Interpolation of Operators: Riesz-Thorin, Marcinkiewicz

INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ

MATT ZIEMKE

If we notice that, for 1 < p < r < q < ∞,

Lp ∩ Lq ⊆ Lr ⊆ Lp + Lq then, if we have a linear operator T which is bounded on Lp and on Lq, it’s natural to wonder whether or not T must be bounded on Lr. The theorems of Riesz-Thorin and Marcinkiewicz will provide us with an answer to this question. In this project, we will discuss the interpolation theorems of Riesz-Thorin and Marcinkiewicz along with their proofs and several applications. Our ﬁrst goal is to prove the Riesz-Thorin Theorem but we will need to prove several lemmas beforehand. The Riesz-Thorin Theorem pertains to complex valued functions so this is assumed for all lemmas along with the theorem. Lemma 1.1 (Hadamard’s Three-Line Theorem). Suppose f(z) is a bounded and continuous function on 0 ≤

Mθ = sup |f(θ + iy)| y∈R 1−θ θ Then Mθ ≤ M0 M1 for 0 ≤ θ ≤ 1.

Proof. First, we may assume that M0,M1 > 0 else f = 0 and the proof is trivial. Case 1: M0 = M1 = 1. Since f(z) is bounded on 0 ≤ 0 so that |f(z)| ≤ k on 0 ≤ 0 and deﬁne f(z) F = . 1 + z −k k Further, let Ω = {z : 0 <

Date: December 2011. 1 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE

So, for a ﬁxed > 0, |F(z)| ≤ 1 on δR. So, by the maximum modulus principle, maxz∈R |F(z)| ≤ 1. k Also for z = x + iy ∈ Ω − R, we have |y| > , so, by (1), k k |F (z)| ≤ ≤ ≤ 1. |y| k | |

Hence |F(z)| ≤ 1 for all z ∈ Ω. Then we have that |f(z)| ≤ |1 + z| ≤ 1 + |z|. Since this is true 1−θ θ for all > 0, we have that |f(z)| ≤ 1. So, for 0 ≤ θ ≤ 1, we have Mθ ≤ 1 = M0 M1 . Case 2: M0,M1 > 0. Let f(z) G(z) = 1−

Lemma 1.2. If f ∈ Lp and 1 ≤ p < ∞ then

kfkp = sup |hf, gi| khkp0 =1 R 1 1 where hf, gi = U fgdµ, p + p0 = 1 and h is a simple function with compact support. Proof. First we will establish the validity of the statement when the supremum is taken over all g ∈ Lp0 where kgkp0 = 1. If we let g be such then Z |hf, gi| ≤ |fg|dµ U 1/p 1/p0 Z Z 0 ≤ |f|pdµ |g|p dµ Holder U U = kfkpkgkp0

= kfkp 2 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ

So sup khf, gi| ≤ kfk . Let kgkp0 =1 p

e−i arg f(x)|f(x)|p−1 h(x) = p−1 kfkp

Well,

p0 −i arg f(x) p−1 0 Z e |f(x)| khkp = dµ p0 p−1 U kfkp

1 Z 0 = |f(x)|p (p−1)dµ p0(p−1) kfkp U Z 1 p 0 = p |f(x)| dµ since p = p (p − 1) kfkp U = 1

Also,

Z e−i arg f(x)|f(x)|p−1 hf, hi = f(x) p−1 dµ U kfkp Z |f(x)||f(x)|p−1 = p−1 kfkp Z 1 p = p−1 |f(x)| dµ kfkp U = kfkp

So we have that kfk = sup |hf, gi|. Now we want to show p kgkp0 =1

sup |hf, gi| = sup |hf, gi| khkp0 =1 kgkp0 =1 where g ∈ Lp0 and h is simple with compact support. Clearly

sup |hf, gi| ≤ sup |hf, gi| khkp0 =1 kgkp0 =1

since h ∈ Lp0 for all h. For the other direction, let > 0. Then there exists g0 ∈ Lp0 so that kg0k = 1 and sup |hf, gi| ≤ |hf, g i| + 0 2 kgkp0 =1 3 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE

We are now ready to prove the Riesz-Thorin Interpolation Theorem.

