A sharp Trudiger-Moser type iequality for ubouded domais i R Yuxiag Li ad Berhard Ruf Abstract The Trudiger-Moser iequality states that for fuctios u H, 0 (Ω) (Ω R a bouded domai) with Ω u dx oe has Ω (eα u )dx c Ω, with c idepedet of u. Recetly, the secod author has show that for = 2 the boud c Ω may be replaced by a uiform costat d idepedet of Ω if the Dirichlet orm is replaced by the Sobolev orm, i.e. requirig Ω ( u + u )dx. We exted here this result to arbitrary dimesios > 2. Also, we prove that for Ω = R the supremum of (e α u R fuctios is attaied. The proof is based o a blow-up procedure. )dx over all such Keywords: Trudiger-Moser iequality, blow-up, best costat, ubouded domai. Mathematics subject classificatio (2000): 35J50, 46E35 Itroductio Let H,p 0 (Ω), Ω R, be the usual Sobolev space, i.e. the completio of C0 ( ) u H,p (Ω) = ( u p + u p p )dx. Ω It is well-ow that H,p p 0 (Ω) L p (Ω) if p < H,p 0 (Ω) L (Ω) if < p The case p = is the it case of these embeddigs ad it is ow that H, 0 (Ω) L q (Ω) for q < +. (Ω) with the orm Whe Ω is a bouded domai, we usually use the Dirichlet orm u D = ( u dx) i place of H,. I this case, we have the famous Trudiger-Moser iequality (see [P], [T], [M]) for the it case p = which states that { sup (e α u < + whe α α )dx = c(ω, α) (.) u D Ω = + whe α > α where α = ω, ad ω is the measure of the uit sphere i R. The Trudiger-Moser result has bee exteded to Sobolev spaces of higher order ad Soboleve spaces over compact
maifolds (see [A], [Fo]). Moreover, for ay bouded Ω, the costat c(ω, α ) ca be attaied. For the attaiability, we refer to [C-C], [F], [Li], [L], [L2], [d-d-r], [L3]. Aother iterestig extesio of (.) is to costruct Trudiger-Moser type iequalities o ubouded domais. Whe = 2, this has bee doe by B. Ruf i [R]. O the other had, for a ubouded domai i R, S. Adachi ad K. Taaa ([A-T]) get a weaer result. Let S. Adachi ad K. Taaa s result says that: 2 Φ(t) = e t t j j!. Theorem A For ay α (0, α ) there is a costat C(α) such that j= u u Φ(α( ) L )dx C(α) (R ) R u L (R ) u L (R ) I this paper, we shall discuss the critical case α = α. followig:, for u H, (R ) \ {0}. (.2) More precisely, we prove the Theorem.. There exists a costat d > 0, s.t. for ay domai Ω R, sup Φ(α u )dx d. (.3) u H, (Ω), u H, (Ω) Ω The iequality is sharp: for ay α > α, the supremum is +. We set Further, we will prove S = Theorem.2. S is attaied. u H, (R ) =, s.t. sup u H, (R ), u H, (R ) R Φ(α u )dx. I other words, we ca fid a fuctio u H, (R ), with S = R Φ(α u )dx. The secod part of Theorem. is trivial: Give ay fixed α > α, we tae β (α, α). By (.) we ca fid a positive sequece {u } i {u H, 0 (B ) : u dx = }, B such that + e βu B = +. By Lio s Lemma, we get u 0. The by the compact embeddig theorem, we may assume u L p (B ) 0 for ay p >. The, R ( u + u )dx, ad u α( u H, ) 2 > βu
