SCHUR S LEMMA AND BEST CONSTANTS IN WEIGHTED NORM INEQUALITIES. Gord Sinnamon The University of Western Ontario. December 27, 2003

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1 SCHUR S LEMMA AND BEST CONSTANTS IN WEIGHTED NORM INEQUALITIES Gord Sinnamon The University of Western Ontario December 27, 23 Abstract. Strong forms of Schur s Lemma and its converse are roved for mas taking non-negative functions to non-negative functions and having formal adjoints. These results are alied to give best constants in a large class of weighted Lebesgue norm inequalities for non-negative integral oerators. Since general measures are used, norms of non-negative matrix oerators may be calculated by the same method. 1. Introduction Schur s Lemma is rimarily a way of establishing the boundedness of integral oerators with non-negative kernels, or of matrix oerators with non-negative entries, between Lebesgue saces. As a sufficient condition [1, 3, 4, 7, 11] it has been roved in many different forms and alied in a great many situations. In a tyical alication, the lemma deduces the boundedness of the oerator in question from the hyothesis that there exist a ositive function, or sequence, satisfying a certain inequality. The clever choice of such a function, or sequence, then finishes the roof. After a great many clever choices have been made, one begins to susect that there is always some choice that will serve. The converses to Schur s Lemma [2, 3, 5, 7, 8, 9, 1, 13] assert just that. Naturally enough, Schur s Lemma rovides not only boundedness but also an estimate of the norm of the integral or matrix oerator. The various converses generally show that the actual norm can be aroximated arbitrarily closely by such estimates. The question of whether or not the oerator norm can be reached has been addressed for sequences in [2, 9] and for certain integral oerators in [3, 4, 7] Mathematics Subject Classification. Primary 26D15; Secondary 26A33. Key words and hrases. Hardy inequality, Weight, Averaging oerator. Suort from the Natural Sciences and Engineering Research Council of Canada is gratefully acknowledged. 1 Tyeset by AMS-TEX

2 2 GORD SINNAMON In this aer we rove strong forms of Schur s Lemma and its converse for mas between non-negative functions on general measure saces. The mas are only required to have formal adjoints so they include integral oerators with non-negative kernels, non-negative matrix oerators, comosition oerators, and multilication oerators. These results are given in Section 2 but some of the roofs have been deferred to Section 4 where we also give an iterative rocedure for aroximating the norm of such an oerator. For matrix oerators, the rocedure is easily imlemented on a comuter and convergence is raid. The delicate iteration introduced in the roofs of Lemma 4.3 and Theorem 2.5 may be of indeendent interest as a fixed oint result. In Section 3, Schur s Lemma and its converse are alied to establish best constants for a large class of weighted Lebesgue norm inequalities, including essentially all such inequalities for non-negative integral oerators when the Lebesgue index in the domain sace is larger than the index in the codomain. The method of generating inequalities with best constants is quite simle and the calculations can be readily carried out by hand, with a comuter algebra system, or numerically. Several examles are given. Although these secific examles may be of direct interest to the secialist, they are included here simly as alications of Theorem 3.1 and illustrations of the method. The reader is encouraged to start with a favorite ositive oerator, select a domain sace weight, fix a ositive function, and see how simle it is to generate a weighted norm inequality with best constant. A great deal of rogress has been made recently in the understanding of weighted norm inequalities but the focus has been on establishing the finiteness of the constant rather than its best (smallest) ossible value. Boyd s work [3, 4] as well as that of Howard and Sche [7] are excetions to this and do find best constants. There is some overla between Boyd s aroach and this one but the methods and objectives are quite different. Some of the standard roofs of Schur and Gagliardo aear again here in no greater generality than in Howard and Sche s work but in a somewhat different context. Throughout the aer we work with the extended real numbers under the convention that =. The dual of the Lebesgue index is denoted so that 1/ + 1/ = 1. Schur s Lemma and its converse Let L + µ and L + denote the non-negative, extended real valued functions on the measure saces (X, dµ) and (Y, d) resectively. We say that a ma T : L + L + µ has a formal adjoint T : L + µ L + rovided (T f)g dµ = f(t g) d for all f L + T by and g L + µ. For fixed indices and q with 1 < q < we define T = su{ T f L q µ : f L +, f L 1}

