Vietnam Journal of Mathematics 33:2 (2005) 207–213 A Gagliardo-Nirenberg Inequality for Lorentz Spaces * Mai Thi Thu Ca Mau Pedagogical College, Nguyen Tat Thanh R oad, Ca Mau City, Vietnam Received July 23, 2004 Revised November 26, 2004 Abstract. In this paper, essentially developing the method of [1] and [10], we give an extension of the Gagliardo-Nirenberg inequality to Lorentz spaces. Let Φ : [0, ∞) → [0, ∞) be a non-zero concave function, which is non-decreasing and Φ(0+) = Φ(0) = 0. We put Φ(∞) = lim t→∞ Φ(t). For an arbitrary mea- surable function f we define f N Φ = ∞ 0 Φ λ f (y) dy, where λ f (y)=mes{x ∈ R n : |f(x)| >y} ,(y ≥ 0). The space N Φ (R n ) consisting of measurable functions f such that f N Φ < ∞ is a Banach space. Denote by M Φ (R n ) the space of measurable functions g such that g M Φ =sup 1 Φ(mes Δ) Δ |g(x)|dx :Δ⊂ R n , 0 < mes Δ < ∞ < ∞. Then M Φ (R n ) is a Banach space, see [7 - 9]. We have the following results [8 - 9]: ∗ This work was supported by the Natural Science Council of Vietnam. 208 Mai Thi Thu Lemma 1. If f ∈ N Φ (R n ), g ∈ M Φ (R n ) then fg ∈ L 1 (R n ) and R n |f(x)g(x)|dx ≤f N Φ g M Φ . Lemma 2. If f ∈ N Φ (R n ) then f N Φ =sup g M Φ ≤1 R n f(x)g(x)dx . Let ≥ 2. Denote by W ,∞ (R n ) the set of all measurable functions f such that f and its generalized derivatives D β f,0< |β|≤,belongtoL ∞ (R n ). The following is the well-known Gagliardo-Nirenberg inequality: Lemma 3. [6] For fixed α, 0 < |α| <, there is the best constant C α, such that D α f ∞ ≤ C α, f 1− |α| ∞ |β|= D β f ∞ |α| , for any f ∈ W ,∞ (R n ). The following result is an extension of the Gagliardo-Nirenberg inequality ([2 - 6]) to Lorentz spaces. Note that the Gagliardo-Nirenberg inequality has applications to partial differential equations and interpolation theory. Theorem 1. Let ≥ 2, f and its generalized derivatives D β f, |β| = be in N Φ (R n ).ThenD α f ∈ N Φ (R n ) for all α, 0 < |α| <and D α f N Φ ≤ C α, f 1− |α| N Φ ( |β|= D β f N Φ ) |α| , (1) where the constant C α, is defined in Lemma 3. Proof. We begin to prove (1) with the assumption that D α f ∈ N Φ (R n ), 0 ≤ |α|≤. Fix 0 < |α| <. By Lemma 2 we have D α f N Φ =sup v M Φ ≤1 R n D α f(x)v(x)dx . Let >0. We choose a function v ∈ M Φ (R n ) such that v M Φ =1and R n f(x)v (x)dx ≥f N Φ − /2. By Lemma 1, there is H := [−H, H] n such that R n f(x)v(x)dx ≥f N Φ − , (2) where v = v(H,):=χ H v and χ H is the characteristic function of H.Put A Gagliardo-Nirenber g Inequality for Lorentz Spaces 209 F (x)= R n f(x + y)v(y)dy. Then F ∈ L ∞ (R n ) by virtue of Lemma 1, and it is easy to check that D β F ε (x)= R n D β f(x + y)v(y)dy, 0 ≤|β|≤ (3) in the distribution sense. For all x ∈ R n , clearly, |D β F ε (x)|≤D β f(x + ·) N Φ v M Φ ≤D β f N Φ . (4) Now we prove the continuity of D β F ε on R n (0 ≤|β|≤). We show this for β = 0. Clearly, it suffices to prove that for any x ∈ R n , lim t→0 χ H (·) f(x + t + ·) − f(x + ·) N Φ =0. Assume the contrary that for some δ>0, point x 0 and sequence t m → 0, χ H (·) f(x 0 + t m + ·) − f(x 0 + ·) N Φ ≥ δ, m ≥ 1. (5) For simplicity of notation we suppose