Theorem 1.3 (Riesz-Thorin). Let T be a linear operator and let 1 ≤ p0, p1, q0, q1 ≤ ∞ where p0 6= p1 and q0 6= q1. Suppose T : Lp0 (µ) −→ Lq0 (ν) is bounded with norm M0 and T : Lp1 (µ) −→ L (ν) is bounded with norm M . Let 1 = 1−θ + θ and 1 = 1−θ + θ where 0 < θ < 1. Then q1 1 p p0 p1 q q0 q1 1−θ θ T : Lp(µ) −→ Lq(ν) is bounded with norm M ≤ M0 M1 .

Proof. Case 1: 1 ≤ p0, p1, q0, q1 < ∞ 1−θ θ Let f ∈ Lp where kfkp = 1. We want to prove that kT (f)kq ≤ M0 M1 . By lemma 1.2, it suffices R 1−θ θ to show, for any simple function g with finite support, | U T (f)gdµ| ≤ M0 M1 . Case 1: f is simple with compact support. Pn Pm Let f = j=1 aj1Aj and let g = k=1 bk1Bk where aj, bj ∈ C and (An) and (Bn) are both disjoint collections of subsets of U(not necessarily disjoint from each other) where µ(Aj) < ∞ and ν(B ) < ∞ for all j, k. Let α(z) = 1−z + z and let β(z) = 1−z + z . For any z = x + iy where k p0 p1 q0 q1 0 ≤ x ≤ 1, define

α(z) i arg f(t) fz(t) = |f(t)| α(θ) e and

1−β(z) i arg g(t) gz(t) = |g(t)| 1−β(θ) e

Also, let Z F (z) = T (fz)gzdν V 4 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ

. Claim: F (z) is analytic on 0 <

 n  m ! Z α(z) 1−β(z) X α(θ) i arg aj X 1−β(θ) i arg bk =  |aj| e T (1Aj ) |bk| e 1Bk dν V j=1 k=1 n m α(z) 1−β(z) Z X X α(θ) 1−β(θ) i(arg aj +arg bk) = |aj| |bk| e T (1Aj )1Bk dν j=1 k=1 V

If we recall how α and β are deﬁned then we see that the above is just a linear combination of exponential functions so we actually have that F (z) is entire. From this, we also see that F (z) is bounded on 0 ≤

since T is bounded from Lp0 to Lq0 with norm M0. Further,

Z α(iy) p0 p0 α(θ) i arg f(t) kfiykp0 = |f| e dµ U Z α(iy) p0

= |f| α(θ) dµ U Z = |f|pdµ (∗) U = 1 where we get (*) because

α(iy) p0 α(iy) α(iy) p0 α(iy) <( ) i=( ) <( )p0 |f(z)| α(θ) = |f(z)| α(θ) |f(z)| α(θ) = |f(z)| α(θ)

5 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE and " # α(iy) 1 − iy iy 1 − θ θ −1 < = < + + α(θ) p0 p1 p0 p1 p − ip y + ip y 1 − θ θ −1 1−1 = < 1 1 0 p since + = = p p0p1 p0 p1 p p p = 1 p0p1 p = p0

α(iy) p So < p = p = p. Similarly, kg k 0 = 1 so we have that |F (iy)| ≤ M . Also, α(θ) 0 p0 0 iy q0 0 Z

|F (1 + iy)| = T (f1+iy)g1+iydν V ≤ kT (f )k kg k 0 Holder 1+iy q1 1+iy q1 ≤ kT k kf k kg k 0 p1→q1 1+iy p1 1+iy q1 = M kf k kg k 0 (2) 1 1+iy p1 1+iy q1 Further, q0 q0 Z 1−β(1+iy) 1 1 1−β(θ) kg1+iykq0 = |g(t)| dν 1 V Z < 1−β(1+iy) p0 = |g(t)| 1−β(θ) 1 dν V Z q0 = |g(t)| 1 dν (∗) V 0 q1 = kgk 0 q1 = 1 where we get (*) because " # 1 − β(1 + iy) 1 − (1 + iy) (1 + iy) 1 − θ θ −1 < = < 1 − − 1 − ( + ) 1 − β(θ) q0 q1 q0 q1

