A remark on the relationship 1/p = 1/p_1 + 1/p_2 for boundedness of bilinear pseudo-differential operators with exotic symbols
aa r X i v : . [ m a t h . C A ] J a n A REMARK ON THE RELATIONSHIP /p = 1 /p + 1 /p FORBOUNDEDNESS OF BILINEAR PSEUDO-DIFFERENTIALOPERATORS WITH EXOTIC SYMBOLS
TOMOYA KATO AND NAOTO SHIDA
Abstract.
We consider the bilinear pseudo-differential operators with symbolsin the bilinear H¨ormander classes BS mρ,ρ , 0 < ρ <
1. In this paper, we show thatthe condition 1 /p = 1 /p + 1 /p is necessary when we consider the boudnednessfrom H p × H p to L p of those operators for the critical case. Introduction
For m ∈ R and 0 ≤ ρ, δ ≤
1, the bilinear H¨ormander symbol class BS mρ,δ consistsof all functions σ ( x, ξ , ξ ) ∈ C ∞ ( R n × R n × R n ) such that | ∂ αx ∂ β ξ ∂ β ξ σ ( x, ξ , ξ ) | ≤ C α,β ,β (1 + | ξ | + | ξ | ) m + δ | α |− ρ ( | β | + | β | ) for all multi-indices α, β , β ∈ N n = { , , , . . . } n . For a symbol σ ∈ BS mρ,δ , thebilinear pseudo-differential operator T σ is defined by T σ ( f , f )( x ) = 1(2 π ) n Z ( R n ) e ix · ( ξ + ξ ) σ ( x, ξ , ξ ) b f ( ξ ) b f ( ξ ) dξ dξ for f , f ∈ S ( R n ).Let X , X and Y be function spaces on R n equipped with quasi-norms k · k X , k · k X and k · k Y . If there exists a constant C > k T σ ( f , f ) k Y ≤ C k f k X k f k X , f ∈ S ∩ X , f ∈ S ∩ X holds, then, with a slight abuse of terminology, we say that the operator T σ isbounded from X × X to Y . The smallest constant C of (1.1) is denoted by k T σ k X × X → Y . Op( BS mρ,δ ) denotes the class of all bilinear pseudo-differential opera-tors T σ with σ ∈ BS mρ,δ . If all T σ with σ ∈ BS mρ,δ are bounded from X × X to Y ,then we write Op( BS mρ,δ ) ⊂ B ( X × X → Y ).The boundedness properties of the bilinear pseudo-differential operators withsymbols in the bilinear H¨ormander classes have been studied in many researches.In the case ρ = 1 and δ <
1, bilinear pseudo-differential operators with symbolsin BS ,δ can be regarded as bilinear Calder´on-Zygmund operators in the sense ofGrafakos-Torres [9], and they are bounded from L p × L p to L p , 1 < p , p < ∞ ,1 /p = 1 /p + 1 /p . For details, see Coifman-Meyer [6], B´enyi-Torres [3] and B´enyi-Maldonado-Naibo-Torres [2]. We remark that the assumption 1 /p = 1 /p + 1 /p isreasonable since a symbol σ ≡ BS ,δ and the corresponding bilinear operator T σ is the product of two functions. Date : January 27, 2021.2020
Mathematics Subject Classification.
Key words and phrases.
Bilinear pseudo-differential operators, bilinear H¨ormander symbolclasses.This work was partially supported by JSPS KAKENHI Grant Numbers JP20K14339 (Kato).