x 0 =0.Sincef ∈ N Φ (R n ), f ∈ L 1,oc (R n ). So, it is known that H |f(x + t m ) − f(x)|dx → 0asm →∞. Therefore, there exists a subsequence {t m j }, which we still denote by {t m },such that f (· + t m ) → f a.e. on H. Define g n (x)= inf m≥n |f(x + t m )|,x∈H; then {g n } is a non-decreasing sequence and g n →|f| a.e. on H.Itiseasytosee that λ χ H g n (t) → λ χ H |f | (t)asn →∞, for every t>0. We have Φ(λ χ H |f | (t)) = lim m→∞ Φ(λ χ H |g m | (t)) ≤ lim m→∞ Φ(λ χ H |f (·+t m )| (t)),t>0. (6) It follows from the definition of Φ that Φ(a + b) ≤ Φ(a)+Φ(b)fora, b ≥ 0. Observing that, for any f, g ∈ N Φ (R n )andt>0wehaveλ χ H (f +g) (2t) ≤ λ χ H f (t)+λ χ H g (t), then Φ(λ χ H |f (·+t m )−f | (2t)) ≤ Φ(λ χ H |f (·+t m )| (t)) + Φ(λ χ H |f | (t)). Hence for all m ≥ 1, 0 ≤ Φ(λ χ H |f (·+t m )| (t)) + Φ(λ χ H |f | (t)) − Φ(λ χ H |f (·+t m )−f | (2t)), ∀t>0. 210 Mai Thi Thu It is easy to check that χ H f(· + t m ) N Φ = χ H f N Φ , ∀m ≥ 1. Applying Fatou’s lemma to the sequence {Φ(λ χ H |f (·+t m )| (t)) + Φ(λ χ H |f | (t)) − Φ(λ χ H |f (·+t m )−f | (2t))}, we obtain ∞ 0 lim m→∞ Φ(λ χ H |f (·+t m )| (t)) + Φ(λ χ H |f | (t)) − Φ(λ χ H |f (·+t m )−f | (2t)) dt ≤ lim m→∞ ∞ 0 Φ(λ χ H |f (·+t m )| (t)) + Φ(λ χ H |f | (t)) − Φ(λ χ H |f (·+t m )−f | (2t) dt =2 ∞ 0 Φ(λ χ H |f | (t))dt − 1 2 lim m→∞ ∞ 0 Φ(λ χ H |f (·+t m )−f | (t))dt. (7) On the other hand, λ χ H |f (·+t m )−f | (t)=mes{x ∈H: |f (x + t m ) − f(x)| >t}. Therefore, since f(· + t m ) → f a.e. on H,wehave lim m→∞ λ χ H |f (·+t m )−f | (t)=0 and then lim m→∞ Φ(λ χ H |f (·+t m )−f | (t)) = 0. So, by (6) we get for any t>0 2Φ(λ χ H |f | (t)) = lim m→∞ Φ(λ χ H |g m | (t)) + Φ(λ χ H |f | (t)) − lim m→∞ Φ(λ χ H |f (·+t m )−f | (2t)) ≤ lim m→∞ Φ(λ χ H |f (·+t m )| (t)) + Φ(λ χ H |f | (t)) − Φ(λ χ H |f (·+t m )−f | (2t)) . (8) From (7) and (8), we have 2 ∞ 0 Φ(λ χ H |f | (t))dt ≤ 2 ∞ 0 Φ(λ χ H |f | (t))dt − 1 2 lim m→∞ ∞ 0 Φ(λ χ H |f (·+t m )−f | (t))dt. Hence ∞ 0 Φ(λ χ H |f (·+t m )−f | (t))dt → 0asm →∞, i.e., lim m→∞ χ H f(· + t m ) − f N Φ =0, A Gagliardo-Nirenber g Inequality for Lorentz Spaces 211 which contradicts (5). The cases 1 ≤|β|≤ are proved similarly. The continuity of D α F ε , 0 ≤ |β|≤ has been proved. The functions D β F ε , 0 ≤|β|≤ are continuous and bounded on R n .There- fore,itfollowsfromLemma3and(2)-(3)that D α f|| N Φ − ≤|D α F ε (0)|≤D α F ε ∞ ≤ ≤ C α, F ε 1− |α| ∞ |β|= D β F ε ∞ |α| , which together with (4) implies ||D α f|| N Φ − ≤ C α, f 1− |α| N Φ |β|= D β f N Φ |α| . By letting → 0wehave(1). Step 2. To complete the proof, it remains to show that D α f ∈ N Φ (R n ), ∀α : 0 < |α| <if f, D α f ∈ N Φ (R n ), |α| = .Sincef ∈ L 1,oc (R n )and D α f ∈ L 1,oc (R n ), |α| = ,wegetD α f ∈ L 1,oc (R n ), 0 < |α| <(see [5], p. 7). Let ψ(x) ∈ C ∞ 0 (R n ),ψ(x) ≥ 0,ψ(x)=0for |x|≥1and R n ψ(x)dx = 1. We put ψ λ (x)= 1 λ n ψ( x λ ),λ>0andf λ = f ∗ ψ λ .ThenD α f λ = f ∗ D α ψ λ , |α|≥0 and D α f λ = D α f ∗ ψ λ , 0 ≤|α|≤ in the D (R n ) sense. Actually, for every ϕ ∈ C ∞ 0 (R n ), D α f λ (x),ϕ(x) =(−1) |α| f λ (x),D α ϕ(x) =(−1) |α| R n R n f(x − y)ψ λ (y)dy D α ϕ(x)dx =(−1) |α| R n ψ λ (y) R