Properties of BMO functions whose reciprocals are also BMO


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1 Properties of BMO functions whose reciprocals are also BMO R. L. Johnson and C. J. Neugebauer The main result says that a nonnegative BMOfunction w, whose reciprocal is also in BMO, belongs to p> A p,and that an arbitrary u BMO can be written as u = w /w, for w as above. This leads then to some observations concerning the JohnNirenberg distribution inequality for F u, u BM O and F Lip α. Key words: Bounded mean oscillation, A p weights AMS subject classifications: 42B25. Introduction We will consider the question of when a function w and its reciprocal /w are in BMO. If we assume that w : R n R + and consider this question for various spaces X, we obtain distinct results. The answer for L p (R n ) is that if w, /w L p (R n ), then p = while w, /w L implies that w which is also equivalent to the fact that w, /w A (for the precise definition of the A p classes see below). It is known that BMO is the right space to consider in place of L p as p in a number of situations and we will give the answer to this question for BMO in this paper. The definition of BMO is that f BMO if sup f(x) f dx = f < + where f = f(x)dx, and is a cube with sides parallel to the coordinate axes. important to know that the L norm can be replaced by the L p norm for 0 < p <, ( /p sup f(x) f dx) p = f,p f. It is We need also to recall the JohnNirenberg lemma, the reason for the above result, for functions of bounded mean oscillation. If f BMO, there are constants c, c 2 > 0 independent of f and such that t : f(t) f > λ} c e c 2λ/ f, for all λ > 0. Of course, bounded functions are in BMO and ln / x is an unbounded function in BMO. The precise space we will study is BMO = w : R n R + : w, /w BMO}.
2 We need to recall the A p weights which are defined by the condition A p (w) = sup ( ) ( ) p w w p < +, where is again a cube. The A p weights solve the problem of characterizing when the Hardy Littlewood maximal function maps L p w into L p w, where Mf(x) = sup x f(y) dy, and the result is Mf(x) p w(x) dx C p f(x) p w(x) dx w A p. We will also need to consider A = w Mw(x) Cw(x)}, with the smallest such C being denoted A (w) and A = p> A p. Since the A p constants decrease by Hölder s inequality, we can set A (w) = lim p A p (w). We have the set inclusions A A p A q A, where p q. The A p weights also solve the corresponding problem for the Hilbert transform f(y) Hf(x) = lim ɛ 0 ɛ< x y </ɛ x y dy. It is known that if w, /w A p, then w A 2, and we may limit our study to the case p 2 by the inclusion properties of A p. It is also known that [, p. 474] w, /w p> A p ln w clos BMO L. () We say that w RH p0 (reverse Hölder) if ( ) /p0 w p C 0 w, and we abbreviate by RH p0 (w) the infimum of all such C. We will use the fact, due to Strömberg and Wheeden, that w RH p0 if and only if w p 0 A. An alternate proof of this fact can be found in [3, Lemma 3.]. 2. Preliminary results Our first result shows that Hölder continuous functions operate on BM O. Lemma : If F is Hölder continuous of order α, where 0 < α and f BMO, then F f BMO and F f 2 F Lip α f α. Proof. If there is a constant c such that f(x) c dx A, then it is well known that f 2A. We compute ( /p ( /p F (f(x)) F (f ) dx) p F p Lip α f(x) f dx) αp. 2