whe are sufficietly large. So, we get u Φ(α( + R u H, ) )dx + (e βu B )dx = +. The first part of Theorem. ad Theorem.2 will be proved by blow up aalysis. We will use the ideas from [L] ad [L2] (see also [A-M] ad [A-D]). However, i the ubouded case we do ot obtai the strog covergece of u i L (R ), ad so we eed more techiques. Cocretely, we will fid positive ad symmetric fuctios u H, 0 (B R ) which satisfy B R ( u + u )dx = ad Φ(β u )dx = sup B R B ( v + v )=, v H, R 0 (B R ) B R Φ(β v )dx. Here, β is a icreasig sequece tedig to α, ad R is a icreasig sequece tedig to +. Furthermore, u satisfies the followig equatio: div u 2 u + u = u Φ (β u ), λ where λ is a Lagrage multiplier. The, there are two possibilities. If c = max u is bouded from above, the it is easy to see that + R (Φ(β u where u is the wea it of u. R Φ(α u )dx, or ) β u ( )! )dx = (Φ(α u α ) u R ( )! )dx It the follows that either R Φ(β u S α ( )!. If c is ot bouded, the ey poit of the proof is to show that β c (u (r x) c ) log( + c r ), locally for a suitably chose sequece r (ad with c = ( ω ) ), ad that c u G, )dx coverges to o ay Ω R \ {0}, where G is some Gree fuctio. This will be doe i sectio 3. The, we will get i sectio 4 the followig 3
Propositio.3. If S ca ot be attaied, the where A = r 0 (G(r) + α log r ). S mi{ α ( )!, ω eαa++/2+ +/() }, So, to prove the attaiability, we oly eed to show that S > mi{ α ( )!, ω eαa++/2+ +/() }. I sectio 5, we will costruct a fuctio sequece u ɛ such that Φ(α uɛ )dx > ω eαa++/2+ +/() R whe ɛ is sufficietly small. Ad i the last sectio we will costruct, for each > 2, a fuctio sequece u ɛ such that for ɛ sufficietly small Φ(α uɛ )dx > α R ( )!. Thus, together with Ruf s result of attaiability i [R] for the case = 2, we will get Theorem.2. 2 The maximizig sequece Let {R } be a icreasig sequece which diverges to ifiity, ad {β } a icreasig sequece which coverges to α. By compactess, we ca fid positive fuctios u H, 0 (B R ) with B R ( u + u )dx = such that Φ(β u )dx = sup Φ(β v )dx. B R B ( v + v )=, v H, B R R Moreover, we may assume that R Φ(β u 0 (B R ) )dx = B R Φ(β u Lemma 2.. Let u as above. The a) u is a maximizig sequece for S; b) u may be chose to be radially symmetric ad decreasig. )dx is icreasig. Proof. a) Let η be a cut-off fuctio which is o B ad 0 o R \ B 2. The give ay ϕ H, (R ) with R ( ϕ + ϕ )dx =, we have τ (L) := ( η( x R L )ϕ + η( x L )ϕ )dx, as L +. Hece for a fixed L ad R > 2L ϕ Φ(β τ(l) )dx B 2L Φ(β η( x L )ϕ τ(l) )dx 4 B R Φ(β u )dx
By the Levi Lemma, we the have ϕ Φ(α τ(l) )dx + The, lettig L +, we get Hece, we get + R Φ(α ϕ )dx + Φ(β u )dx = sup R R Φ(β u R Φ(β u R ( v + v )=, v H, (R ) )dx. )dx. R Φ(α v )dx. b) Let u be the radial rearragemet of u, the we have τ := ( u + u )dx ( u + u )dx =. B R B R It is well-ow that τ = iff u is radial. Sice Φ(β u )dx = Φ(β u B R B R we have Φ(β ( u ) )dx B R τ B R Φ(β u ad = holds iff τ =. Hece τ = ad Φ(β u )dx = sup B R B ( v + v )=, v H, R 0 (B R ) )dx, )dx, B R Φ(β v )dx. So, we ca assume u = u ( x ), ad u (r) is decreasig. Assume ow u u. The, to prove Theorem. ad.2, we oly eed to show that Φ(β u + )dx = Φ(α u )dx. R R 3 Blow up aalysis By the defiitio of u we have the equatio div u 2 u + u where λ is the costat satisfyig λ = First, we eed to prove the followig: u B R = u Φ (β u Φ (β u ), (3.) λ )dx. 5