3 and the ma S : L + L + by SCHUR S LEMMA AND BEST CONSTANTS 3 Sϕ = (T ((T ϕ) q 1 )) 1. An extremal function for T from L to L q µ is a non-zero function f L + satisfying T f L q µ = T f L <. Although the results of this section aly to any ma T having a formal adjoint we are most interested in the class of non-negative integral oerators. If k(x, t) is a non-negative, extended real valued, µ -measurable function then the mas T k : L + L + µ and Tk : L+ µ L + defined by T k f(x) = Y k(x, y)f(y) d(y) and Tk g(y) = X k(x, y)g(x) dµ(x) are easily seen to be formal adjoints. Moreover, T is just the usual norm of T k considered as a linear transformation from L to L q µ. The (non-linear) oerator S k corresonding to S is S k ϕ(y) = ( Y ( ) q 1 k(x, y) k(x, z)ϕ(z) d(z)) 1 dµ(x). X Mas having formal adjoints inherit ositive homogeneity, additivity, monotonicity, and Hölder s inequality from integration. Lemma 2.1. Suose that T : L + L + µ has a formal adjoint, that a, and that f 1, f 2 L +. Then T (af 1 ) = at f 1, T (f 1 + f 2 ) = T f 1 + T f 2, T f 1 T f 2 whenever f 1 f 2, and if 1 < q < then T (f 1 f 2 ) [T (f q 1 )]1/q [T (f q 2. Also, the formal )]1/q adjoint of T is unique. Proof. Standard arguments show that if g 1, g 2 L + µ and g 1 g dµ g 2 g dµ for all g L + µ then g 1 g 2 µ-almost everywhere. Consequently, if g 1, g 2 L + µ and g1 g dµ = g 2 g dµ for all g L + µ then g 1 = g 2 µ-almost everywhere. Let T be a formal adjoint of T. For all g L + µ, T (af 1 )g dµ = af 1 T g d = a f 1 T g d = at f 1 g dµ so we have T (af 1 ) = at f 1. In just the same way we show that T (f 1 + f 2 ) = T f 1 + T f 2. If f 1 f 2 and g L + µ then T f 1 g dµ = f 1 T g d f 2 T g d = T f 2 g dµ so T f 1 T f 2. The analogue of Hölder s inequality is roved using the ositive homogeneity and additivity of T and the well-known inequality AB (1/q)A q + (1/q )B q for

4 4 GORD SINNAMON A, B but first we must disense with the case where the right hand side is zero. Let g L + µ be suorted on the set where [T (f q 1 )]1/q [T (f q 2 vanishes and write )]1/q g = g 1 + g 2 where T (f q 1 )g 1 = and T (f q 2 )g 2 =. We have = T (f q 1 )g 1 dµ = f q 1 T g 1 d and hence f 1 f 2 T g 1 d =. Similarly, f 1 f 2 T g 2 d =. Putting these together yields T (f 1 f 2 )g dµ = T (f 1 f 2 )g 1 dµ + T (f 1 f 2 )g 2 dµ = f 1 f 2 T g 1 d + f 1 f 2 T g 2 d =. Since the only restriction on g was its suort, we see that T (f 1 f 2 ) vanishes whenever [T (f q 1 )]1/q [T (f q 2 does. )]1/q It remains to establish the analogue of Hölder s inequality on the set where both T (f q q 1 ) and T (f2 ) are ositive. If either is infinite then there is nothing to rove so Now, α 1 [T (f q 1 )] 1/q (, ) and α 2 [T (f q 2 )] 1/q (, ). α 1 α 2 T (f 1 f 2 ) = T (α 1 f 1 α 2 f 2 ) T ((α 1 f 1 ) q /q + (α 2 f 2 ) q /q ) = (α q 1 /q)t (f q 1 ) + (αq 2 /q )T (f q 2 ) = 1/q + 1/q = 1 and we have T (f 1 f 2 ) [T (f q 1 )]1/q [T (f q 2 )]1/q as desired. If T # is also a formal adjoint of T and g L + µ then for all f L +, ft # g d = T fg dµ = ft g d. Thus T # g = T g and so the formal adjoint is unique. Our first version of Schur s Lemma is in the next theorem. Although it alies to any ma having a formal adjoint it is essentially the standard result: If the ositive function ϕ satisfies the aroriate inequality then the ma T is bounded and an estimate of the norm of T is given. Theorem 2.2. Let 1 < q <. Suose that T : L + L + µ has a formal adjoint T : L + µ L + and that there exist an A > and a ositive, -measurable function ϕ which is finite -almost everywhere and satisfies Sϕ Aϕ -almost everywhere in Y. If = q then T A ( 1)/.