3 Thus we obtain with p = /α, F f 2 F Lip α f α. This has been, at least partially, observed by many people. If f BMO, then f α BMO, for 0 < α and maxf, g} and minf, g} are in BMO if f, g are in BMO. We haven t noticed the converse observed, but it is true. If F f A f α, then F Lip α. The proof may be found in [2], but as this is not generally available, we give the proof here. Without loss of generality, we may assume F (0) = 0 and consider only cubes centered at the origin since BMO is translation invariant. Suppose that = [ d 2, d 2 ]n and that x on the double of f(x) = 0 outside the double of. One checks that F (f(x)) = F (x ) for x 2, 0 outside the double of. and since f f d 2, one finds d d 2 d 2 F (x ) F dx Ad α, where is the onedimensional cube [ d 2, d 2 ], and by the CampanatoMeyer theorem [4], this proves the result. We can use the lemma to show that there is a close connection between BMO and BMO. Theorem : A real valued function u is in BMO there exists a w BMO such that u = w /w and w + /w u. Proof. If u admits the decomposition, it is clear that u BMO. If we are given a u BMO, it is easy to see that the equation for w leads to a quadratic equation with a solution of w = 2 (u + u 2 + 4). The function F (x) = 2 (x + x 2 + 4) is everywhere differentiable with derivative bounded by. By Lemma, w BMO. Remark. We note that the same proof proves the corresponding result for functions of vanishing mean oscillation, which are defined as is BMO but when the sup is taken over cubes of side r, and the resulting sup goes to 0 as r 0+. Another application of Lemma is to the determination of conditions under which the square of a function belongs to BMO. By Lemma with F (x) = x, it follows that such a function belongs to BMO. We show that more is true. Lemma 2: If f = F (u), F Lip α, u BMO, then x : f(x) F (u ) > λ} c e c 2λ /α / F /α Lip α u. Proof. Because u BMO, by the JohnNirenberg lemma, there are constants c and c 2 such that t : u(t) u > λ} c e c 2λ/ u. Hence, since ( ) /α t : f(t) F (u ) > λ} t : u(t) u λ > F Lip α, we have the inequality t : f(t) F (u ) > λ} c e c 2( λ ) F /α / u Lip α, 3
4 which is the desired result. Corollary : For any ɛ < c 2, e ( ) Lipα u c2 ɛ dx c. ɛ (c 2 ɛ) f(x) F (u ) /α F /α Proof. Let φ(x) = e Ax/α, which is increasing with φ (x) = A α x/α e Ax/α. As long as A is positive, e A f(x) F (u) /α dx = A x : f(x) F (u ) > λ} λ /α e Aλ/α dλ α 0 ( ) A α c 0 e c 2 F /α Lip α u A If we choose A less than the fraction, we can use the fact that α to obtain the above estimate. 0 e ɛλ/α ɛλ /α dλ = 0 e u du = λ /α λ /α dλ. If we modify the choice of φ slightly by putting ψ(x) = e Ax/α, we see that for α, ψ is convex and we can apply Jensen s formula to, p =, f = f(x) F (u ) and if we note that f F (u ) = (f(x) F (u )) f(x) F (u ), we can make the estimate ( ) ψ(2 f F (u ) ) ψ 2 f(x) F (u ) e A2/α f(x) F (u ) /α. We now combine this with Corollary and obtain e A f(x) f /α = e A f(x) F (u )+F (u ) f /α e A2/α f(x) F (u ) /α +A2 /α f F (u ) /α = e A2/α f F (u ) /α e A2/α f(x) F (u ) /α. If we choose A2 /α = (c 2 ɛ)/( F /α Lip α u ), we can estimate this and ( ) e A f(x) f /α c ( c2 ɛ ɛ 4 ) + ψ(2 f F (u ) )
