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The Schur Lemma

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The Schur Lemma Appendix A The Schur Lemma Schur’s lemma provides sufficient conditions for linear operators to be bounded on Lp. Moreover, for positive operators it provides necessary and sufficient such condi- tions. We discuss these situations. A.1 The Classical Schur Lemma We begin with an easy situation. Suppose that K(x,y) is a locally integrable function on a product of two σ-finite measure spaces (X, μ) × (Y,ν), and let T be a linear operator given by T( f )(x)= K(x,y) f (y)dν(y) Y when f is bounded and compactly supported. It is a simple consequence of Fubini’s theorem that for almost all x ∈ X the integral defining T converges absolutely. The following lemma provides a sufficient criterion for the Lp boundedness of T. Lemma. Suppose that a locally integrable function K(x,y) satisfies sup |K(x,y)|dν(y)=A < ∞, x∈X Y sup |K(x,y)|dμ(x)=B < ∞. y∈Y X Then the operator T extends to a bounded operator from Lp(Y) to Lp(X) with norm − 1 1 A1 p B p for 1 ≤ p ≤ ∞. Proof. The second condition gives that T maps L1 to L1 with bound B, while the first condition gives that T maps L∞ to L∞ with bound A. It follows by the Riesz– − 1 1 Thorin interpolation theorem that T maps Lp to Lp with bound A1 p B p . This lemma can be improved significantly when the operators are assumed to be positive. A.2 Schur’s Lemma for Positive Operators We have the following necessary and sufficient condition for the Lp boundedness of positive operators. L. Grafakos, Modern Fourier Analysis, Graduate Texts in Mathematics 250, 589 DOI 10.1007/978-1-4939-1230-8, © Springer Science+Business Media New York 2014 590 A The Schur Lemma Lemma. Let (X, μ) and (Y,ν) be two σ-finite measure spaces, where μ and ν are positive measures, and suppose that K(x,y) is a nonnegative measurable function on X ×Y.Let1 < p < ∞ and 0 < A < ∞. Let T be the linear operator T( f )(x)= K(x,y) f (y)dν(y) Y and T t its transpose operator T t (g)(y)= K(x,y)g(x)dμ(x). X To avoid trivialities, we assume that there is a compactly supported, bounded, and positive ν-a.e. function h1 on Y such that T(h1) > 0 μ-a.e. Then the following are equivalent: (i) T maps Lp(Y) to Lp(X) with norm at most A. (ii) For all B > A there is a measurable function h on Y that satisfies 0 < h < ∞ ν-a.e., 0 < T(h) < ∞ μ-a.e., and such that p p T t T(h) p ≤ Bp h p . (iii) For all B > A there are measurable functions u on X and v on Y such that 0 < u < ∞ μ-a.e., 0 < v < ∞ ν-a.e., and such that T(up ) ≤ Bvp , T t (vp) ≤ Bup. Proof. First we assume (ii) and we prove (iii). Define u,v by the equations vp = T(h) and up = Bh and observe that (iii) holds for this choice of u and v. Moreover, observe that 0 < u,v < ∞ a.e. with respect to the measures μ and ν, respectively. Next we assume (iii) and we prove (i). For g in Lp (X) we have v(x) u(y) T( f )(x)g(x)dμ(x)= K(x,y) f (y)g(x) dν(y)dμ(x). X X Y u(y) v(x) We now apply Holder’s¨ inequality with exponents p and p to the functions v(x) u(y) f (y) and g(x) u(y) v(x) with respect to the measure K(x,y)dν(y)dμ(x) on X ×Y. Since p 1 v(x) p 1 ( )p ( , ) μ( ) ν( ) ≤ p f y p K x y d x d y B f Lp(Y) Y X u(y) A.2 Schur’s Lemma for Positive Operators 591 and 1 p 1 u(y) p ( )p ( , ) ν( ) μ( ) ≤ p , g x K x y d y d x B g Lp (X) X Y v(x)p we conclude that 1 1 + ( )( ) ( ) μ( ) ≤ p p . T f x g x d x B f Lp(Y) g Lp (X) X Taking the supremum over all g with Lp (X) norm 1, we obtain ( ) ≤ . T f Lp(X) B f Lp(Y) Since B was any number greater than A, we conclude that ≤ , T Lp(Y)→Lp(X) A which proves (i). We finally assume (i) and we prove (ii). Without loss of generality, take