5 SCHUR S LEMMA AND BEST CONSTANTS 5 If q < then T A ( 1)/q ϕ ( q)/q. L Proof. Let f L +. The hyotheses on ϕ ensure that ϕ 1/q ϕ 1/q = 1 -almost everywhere. Using the analogue of Hölder s inequality given in Lemma 2.1, we have T f q L = [T (fϕ 1/q ϕ 1/q )] q dµ T (f q ϕ 1 q )(T ϕ) q 1 dµ. Introducing the formal adjoint, T, the last exression becomes f q ϕ 1 q T ((T ϕ) q 1 ) d = f q ϕ 1 q (Sϕ) 1 d A 1 f q ϕ q d. Here we have used the definition of S and the hyothesis Sϕ Aϕ. If q = this simlifies to T f L A 1 f and, since f was arbitrary, µ L T A ( 1)/ as required. If q < we aly Hölder s inequality with indices /q and /( q) to get T f q L q µ A 1 ( ) q/ ( f d ) ( q)/ ϕ d = A 1 ϕ q L f q L and, since f was arbitrary, we conclude that T A ( 1)/q ϕ ( q)/q to comlete L the roof. In our second version we make stronger hyotheses and get a stronger conclusion. Theorem 2.3. Let 1 < q <. Suose that T : L + L + µ has a formal adjoint T : L + µ L + and that there exist an A > and a ositive, -measurable function ϕ L which satisfies Sϕ = Aϕ -almost everywhere in Y. Then (2.1) T = A ( 1)/q ϕ ( q)/q, L the constant multiles of ϕ are extremal functions for T from L to L q µ and if q < they are the only ones. Proof. Since ϕ L it is necessarily finite -almost everywhere. Thus Theorem 2.2 alies and we have T A ( 1)/q ϕ ( q)/q for 1 < q <. (Since ϕ L L the second factor vanishes in the case q =.) To rove the reverse inequality we use the definitions of S and T. A 1 ϕ L = ϕ(aϕ) 1 d = ϕ(sϕ) 1 d = ϕt ((T ϕ) q 1 ) d = T ϕ(t ϕ) q 1 dµ = (T ϕ) q dµ T q ϕ q L.

6 6 GORD SINNAMON This may be rearranged to yield T A ( 1)/q ϕ ( q)/q. Not only do we have L (2.1) but we also see that the inequality in the above calculation is an equality. That is, T ϕ L q µ = T ϕ L, and therefore ϕ (and its constant multiles) are extremal functions for T from L to L q µ. If q < and f is an extremal function for T from L to L q µ then we consider the argument of Theorem 2.2 alied to f. Since f is extremal, all inequalities necesarily become equalities. In articular, the alication of Hölder s inequality that gave ( f q ϕ q d ) q/ ( f d ) ( q)/ ϕ d is an equality. This occurs only when f is a constant multile of ϕ. The next two theorems exlore the extent to which these versions of Schur s Lemma are reversible. First we note that if (Y, ) is not σ-finite then there is no ositive function in L so, with the ossible excetion of the case q = in Theorem 2.2, the above results hold vacuously and no converse is to be exected. It may be ossible to give a artial converse to Theorem 2.2 in the case q = with (Y, ) non-σ-finite but at very least it would have to be assumed that Y contain no infinite atoms. We will not investigate this ossibility further here. The iteration used to rove Theorem 2.4 is essentially given in [5, 13]. A refinement of this iteration scheme is used to rove Theorem 2.5. Both roofs will be deferred to Section 4. Theorem 2.4. Suose that (Y, ) is σ-finite, 1 < q = <, T : L + L + µ has a formal adjoint T : L + µ L + and T <. Then for every ε > there exists an A > and a ositive ϕ L such that Sϕ Aϕ and ε + A ( 1)/ T A ( 1)/. The next theorem gives the converse of Theorem 2.2 in the case q < and shows that A may be taken to be 1. It also gives the converse to Theorem 2.3 in the case q <. In Examle 2.7 it is shown that a converse to Theorem 2.3 is not ossible when q =. In addition to the σ-finiteness of (Y, ) there is another mild but necessary assumtion for a converse to Theorem 2.3: (2.2) T g > whenever g >. For the integral oerator T k, this asserts that k(x, y) does not vanish identically on any tube X Y with (Y ) > so that the domain of T k is not artificially too large. Stated in terms of the ma T, (2.2) becomes: There is no set of ositive measure such that T f = for all f L + suorted there.