5 by using Corollary and now we apply Jensen s inequality to get ( ( ) ) e A f(x) f /α c2 ɛ c + ɛ ( ( ) c2 ɛ 2 c + ). ɛ We can now state and prove the following. e A2/α f(x) F (u ) /α Theorem 2: Consider the set of f = F (u), u BMO, 0 < α. The following two statements are equivalent. (i) F Lip α (ii) there exists 0 < c, c 2 <, 0 < A <, independent of, u BMO such that and then A F /α Lip α. x : f(x) f > λ} c e c 2 λ /α A u. Proof. We will first prove that (i) implies (ii). By restricting the range of integration in the inequality derived after Corollary, we see that E λ x : f(x) f > λ} e Aλ/α e A f(x) f /α dx, since e Aλ/α e A f(x) f /α > on E λ. This is the desired result if we choose ɛ = c 2 2 and A as above. We next show that (ii) implies (i). We first observe that (ii) implies that for some constants 0 < c 3, c 4 < ( c3 f(x) f /α ) exp c 4. A u This implies that Hölder now gives us, since /α, Hence L c 3 f(x) f /α c 4. A u L ( c α 3 f(x) f ) /α A α u α. f(x) f CA α u α. The proof is now completed by an application of [2]; see the argument after Lemma. Corollary 2: If b k BMO, then x : b(x) b > λ} c e c 2λ k / b k. 5
6 Proof. Apply the above theorem with u(x) = b k, F (x) = x /k which is Lipschitz continuous of order /k with Lipschitz constant. Remark. The argument actually shows that if x : u(x) u > λ} c e c 2λ k, then x : f(x) f > λ} (c + ) 2 e c 2 2/α F /α Lip α λ k/α. Our main result connects the behavior of functions in BMO with the A p classes. Theorem 3: The set of nonnegative functions which are BMO along with their reciprocals is contained in the intersection of all the A p classes for p >, i.e. BMO p> A p. Remarks.() Of course, if b BMO, then /b BMO p> A p and () above implies ln b clos BMO L. (2) The class BMO is nonempty. For example, b (x) = max(ln / x, e) BMO and /b L BMO. Moreover, if we take b 2 (x) = max(ln / x, / ln( x e 2 )) we get an example of a function which is unbounded and whose inverse is unbounded, yet both b 2, /b 2 BMO. (3) The result is sharp in the sense that the function b in the theorem cannot be in A since if it were, /b would also be in A and then by a result of Johnson and Neugebauer [3, Lemma 2.2], b. (4) The converse is, however, not true because with the same function b as above, b 2 satisfies /b 2 L and ln b 2 = 2 ln b clos BMO L and therefore b 2 p> A p and b 2 p> A p, but b 2 / BMO. We will prove Theorem 3 as a special case of a more general result, but let us indicate how it can be proved directly. The first step is a lemma. Lemma 3: Let us denote by then we have f = f(x)dx, (fg) f g = (f(x) f )(g(x) g )dx. Proof. Compute and use the fact that g g has mean value zero. We are ready for the first step in this version of the proof of Theorem 3. Theorem 4: Suppose b BMO, then b is in A 2. 6
7 Proof. Apply Lemma 3 to b and /b which gives b (/b) = (b(x) b )(/b(x) (/b) )dx and allows us to make the estimate b (/b) b /b. Hölder s inequality shows that b (/b) and the above becomes b (/b) + b /b. Theorem 5: If b BMO, then b A 3/2. For the proof of this statement we have to estimate ( ) ( ) /2 b b 2. First we require another lemma. Lemma 4: With the same notation as in Lemma 3, we have (f(t) f )(g(t) g )(h(t) h )(l(t) l )dt = (fghl) f (ghl) g (fhl) h (fgl) l (fgh) + f g (hl) +f h (gl) + f l (gh) + g h (fl) + g l (fh) +h l (fg) 3f g h l. Proof. We expand the integrand and compute the resulting terms. Take f = h = b, g = l = b. We obtain b ( b ) ( b ) b b ( b ) ( b ) b + b ( b ) + (b ) 2 ( b 2 ) + b ( b ) + ( b ) b + (( b ) ) 2 (b 2 ) + b ( } b ) This allows us to estimate which means that + (b ) 2 ( b 2 ) + + (b ) 2 ( b 2 ) + In particular, b ( b 2 ) /2 3(b ) 2 (( b ) ) 2 = ( 2 (b(t) b ) ) 2 b (t) ( b ) dt. ( ( ) 2 ( b ) (b 2 ) 3(b ) 2 ( ) 2 b ) b 2,4 b 2,4 ( ( ) 2 ( b ) (b 2 ) 3(b ) 2 ( ) 2 b ) + b 2,4 b 2,4. b b + 3A 2 (b), which proves that A 3/2 (b) 3 + ( 3 + ) b b. The remainder of the direct proof of Theorem 3 proceeds like this. To prove that b, /b are in A 4/3 do the corresponding formula with 8 terms of which 4 are b and 4 are /b, etc