1 1 −1 = (1 − )( 0 ) q1 q q0 = 0 q1

|F (θ)| = T (f)gdν V Then Z 1−θ θ T (f)gdν ≤ sup |F (θ + iy)| ≤ M0 M1 V y∈R Since this is true for any simple function g with compact support, we have then that

1−θ θ M = kT kp→q ≤ M0 M1

. Case 2: f ∈ Lp. Since f ∈ Lp there exists a sequence of simple functions with compact support (fn)n≥1 so that |fn| ≤ |f| and fn → f pointwise. Let E = {x : |f(x)| > 1}, let g = fχE and gn = fnχE. Further, let h = f − g and hn = fn − gn. Without loss of generality assume p0 < p1. Then Z p0 p0 kgkp0 = |g| dµ U Z p0 = |f| χEdµ U Z p ≤ |f| χEdµ U Z ≤ |f|pdµ U < ∞

So h ∈ Lp1 . Since fn → f a.e., |fn| ≤ |f| and |f| ∈ Lp, by the dominated convergence theorem we have kfn − fkp → 0 and so kfnkp → kfkp. Similarly, by the dominated convergence theorem we 7 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE

also have that kgn − gkp0 → 0 and lastly, since

|hn| = |fn − gn|

= |fn(1 − χE)|

≤ |f|(1 − χE)

= |f − fχE| = |h| hn → h a.e, and h ∈ Lp1 we have that khn − hkp1 → 0. Since T is bounded from Lp0 to Lq0 and from Lp1 to Lq1 , we then have that kT (gn) − T (g)kq0 → 0 and kT (hn) − T (h)kq1 → 0 and so T (gn) converges to T (g) in measure and T (hn) converges to T (h) in measure. So there exists a subsequence T (gnk ) of T (gn) so that T (gnk ) → T (g) a.e. and there exists a subsequence T (hmk ) of

T (hm) so that T (hmk ) → T (h) a.e. Then

T (fn) = T (gn) + T (hn) → T (h) + T (g) = T (f) so, by Fatou’s lemma, we have 1−θ θ 1−θ θ kT (f)kq ≤ lim inf kT (fn)kq ≤ lim inf M M kfnkq ≤ M M kfkq n→∞ n→∞ 0 1 0 1

Case 2: Some of p0, p1, q0, q1 are ∞. The proof of Case 1 works here simply by interchanging some integrals with supremums. To see the signiﬁcance of the Reisz-Thorin theorem, we will now look at several of its applications. For the ﬁrst one we are using the notation that Z f ∗ g(x) = f(y)g(x − y)dy R .

1 1 1 Theorem 1.4 (Young’s Inequality). Suppose 1 ≤ p, q, r ≤ ∞ and p + q = r + 1. If f ∈ Lp and g ∈ Lq then f ∗ g ∈ Lr and kf ∗ gkr ≤ kfkpkgkq Proof. Case 1: p = 1 and r = q. Then,

kf ∗ gkr = kf ∗ gkq Z = k f(y)g(· − y)dykq R Z ≤ kf(y)g(· − y)kqdy Minkowski R Z = |f(y)|kg(· − y)kqdy R Z = kg(· − y)kq |f(y)|dy R = kgkqkfk1 8 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ

q Case 2: p = q−1 and r = ∞. Then,

kf ∗ gkr = kf ∗ gk∞ Z

= sup f(y)g(x − y)dy x∈R R Z ≤ sup |f(y)g(x − y)|dy x∈R R ≤ sup kfkpkgkp0 x∈R 0 = kfkpkgkq since q = p Case 3: General case. Fix q so that 1 ≤ q ≤ ∞ and ﬁx g ∈ Lq. Let T be the linear operator deﬁned by T (f) = f ∗ g Then, by case 1,

kT (f)k ≤ kgkqkfk1 and, by case 2, we have

kT (f)k∞ ≤ kfkpkgkp0 0 In reference to the statement of the Riesz-Thorin theorem we have p0 = 1, q0 = q, p1 = q , q1 = ∞,M0 = kgkq, and M1 = kgkq. So, by the Riesz-Thorin Theorem we have that