Lemma 3.. if λ > 0. Proof. Assume λ 0. The u dx C R u R Φ (β u )dx Cλ 0. Sice u ( x ) is decreasig, we have u (L) u, ad the Set ɛ = u (L) ω L. (3.2) ω L. The u (x) ɛ for ay x /, ad hece we have, usig the form of Φ, that Φ(β u )dx C u dx Cλ 0. R \ R \ Ad o, sice u 0 i L q ( ) for ay q >, we have by Lebesgue [ Φ(β u + )dx Cu + Φ (β u )dx + Cλ + Φ(0)dx + = 0. {x :u (x) } Φ(β u )dx ] This is impossible. We deote c = max u = u (0). The we have Lemma 3.2. If sup c < +, the i) Theorem. holds; ii) if S is ot attaied, the S α ( )!. Proof. If sup c < +, the u u i C loc (R ). By (3.2), we are able to fid L s.t. u (x) ɛ for x /. The R \ (Φ(β u Lettig ɛ 0, we get Hece + + ) β u ( )! )dx C R (Φ(β u Φ(β u ) = R 2 u R \ ) β u ( )! )dx = R Φ(α u )dx + 6 dx Cɛ 2 R u (Φ(α u α ) u R ( )! )dx. α ( )! + 2 dx Cɛ. R (u u )dx. (3.3)
Whe u = 0, we ca deduce from (3.3) that Now, we assume u 0. Set τ = By the Levi Lemma, we have τ. Let ũ = u( x τ ). The, we have ũ dx = R ad The ũ dx = τ R R ( ũ + ũ )dx S + α ( )!. R u dx R u dx = + u dx R. u dx R + + R u dx, R u dx. ( u + u )dx =. R Hece, we have by (3.3) S Φ(α ũ )dx R = τ Φ(α u )dx R [ = Φ(α u )dx + (τ α ] ) R R ( )! u dx = Φ(β u + )dx + (τ ) (Φ(α u ) R R = S + (τ ) (Φ(α u α ) R ( )! u )dx + (τ ) α ( )! u )dx R (Φ(α u ) α ( )! u )dx Sice Φ(α u ) α ()! u > 0, we have τ =, ad the So, u is a extremal fuctio. S = R Φ(α u )dx. From ow o, we assume c +. We perform a blow-up procedure: We defie r = c λ e β c. 7
By (3.2) we ca fid a sufficietly large L such that u o R \. The (u u (L)) + dx ad hece, by (.), we have e α[(u u (L))+ ] C(L). Clearly, for ay p < α we ca fid a costat C(p), s.t. pu α [(u u (L)) + ] + C(p), ad the we get Hece, λ e β 2 c = e β 2 c C [ e pu u dx e R \ u R \ β 2 c + dx < C = C(L, p). Φ (β u )dx + u e β 2 u u dx. Φ (β u )dx ] Sice u coverges strogly i L q ( ) for ay q >, we get λ Ce β 2 c Now, we set r Ce β 2 c. v (x) = u (r x), w (x) = β c (v c ),, ad hece where v ad w are defied o Ω = {x R : r x B }. Usig the defiitio of r ad (3.) we have div w 2 w = v ( β ) e β (v c ) + O(r c ). c By Theorem 7 i [S], we ow that osc BR ω C(R) for ay R > 0. The from the result i [T] (or [D]), it follows that w C,δ (B R ) < C(R). Therefore w coverges i Cloc ad v c 0 i Cloc. Sice we get β (v v c = c ( + v c ) c ) w i Cloc 0, ad so we have = c ( + v c + O( c c 2 )), div w 2 w = ( α ) e w, (3.4) 8