here A = 1 and B > 1. Define a map S : Lp(Y) → Lp(Y) by setting p p S( f )(y)= T t T( f ) p p (y). We observe two things. First, f1 ≤ f2 implies S( f1) ≤ S( f2), which is an easy con- sequence of the fact that the same monotonicity is valid for T. Next, we observe that f Lp ≤ 1 implies that S( f )Lp ≤ 1 as a consequence of the boundedness of T on Lp (with norm at most 1). Construct a sequence hn, n = 1,2,..., by induction as follows. Pick h1 > 0on Y as in the hypothesis of the theorem such that T(h1) > 0 μ-a.e. and such that −p p h1Lp ≤ B (B − 1). (The last condition can be obtained by multiplying h1 by a small constant.) Assuming that hn has been defined, we define 1 h + = h + S(h ). n 1 1 Bp n We check easily by induction that we have the monotonicity property hn ≤ hn+1 and the fact that hnLp ≤ 1. We now define h(x)=suph (x)= lim h (x). n →∞ n n n Fatou’s lemma gives that hLp ≤ 1, from which it follows that h < ∞ ν-a.e. Since h ≥ h1 > 0 ν-a.e., we also obtain that h > 0 ν-a.e. 592 A The Schur Lemma Next we use the Lebesgue dominated convergence theorem to obtain that hn → h p p p in L (Y). Since T is bounded on L ,itfollowsthatT(hn) → T(h) in L (X).It p p p t p follows that T(hn) p → T(h) p in L (X). Our hypothesis gives that T maps L (X) p p p t t p to L (Y) with norm at most 1. It follows T T(hn) p → T T(h) p in L (Y). p ( ) → ( ) p( ) Raising to the power p , we obtain that S hn S h in L Y . ( ) → ( ) It follows that for some subsequence nk of the integers we have S hnk S h a.e. in Y. Since the sequence S(hn) is increasing, we conclude that the entire sequence S(hn) converges almost everywhere to S(h). We use this information in conjunction 1 with hn+ = h + S(hn). Indeed, letting n → ∞ in this identity, we obtain 1 1 Bp 1 h = h + S(h). 1 Bp p Since h1 > 0 ν-a.e. it follows that S(h) ≤ B h ν-a.e., which proves the required estimate in (ii). p It remains to prove that 0 < T(h) < ∞ μ-a.e. Since hLp ≤ 1 and T is L bounded, it follows that T(h)Lp ≤ 1, which implies that T(h) < ∞ μ-a.e. We also have T(h) ≥ T(h1) > 0 μ-a.e. A.3 An Example Consider the Hilbert operator ∞ f (y) T( f )(x)= dy, 0 x + y where x ∈ (0,∞). The operator T takes measurable functions on (0,∞) to measurable functions on (0,∞). We claim that T maps Lp(0,∞) to itself for 1 < p < ∞; precisely, we have the estimate ∞ π ( )( ) ( ) ≤ . T f x g x dx f Lp(0,∞) g Lp (0,∞) 0 sin(π/p) To see this we use Schur’s lemma. We take − 1 u(x)=v(x)=x pp . We have that 1 1 ∞ − ∞ − 1 y p − 1 t p 1 −1 1 − T(up )(x)= dy = x p dt = v(x)p (1 − s) p s p 1 ds, 0 x + y 0 1 +t 0 A.3 An Example 593 where last identity follows from the change of variables s =(1 +t)−1. Now an easy calculation yields 1 1 −1 1 − π ( − ) p p 1 = 1 , 1 = , 1 s s ds B p p 0 sin(π/p) π p→ p ≤ so the lemma in Appendix I.2 gives that T L L sin(π/p) . The sharpness of this constant follows by considering the sequence of functions − 1+ε ( )= p χ ( ) hε t t (1,∞) t for ε > 0. To verify the last assertion notice that for x > 1 and 0 < ε < p−1wehave − 1+ε − 1+ε ∞ t p 1 t p T(hε )(x)= dt − dt 0 x +t 0 x +t +ε +ε ∞ − 1 1 − 1 − 1+ε t p − 1+ε x t p = x p dt − x p dt 0 1 +t 0 1 +t +ε ∞ − 1 1 − 1+ε t p − 1+ε x x − 1+ε ≥ x p dt − x p t p dt 0 1 +t x + 1 0 +ε ∞ − 1 − 1+ε t p 1 p = x p dt − . 0 1 +t x + 1 p − 1 − ε Notice that the expression directly after the ≥ sign is nonnegative, and so is the last expression. It follows that − 1+ε ∞ p t p 1 ( ε ) ≥ ε p − . T h Lp(1,∞) dt h L (1,∞) (·)+1 Lp(1,∞) 0 1 +t p − 1 − ε = = ε−1/p Dividing both sides of this inequality by hε Lp(0,∞) hε Lp(1,∞) , and letting ε → 0 we obtain ∞ − 1 ( ε ) p p π T h L (0,∞) t 1 1 liminf ≥ dt = B , = .
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