n f(x − y)D α ϕ(x)dx dy = R n ψ λ (y) R n D α f(x − y)ϕ(x)dx dy = R n R n D α f(x − y)ψ λ (y)dy ϕ(x)dx = R n D α f(x − y)ψ λ (y)dy , ϕ(x) . TakingLemma1intoaccount,wegeteasilyD β f λ = f ∗ D β ψ λ ∈ N Φ (R n ), 0 ≤ |β|≤ and f λ N Φ = f ∗ ψ λ N Φ ≤f(x −·) N Φ ψ λ 1 = f N Φ , (9) D α f λ N Φ = D α f ∗ ψ λ N Φ ≤D α f(x −·) N Φ ψ λ 1 = D α f N Φ . (10) 212 Mai Thi Thu Applying (1) for f λ , we have from (9) - (10) for every α (0 < |α| <) ||D α f λ || N Φ ≤ C α, f λ 1− |α| N Φ |β|= D β f λ N Φ |α| ≤ C α, f 1− |α| N Φ |β|= D β f N Φ |α| . (11) Fix α (0 < |α| <). Since D α f ∈ L 1,oc (R n ), for each j =1, 2, , we have D α f ∗ ψ λ → D α f in L 1 ([−j, j] n )asλ → 0. Therefore, there is a sequence {λ j k } ∞ k=1 ,λ j k 0 such that D α f λ j k → D α f a.e. in [−j, j] n as k →∞.So,by the diagonal process, there exists a subsequence denoted by {λ m } ∞ 1 : λ m → 0 such that lim m→∞ D α f λ m (x)=D α f(x) (12) a.e. in R n . For each function v ∈ M Φ (R n ), v M Φ ≤ 1andm ≥ 1, by (11) - (12) and the definition of the Lorentz norm we get R n (D α f λ m )(x)v(x) dx ≤ C α, f 1− |α| N Φ |β|= D β f N Φ |α| . (13) Therefore, using Fatou’s lemma, (12) and (13), we obtain R n D α f(x)v(x)dx ≤ R n lim inf m→∞ D α f λ m (x)v(x) dx ≤ lim inf m→∞ R n (D α f λ m )(x)v(x) dx ≤ C α, f 1− |α| N Φ |β|= D β f N Φ |α| . (14) Because (14) is true for all v ∈ M Φ (R n ), v M Φ ≤ 1, by the definition of · N Φ we have ||D α f|| N Φ ≤ C α, f 1− |α| N Φ |β|= D β f N Φ |α| < ∞, 0 < |α| <. The proof is complete. By Theorem 1, we have Theorem 2. Let ≥ 2, f and its generalized derivatives D β f, |β| = be in N Φ (R n ).ThenD α f ∈ N Φ (R n ) for all α, 0 < |α| = r<and |α|=r D α f N Φ ≤ C r, f 1− r N Φ |β|= D β f N Φ r . A Gagliardo-Nirenber g Inequality for Lorentz Spaces 213 Corollary 1. Let ≥ 2, f and its generalized derivatives D β f, |β| = be in N Φ (R n ). Then D α f ∈ N Φ (R n ) for all α, 0 < |α| = r<and |α|=r D α f N Φ ≤ Ch − r −r f N Φ + Ch |β|= D β f N Φ , for all h>0 and C does not depend on f . Remark 1. Note that the Gagliardo-Nirenberg inequality for Orlicz spaces L Φ (R n ) was proved in [1] but the techniques used there cannot be applied to Lorentz spaces N Φ (R n ). In conclusion the author would like to thank professor Ha Huy Bang for the helpful suggestions. References 1. H. 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Steigerwalt, Some Banach Function Spaces Related to Orlicz Spaces,Uni- versity of Aberdeen Thesis, 1967. 10. E. M. Stein, Functions of exponential type, Ann. Math. 65 (1957) 582–592. . W ,∞ (R n ). The following result is an extension of the Gagliardo-Nirenberg inequality ([2 - 6]) to Lorentz spaces. Note that the Gagliardo-Nirenberg inequality has applications to partial differential. of the Gagliardo-Nirenberg inequality to Lorentz spaces. Let Φ : [0, ∞) → [0, ∞) be a non-zero concave function, which is non-decreasing and Φ(0+) = Φ(0) = 0. We put Φ(∞) = lim t→∞ Φ(t). For an. well-known Gagliardo-Nirenberg inequality: Lemma 3. [6] For fixed α, 0 < |α| <, there is the best constant C α, such that D α f ∞ ≤ C α, f 1− |α| ∞ |β|= D β f ∞ |α| , for any