Lecture notes 4.

INTERPOLATION OF OPERATORS - RIESZ-THORIN

SHOBHA MADAN

1. Riesz-Thorin Interpolation Theorem

The setting: (X,µ) and (Y, ν) are two measure spaces, on whichwe consider the (complex-valued) function spaces Lp(X) and Lp(Y )for 1 ≤ p ≤ ∞. Let D denote a dense subspace of Lp(X) and letT : D → M(Y ) be a linear mapping, where M(Y ) is the space ofν-measurable functions on Y . In the following, we will let D be thelinear space of simple functions on X

Theorem 1.1. Let 1 ≤ pj, qj ≤ ∞, j = 1, 2, and suppose there existconstants C1, C2 > 0 such that

‖Tf‖qj ≤ Cj‖f‖pj , j = 1, 2

Then for every t ∈ [0, 1], if pt and qt are given by 1/pt = (1−t)/p1+t/p2and 1/qt = (1− t)/q1 + t/q2, we have

‖Tf‖qt ≤ Ct‖f‖pt , ∀f ∈ D

where Ct is a constant satisfying Ct ≤ C1−t1 Ct

2. In other words, as afunction of t, the operator norm is a convex function.

First, some remarks:

(1) (Historical.) This theorem was first proved by M. Riesz in1926 for real Lp spaces, using real variable methods, but withthe restriction that pj ≤ qj. Thorin (1939) extended the resultas stated above for complex spaces. (Littlewood called Thorin’sproof ”the most impudent idea in Mathematics”. Later, in 1964,Calderon developed the method of Thorin into the ComplexMethod of Interpolation. (I strongly recommend the chapteron this in Katznelson’s book).

(2) The theorem also holds for sublinear operators. Svante Jansonused the Hahn-Banach Theorem to linearize the operators, weexplain the idea in these notes.

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2 SHOBHA MADAN

(3) The proof of the Riesz-Thorin theorem will use a classical the-orem from complex analysis called Hadamard’s three lines the-orem. This is one of a class of results called the Phragmen-Lindelof theorems, which are analogues of the maximum mod-ulus theorem for unbounded domains.

Theorem 1.2 (Hadamard’s three lines theorem). Let Ω = z ∈C : 0 < |z| < 1. Suppose f : Ω → C is a bounded continuousfunction which is holomorphic on Ω. Let M : [0, 1] → R be given byM(t) = sup|f(t+ iy)| : y ∈ R, then logM(t) is a convex function oft. In particular,

M(t) ≤M(0)1−tM(1)t,

Proof outline: Without loss of generality, we can choose a = 0, b = 1,and by passing to f1(z) = M(0)z−1M(1)−zf(z), we may also assumethat M(0) = M(1) = 1. Next, use an auxiliary function such as

hε(z) = eεz2

and let Fε(z) = f1(z)hε(z). Then Fε(z) → f1(z) as ε → 0

and |Fε(z)| = |f1(z)|eε(x2−y2) → 0 as |y| → ∞. Now for large y, usethe Maximum Modulus Principle for bounded domains to conclude that|Fε(z)| ≤ 1. Then we get |f(z)| ≤ 1 and so |f(x+iy)| ≤M(0)1−xM(1)x.

Proof of Riesz-Thorin Interpolation Theorem: Let f =∑n

1 akχEk

be a simple function on X, where Ek’s are mutually disjoint sets, eachof finite measure. We need to estimate ‖Tf‖Lqt (Y ). We do this byduality; take a simple function g =

∑m1 bkχFk

on Y . We will provethat ∣∣ ∫

Y

Tf(y)g(y)dν(y)∣∣ ≤ C1−t

1 Ct2‖f‖Lpt (X)‖g‖Lq′t (Y )

.

where 1/qt + 1/q′t = 1. Without loss of generality, we assume that‖f‖Lpt (X), ‖g‖Lq′t (Y )

≤ 1. Define two functions on the strip Ω by

φ(z) =1− zp1

+z

p2, ψ(z) =

1− zq1

+z

q2

Note that φ(t) = 1/pt and ψ(t) = 1/qt. We consider four cases:

Case 1. pt = ∞, qt = 1. In this case one of the pairs (p1, q1) or(p2, q2) has to be (∞, 1), and so there is nothing to prove.

Case 2. pt < ∞, qt > 1. This is the heart of the matter. We willdefine a holomorphic function on Ω, such that

∫YTf(y)g(y)dν(y) is one

of its values. To get an estimate on the modulus of its value, we useHadamard’s three lines theorem, given the bounds on the boundary ofΩ.

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Define

fz =n∑1

|aj|φ(z)pteiθjχEj

gz =m∑1

|bj|(1−ψ(z))q′teiηjχFj

where θj = arg aj and ηj = arg bj. Let

F (z) =

∫Y

Tfz(y)gz(y)dν(y), z ∈ Ω

Clearly F is a bounded continuous function on Ω and is also holomor-phic on Ω. Further, for s ∈ R, we have the estimate

|F (is)| =∣∣ ∫

Y

Tfis(y)gis(y)dν(y)∣∣

≤ ‖Tfis‖Lq1 (Y )‖gis‖Lq′1 (Y )

≤ C1‖fis‖Lp1 (X)‖gis‖Lq′1 (Y )

Since the Ej’s are disjoint, we have

|fis(y)| =n∑1

|aj|pt/p1χEj(y) = |f(y)|pt/p1

and similarly|gis| = |g(y)|q′t/q′1

It follows that‖fis‖p1Lp1 (X) = ‖f‖ptLpt (X) ≤ 1

and‖gis‖

q′1

Lq′1 (Y )= ‖f‖q

′t

Lq′t (Y )≤ 1

Together this implies that

|Fis| ≤ C1 ∀s ∈ RA similar computation will give that

|F1+is| ≤ C2 ∀s ∈ RThe three lines theorem immediately yields

|F (t)| ≤ C1−t1 Ct

2.