8 3. A p weights whose reciprocals are A p weights We will now obtain Theorem 3 as a special case of the next result. Theorem 6: Suppose < p 0 2. Then the following are equivalent. w, /w A p0 (2) L = w w p 0 w ( w ) p 0 c < +. (3) Proof. Suppose (2) holds. Let r = p 0. Note that L w r (w ) r w r ( w )r + (w r ) ( w )r + w r ( w r ) + w r ( w )r + A p0 ( w ) + A p 0 (w) + A 2 (w) r c < +, because w A p0 implies w A 2. Conversely, if (3) holds, then we first note that w A 2. This follows from the next sequence of inequalities: c /r L /r w w w ( w ) (w w )(( w ) w ) = w ( w ) w ( w ) + w ( w ). We use the fact that if r, then a r b r ar b 2 r. Write r (w w )( w ( w ) ) = w( w ) + w w (w ( w ) + ) which allows us to estimate the integrand below by (w w )( w ( w ) r ) 2 r w( w ) + r ( } w w w ( r w ) + ), 2 r w r ( w )r + } ( w r wr w ( r w ) + ). Now we take the average of this over which gives 2 r (w r ) ( w )r + ( } w r ) w r c + (A 2 (w) + ) r, and we conclude that w, w A p 0. Theorem 3 follows from this result; in fact, we obtain the estimate L w p 0,p(p 0 ) w p 0,p (p 0 ) and as BMO is characterized by f,p for any p > 0, we can have any p 0 > which proves the result. Although we proved Theorem 6 for A p, it immediately implies a result about RH r. 8
9 Theorem 7: The following statements are equivalent for r < : w, /w RH r (4) w, /w A +/r (5) w r A 2. (6) Proof. (4) (5). Since w r, /w r A, we have that w r, /w r A 2, and hence w, /w A +/r. (5) (4). w A +/r /w RH r. Similarly, w RH r. (4) (6). Since w r, /w r A, we have that w r A 2 as above. (6) (4). Since w r A 2, w A +/r /w RH r. From the fact that w r A 2, it follows that w r A 2 and this implies that we can apply the above remark to /w. Theorem 8: Suppose u BMO and α > 0. Then u 2 + α p> A p. Proof. For any λ > 0, write λu = w λ /w λ, for some w λ BMO. Then λ 2 u 2 = wλ wλ 2 By Theorem 7, wλ 2 A and since w λ p> A p, by Lemma 2.4 in [3], wλ 2 p> A p and a similar result holds for. This shows that λ 2 u wλ 2 p> A p and hence, u λ 2 p> A p. Since λ is an arbitrary positive number, the result follows. References [] GarcíaCuerva, J. and Rubio de Francia, J. L.: Weighted norm inequalities and related topics, (North Holland Math. Studies: Vol. 6), Amsterdam: NorthHolland, 985. [2] Johnson, R. L.: Behaviour of BMO under certain functional operations. Preprint University of Maryland, September 975. [3] Johnson, R. L. and Neugebauer, C. J.: Homeomorphisms preserving A p, Rev. Mat. Iberoamericana, 3(2), 987, [4] Meyers, N. G.: Mean oscillation over cubes and Hölder continuity, Proc. Amer. Math. Soc., 5(964), R. L. Johnson Supported by a grant from the NSF Department of Mathematics University of Maryland College Park, Maryland and C. J. Neugebauer Department of Mathematics Purdue University West Lafayette, Indiana June 2, 99 9
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