7 SCHUR S LEMMA AND BEST CONSTANTS 7 The assumtion is necessary because if there were such a set then for all f suorted there and all g L + µ we would have ft g d = T fg dµ = so T g = on the set. If Sϕ = Aϕ for some ositive real number A then ϕ = A 1 (T ((T ϕ) q 1 )) 1 would be zero on the set and hence ϕ could not be ositive and there could be no converse to Theorem 2.3. We view the condition (2.2) as a non-degeneracy condition on the ma T. Theorem 2.5. Suose that (Y, ) is σ-finite, 1 < q < <, T : L + L + µ has a formal adjoint T : L + µ L + and T <. Then for every ε > there exists a ositive ϕ, finite -almost everywhere, such that Sϕ ϕ and ε + ϕ ( q)/q L T ϕ ( q)/q. L If (2.2) is satisfied then there exists a ositive function ϕ such that Sϕ = ϕ and T = ϕ ( q)/q. L Combining the last statement of Theorem 2.5 with the last statement of Theorem 2.3 gives the following interesting result. Corollary 2.6. Suose 1 < q < < and (Y, ) is σ-finite. Every non-negative integral oerator satisfying (2.2) that is bounded from L to L q µ has a unique extremal function u to constant multiles. This function is never zero nor does it change sign. Examle 2.7. Hardy s inequality, [6, Theorem 327], is a strict inequality. If > 1 and f then ( ( 1 x ) 1/ ( ) 1/ f(y) dy) dx < f(y) dy x unless f. The constant is best ossible. This inequality has no ositive extremal function so Theorem 2.5 and Corollary 2.6 do not extend to the case q =. Remark. The case q = in Theorem 2.3 deserves further study. The hyothesis that ϕ L is too restrictive. It is not required to get an uer bound on the norm of the oerator as we saw in Theorem 2.2. If this assumtion can be weakened (at least finiteness -almost everywhere is required) then there may be a converse of a sort. The results of [2, 9] show that for matrix oerators there is always a ositive extremal sequence but that equality in Sϕ Aϕ is not necessarily achieved. The difficulty arises with oerators having a direct sum decomosition. Perron showed that a (finite) matrix with ositive entries has a unique ositive eigenvalue

8 8 GORD SINNAMON but to extend the result to matrices with non-negative entries Frobenius had to imose the condition that the matrix be indecomosable. This observation is at the root of the difficulty with the case q =. We seem to be faced with two choices. Either restrict our attention to oerators with no direct sum decomosition, or give u the uniqueness of the extremal and the equality in Sϕ Aϕ. We hoe to return to this dilemma in future work. 3. Weighted Norm Inequalities A weighted norm inequality for the ma K : L + η L + ξ form (3.1) ( 1/q ( (KF ) q u dξ) C F v dη) 1/. is an inequality of the Here (X, ξ) and (Y, η) are measure saces, F L + η, and the weights u and v are fixed functions in L + ξ and L+ η reectively. The best constant in (3.1) is the smallest constant C, finite or infinite, such that the inequality holds for all F L + η. It may seem redundant to include the weight functions u and v as well as the measures ξ and η in inequality (3.1). After all, multilying a measure by a weight function just gives another measure. However, the measures ξ and η rovide the inner roducts necessary for the definition of the formal adjoint and the weights allow us to consider the action of the oerator on Lebesgue saces with different measures than those used for the inner roducts. In order to reserve the integrability of the functions in L + ξ and L+ η with resect to different measures we naturally exect the new measures to be absolutely continuous with resect to the old ones. The Radon-Nikodym Theorem leads us to the weighted measures uξ and vη. To avoid certain excetional cases we assume throughout this section that the weight v is ositive. If this is not the case we simly relace the sace Y underlying η by the suort of v. Taken together, the first two theorems of this section give a formula for the best constant in all weighted norm inequalities for non-degenerate mas with formal adjoints in the case 1 < q < <. This includes weighted norm inequalities for integral oerators with non-negative kernels. In the case q =, best constants are given for a large class of weighted norm inequalities but Examle 2.7 has shown that not all weighted norm inequalities are included. So that we may easily transfer the results of the revious section to the investigation here, we make the following identifications: If u, v, ξ, η, K, and F are as above and K is a formal adjoint of K then (3.2) F = fv 1, µ = uξ, = v 1 η, T f = K(fv 1 ), and T g = K (gu). Note that T : L + L + µ and T : L + µ L + are formal adjoints because (T f)g dµ = K(fv 1 )gu dξ = fv 1 K (gu) dη = ft (g) d,