1−t t kT (f)kqt ≤ M0 M1kfkpt = kgkqkfkpt , ∀f ∈ Lpt where 1 1 − t t 1 1 − t t = + , = + ; 0 < t < 1 pt p0 p1 qt q0 q1 And, in our case, this is 1 t 1 1 − t = 1 − t + 0 , = pt q qt q 1 1 1 We want qt = r. If this is the case then the above gives that p + q = r + 1.

For the next application of the Riesz-Thorin theorem we need to ﬁrst have a deﬁnition.

2πikx Deﬁnition 1.5. Let f ∈ L2[0, 1]. The set B = {Ek = e : k ∈ Z} is an orthonormal basis for L2[0, 1] so ∞ X f = hf, EkiEk k=−∞ ∞ ˆ Then the sequence {hf, Eki}k=−∞ is the Fourier Transform of f, denoted f. Note: Z 1 Z 1 2πikx −2πikx hf, Eki = f(x)e dx = f(x)e dx 0 0 9 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE

Theorem 1.6 (Hausdorﬀ-Young Theorem). Suppose p and q are conjugate exponents and 1 ≤ p ≤ 2. If f ∈ Lp[0, 1] then fˆ ∈ `q and kfˆkq ≤ kfkp. Proof. First, a minor discussion is in order. The Riesz-Thorin theorem is about operators from Lp(X, M, µ) to Lq(Y, N, ν) so how can we talk about `q instead? Let ν be the counting measure on Y = Z, i.e., ν(E) = card(E) for all E ⊆ Z. In this sense then we have that Lq = `q. Let T be the ˆ linear operator deﬁned by T (f) = f. First, consider T : L1 −→ `∞. Let f ∈ L1 and k ∈ Z. Then Z 1 Z 1 Z 1 −2πikx −2πikx |hf, Eki| = f(x)e dx ≤ |f(x)||e |dx = |f(x)|dx = kfk1 0 0 0 This is true for all k ∈ Z so

k{hf, Eki}k∈Zk∞ = sup |hf, Eki| ≤ kfk1 k∈Z

So we have that kT (f)k∞ ≤ M0kfk1 where M0 ≤ 1. Further, consider T : L2 −→ `2. Let f ∈ L2. Then !1/2 !1/2 X ˆ 2 X 2 |f(n)| = |hf, Eki| = kfk2 n∈Z n∈Z by Parseval. So we have that kT (f)k2 ≤ M1kfk2 where M1 = 1. By Riesz-Thorin, for 0 ≤ θ ≤ 1, 1−θ θ T : Lp −→ `q is bounded with norm M ≤ M0 M1 ≤ 1 where 1 θ 1 θ = 1 − θ + , = ; 0 ≤ θ ≤ 1 p 2 q 2 1 1 i.e., where 1 ≤ p ≤ 2 and p + q = 1. The next topic of discussion is the Marcenkiewicz Interpolation Theorem along with its proof and an application. First we need to look at several deﬁnitions along with some lemmas. Also, Marcenkiewicz’s theorem is about real-valued Lp spaces so this is assumed for the remainder of this paper.

Deﬁnition 2.1. If f is a measurable function on (X, M, µ), we deﬁne its distribution function λf : (0, ∞) −→ [0, ∞] by λf (α) = µ({x : |f(x)| > α}).