In this paper, we shall consider the case 0 ≤ ρ = δ <
1. In the linear case, the cel-ebrated Calder´on-Vaillancourt theorem [5] states that the linear pseudo-differentialoperator with the symbol satisfying the estimate | ∂ αx ∂ βξ σ ( x, ξ ) | ≤ C α,β (1 + | ξ | ) ρ | α |− ρ | β | is bounded on L . However, in the bilinear settings, it was pointed out in B´enyi-Torres [4] that for any 1 ≤ p, p , p < ∞ satisfying 1 /p = 1 /p + 1 /p , there existsa symbol σ ∈ BS ρ,ρ such that T σ is not bounded from L p × L p to L p . Thus, thereis certain difference between linear and bilinear cases.For 0 ≤ ρ < < p , p ≤ ∞ , we define m ρ ( p , p ) = (1 − ρ ) m ( p , p ) ,m ( p , p ) = − n (cid:18) max (cid:26) , p , p , − p − p , p + 1 p − (cid:27)(cid:19) . We see that the explicit values of m ( p , p ) are m ( p , p ) = n/p + n/p − n if (1 /p , /p ) ∈ J ; − n/ /p , /p ) ∈ J ; − n/p if (1 /p , /p ) ∈ J ; − n/p if (1 /p , /p ) ∈ J ; n/ − n/p − n/p if (1 /p , /p ) ∈ J , where J , . . . , J are the following 5 regions; 1 /p /p / / J J J J J Then, Miyachi-Tomita [14, 15, 16] proved the following theorem.
Theorem A.
Let ≤ ρ < , < p , p , p ≤ ∞ and /p = 1 /p + 1 /p . Then allbilinear pseudo-differential operators with symbols in BS m ρ ( p ,p ) ρ,ρ are bounded from H p × H p to L p , where L p should be replaced by BM O if p = p = p = ∞ . For the definition of the Hardy space H p and the space BM O , see Section 2. In[14], it was also proved that, for 0 < p , p , p ≤ ∞ with 1 /p + 1 /p = 1 /p , the class BS mρ,ρ with m > m ρ ( p , p ) does not assure the boundedness from H p × H p to L p .In this sense, the class BS m ρ ( p ,p ) ρ,ρ can be regarded as a critical class. We remarkthat the sharper boundedness was proved in [14] for the case ρ = 0 and in Kato [10]for the case 0 ≤ ρ < /p , /p ) ∈ J , 1 /p = 1 /p + 1 /p . We also noticethat the boundedness for the case 0 ≤ ρ < / p = p = p = ∞ was also ILINEAR PSEUDO-DIFFERENTIAL OPERATORS 3 obtained in Naibo [17]. For the subcritical case m < m ρ ( p , p ), see B´enyi-Bernicot-Maldonado-Naibo-Torres [1] and Michalowski-Rule-Staubach [13].Now, we focus on the condition 1 /p = 1 /p + 1 /p , which is often assumed whenthe L p × L p → L p boundedness is considered. It may be difficult to determinewhether the assumption 1 /p = 1 /p + 1 /p in Theorem A is valid or not since thesymbol σ ≡ BS m ρ ( p ,p ) ρ,ρ for any 0 ≤ ρ < < p , p ≤ ∞ . Recently,Kato-Miyachi-Tomita [11, 12] proved by using L -based amalgam spaces that if(1 /p , /p ) ∈ J , 1 ≤ p ≤ /p ≤ /p + 1 /p , then all bilinear operators T σ with σ ∈ BS m ( p ,p )0 , = BS − n/ , are bounded in L p × L p → L p . From this result,we conclude that the condition 1 /p = 1 /p + 1 /p is not always necessary when wediscuss the boundedness of bilinear pseudo-differential operators with symbols inthe critical class of ρ = 0.The purpose of this paper is to show that the condition 1 /p = 1 /p +1 /p is essen-tial to have the boundedness of bilinear pseudo-differential operators with symbolsin BS m ρ ( p ,p ) ρ,ρ , 0 < ρ <
1. The following is the main result of this paper.
Theorem 1.1.
Let < ρ < and < p , p , p ≤ ∞ . If all bilinear pseudo-differential operators with symbols in BS m ρ ( p ,p ) ρ,ρ are bounded from H p ( R n ) × H p ( R n ) to L p ( R n ) when p < ∞ or to BM O ( R n ) when p = ∞ , then /p = 1 /p + 1 /p . The contents of this paper are as follows. In Section 2, we recall some basicnotations and preliminary facts. In Section 3, we prove Theorem 1.1.2.