9 SCHUR S LEMMA AND BEST CONSTANTS 9 whenever f L + and g L + µ. The assumtion that v does not vanish enables us to show that for each C >, (3.1) holds for all F L + η if and only if (3.3) ( 1/q ( (T f) dµ) q C ) 1/ f d holds for all f L +. We conclude that the best constant in (3.1) and the best constant in (3.3) coincide. Theorem 3.1. Suose that 1 < q <, K : L + η L + ξ has a formal adjoint K : L + ξ L + η, and < v L + η. Let h be in L + ξ, set ϕ = [K h] 1 and u = [K(ϕv 1 )] 1 q h. If < ϕ L (v 1 )η then the best constant in (3.1) is ( C = ϕ v 1 dη) ( q)/(q). Proof. The best constant in (3.1) is also the best constant in (3.3) which was earlier denoted T. Therefore the desired result will follow from Theorem 2.3 once we show that Sϕ = ϕ: Sϕ = [T ([T ϕ] q 1 )] 1 = [K ([K(ϕv 1 )] q 1 u)] 1 = [K h] 1 = ϕ. For the converse, in the case q <, we imose non-degeneracy conditions similar to (2.2) on both K and K : (3.4) K(fv 1 ) > whenever f >, and K (gu) > whenever g >. Theorem 3.2. Suose that 1 < q < <, (Y, η) is σ-finite, K : L + η L + ξ has a formal adjoint K : L + ξ L+ η, u L + ξ, < v L+ η, and (3.4) holds. If the best constant in (3.1) is finite then there exists a function h L + ξ such that < ϕ = (K h) 1 L and u = )] 1 q h. (v 1 )η [K(ϕv1 Proof. The best constant in (3.1), and hence in (3.3), is finite which means that T <. Note that since v >, the σ-finiteness of (Y, η) imlies that (Y, ) is also σ-finite. Also note that (3.4) imlies (2.2). By Theorem 2.5 there exists a ositive ϕ L (v 1 )η with Sϕ = ϕ. Set h = [K(ϕv1 )] q 1 u and we immediately have ϕ = Sϕ = [T ([T ϕ] q 1 )] 1 = [K ([K(ϕv 1 )] q 1 u)] 1 = (K h) 1. If < K(ϕv 1 ) < then we may divide by it in the definition of h to see that u = [K(ϕv 1 )] 1 q h. Since ϕ >, (3.4) imlies K(ϕv 1 ) > and since

10 1 GORD SINNAMON ϕ L we have (v 1 ϕv1 L )η vη and the finiteness of the constant in (3.1) shows that K(ϕv 1 ) L q uξ. Thus, on the set where K(ϕv1 ) is infinite u vanishes ξ- almost everywhere and so in this case, too, u = (K(ϕv 1 )) 1 q h. Theorem 3.1 gives a simle method of generating inequalities with best constants. We give several examles to illustrate the ease and versatility of the method. The details of calculating the weight u and the norm of ϕ have been omitted. We begin with a one arameter family of weighted Hardy inequalities. Examle 3.3. Suose 1 < q < and fix α >. The inequality ( ( x holds with best constant q ) 1/q f(y) dy) [log(1 + x)] α(1 q) (x + 1) dx ( C ) 1/ f(y) [log(1 + y)] (α 1)(1 ) dy C = ( 1) 1/q+α(1/q 1/) α 1/q 1 Γ(α) 1/q 1/. Proof. Take both (X, ξ) and (Y, η) to be the half line [, ) with Lebesgue measure, take v(y) = [log(1 + y)] (α 1)(1 ), and let h(x) = (x + 1). Sometimes simler is better. Examle 3.4. Suose 1 < <. For any non-negative f, ( 1 (1 + x) log(1 + x) The constant is best ossible. x Proof. Take = q and α = 1 in the revious examle. Here is an inequality for a Steklov oerator. ) f(y) dy log(1 + x) dx 1 f(y) dy. 1 Examle 3.5. Suose 1 < q <. The best constant in the inequality ( 2 2 ( x+1 x 1 is C = 4 1/q /2 1/. q 1/q ( 1 f(y)χ [ 1,1] (y) dy) (2 x ) dx) 1 q C 1 ) 1/ f(y) dy Proof. Take (X, ξ) to be [ 2, 2] with Lebesgue measure, take (Y, η) to be [ 1, 1] with Lebesgue measure, take v = 1 and h = 1. By choosing the measures ξ and η aroriately it is simle to mix integrals and sums. The best constant is exressed in terms of the Riemann zeta function.

11 SCHUR S LEMMA AND BEST CONSTANTS 11 Examle 3.6. Suose 1 < q. The best constant in the inequality ( ( ) q 1/q ( n e ny f(y) f(y) dy) C y (q 1)( 1) (e y 1) dy is n=1 C = Γ(q ) 1/ ζ(q ) 1/q 1/. ) 1/ Proof. Take (X, ξ) to be {1, 2,... } with counting measure, take (Y, η) to be [, ) with Lebesgue measure, take, v(y) = (y (q 1)( 1) (e y 1)) 1 and h = 1. Estimates for the norm of the Hardy oerator on weighted saces with q < have been available for some time. We can now comare them with best constants. We use estimates from [12]. Examle 3.7. Suose 1 < q. The best constant in the inequality ( 1 ( 1 x q 1/q ( 1 f(y) dy) x dx) C f(y) (1 y) dy x ) 1/ is C = 2 1/ 1/q. With q = 2 and = 3 the uer and lower estimates for C given in [12] are not very close, giving aroximately.359 <.899 < Proof. Take (X, ξ) to be the [, 1] with Lebesgue measure, take (Y, η) to be [, 1] with Lebesgue measure, take v = 1 y and h = 1. We end this section with two inequalities for oerators with formal adjoints that are not, strictly seaking, integral oerators. Examle 3.8. Suose 1 < q. The inequality ( 1 ( 1 q 1/q ( 1 f(x) + f(y) dy) dx) 2 holds for all non-negative f. The constant is best ossible. ) 1/ f(y) dy Proof. Take (X, ξ) and (Y, η) to be [, 1] with Lebesgue measure, take v = 1 and h = 1. Examle 3.9. Suose 1 < q. If F is continuously differentiable and F () = then ( 1 ) 1/q F (x) + F (x) q (1 + x) 1 q dx The constant is best ossible. ( 1 1/ (3/2) 1/q 1/ F (y) (2 y) dy). Proof. Take (X, ξ) and (Y, η) to be [, 1] with Lebesgue measure, take v(x) = 2 x and h = 1. The oerator used here is Kf(x) = f(x) + x f(y) dy alied to F.