Definition 2.2. If f is a measurable function on X and 0 < p ≤ ∞, we define 1/p p [f]p = sup α λf (α) , 0 < p < ∞;[f]∞ = kfk∞ α>0 and we define weak Lp to be the set of all f ∈ (X, M, µ) where [f]p < ∞. Note: (1) [f]p ≤ kfkp so Lp ⊆ weak Lp. (2) For 0 < p < ∞, [·]p is not a norm as it fails the triangle inequality.

Deﬁnition 2.3. If U and V are vector spaces, a map T : U −→ V is called sublinear if

|T (cf)(x)| = |c||T (f)(x)| for all c ∈ R; and |T (f + g)(x)| ≤ |T (f)(x)| + |T (g)(x)| . 10 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ

Deﬁnition 2.4. Let T be a map from some vector space V of measurable functions on (X, M, µ) to the space of all measurable functions on (Y, N, ν). A sublinear map T is said to be strong type (p, q), for 1 ≤ p, q ≤ ∞, if Lp(µ) ⊆ V , T maps Lp(µ) into Lq(ν) and there exists a constant c > 0 so that

kT (f)kq ≤ ckfkp for all f ∈ Lp(µ). A linear map T is said to be weak type (p, q), for 1 ≤ p ≤ ∞, 1 ≤ q < ∞, if Lp(µ) ⊆ V , T maps Lp(µ) into weak Lp, and there exists a constant c > 0 such that

[T (f)]q ≤ ckfkp for all f ∈ Lp(µ).

Lemma 2.5. If f ∈ Lp, where 1 ≤ p < ∞, then p Z ∞ kfkp p p−1 λf (α) ≤ p , α > 0; and kfkp = p t λf (t)dt α 0 Proof. Well, Z p p kfkp = |f(x)| dµ U Z ≥ |f(x)|pdµ {x∈U:|f(x)|>α} Z ≥ αpdµ {x∈U:|f(x)|>α} = αpµ({x ∈ U : |f(x)| > α}) p = α λf (α)

p kfkp p So, λf (α) ≤ αp . Let A = {(x, s): |f(x)| > s}. Then, Z Z Z |f(x)|p |f(x)|pdµ = dsdµ U U 0 Z = 1A(x, s)dsdµ U×[0,∞) Z = 1A(x, s)dµds Fubini U×[0,∞) Z ∞ = µ({x ∈ U : |f(x)|p > s})ds 0 Z ∞ = µ({x ∈ U : |f(x)|p > tp})ptp−1dt wheres = tp 0 Z ∞ = p tp−1µ({x ∈ U : |f(x)| > t})dt 0 Z ∞ p−1 = p t λf (t)dt 0 11 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE

Lemma 2.6. If f ∈ Lp, where 1 ≤ p < ∞, then (a) Z Z ∞ p p p−1 |f| dµ = λf (α)α + p t λf (t)dt; and {x:|f(x)|>α} α (b) Z Z α p p p−1 |f| dµ = −λf (α)α + p t λf (t)dt {x:|f(x)|≤α} 0 Proof. (a) Let A = {(x, s): |f(x)|p > s} Z Z Z |f(x)|p |f|pdµ = dsdµ {x:|f(x)|>α} {x:|f(x)|>α} 0 Z Z αp Z Z |f(x)|p = dsdµ + dsdµ {x:|f(x)|>α} 0 {x:|f(x)|>α} αp Z Z p = α dµ + 1A(x, s)dsdµ {x:|f(x)|>α} U×[αp,∞) Z p = α µ({x : |f(x)| > α}) + 1A(x, s)dµds U×[αp,∞) Z p p = α λf (α) + µ({x : |f(x)| > s})ds αp Z p p p p−1 p = α λf (α) + µ({x : |f(x)| > t })pt dt where s = t α Z ∞ p p−1 = α λf (α) + p t λf (t)dt α (b) Z Z Z |f|pdµ = |f|pdµ − |f|pdµ {x:|f(x)|≤α} U {x:|f(x)|>α} Z ∞ Z ∞ p−1 p p−1 = p t λf (t)dt − α λf (α) − p t λf (t)dt by Lemma 2.4 and (a) 0 α Z α p p−1 = −λf (α)α + p t λf (t)dt 0 We are now ready to state and prove The Marcenkiewicz Interpolation Theorem. For this, we will denote the set of all µ-measurable functions on a vector space U by M(U, µ).