Preliminaries
For two nonnegative quantities A and B , the notation A . B means that A ≤ CB for some unspecified constant C >
0, and A ≈ B means that A . B and B . A .For 1 ≤ p ≤ ∞ , p ′ is the conjugate exponent of p , that is, 1 /p + 1 /p ′ = 1.Let S ( R n ) and S ′ ( R n ) be the Schwartz space of rapidly decreasing smooth func-tions on R n and its dual, the space of tempered distributions, respectively. We definethe Fourier transform F f and the inverse Fourier transform F − f of f ∈ S ( R n ) by F f ( ξ ) = b f ( ξ ) = Z R n e − ix · ξ f ( x ) dx and F − f ( x ) = 1(2 π ) n Z R n e ix · ξ f ( ξ ) dξ. For a measurable subset E ⊂ R n , the Lebesgue space L p ( E ), 0 < p ≤ ∞ , is the setof all those mesurable functions f on E such that k f k L p ( E ) = ( R E | f ( x ) | p dx ) /p < ∞ if 0 < p < ∞ or k f k L ∞ ( E ) = ess sup x ∈ E | f ( x ) | < ∞ if p = ∞ . We also write k f k L p ( E ) = k f ( x ) k L px ( E ) when we want to indicate the variable explicitly.We recall the definition of Hardy spaces and the space BM O on R n . Let φ ∈S ( R n ) be such that R R n φ ( x ) dx = 1. For 0 < p ≤ ∞ , the Hardy space H p ( R n )consists of all f ∈ S ′ ( R n ) such that k f k H p = k sup t> | φ t ∗ f |k L p < ∞ , where φ t ( x ) = t − n φ ( t − x ). It is known that the definition of H p ( R n ) is independentof the choice of the function φ , and that H p ( R n ) = L p ( R n ) for 1 < p ≤ ∞ . Thespace BM O ( R n ) consists of all locally integrable functions f on R n such that k f k BMO = sup Q | Q | Z Q | f ( x ) − f Q | dx < ∞ , T. KATO AND N. SHIDA where the supremum is taken over all cubes in R n and f Q = | Q | − R Q f . It is knownthat the dual space of H ( R n ) is BM O ( R n ).3. Proof of Theorem 1.1
In this section, we shall prove Theorem 1.1. For 0 < ρ < ≤ p , p ≤ ∞ ,we set e m ρ ( p , p ) = (1 − ρ ) e m ( p , p ) , e m ( p , p ) = − n (cid:18) max (cid:26) , p , p , p + 1 p − (cid:27)(cid:19) = − n/ /p , /p ) ∈ I ; − n/p if (1 /p , /p ) ∈ I ; − n/p if (1 /p , /p ) ∈ I ; n/ − n/p − n/p if (1 /p , /p ) ∈ I , where I , . . . , I are the following ranges; 1 /p /p / / I I I I We now introduce the following fact proved in Wainger [18, Theorem 10]. Let0 < a < < b < n . For t >
0, we define e f a,b,t ( x ) = X k ∈ Z n \{ } e − t | k | | k | − b e i | k | a e ik · x . Then, the limit e f a,b ( x ) = lim t → +0 e f a,b,t ( x ) exists for all x ∈ R n \ (2 π Z ) n . Furthermore,if 1 ≤ p ≤ ∞ and b > n/ − an/ − n/p + an/p , then e f a,b ∈ L p ( T n ).From this, the following fact was given. Lemma 3.1 ([14, Lemma 6.1]) . Let < a < , < b < n and ϕ ∈ S ( R n ) . For t > , we set f a,b,t ( x ) = X k ∈ Z n \{ } e − t | k | | k | − b e i | k | a e ik · x ϕ ( x ) . If ≤ p ≤ ∞ and b > n − na/ − n/p + na/p , then sup t> k f a,b,t k L p ( R n ) < ∞ . The following lemma plays an essential role in the proof of Theorem 1.1
Lemma 3.2.
Let < ρ < , < p , p < ∞ and < p < ∞ . If all bilinear pseudo-differential operators with symbols in BS e m ρ ( p ,p ) ρ,ρ are bounded from L p ( R n ) × L p ( R n ) to L p ( R n ) , then /p = 1 /p + 1 /p . ILINEAR PSEUDO-DIFFERENTIAL OPERATORS 5
Proof.