12 12 GORD SINNAMON Two Iterations The roofs of a converse to Schur s Lemma given in [5, 13] use a clever iteration which is imitated here to rove Theorem 2.4. Although the same iteration can be used to get the conclusion Sϕ ϕ in Theorem 2.5, we introduce a variation of the argument in Lemma 4.2 which is only valid in the case q <. The main advantage of the modification is that it ermits us to exercise greater control over the outut of the rocess. This is imortant because an instance of the first iteration is used as the recursion ste in the second iteration, a more delicate argument which yields a ositive fixed oint of the ma S to comlete the roof of Theorem 2.5. There is another advantage to the iteration given in Lemma 4.2. In order to roceed, the first iteration requires that T be known or at least that an uer bound for T be known. As we see in Corollary 4.4, it is ossible to carry out the modified iteration when no uer bound for T is known, even if T =. This result is of indeendent interest and may be imortant in comutations. Before resenting the roofs of Theorem 2.4 we need to understand the action of the ma S. For convenience, the non-negative functions in the closed ball of radius σ in L will be denoted by B + σ (L ). Lemma 4.1. Suose 1 < q <, T : L + L + µ has a formal adjoint T : L + µ L +, and T is finite and set α = (q 1)/( 1). Then for any λ > the oerator S given by Sϕ = (T ((T ϕ) q 1 )) 1 is a continuous, order reserving ma from B + λ (L ) to B + λ α T αq (L ). Also, S(λϕ) = λ α Sϕ for every ϕ L +. Proof. Let E s denote the (non-linear) oerator on non-negative functions defined by E s f(t) = f(t) s. It is straightforward to check that if s and σ are ositive then E s : B σ + (L ) B σ + s(l/s ) is continuous and order-reserving. The definition of T shows that T f L q µ T f L for all f L + and a routine calculation shows that T g L T g L q for all g L + µ. It follows µ that T : B σ + (L ) B + σ T (Lq µ) and T : B σ + (L q µ ) B + σ T (L ) are continuous. (A minor modification of the roof that bounded, linear oerators are continuous is needed here.) Lemma 2.2 shows that mas with formal adjoints reserve order so both T and T are order reserving. The diagram, B + λ (L ) T B + λ T (Lq µ) E q 1 B + λ q 1 T (L q q 1 T µ ) B + λ q 1 T q (L ) E 1 B + λ α T αq (L ) shows that the ma S = E 1T E q 1 T is a continuous, order reserving ma from