with w(0) = 0 = max w. Sice ω is radially symmetric ad decreasig, it is easy to see that (3.4) has oly oe solutio. We ca chec that w(x) = log( + c x ), ad e w dx =, R where c = ( ω ). The, L + + u e β u r λ dx = L + e w dx =. (3.5) For A >, let u A = mi{u, c A }. We have Lemma 3.3. For ay A >, there holds sup ( u A + u A )dx + R A. (3.6) Proof. Sice {x : u c A } c A {u c A } u, we ca fid a sequece ρ 0 s.t. {x : u c A } B ρ. Sice u coverges i L p (B ) for ay p >, we have u A + {u > c p dx A } + u p {u > c dx = 0, A } ad + R (u c A )+ u p dx = 0 for ay p > 0. Hece, testig equatio (3.) with (u c A ) +, we have R ( (u c A )+ + (u c ) A )+ u dx = (u c R A )+ u λ e β u dx + o() (u c u r A )+ e β u dx + o() λ v c /A = ( v c + ) e w +o() dx + o(). c c Hece if + R ( (u c A )+ + (u c ) A )+ u dx A e w dx. A 9
Lettig L +, we get if + R Now observe that ( u A + u A )dx = R ( (u c A )+ + (u c ) A )+ u dx A A. + R ( (u c A )+ + (u c ) A )+ u dx (u c R A )+ u dx ( A ) + o(). u {u > c dx + A } u A {u > c dx A } Hece, we get this Lemma. Corollary 3.4. We have + R \B δ ( u + u )dx = 0, for ay δ > 0, ad the u = 0. Proof. Lettig A +, the for ay costat c, we have ( u + u )dx 0. {u c} So we get this Corollary. Lemma 3.5. We have Φ(β u + )dx R ad cosequetly Proof. We have Φ(β u dx) R L + + λ c +, ad sup {u c A } (e β u r Φ(β u )dx + R Φ(β (u A ) )dx + A )dx = sup c λ, (3.7) c < +. (3.8) λ {u > c A } c λ Φ (β u u R λ )dx Φ (β u )dx. 0
Applyig (3.2), we ca fid L such that u o R \. The by Corollary 3.4 ad the form of Φ, we have Φ(pβ (u A ) )dx C(p) u dx = 0 (3.9) + R \ R \ for ay p > 0. Sice by Lemma 3.3 sup R ( u A + + u A )dx A < whe A >, it follows from (.) that sup dx < + for ay p < A. Sice for ay p < p we have p(u A ) sup e p β ((ua u (L))+ ) p ((u A u (L)) + ) + C(p, p ), Φ(pβ (u A ) )dx < + (3.0) for ay p < A. The o BL, by the wea compactess of Baach space, we get + Φ(β (u A ) )dx = Φ(0)dx = 0. Hece we have + Φ(β u )dx R A L + = A + c λ c λ + Cɛ. u λ Φ (β u )dx + Cɛ As A ad ɛ 0 we obtai (3.7). If λ c was bouded or sup c λ = +, it would follow from (3.7) that sup R ( v + v )dx=,v H, (R ) R Φ(α v )dx = 0, which is impossible. λ u Lemma 3.6. We have that c we have + Φ (β u ) coverges to δ 0 wealy, i.e. for ay ϕ D(R ) R ϕ c u λ Φ (β u )dx = ϕ(0).
Proof. Suppose supp ϕ B ρ. We split the itegral B ρ ϕ c u λ Φ (β u )dx + {u c A }\r + r {u < c A } We have I A ϕ C 0 u R \r λ = I + I 2 + I 3. Φ (β u )dx = A ϕ C 0( e w +o() dx), ad I 2 = ϕ(r x) c (c + (v c )) B L c By (3.9) ad (3.0) we have e w+o() dx = ϕ(0) R Φ(pβ u A )dx < C for ay p < A. We set q + p =. The we get by (3.8) e w dx + o() = ϕ(0) + o(). I 3 = {u c A } ϕ c u λ Φ (β u )dx c λ ϕ C 0 u L q (R ) e β u A L p (R ) 0. Lettig L +, we deduce ow that + R ϕ c u λ Φ (β u )dx = ϕ(0). Propositio 3.7. O ay Ω R \{0}, we have that c G C,α loc (R \ {0}) satisfies the followig equatio: Proof. We set U = c u coverges to G i C (Ω), where div G 2 G + G = δ 0. (3.) u, which satisfy by (3.) the equatios: div U 2 U + U For our purpose, we eed to prove that = c u λ B R U q dx C(q, R), 2 Φ (β u ). (3.2)