Theorem 2.7 (Marcenkiewicz). Let 1 ≤ p0 < p1 ≤ ∞ and assume that T : M(U, µ) −→ M(V, ν) is a sublinear operator of weak type (p0, p0) and (p1, p1). Then, for each p0 < p < p1, T is of strong type (p, p).

Proof. Let f ∈ Lp and deﬁne

f(x) if |f(x)| ≤ α g (x) = α 0 if |f(x)| > α 12 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ and 0 if |f(x)| ≤ α h (x) = α f(x) if |f(x)| > α

Claim: gα ∈ Lp1 and hα ∈ Lp0 . Well,

Z Z g p1 Z g p Z kg kp1 = |g |p1 dµ = αp1 α dµ ≤ αp1 α dµ ≤ αp1−p |f|pdµ < ∞ α p1 α U U α U α U Similarly, Z Z h p0 Z h p Z p0 p0 p0 α p0 α p0−p p khαkp0 = |hα| dµ = α dµ ≤ α dµ ≤ α |f| dµ < ∞ U U α U α U

Then f(x) = gα(x) + hα(x) and, since T is sublinear,

|T (f(x)| ≤ |T (gα(x))| + |T (hα(x))| Suppose x ∈ U and |T (f(x))| > α. Then

|T (gα(x))| + |T (hα(x))| > α so either α α |T (g (x))| > or |T (h (x))| > α 2 α 2 so α α {x : |T (f(x))| > α} ⊆ {x : |T (g (x)| > } ∪ {x : |T (h (x)| > } α 2 α 2 and therefore α α λ (α) ≤ λ ( ) + λ ( ) T f T gα 2 T hα 2

Since T is weak (p1, p1), there exists a constant c > 0 so that [T (gα)]p1 ≤ ckgαkp1 , i.e., 1/p Z 1/p 0p1 0 p1 sup α λT gα (α ) ≤ c |g| dµ α0>0 U

αp1 α Z p1 ⇒ λT gα ≤ c1 |g| dµ 2 2 U

α 2 p1 Z p1 ⇒ λT gα ≤ c1 |g| dµ 2 α U 2 p1 Z p1 = c1 |f| dµ α {x:|f(x)|≤α}

Similarly, since T is weak (p0, p0), α 2 p1 Z p0 λT hα ≤ c2 |f| dµ 2 α {x:|f(x)|>α} 13 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE

Then Z ∞ p p−1 kT (f)kp = p α λT f (α)dα 0 Z ∞ Z ∞ p−1 α p−1 α ≤ p α λT gα ( )dt + p α λT hα ( )dα 0 2 0 2 Z ∞ αp−1 Z Z ∞ αp−1 Z p1 p1 p0 p0 ≤ c1p2 p |f| dµdα + c2p2 p |f| dµdα 0 α 1 {x:|f(x)|≤α} 0 α 0 {x:|f(x)|>α} Z ∞ Z α p1 p−p1−1 p1 p1−1 = c1p2 α −λf (α)α + p1 t λf (t)dt dα 0 0 Z ∞ Z ∞ p0 p−p0−1 p0 p0−1 + c2p2 α λf (α)α + p0 t λf (t)dt dα 0 α Z ∞ Z α Z ∞ p1 p−p1−1 p1−1 p0 p−1 ≤ c1p1p2 α t λf (t)dtdα + pc22 α λf (α)dα 0 0 0 Z ∞ Z ∞ p0 p−p0−1 p0−1 + pp0c22 α t λf (t)dtdα 0 α and therefore, Z ∞ Z ∞ p p1 p−p1−1 p1−1 p0 p kT (f)kp = c1p1p2 α t λf (t)dαdt + c22 kfkp 0 t Z ∞ Z t p0 p−p0−1 p0−1 + pp0c22 α t λf (t)dαdt 0 0 Z ∞ Z ∞ p0 p p1 p1−1 p−p1−1 = c22 kfkp + c1p1p2 t λf (t) α dαdt 0 t Z ∞ Z t p0 p0−1 p−p0−1 + pp0c22 t λf (t) α dαdt 0 0 Z ∞ 1 Z ∞ 1 p0 p p1 p1−1 p0 p0−1 = c22 kfkp + c1p1p2 t λf (t)dt + pp0c22 t λf (t)dt 0 p1 − p 0 p − p0 p p p = c3kfkp + c4kfkp + c5kfkp p = c6kfkfp