In this proof, we simply write e m ρ ( p , p ) and e m ( p , p ) as m ρ and m , re-spectively.We first show that 1 /p ≤ /p + 1 /p . If the symbol σ is independent of x ,namely, σ = σ ( ξ , ξ ), then T σ is called a bilinear Fourier multiplier operator. Wewill use the following fact proved in [8, Proposition 7.3.7]: If a nonzero bilinearFourier multiplier operator is bounded from L p × L p to L p , 0 < p , p , p < ∞ , then1 /p ≤ /p + 1 /p . Now, let σ = σ ( ξ , ξ ) ∈ S ( R n ). Then, since σ ∈ BS e mρ,ρ for any e m ≤
0, it follows from our assumption and the above fact that 1 /p ≤ /p + 1 /p .Next, we prove 1 /p ≥ /p + 1 /p . The proof below is based on the idea givenby [14, Proof of Lemma 6.3]. We first give a rough argument omitting necessarylimiting argument and then we incorporate necessary details at the last part of theproof. By our assumption and the closed graph theorem, there exists a positiveinteger M such that k T σ k L p × L p → L p . max | α | , | β | , | β |≤ M k (1 + | ξ | + | ξ | ) − m ρ − ρ ( | α |−| β |−| β | ) ∂ αx ∂ β ξ ∂ β ξ σ ( x, ξ , ξ ) k L ∞ . for all σ ∈ BS m ρ ρ,ρ (see [1, Lemma 2.6]).Let ϕ, e ϕ ∈ S ( R n ) be such thatsupp ϕ ⊂ [ − / , / n , |F − ϕ ( x ) | ≥ − , n , supp e ϕ ⊂ [ − / , / n , e ϕ = 1 on [ − / , / n . For ǫ > < a , a <
1, we set b i = n − na i − np i + na i p i + ǫ, i = 1 , . For sufficiently large j ∈ N , we define σ ( ξ , ξ ) = X ( k ,k ) ∈ D j c k ,k (1 + | k | + | k | ) m ϕ (2 − jρ ξ − k ) ϕ (2 − jρ ξ − k ) , [ f a k ,b k ( ξ k ) = X ℓ k ∈ Z n \{ } | ℓ k | − b k e i | ℓ k | ak e ϕ (2 − jρ ξ k − ℓ k ) , k = 1 , , where { c k ,k } is a sequence satisfying sup k ,k ∈ Z n | c k ,k | ≤
1, and D j = (cid:8) ( k , k ) ∈ Z n × Z n : | k | ≈ | k | ≈ | k + k | ≈ j (1 − ρ ) (cid:9) . Then, since | ξ | , | ξ | ≈ j for ( ξ , ξ ) ∈ supp σ , we have | ∂ β ξ ∂ β ξ σ ( ξ , ξ ) | . j (1 − ρ ) m − jρ ( | β | + | β | ) ≈ (1 + | ξ | + | ξ | ) m ρ − ρ ( | β | + | β | ) , and thus, this σ belongs to BS m ρ ρ,ρ . Here, it should be emphasized that the implicitconstants are independent of { c k ,k } . It follows from Lemma 3.1 that k f a ,b k L p = (cid:13)(cid:13)(cid:13) X ℓ ∈ Z n \{ } | ℓ | − b e i | ℓ | a e i jρ x · ℓ jρn e Φ(2 jρ x ) (cid:13)(cid:13)(cid:13) L p x = 2 jρn (1 − /p ) (cid:13)(cid:13)(cid:13) X ℓ ∈ Z n \{ } | ℓ | − b e i | ℓ | a e ix · ℓ e Φ( x ) (cid:13)(cid:13)(cid:13) L p x ≈ jρn (1 − /p ) , where e Φ = F − e ϕ . To have the equivalence above, we need a slight modificationfor the definition of f a ,b (see the last part of this proof). Similarly, k f a ,b k L p ≈ jρn (1 − /p ) . T. KATO AND N. SHIDA