where C(q, R) does ot deped o. We use the idea i [St] to prove this statemet. Set Ω t = {0 U t}, U t = mi{u, t}. The we have ( U t + U t )dx Ω t ( U t U + U t U ) = R R U t c u λ Φ (β u )dx 2t. Let η be a radially symmetric cut-off fuctio which is o B R ad 0 o B c 2R. The, B 2R ηu t dx C (R) + C 2 (R)t. The, whe t is bigger tha C (R) C 2 (R), we have B 2R ηu t dx 2C 2 (R) t. Set ρ such that U (ρ) = t. The we have { } if v dx : v H, 0 (B 2R ) ad v Bρ = t 2C 2 (R) t. B 2R O the other had, the if is achieved by t log x 2R / log 2R ρ. By a direct computatio, we have ad hece for ay t > C (R) C 2 (R) ω t (log 2R ρ ) 2C 2(R), {x B 2R : U t} = B ρ C 3 (R)e A(R)t, where A(R) is a costat oly depedig o R. The, for ay δ < A, B R e δu dx µ({m U m + })e δ(m+) m=0 e (A δ)m e δ C. m=0 The, testig the equatio (3.2) with the fuctio log +2(U U (R)) + +(U U (R)) + U ( + U U (R))( + 2U 2U (R)) dx B R log 2 B R c u λ Φ (β u )dx, we get U log + 2(U U (R)) B R + (U U (R)) dx C. Give q <, by Youg s Iequality, we have [ U q U dx B R B R ( + U U (R))( + 2U 2U (R)) + (( + U )( + 2U )) [ U ] ( + U U (R))( + 2U 2U (R)) + CeδU dx. B R 3 q ] dx
Hece, we are able to assume that U coverges to a fuctio G wealy i H,p (B R ) for ay R ad p <. Applyig Lemma 3.6, we get (3.). Hece U is bouded i L q (Ω) for ay q > 0. By Corollary 3.4 ad Theorem A, e β u is also bouded i L q (Ω) for ay q > 0. The, applyig Theorem 2.8 i [S], ad the mai result i [T] (or [D]), we get U C,α (Ω) C. So, U coverges to G i C (Ω). For the Gree fuctio G we have the followig results: Lemma 3.8. G C,α loc (R \ {0}) ad ear 0 we ca write G = α log r + A + O(r log r) ; (3.3) here, A is a costat. Moreover, for ay δ > 0, we have ( c + u + (c u ) )dx = ( G + G )dx R \B δ R \B δ = G(δ)( G dx). B δ Proof. Slightly modifyig the proof i [K-L], we ca prove G = α log r + A + o(). Oe ca refer to [L2] for details. Further, testig the equatio (3.2) with, we get ω G (r) r 2 G = G B r = G dx = + O(r log r). B r The, we get (3.3). We have u R \B δ Φ (β u )dx C Recall that U H, 0 (B R ). By equatio (3.2) we get R \B δ ( U + U )dx = c By (3.4) ad (3.8) we the get + λ u R \B δ Φ (β u u dx 0. (3.4) R \B δ ( U + U )dx = R \B δ + = G(δ) )dx B δ U U 2 U ds. B δ U U 2 U ds G 2 ds = G(δ)( G dx). B δ 4 B δ G
We are ow i the positio to complete the proof of Theorem.: We have see i (3.9) that Φ(β u )dx C. R \B R So, we oly eed to prove o B R, e β u B R dx < C The classical Trudiger-Moser iequality implies that dx < C = C(R). By Propositio 3.7, u (R) = O( B R e β ((u u (R))+ ) c ), ad hece we have u ((u u (R)) + + u (R)) ((u u (R)) + ) + C, The, we get e β u B R C. 4 The proof of Propositio.3 We will use a result of Carleso ad Chag (see [C-C]): Lemma 4.. Let B be the uit ball i R. Assume that u is a sequece i H, 0 (B) with B u dx =. If u 0, the sup (e α u )dx B e +/2+ +/(). + Proof of Propositio.3: Set u (x) = (u (x) u (δ)) + u L (Bδ ) of Carleso ad Chag, we have sup + By Lemma 3.8, we have ( c R \B δ B e β u B δ u + (c which is i H, 0 (B δ ). The by the result B δ ( + e +/2+ +/() ). u ) )dx G(δ)( B δ G dx), 5