Case 3. pt <∞, qt = 1. In this case take gz = g, ∀z ∈ Ω.

Case 4. pt =∞, qt > 1. Now take fz = f, ∀z ∈ Ω.

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2. Some Applications

2.1. The Hausdorff-Young Inequality. Recall that the Fourier trans-form of a function f ∈ L1(Rn) is defined as

F(f)(ξ) ≡ f(ξ) =

∫Rn

f(x)e2πix.ξdx.

The operator f → Ff is well-defined for all f ∈ L1 ∩ L∞(Rn), whichis a dense subspace in each Lp(Rn), 1 ≤ p <∞. Clearly,

‖F(f)‖∞ ≤ ‖f‖1and by Plancherel Theorem,

‖F(f)‖2 ≤ ‖f‖2, ∀f ∈ L1 ∩ L2(Rn)

So by interpolating between the pairs (1,∞) and (2, 2), we see that if1 ≤ p ≤ 2, then

‖F(f)‖p′ ≤ ‖f‖p, ∀f ∈ L1 ∩ L2(Rn)

where 1/p + 1/p′ = 1. Hence the Fourier transform extends as abounded operator on Lp(Rn) for 1 ≤ p ≤ 2 to Lp

′(Rn).

2.2. Young’s Inequality. Using Minkowski’s integral inequality andHolder’s inequality, we get the following two convolution inequalities

‖f ∗ g‖p ≤ ‖f‖1‖g‖p, f ∈ L1, g ∈ Lp

‖f ∗ g‖∞ ≤ ‖f‖p‖g‖p′ , f ∈ Lp, g ∈ Lp′

Now if p > 1 is fixed, then each g ∈ Lp(Rn) defines a convolutionoperator by Tg(f) = f ∗ g, which is bounded from L1 → Lp and alsofrom Lp

′ → L∞. By interpolation, if 1 < pt < p′ is given by 1/pt =(1− t) + t/p′ = 1− t/p,, then

‖Tg(f)‖rt = ‖f ∗ g‖rt ≤ ‖f‖pt‖g‖p, f ∈ L1, g ∈ Lp

where rt is given by 1/rt = (1− t)/p. So 1/rt = 1/p+ 1/pt − 1.

2.3. The Conjugation Operator. We can now prove

Theorem 2.1 (M.Riesz). The operator H (Hilbert transform) is boundedon Lp(T), ∀1 < p <∞

Proof. We have already proved the inequalities for every even integervalue of p, and so by the Riesz-Thorin interpolation, the result holdsfor ever p ≥ 2. For p ≤ 2, we use duality to estimate the norms.

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3. The sublinear case

We will now show that the Riesz-Thorin theorem also holds for sub-linear operators. The proof uses the Hahn-Banach theorem to linearize.To understand this, we first assume that the measure space (Y, ν) iscountable with ν a discrete measure. Consider the theorem with theword ’linear’ replaced by ’sublinear’ for the operator T . Fix an f0 ∈ D.We need to estimate ‖Tf0‖qt . For each y ∈ Y , consider a seminorm onD defined by

f → |Tf(y)|By Hahn-Banach Theorem, there exists a linear functional φy on Dsuch that

|〈f, φy〉| ≤ |Tf(y)|for all f ∈ D, and

|〈f0, φy〉| = |Tf0(y)|Now define a linear operator U on D by

Uf(y) = 〈f, φy〉Observe that U satisfies the hypothesis of Riesz-Thorin Theorem, and‖Tf0‖qt = ‖Uf0‖qt .

For general measure spaces, evaluation of a function at a point is notmeaningful. We can do the following:

Fix f0 ∈ D and let ε > 0. Let

Ej = y ∈ Y : jε ≤ |Tf0(y)| < (j + 1)ε, j = 0, 1, 2, ...

The Ej’s are mutually disjoint measurable subsets of Y . For each j,the map

f →∫Ej

|Tf(y)|dν(y)

defines a seminorm on D, so as before, there exists a linear functionalφj such that

|〈f, φj〉| ≤∫Ej

|Tf(y)|dν(y)

|〈f0, φj〉| =

∫Ej

|Tf0(y)|dν(y)

Now define a linear operator U by

Uf =∑j

1

ν(Ej)〈f, φj〉χEj

.

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Note that this is a finite sum. Then

|Uf0(z)| =∑j

1

ν(Ej)(

∫Ej

|Tf0(y)|dν(y))χEj(z)

= E(|Tf0|/FRj)Hence

‖Uf0‖qt = ‖E(|Tf0|/FRj)‖qt → ‖Tf0‖qtas ε→ 0, by standard approximation arguments.

Department of Mathematics and Statistics, I.I.T. Kanpur, India