On the other hand, since ϕ ( · − k ) e ϕ ( · − ℓ ) = ϕ ( · − k ) if k = ℓ , and 0 otherwise, wehave T σ ( f a ,b , f a ,b )( x ) = X ( k ,k ) ∈ D j c k ,k (1 + | k | + | k | ) m | k | − b | k | − b × e i | k | a e i | k | a e i jρ x · ( k + k ) { jρn Φ(2 jρ x ) } with Φ = F − ϕ . Now, let { r k ( ω ) } k ∈ Z n be a sequence of the Rademacher functionson ω ∈ [0 , n . We choose { c k ,k } as c k ,k = r k + k ( ω ) e − i | k | a e − i | k | a . Then, T σ ( f a ,b , f a ,b )( x )= X ( k ,k ) ∈ D j r k + k ( ω )(1 + | k | + | k | ) m | k | − b | k | − b e i jρ x · ( k + k ) { jρn Φ(2 jρ x ) } = { jρn Φ(2 jρ x ) } X | k |≈ j (1 − ρ ) r k ( ω ) e i jρ x · k d k , where d k = X k : | k |≈| k − k |≈ j (1 − ρ ) (1 + | k | + | k − k | ) m | k | − b | k − k | − b . Here, we see from the way to generate d k that the cardinality of the set { k ∈ Z n : | k | ≈ | k − k | ≈ j (1 − ρ ) } for | k | ≈ j (1 − ρ ) is equivalent to 2 j (1 − ρ ) n . By a change ofvariables and the assumption | Φ | ≥ − , n , we have k T σ ( f a ,b , f a ,b ) k L p = 2 jρn (2 − /p ) (cid:13)(cid:13)(cid:13) { Φ( x ) } X | k |≈ j (1 − ρ ) r k ( ω ) e ix · k d k (cid:13)(cid:13)(cid:13) L px ≥ jρn (2 − /p ) (cid:13)(cid:13)(cid:13) X | k |≈ j (1 − ρ ) r k ( ω ) e ix · k d k (cid:13)(cid:13)(cid:13) L px ([ − , n ) . Thus, our assumption implies that2 jρn (1 /p − /p − /p ) & (cid:13)(cid:13)(cid:13) X | k |≈ j (1 − ρ ) r k ( ω ) e ix · k d k (cid:13)(cid:13)(cid:13) L px ([ − , n ) . (3.1)It should be emphasized that the implicit constant in this inequality does not dependon ω ∈ [0 , n . Raising (3.1) to a p th-power and integrating over ω , we obtain(2 jρn (1 /p − /p − /p ) ) p & Z [ − , n Z [0 , n (cid:12)(cid:12)(cid:12) X | k |≈ j (1 − ρ ) r k ( ω ) e ix · k d k (cid:12)(cid:12)(cid:12) p dωdx. By Khintchine’s inequality (see, e.g., [7, Appendix C]), the right hand side is equiv-alent to Z [ − , n (cid:16) X | k |≈ j (1 − ρ ) | e ix · k d k | (cid:17) p/ dx ≈ (cid:16) X | k |≈ j (1 − ρ ) | d k | (cid:17) p/ . Since | d k | ≈ j (1 − ρ )( m − b − b ) (cid:16) X k : | k |≈| k − k |≈ j (1 − ρ ) (cid:17) ≈ j (1 − ρ )( m − b − b + n ) ILINEAR PSEUDO-DIFFERENTIAL OPERATORS 7 for | k | ≈ j (1 − ρ ) , we obtain (cid:16) X | k |≈ j (1 − ρ ) | d k | (cid:17) p/ ≈ j (1 − ρ )( m − b − b + n ) p (cid:16) X | k |≈ j (1 − ρ ) (cid:17) p/ ≈ j (1 − ρ )( m − b − b +3 n/ p . Combining the above estimates, we have2 jρn (1 /p − /p − /p ) & j (1 − ρ )( m − b − b +3 n/ , and since j is arbitrarily large, this implies that ρn (cid:18) p − p − p (cid:19) ≥ (1 − ρ ) (cid:26) m + (cid:18) np + np − n (cid:19) + a (cid:18) n − np (cid:19) + a (cid:18) n − np (cid:19) − ǫ (cid:27) . (3.2)As mentioned at the beginning of this section, we have m = − n/ /p , /p ) ∈ I ; − n/p if (1 /p , /p ) ∈ I ; − n/p if (1 /p , /p ) ∈ I ; n/ − n/p − n/p if (1 /p , /p ) ∈ I , and hence, we see that the right hand side of (3.2) converges to 0 as ǫ → a → , a → /p , /p ) ∈ I ; a → , a → /p , /p ) ∈ I ; a → , a → /p , /p ) ∈ I ; a → , a → /p , /p ) ∈ I . Thus, since ρ >
0, we obtain 1 /p − /p − /p ≥