ad therefore we get u dx = B δ ( u + u )dx R \B δ u dx = G(δ) + ɛ (δ), (4.) B δ c where ɛ (δ) = 0. δ 0 + By (3.9) i Lemma 3.5 we have L + + e β u B ρ\r dx = B ρ, for ay ρ < δ. Furthermore, o B ρ we have by (4.) (u ) u ( G(δ)+ɛ (δ) c ) = u ( + G(δ) + ɛ (δ) + O( c c 2 = u + G(δ)(u ) + O(c c ) )) u log δ ( )α. The we have L + + e β u B ρ\r dx O(δ ) L + + e β u B ρ\r dx B ρ O(δ ). Sice u 0 o B δ \ B ρ, we get (e βu + B δ \B ρ the Lettig ρ 0, we get 0 L + + L + + B δ \r (e β u B δ \r (e β u )dx = 0, )dx B ρ O(δ ). )dx = 0. So, we have L + + r (e β u )dx e +/2+ +/() B δ. 6
Now, we fix a L. The for ay x r, we have u β u = β ( u ) L ( u (Bδ ) dx) B δ = β (u + u (δ) u ) L ( u (Bδ ) dx) B δ ( usig that u (δ) = O( ) ad u L (B δ ) = + O( = β ( u + u (δ) + O( + c = β u ( + u (δ) + O( u [ c = β u It is easy to chec that c ) ) ( B δ u dx ) 2 + u (δ) u ) ) ( G(δ)+ɛ (δ) c ] G(δ)+ɛ (δ) c + O( 2 c ) u (r x) c, ad ( u (r x) ) u (δ) G(δ). ). c ) ) So, we get L + + (e β u r )dx = L + + eαg(δ) r (e β u )dx e αg(δ) δ ω e +/2+ +/() = e α( α log δ +A+O(δ log δ)) δ ω e +/2+ +/(). Lettig δ 0, the the above iequality together with Lemma 3.2 imply Propositio.3. 5 The test fuctio I this sectio, we will costruct a fuctio sequece {u ɛ } H, (R ) with u ɛ H, = which satisfies Φ(α u ɛ ω )dx > eαa++/2+ +/(), R for ɛ > 0 sufficietly small. Let C () log(+c x ɛ )+Λ ɛ x Lɛ u ɛ = α C G( x ) C x > Lɛ, where Λ ɛ, C ad L are fuctios of ɛ (which will be defied later, by ( 5.), (5.2), (5.5) ) which satisfy 7
ad i) L +, C +, ad Lɛ 0, as ɛ 0 ; ii) C iii) log L C () log(+cl )+Λ ɛ α C 0, as ɛ 0. = G(Lɛ) C ; We use the ormalizatio of u ɛ to obtai iformatio o Λ ɛ, C ad L. We have ( u ɛ + u ( ɛ )dx = R \ɛ C G dx + G dx ) Bc Lɛ BLɛ c 2 G = C G(Lɛ) G ɛ ds G(Lɛ) G(Lɛ) G dx B = Lɛ. ɛ u ɛ dx = = α C α C cl 0 cl 0 C u ( + u) du (( + u) ) ( + u) du where we used the fact = α C + α C 2 =0 C ( ) log( + c L ) + O( L ) C = ( ) + /2 + /3 + + /( ) α C + α C log( + c L ) + O( L ), C 2 =0 C ( ) = + 2 + +. It is easy to chec that ad thus we get R ( u ɛ + u ɛ )dx = ɛ u ɛ dx = O((Lɛ) C log L), { ( ) ( + /2 + + /( ) ) + α α C A } +( ) log( + c L ) log(lɛ) + φ, 8
where ) φ = O ((Lɛ) C log L + (Lɛ) log Lɛ + L. Settig R ( u ɛ + u ɛ )dx =, we obtai α C = ( ) ( + /2 + + /( ) ) + α A + log (+cl ) L log ɛ + φ By ii) we have ad hece this implies that = ( ) ( + /2 + + /( ) ) + α A + log ω log ɛ + φ. α C ( ) log( + c L ) + Λɛ = αg(lɛ) ( ) ( + /2 + + /( ) ) + α A log (Lɛ) + φ + Λ ɛ = αg(lɛ) ; Next, we compute (5.) Λ ɛ = ( )( + /2 + + /( )) + φ. (5.2) e α uɛ ɛ dx. Clearly, ϕ(t) = t + t is icreasig whe 0 t ad decreasig whe t 0, the t t, whe t <. Thus we have by ii), for ay x ɛ The we have α uɛ = α C ( ) log( + c x ɛ ) + Λɛ α C α C ( ) log( + c x ( ɛ ) + Λɛ ). α C ɛ e α uɛ dx e αc log(+c x ɛ ) Λɛ ɛ = e αc Λɛ ɛ ( + c x dx ) = e αc Λɛ ( )ɛ cl = e αc Λɛ ( )ɛ cl 0 0 = e αc Λɛ ɛ ( + O(L )) u 2 ( + u) du ((u + ) ) 2 ( + u) du (5.3) = ω eαa++/2+ +/() ) + O ((Lɛ) C log L + L + (Lɛ) log Lɛ. 9