13 SCHUR S LEMMA AND BEST CONSTANTS 13 B + λ (L ) to B + λ α T αq (L ). Since T and T are ositive homogeneous by Lemma 2.1 it is easy to see that S(λϕ) = λ α Sϕ. Proof of Theorem 2.4. If T = then T is the zero ma so we may rove the result by choosing A = ε and taking any ositive ϕ L. Suose, then, that T > and choose δ (, 1) such that A = (1 δ) 1 T satisfies ε + A ( 1)/ T A ( 1)/. Note that A is ositive. Since (Y, ) is assumed to be σ-finite we may choose a ositive function ψ with ψ L = δ. Define a sequence in L + recursively by setting ψ n+1 = ψ + A 1 Sψ n. Using induction we see that the sequence is non-decreasing since ψ 1 ψ = A 1 Sψ and if ψ n ψ n 1 then ψ n+1 ψ n = A 1 (Sψ n Sψ n 1 ). Moreover, ψ L = δ 1 and if ψ n L 1 then by Lemma 4.1 with q = and λ = 1, ψ n+1 L δ + A 1 Sψ n L δ + A 1 T = 1. By induction, ψ n L 1 for all n. This is a non-decreasing sequence whose terms are bounded above in L. Therefore, it converges in L to its ointwise limit ϕ and ϕ L. The continuity of S shows that ϕ = ψ + A 1 Sϕ > A 1 Sϕ so we have ϕ > and Sϕ Aϕ. The next lemma uses a variation of the iteration above to reare for the recursion ste in the roof of Theorem 2.5. Lemma 4.2. Suose that 1 < q < <, T : L + L + µ has a formal adjoint T : L + µ L +, and T <. Let C = T q/( q) and α = (q 1)/( 1). If ψ is a ositive function in B + λ 2 C (L ) for some λ > 1 then there exists a ositive function ϕ B + λc (L ) satisfying ϕ = ((λ λ α )/λ 2 )ψ + Sϕ. If Sψ ψ then ϕ λψ. Proof. Set ψ = ((λ λ α )/λ 2 )ψ. For n define ψ n+1 = ψ + Sψ n. Since S is order reserving, ψ 1 ψ = Sψ and if ψ n ψ n 1 then ψ n+1 ψ n = Sψ n Sψ n 1. Induction shows that ψ, ψ 1,... is a non-decreasing sequence. We define ϕ to be the ointwise limit of this sequence. Now ψ B + λ 2 C (L ) so ψ L (λ λ α )C λc. If ψ n L λc then Sψ n L λ α C by Lemma 4.1 so ψ n+1 L ψ L + Sψ n L (λ λ α )C + λ α C = λc. By induction ψ n L λc for all n and so the sequence converges in L and ϕ L λc. By the continuity of S and the hyothesis that ψ is ositive we see that ϕ = ψ + Sϕ > Sϕ and that ϕ is ositive. To rove the last statement of the lemma we suose that Sψ ψ. Since λ > 1, ψ = ((λ λ α )/λ 2 )ψ (λ λ α )ψ λψ. If ψ n λψ for some n then ψ n+1 = ψ + Sψ n (λ λ α )ψ + S(λψ) (λ λ α )ψ + λ α ψ = λψ. By induction ψ n λψ for all n and in the limit we have ϕ λψ. This comletes the roof. Proof of Theorem 2.5. Define C and α as in Lemma 4.2 and note that < α < 1. Let ϕ be a ositive function with ϕ L 4C. Such a function exists because (Y, ) is σ-finite. For n =, 1,... aly Lemma 4.2 with ψ = ϕ n and λ = 2 2 n to

14 14 GORD SINNAMON recursively roduce ϕ n+1 satisfying ϕ n+1 L 2 2 n C and ϕ n+1 = ((2 2 n 2 α2 n )/2 21 n )ϕ n + Sϕ n+1. Since Sϕ n+1 ϕ n+1, Theorem 2.2 yields C ϕ n+1 L so we have T ϕ n+1 ( q)/q L (2 2 n ) ( q)/q T and we see that ϕ n ( q)/q is eventually less than T + ε for any ε >. This L roves the first art of the theorem. Alying the last statement of Lemma 4.2, we see that for n 1, Sϕ n ϕ n so ϕ n n ϕ n. Thus we have a decreasing sequence of ositive L functions 2ϕ n ϕ n 2 21 (n+1) ϕ n+1. We denote the ointwise limit of this sequence by ϕ and aly the Dominated Convergence Theorem and the above estimates of ϕ n+1 L to get ϕ ( q)/q L = lim ϕ n 22 n n+1 ( q)/q = T. L The continuity of S yields ϕ = lim ϕ n 22 n n+1 = lim (2 n 2 2 n 2 n 2 α2 n )ϕ n n Sϕ n+1 = Sϕ in L because 2 2 n 2 α2 n as n and the ϕ n are uniformly bounded in L norm. To comlete the roof of the second art we have to show that ϕ is ositive - almost everywhere. Since ϕ is the ointwise limit of a sequence of ositive functions, it is clearly non-negative. Thus it is enough to show that {y Y : ϕ(y) = } has measure zero. That is the object of the next lemma. Lemma 4.3. Suose 1 < q < <, Sϕ = ϕ, and ϕ L = T q/( q). If (2.2) holds then ϕ is ositive -almost everywhere. Proof. Let Y be the set where ϕ is zero and Y 1 be its comlement in Y. Let X be the set where T ϕ is zero and X 1 be its comlement in X. If f is suorted on Y then T f(t ϕ) q 1 dµ = ft ((T ϕ) q 1 ) d = f(sϕ) 1 d = fϕ 1 d which is zero because f and ϕ have disjoint suorts. Since (T ϕ) q 1 is ositive on X 1 we see that T f is suorted on X.