0, that is, 1 /p ≥ /p + 1 /p .We notice that the above argument is not rigorous since f a ,b and f a ,b are notin S . In order to correct this point, we replace f a k ,b k by f a k ,b k ,t satisfying \ f a k ,b k ,t ( ξ k ) = X ℓ k ∈ Z n \{ } e − t | ℓ k | | ℓ k | − b k e i | ℓ k | ak e ϕ (2 − jρ ξ k − ℓ k ) , k = 1 , , where a k and b k are the same as above. Then, f a k ,b k ,t ∈ S , and Lemma 3.1 impliesthat sup t> k f a k ,b k ,t k L pk ≈ jρn (1 − /p k ) , k = 1 ,
2. In the same way as above, we have T σ ( f a ,b ,t , f a ,b ,t )( x ) = { jρn Φ(2 jρ x ) } X | k |≈ j (1 − ρ ) r k ( ω ) e i jρ x · k d k,t with d k,t = X k : | k |≈| k − k |≈ j (1 − ρ ) (1 + | k | + | k − k | ) m e − t ( | k | + | k − k | ) | k | − b | k − k | − b . Since this is the sum of a finite number of terms, we have d k,t → d k as t →
0. Therest of the proof is the same as above. (cid:3)
In the rest of this section, we will complete the proof of Theorem 1.1. Let J , . . . , J be the same region as in Section 1. The following proof is also based on the idea by[14, Theorem 6.4].First, we assume p < ∞ . We consider the case (1 /p , /p ) ∈ (0 , . Since H p = L p , 1 < p ≤ ∞ , the desired results for (1 /p , /p ) ∈ S k =1 J k ∩ (0 , followfrom Lemma 3.2. In the case (1 /p , /p ) ∈ J ∩ (0 , , we will use a duality T. KATO AND N. SHIDA argument. It should be remarked that p is restricted to 2 ≤ p < ∞ in this case,since 1 /p ≤ /p + 1 /p ≤ / /p , /p ) ∈ J ∩ (0 , (for the first inequality,see the argument in the proof of Lemma 3.2). It is known that T σ is bounded from L p × L p to L p if and only if T σ ∗ is bounded from L p ′ × L p to L p ′ , where σ ∗ isgiven by Z R n T σ ( f , f )( x ) g ( x ) dx = Z R n T σ ∗ ( g, f )( x ) f ( x ) dx. It is further known that if σ is in BS mρ,ρ , then the transposition σ ∗ is also in BS mρ,ρ (see [2, Theorem 2.1]). Thus, it follows from these facts that if all the opera-tors in Op( BS mρ,ρ ) are bounded from L p × L p to L p , then all the operators inOp( BS mρ,ρ ) are bounded from L p ′ × L p to L p ′ . Now, since we have BS − (1 − ρ ) n/p ′ ρ,ρ ⊂ BS − (1 − ρ ) n (1 − /p − /p ) ρ,ρ by recalling 1 /p ≤ /p + 1 /p , our assumption and the aboveargument yield that all T σ with σ ∈ BS − (1 − ρ ) n/p ′ ρ,ρ are bounded from L p ′ × L p to L p ′ . Since (1 /p ′ , /p ) ∈ J ∩ (0 , and 1 < p ′ <
2, by Lemma 3.2, we have1 /p ′ = 1 /p ′ + 1 /p , that is, 1 /p = 1 /p + 1 /p .For the case (1 /p , /p ) ∈ [0 , ∞ ) \ (0 , , we first consider the case (1 /p , /p ) ∈ J \ (0 , . The same argument works for the other cases J k ∩ (0 , , k = 1 , , BS − (1 − ρ ) n/p ρ,ρ are bounded from H p × H p to L p . Since(3.3) Op( BS − (1 − ρ ) n/ ρ,ρ ) ⊂ B ( L × L → L )(see [15]), by interpolation, all T σ with σ ∈ BS − (1 − ρ ) n/ e p ρ,ρ are bounded from H e p × H e p to L e p with (1 / e p , / e p , / e p ) = (1 − θ )(1 / , / ,