Here, we used the fact The ɛ Φ(α u ɛ m =0 ( ) m m + C m = m +. )dx ω ) eαa++/2+ +/() +O ((Lɛ) C log L + L + (Lɛ) log Lɛ. Moreover, o R \ ɛ we have the estimate Φ(α uɛ )dx α R \ɛ ( )! R \ɛ G(x) C dx, ad thus we get Φ(α uɛ )dx ω R + α G(x) ()! R \ɛ C = ω eαa++/2+ +/() [ + α ()! C R \ɛ eαa++/2+ +/() ) dx + O ((Lɛ) C log L + L + (Lɛ) log Lɛ G(x) dx + O ((Lɛ) C + C log L + L ) ] + C (Lɛ) log Lɛ (5.4) We ow set the Lɛ 0 as ɛ 0. equatio (5.). We set L = log ɛ ; (5.5) We the eed to prove that there exists a C = C(ɛ) which solves Sice f(t) = α t ( )( + /2 + + /( )) + α A + log ω log ɛ + φ, for ɛ small, ad for ɛ small, f has a zero i satisfies Therefore, as ɛ 0, we have f(( 2 α log ɛ ) ) = log ɛ + o() + φ < 0 f(( log ɛ ) ) = 2α 2 log ɛ + o() + φ > 0 ( ( 2α log ɛ ), ( 2 α log ɛ ) α C = log ɛ + O(). log L C 0, ). Thus, we defied C, ad it 20
ad the (Lɛ) C + log L + C L + C (Lɛ) log Lɛ 0. Therefore, i), ii), iii) hold ad we ca coclude from (5.4) that for ɛ > 0 sufficietly small Φ(α uɛ )dx > ω eαa++/2+ +/(). R 6 The test fuctio 2 I this sectio we costruct, for > 2, fuctios u ɛ such that u ɛ Φ(α ( ) α )dx > R u ɛ H, ( )!, for ɛ > 0 sufficietly small. Let ɛ = e αc, ad u ɛ = c log x L α c x < Lɛ Lɛ x L 0 L x, where L is a fuctio of ɛ which will be defied later. ad We have The u ɛ Φ(α ( R u ɛ H, u ɛ dx = ω R ) α )dx ( )! R u ɛ =, c (Lɛ) + ω L α c R u ɛ dx + R u ɛ dx + α! ɛ r log rdr. 2 R \ɛ uɛ ( + R u ɛ dx) dx = α ( )! α ( )! + ω c (Lɛ) + ω L α c ɛ r log rdr + α! ( 2 ω L /c () 2 ( α ) 2 + ω c (Lɛ) + ω L α c ɛ r log 2 r ɛ r log rdr ) We ow as that L satisfies c 0, as ɛ 0. (6.) L 2
The, for sufficietly small ɛ, we have α ( )! + α! ( + ω c (Lɛ) + ω L α c 2 ω L /c () 2 ( α ) 2 + ω c (Lɛ) + ω L α c 2 c B L B2 L = c L ( = c L (B 2 2 c L ( 2) c B 2 ) B 2 ), ɛ r log rdr + ɛ r log 2 r ɛ r log rdr ) where B, B 2 are positive costats. Whe > 2, we may choose L = b c 2 ; the, for b sufficietly large, we have L ( 2) B B c 2 = B b ( 2) B 2 > 0, ad (6.) holds. Thus, we have proved that for ɛ > 0 sufficietly small u ɛ Φ(α ( R u ɛ H, (R ) ) )dx > α ( )!. Refereces [A-T] S. Adachi ad K. Taaa: Trudiger type iequalites i R ad their best expoets. Proc. AMS, 28: 205-2057,999. [A] D. R. Adams: A sharp iequality of J. Moser for higher order derivatives. Aals of Math., 28: 385-398, 988. [A-D] Adimurthi ad O.Druet, Blow up aalysis i dimesio 2 ad a sharp form of Trudiger Moser iequality, Comm. PDE 29 No -2, (2004), 295-322. [A-M] Adimurthi ad M. Struwe: Global Compactess properties of semiliear elliptic equatios with critical expoetial growth. J. Fuct. Aal., 75, o :25 67, 2000. [C-C] L. Carleso ad S. Y. A. Chag: o the existece of a extremal fuctio for a iequality of J.Moser. Bull. Sc. Math., 0, 3-27, 986. [d-d-r] D.G. de Figueiredo, J.M. do O, B.Ruf: O a iequality by N.Trudiger ad J.Moser ad related elliptic equatios. Comm. Pure. Appl. Math., 55, 35-52, 2002. [D] E. DiBeedetto: C,α local regularity of wea solutio of degeerate elliptic equatios. Noliear Aalysis. 7 o.8, 827-850, 983. 22
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