15 SCHUR S LEMMA AND BEST CONSTANTS 15 Similarly, if g is suorted on X then ϕt g d = T ϕg dµ = and since ϕ is ositive on Y 1 we see that T g is suorted on Y. Thus, if f is suorted on Y 1 then for all g L + µ, T fχ X g dµ = ft (χ X g) d = which imlies that T fg dµ = T fχ X g dµ + T fχ X1 g dµ = χ X1 T fg dµ. It follows that T f = χ X1 T f so T f is suorted on X 1. These observations give the decomosition (4.1) T f = χ X T (fχ Y ) + χ X1 T (fχ Y1 ) for all f L +. For i =, 1; let µ i = χ Xi µ, i = χ Yi, and define T i : L + i L + µ i and T i : L + µ i L + i by T i f = T f and T i g = T g. It is easy to check that T i and T i are formal adjoints. In view of (4.1) it is a routine calculation to show that T q/( q) = T q/( q) + T 1 q/( q). Now we aly Theorem 2.3 to the oerator T 1. Since ϕ is ositive 1 -almost everywhere and ϕ = (T ((T ϕ) q 1 )) 1 = (T 1 ((T 1 ϕ) q 1 )) 1, we conclude that T 1 = ϕ ( q)/q L 1 = ϕ ( q)/q L = T. It follows that T = and hence T f = whenever f is suorted on Y. The remark following (2.2) shows that Y has measure zero. Our final result is a corollary of Lemma 4.2 and gives an iterative rocedure which converges to T to within any reset tolerance without an a riori bound on T. It rovides uer and lower bounds for the norm of T at each stage of the iteration. Corollary 4.4. Suose 1 < q < <, T : L + L + µ has a formal adjoint T : L + µ L +, Sϕ = (T ((T ϕ) q 1 )) 1, and α = (q 1)/( 1). Fix λ > 1 and a ositive function ϕ L. Define ψ n recursively by ψ = (λ λ α ) T ϕ q/( q) L q µ ϕ /(q ) ϕ L and ψ n+1 = ψ + Sψ n.

16 16 GORD SINNAMON Then ψ n is a non-decreasing sequence, (4.2) [(1/λ) ψ n L ] ( q)/q T [su Sψ n (y)/ψ n (y)] ( 1)/q ψ n ( q)/q L y Y and (4.3) [(1/λ) lim ψ n L n ] ( q)/q T [ lim ψ n L n ] ( q)/q for each n. Proof. The right hand inequality in (4.2) follows from Theorem 2.2 with A = su y Y Sψ n (y)/ψ n (y). To rove the other inequalities we first suose that T <. Set ψ = (λ 2 /(λ λ α ))ψ to get ψ L = λ 2 T ϕ q/( q) ϕ L q µ q/(q ) λ 2 T q/( q). L Lemma 4.2 shows that ψ n is a non-decreasing sequence whose ointwise limit ϕ is ositive and satisfies both Sϕ ϕ and ϕ L λ T q/( q). The latter gives the left hand inequality in (4.2) and the former, together with Theorem 2.2, rovides the right hand inequality in (4.3). If T = then it is enough to show that lim n ψ n L is infinite. The roof of Lemma 4.2 still shows that the sequence ψ n is non-decreasing and we again define ϕ to be its ointwise limit. It is an exercise to show that Sψ n converges ointwise to Sϕ even without assuming continuity of S. We obtain Sϕ ϕ and aly Theorem 2.2 again to see that ϕ L =. The Monotone Convergence Theorem shows that lim n ψ n L = as required. References 1. N. Aronszajn, F. Mulla, and P. Szetycki, On saces of otentials connected with L saces, Ann. Inst. Fourier (Grenoble) 12 (1963), D. Borwein and X. Gao, Matrix oerators on l to l q, Canad. Math. Bull. 37 (1994), D. Boyd, Best constants in a class of integral inequalities, Pac. J. Math. 3 (1969), D. Boyd, Inequalities for ositive integral oerators, Pac. J. Math. 38 (1971), E. Gagliardo, On integral transformations with ositive kernel, Proc. Amer. Math. Soc. 16 (1965), G. Hardy, J. E. Littlewood, and G. Pólya, Inequalities, Second Edition, Cambridge University Press, Cambridge, R. Howard and A. R. Sche, Norms of ositive oerators on L -saces, Proc. Amer. Math. Soc. 19 (199), S. Karlin, Positive oerators, J. Math. Mech. 6 (1959), M. Koskela, A characterization of non-negative matrix oerators on l to l q with > q > 1, Pac. J. Math. 75 (1978), L. A. Ladyženskiĭ, Über ein Lemma von Schur, Latviĭsk Mat. Ežegodnik 9 (1971), I. Schur, Bermerkungen zur Theorie der beschränkten Bilinearformen mit unendlich vielen Veränderlicken, J. reine angew. Math. 14 (1911), G. Sinnamon and V. Steanov, The weighted Hardy inequality: New roofs and the case = 1, J. London Math. Soc. (2) 54 (1996), P. Szetycki, Notes on integral transformations, Dissertationes Mathematicae 231 (1984), Deartment of Mathematics, University of Western Ontario, London, Ontario, N6A 5B7, CANADA address: sinnamon@uwo.ca