The case when p1 = ∞ is similar by taking f(x) if |f(x)| ≤ α g (x) = 2c1 α α sign(f(x)) if |f(x)| > α 2c1 2c1 and

hα(x) = f(x) − gα(x) where c1 is a positive constant so that kT (f)k∞ ≤ c1kfk∞.

The last topic of this paper is then to see an application of Marcinkiewicz’s Theorem but ﬁrst we must recall some things about the Hardy-Littlewood maximal function. 14 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ

Deﬁnition 2.8. Let f ∈ L1. Then we deﬁne the Hardy − Littlewood maximal function Mf by 1 Z Mf(x) = sup |f(y)|dy r>0 µ(Br(x)) Br(x) Recall: The Maximal Theorem states that for all f ∈ L1, c Z µ({x : Mf(x) > α}) ≤ |f(x)|dx α R Theorem 2.9. If f ∈ Lp(R), where 1 < p < ∞, then there exists a constant c > 0 so that kMfkp ≤ ckfkp.

Proof. Let f ∈ L1 and let α > 0. Then, by the Maximal Theorem c Z µ({x : Mf(x) > α}) ≤ |f(x)|dx α R i.e., αλMf (α) ≤ ckfk1 and so [Mf]1 ≤ ckfk1 Also, if f ∈ L∞ then

1 Z kMfk = ess sup sup |f(y)|dy ∞ µ(B (x)) x∈R r>0 r Br(x)

1 ≤ ess sup sup kfk∞µ(Br(x)) x∈R r>0 µ(Br(x)) = kfk∞

So, by Marcinkiewicz, for all 1 < p < ∞, there exists cp > 0 so that kMfkp ≤ cpkfkp for all f ∈ Lp.

Another application of Marcinkiewicz is in showing the Hilbert transform is bounded from Lp to Lp when 1 < p < ∞. First, we will look at the deﬁnition.

Deﬁnition 2.10. For f ∈ Lp where 1 < p < ∞ the Hilbert transform of f, denoted Hf, is given by 1 Z f(y) Hf(x) = lim dy. →0 π |x−y|≥ x − y The proof of the following theorem will not be given in this project, however it can be proven with Marcinkiewicz along with some basic properties of the Hilbert transform.

Theorem 2.11. For all 1 < p < ∞, there exists a constant Cp > 0 such that

kHfkp ≤ Cpkfkp for all f ∈ Lp.

While the interpolation theorems of Riesz-Thorin and Marcinkiewicz are interesting themselves, we also see they are very useful tools. They give us a way to prove certain statements which have possibly inﬁnitely many cases by simply considering two of them.

15 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE References

1. L. Grafakos, Classical Fourier Analysis, Graduate Texts in Mathematics, Vol. 249, Springer, New York, 2008.

2. J. Bergh and J. Lofstrom, Interpolation Spaces: An Introduction, A series of Comprehensive Studies in Mathematics, Vol. 223, Springer-Verlag, Berlin Heidelberg New York, 1976

3. G Folland, Real Analysis: Modern Techniques and their Applications, A Wiley-Interscience Series of Texts, John Wiley & Sons,Inc., New York, 1984