²(Rⁿ) to itself with norm 1. Using the estimates in Appendices B.6 and B.7, conclude that T_λ mapsL¹(Rⁿ)to itself with an appropriate constant when Reλ = (n−1)/2+δ (forδ >0) and then appeal to Theorem 1.3.7.

1.3.6.Under the same hypotheses as in Theorem 1.3.7, prove the stronger conclu- sion

Tz(f)

L^q≤B(z) f

L^p

forzin the open stripS= (0,1)×R, where B(t+is) =exp

sin(πt) 2

Z ∞

−∞

logM₀(y) cosh(π(y−s))−cos(πt) + logM₁(y)

cosh(π(y−s)) +cos(πt)

dy

.

Hint:Apply Theorem 1.3.7 to the analytic familyTe_z=Tz+is.

1.3.7.(Yano [294]) Let(X,µ)and(Y,ν)be two measure spaces with µ(X)<∞ andν(Y)<∞. Let T be a sublinear operator that mapsL^p(X)toL^p(Y)for every 1<p≤2 with norm

T

_Lp→Lp≤A(p−1)^−αfor some fixedA,α>0. Prove that for all f measurable onXwe have

Z

Y

|T(f)|dν≤6A(1+ν(Y))¹² _Z

X

|f|(log⁺₂|f|)^αdµ+C_α+µ(X)¹²

,

whereCα=∑^∞_k=1k^α(2/3)^k. This result provides an example ofextrapolation.

Hint:Write

f =

∞

∑

k=0

fχS_k,

whereS_k={2^k≤ |f|<2^k+1} when k≥1 andS₀={|f|<2}. Using H¨older’s inequality and the hypotheses onT, obtain that

Z

Y

|T(fχS_k)|dν≤2Aν(Y)^k+11 2^kk^αµ(S_k)^k+1k

fork≥1. Note that fork≥1 we haveν(Y)^k+11 ≤max(1,ν(Y))¹² and consider the casesµ(S_k)≥3^−k−1andµ(S_k)≤3^−k−1when summing ink≥1. The term with k=0 is easier.

1.3.8.Prove that for 0<x<1 we have 1

2 Z +∞

−∞

sin(πx)

cosh(πy) +cos(πx)dy =x, 1

2 Z +∞

−∞

sin(πx)

cosh(πy)−cos(πx)dy =1−x,

and conclude that Lemma 1.3.8 is indeed an extension of Lemma 1.3.5.

Hint:In the first integral write cosh(πy) = ¹₂(e^πy+e^−πy). Then use the change of variablesz=e^πy.