1) + θ (1 /p , /p , /p ) and 0 < θ < θ sufficiently close to 0, we have (1 / e p , / e p ) ∈ J ∩ (0 , . By Lemma 3.2,we have 0 = 1 e p − e p − e p = θ (cid:18) p − p − p (cid:19) , and this implies 1 /p = 1 /p + 1 /p . The remaining part is (1 /p , /p ) ∈ J \ (0 , .The case J \ { (0 , ∪ { (0 , }} is similar to the above. In fact, using interpolationwith(3.4) Op( BS − (1 − ρ ) n/ ρ,ρ ) ⊂ B ( L a × L b → L ) , /a, /b > , /a + 1 /b = 1 / , which was also proved in [15], and repeating the same argument just above, we obtainthe desired result from the fact for J ∩ (0 , proved in the previous step. Finally,we consider the case (1 /p , /p ) = (0 , /p , /p ) = (0 , BS − (1 − ρ ) nρ,ρ ) ⊂ B ( L ∞ × L ∞ → L p ) holds. Interpolating this with (3.4), we haveOp( BS − (1 − ρ ) n (1 − / e p − / e p ) ρ,ρ ) ⊂ B ( L e p × L e p → L e p ) , where (1 / e p , / e p , / e p ) = (1 − θ )(1 /a, /b, /
2) + θ (0 , , /p ) and 0 < θ <
1. Thenfrom the fact for J ∩ (0 , , 1 / e p = 1 / e p + 1 / e p holds, which is identical with θ/p = 0. However, this is impossible since we are now assuming p < ∞ . Therefore,we complete the proof for the case p < ∞ .Next, we consider the case p = ∞ , namely, our assumption is(3.5) Op( BS m ρ ( p ,p ) ρ,ρ ) ⊂ B (cid:0) H p × H p → BM O (cid:1) . For the case p = p = ∞ is clear, since 1 / ∞ + 1 / ∞ = 1 / ∞ . In what follows, weassume (1 /p , /p ) ∈ [0 , ∞ ) \ { (0 , } . We first consider the case (1 /p , /p ) ∈ ILINEAR PSEUDO-DIFFERENTIAL OPERATORS 9 [0 , \ { (0 , } . However, this choice is invalid. In fact, we have Op( BS m ρ ( p ,p ) ρ,ρ ) ⊂ B ( H × L p → L p ′ ) by duality of (3.5). Since 1 ≤ p ′ < ∞ , the conclusion for p < ∞ gives 1 + 1 /p = 1 /p ′ , which is identical with 1 /p + 1 /p = 0. However, this isimpossible. Next, we consider the case (1 /p , /p ) ∈ [0 , ∞ ) \ [0 , . However, thiscase is also invalid. In fact, by interpolating between (3.5) in this case and (3.3), wehave Op( BS m ρ ( e p , e p ) ρ,ρ ) ⊂ B ( L e p × L e p → L e p ) , where (1 / e p , / e p , / e p ) = (1 − θ )(1 / , / ,
1) + θ (1 /p , /p ,
0) and 0 < θ <
1. Since1 / e p = 1 − θ , namely, 1 < e p < ∞ , we have 1 / e p = 1 / e p + 1 / e p by the conclusion for p < ∞ . This coincides with θ (1 /p + 1 /p ) = 0. However, this is again impossible.Consequently, in the case p = ∞ , the possible choice is only p = p = ∞ , whichmeans 1 /p = 1 /p + 1 /p .The proof of Theorem 1.1 is complete. Acknowledgement
The authors sincerely express their thanks to Professor Akihiko Miyachi and Pro-fessor Naohito Tomita for valuable discussions and fruitful advices.
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Department of Mathematics, Graduate School of Science, Osaka Uni-versity, Toyonaka, Osaka 560-0043, Japan
Email address , T. Kato: [email protected]
Email address , N. Shida:, N. Shida: