Pacific Journal of Mathematics
ON A SHARP MOSER–AUBIN–ONOFRI INEQUALITY FOR FUNCTIONS ON S 2 WITH SYMMETRY Changfeng Gui and Juncheng Wei
Volume 194
No. 2
June 2000
PACIFIC JOURNAL OF MATHEMATICS Vol. 194, No. 2, 2000
ON A SHARP MOSER–AUBIN–ONOFRI INEQUALITY FOR FUNCTIONS ON S 2 WITH SYMMETRY Changfeng Gui and Juncheng Wei We show that for α ≥ 21 , the following inequality holds: Z Z 1 Z α 1 1 1 2g(x) (1 − x2 )|g 0 (x)|2 dx + g(x)dx − log e dx ≥ 0, 2 −1 2 −1 −1 R1 for every function g on (−1, 1) satisfying kgk2 = −1 (1 − R1 x2 )|g 0 (x)|2 dx < ∞ and −1 e2g(x) xdx = 0. This improves a result of Feldman et al., 1998, and answers a question of Chang and Yang in the axially symmetric case.
1. Introduction. On S 2 let Jα denote the functional on the Sobolev space H 1,2 (S 2 ) defined by Z Z Z 2 Jα (g) = α |∇g| dw + 2 gdw − log e2g dw. S2
S2
S2
Here dw R denotes the Lebesgue measure on the unit sphere, normalized to make S 2 dw = 1. The famous Moser-Trudinger inequality says that J1 is bounded below by a non-positive constant C1 . Later Onofri [6] showed that C1 can be taken to be 0. (Another proof was also given by OsgoodPhillips-Sarnack [7].) On the other hand, if we restrict Jα to the class 2g of to 0, that is R G 2gof functions g for which e has centre of mass equal 1 e ~ x dw = 0, then Aubin in [2] showed that for α ≥ , the functional 2 S2 Jα is again bounded below by a non-positive constant Cα . In [3] and [4] A. Chang and P. Yang showed that Cα = 0 for α close enough to 1. This led them to the following R Conjecture. Let G denote the functions in H 1,2 (S 2 ) for which S 2 e2g ~xdw = 0. If α ≥ 21 , then inf g∈G Jα (g) = 0. In this note, we prove this conjecture in the axially symmetric case. We note that Feldman, Froese, Ghoussoub and Gui [5] proved that the above conjecture holds for the axially symmetric case when α > 16 25 − for some small . They also gave an example which says the inequality is not true if α < 12 . It is also known that Jα (g) ≥ 0 if g is an even function, i.e., g(~x) = g(−~x) on S 2 . (See [7].) 349
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CHANGFENG GUI AND JUNCHENG WEI
Let θ and ϕ denote the usual angular coordinates on the sphere, and define x = cos(θ). Axially symmetric functions depend on x only. For such functions, it is well-known (see [5]) that the functional Jα can be written as Z 1 Z Z α 1 1 1 2g(x) 2 0 2 g(x)dx − log Iα (g) := (1 − x )|g (x)| dx + e dx. 2 −1 2 −1 −1 The set G is then replaced by Z 1 Z 2 0 2 (1 − x )|g (x)| < ∞, Gr := g|
1 2g(x)
e
xdx = 0 .
−1
−1
It is proved in [5, Proposition 3.1] that any critical point g of Iα restricted to Gr satisfies the following differential equation Z 1 2 (1.1) e2g dx. α((1 − x2 )g 0 )0 − 1 + e2g = 0, λ = λ −1 The main result of this note is the following: Theorem 1.1. If α ≥ 21 , then the only critical points of the functional Iα restricted to Gr are constant functions. As a consequence, the above theorem implies that the Conjecture of Chang and Yang is true in the axially symmetric case. Theorem 1.2. If α ≥ 21 , then Iα (g) ≥ 0 for g ∈ Gr . The rest of the paper is devoted to the study of (1.1). To this end, we need some notations and some basic facts. Let g be a solution of (1.1). Following [5], we set G = (1 − x2 )g 0 . Then G satisfies (see [5]) αG0 − 1 +
(1.2)
2 2g e = 0, λ
and (1.3)
(1 − x2 )G00 + α2 G − 2GG0 = 0 G(−1) = G(1) = 0.
We also need some facts about the Legendre’s polynomials. Let Pn (x) be the n−th Legendre polynomial, i.e., Pn satisfies ((1 − x2 )Pn0 )0 + λn Pn = 0, λn = n(n + 1), n = 0, 1, . . . . Note that 1 P0 = 1, P1 = x, P2 = (3x2 − 1), . . . 2
MOSER–AUBIN–ONOFRI INEQUALITY
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Moreover (see [1]) (1.4)
1 ≤ λn , 2
|Pn0 (x)|
Z
1
−1
Pn2 =
2 . 2n + 1
Acknowledgements. The research of the first author is supported by a NSERC grant of Canada, while the second author is supported by an Earmarked Grant from RGC of Hong Kong. This work was done when the first author visited the Chinese University of Hong Kong, to which the first author is very grateful for its hospitality. Both authors would like to thank the anonymous referee for carefully reading the first draft of the paper and giving valuable comments. 2. Proof of Theorem 1.1. In this section, we shall prove Theorem 1.1. Let ∞
X 1 G(x) = βx + a2 (3x2 − 1) + ak Pk (x), 2 k=3
G2 =
∞ X
ak Pk (x)
k=3
and b2k
=
a2k
Z
1
−1
Pk2 , k ≥ 2.
We first derive some equalities: Z 1 Z 1 2 2 0 2 (1 − x )(G ) = (2.1) −1 G2 , α −1 −1 Z 1 2 (2.2) P1 G = β, 3 −1 Z 1 e2g 2 (2.3) (1 − x2 ) = (1 − αβ), λ 3 −1 Z 1 Z 1 e2g 2 (2.4) Pk G = − (1 − x2 )Pk0 , k ≥ 2, αλk −1 λ −1 Z 1 2 2 (2.5) G2 = 6 − β, α 3 −1 Z 1 Z 1 2 2 2 (2.6) β 4β + 7 − −6 = (1 − x2 )(G02 )2 − 6 G22 , 3 α α −1 −1
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CHANGFENG GUI AND JUNCHENG WEI
Z
1
(2.7) −1
(1 − x2 )(G02 )2 − 6
Z
1
−1
G22 =
∞ X
(λk − 6)b2k .
k=3
Proofs of (2.1)-(2.7). Multiplying (1.3) by G and integrating over [−1, 1], we Robtain (2.1). The relation (2.2) follows by definition. Multiplying (1.2) x by −1 Pk (s)ds, k ≥ 1 and integrating over [−1, 1] we obtain (2.3) and (2.4). Multiplying (1.3) by x and integrating from −1 to 1 we obtain (2.5). To show (2.6), we just need to use (2.1), (2.5) and the definition of G2 . The equality (2.7) follows from definition. We will show β = 0, which implies G = 0 by (2.5). Our basic strategy is to show that if β 6= 0, then β=
1 , α
which will lead to a contradiction. Below we assume that β 6= 0. Next we obtain some inequalities. From (2.3) we have 1 − β > 0. α
(2.8) By definition we have b2k
R1 ( −1 GPk )2 = R1 = 2 −1 −1 Pk Z 1 2g 2 2k + 1 2 2 0 e ≤ (1 − x )|Pk | 2 αλk −1 λ 2 2k + 1 2 λk 2 ≤ (1 − αβ) . 2 αλk 2 3 a2k
Z
1
Pk2
Hence we obtain (2.9)
b2k ≤
2(2k + 1) 9
1 −β α
2 , k ≥ 2.
Similarly we obtain 3 1 |a2 | ≤ − β. 5 α
(2.10) From (2.6) (since β > 0),
2 4β + 7 − α
2 − 6 ≥ 0. α
MOSER–AUBIN–ONOFRI INEQUALITY
353
Since α ≥ 0.5, we have 1 β≥ 4
(2.11)
2 7− α
2 6− ≥ 1.5. α
From (2.6) and (2.8), we have 2 2 4 + 7− −6 ≥0 α α α which implies that α ≤ 0.537. From (2.6) we have 2 2 2 β 4β + 7 − −6 3 α α Z 1 Z 1 = (1 − x2 )(G02 )2 − 6 G22 −1
1 ≥ 2
Z
−1
1
−1
(1 − x2 )(G02 )2
Z 1 1 4 2 12 2 2 0 2 ≥ (1 − x )(G ) − β − a2 2 −1 3 5 1 2 2 2 4 2 12 2 ≥ −1 6− β − β − a2 . 2 α α 3 3 5 Hence we obtain 2 5 2 2 1 2 2 (2.12) β + 7− −6 − −1 6− 3 α α α 2 α α 1 6 10 − β − a22 ≥ β 3 α 5 2 10 1 6 25 1 ≥ β −β − × −β 3 α 5 9 α 10 1 1 ≥ 2β − −β . 3 α α Since ( α1 − β) ≥ 0, α ≥ 0.5 and 2β − α1 ≥ 0, we conclude that (since β > 0) 5 2 2 1 2 2 (2.13) 0≤ + 7− −6 − −1 6− ≤1 α α α 2 α α which implies, by a simple computation, that (2.14)
α ≤ 0.52.
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CHANGFENG GUI AND JUNCHENG WEI
Moreover since α ≥ 0.5 and β ≥ 1.5, we obtain from (2.12) and (2.13) that 1 β β −β ≤ ≤ . 1 α 5 5(2β − α )
(2.15)
To obtain better estimates, we fix an integer n ≥ 3. We have by (2.6) and (2.7) 2 2 2 β 4β + 7 − −6 3 α α ∞ X = (λk − 6)b2k k=3
=
≥
n X k=3 n X
(λk −
6)b2k
=
(λk − 6)b2k
k=n+1
(λk − 6)b2k +
∞ λn+1 − 6 X λk b2k λn+1 k=n+1
k=3 n X
+
∞ X
(λk − 6)b2k
k=3
! n λn+1 − 6 2 2 2 4 2 12 2 X + 6− − β − a2 − β −1 λk b2k λn+1 3 α α 3 5 k=3 n X λn+1 − 6 = λk − 6 − λk b2k λn+1 k=3 λn+1 − 6 2 2 2 4 2 12 2 + β −1 6− − β − a2 λn+1 3 α α 3 5 n X λk − λn+1 12 λn+1 − 6 = 6 b2k − a22 λn+1 5 λn+1 k=3 λn+1 − 6 2 2 2 4 2 β −1 6− − β . + λn+1 3 α α 3 Hence we have (2.16)
2 2 2 β 4β + 7 − −6 3 α α λn+1 − 6 2 2 2 4 2 − β −1 6− − β λn+1 3 α α 3 n X λk − λn+1 12 λn+1 − 6 ≥ 6 b2k − a22 . λn+1 5 λn+1 k=3
MOSER–AUBIN–ONOFRI INEQUALITY
355
After some simple computations, the left hand of (2.16) equals to 2 2 1 4β 2 − 12β −2 + −1 6− α λn+1 α α α 2 1 − 4β 1 − −β . λn+1 α Thus we have by (2.9), (2.10) and (2.16) 1 2 2 4β 2 (2.17) 12β − −2 + −1 6− α λn+1 α α α 1 2 12 λn+1 − 6 ≥ 4β 1 − − β − a22 λn+1 α 5 λn+1 2 n X λk − λn+1 2(2k + 1) 1 +6 −β λn+1 9 α k=3 2 20 λn+1 − 6 1 2 1 −β − −β ≥ 4β 1 − λn+1 α 3 λn+1 α 2 n 4 X λn+1 − λk 1 − −β (2k + 1) 3 λn+1 α k=3 2 ≥ 4β 1 − λn+1 20 λn+1 − 6 1 4 1 1 − − β − cn −β −β 3 λn+1 α 3 α α where n X λn+1 − λk cn = (2k + 1). λn+1 k=3
Since 1/2 < α ≤ 1 and λn > 2 for n ≥ 1, we have 4β 2 2 2 8β 1 12β −2 + −1 6− − − α λn+1 α α α λn+1 1 4 1 = 4β −2 3− −1 α λn+1 α ≤ 0. Thus the left hand side of (2.17) satisfies (2.18)
LHS of (2.17) ≤
8β . λn+1
We now claim (2.19)
1 4 −β ≤ , ∀n ≥ 4. α λn
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CHANGFENG GUI AND JUNCHENG WEI
By (2.18), we just need to show that the right hand side of (2.17) satisfies 1 RHS of (2.17) ≥ 2β (2.20) −β . α We prove it by induction. We first prove n = 4. To this end, we iterate the inequality (2.17). Note that the right hand side of (2.17) with n = 3 equals 20 20 − 6 1 2 − (2.21) 4β 1 − −β 20 3 20 α 1 1 4 20 − 12 ×7 −β −β − 3 20 α α 14 1 56 1 1 9 −β − −β −β ≥ 4β − 10 3 α 15 α α 126 1 1 ≥ 3.6β − −β −β 15 α α 126 β 1 ≥ 3.6β − −β (by (2.15)) 15 5 α 1 ≥ 1.92β −β . α By using (2.18) and (2.17) again, we obtain 8 1 1 −β ≤ < 0.25. α 20 1.92 Similarly, by using (2.22), we have 126 1 RHS of (2.17) ≥ 3.6β − × 0.25 −β (by (2.22)) 15 α 1 ≥ 2β −β (since β ≥ 1.5 by (2.11)). α
(2.22)
Thus (2.20) holds for n = 4 and hence (2.19) holds for n = 4. Let us now assume that 1 4 −β ≤ , k = n ≥ 4. α λk We observe that for n ≥ 4 cn =
n X
(2k + 1) −
k=3
=
n X k=3
(2k + 1) −
1 λn+1 1 λn+1
n X
λk (2k + 1)
k=3 n X k=3
k(k + 1)(2k + 1)
MOSER–AUBIN–ONOFRI INEQUALITY
357
1 36 . = λn+1 − 9 + 2 λn+1 Hence we have by (2.17) (2.23)
1 2 2 4β 2 12β − −2 + −1 6− α λn+1 α α α 2 ≥ 4β 1 − λn+1 1 20 λn+1 − 6 4 1 36 1 + − λn+1 − 9 + −β −β . 3 λn+1 3 2 λn+1 α α The right hand of (2.23) satisfies RHS of (2.23) 2 64 1 32 8 λn+1 1 ≥ 4β 1 − −β . + − − λn+1 3 λn λn λn+1 3 λn α To show (2.20), we only need to show 4 16 4 λn+1 32 1 β 1− + + , ≥− λn+1 3 λn λn λn+1 3 λn or β≥
4 λ2n+1 − 8λn+1 + 12 · . 3 λn (λn+1 − 4)
In view of the inductive assumption, it suffices to show 1 4 λn+1 λn+1 − 5 ≥ · . · α 3 λn λn+1 − 4 Because of (2.14), it is easy to verify that the above inequality holds for n ≥ 4. In conclusion, we have obtained (2.19). Finally we can finish the proof of Theorem 1.1. In fact, if we let n → +∞ in (2.19), we obtain 1 −β =0 α which is a contradiction to (2.8). This implies that β = 0 and therfore G ≡ 0. Hence g 0 ≡ 0, and g ≡ Constant.
358
CHANGFENG GUI AND JUNCHENG WEI
References [1] M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Washington, National Bureau of Standards Applied Mathematics, 1964. [2] T. Aubin, Mulleures constantes dans des theoremes d’inclusion de Sobolev at un theoreme de Fredholm non-linearire pour la transformation conforme de la courbure scalaire, J. Funct. Anal., 32 (1979), 149-179. [3] S.-Y.A. Chang and P. Yang, Conformal deformations of metrics on S 2 , J. Diff. Geom., 27 (1988), 215-259. [4]
, Precribing Gaussian curvature on S 2 , Acta Math., 159 (1987), 214-259.
[5] J. Feldman, R. Froese, N. Ghoussoub and C. Gui, An improved Moser-Aubin-Onofri inequality for radially symmetric functions on S 2 , Cal. Var. PDE, 6 (1998), 95-104. [6] E. Onofri, On the positivity of the effective action in a theorem on radom surfaces, Comm. Math. Phys., 86 (1982), 321-326. [7] B. Osgood, R. Phillips and P. Sarnack, Extremals of determinants of Laplacians, J. Funct. Anal., 80 (1988), 148-211. Received September 10, 1998. Department of Mathematics University of British Columbia Vancouver, BC V6T 1Z2 Canada E-mail address:
[email protected] Department of Mathematics The Chinese University of Hong Kong Shatin, Hong Kong China E-mail address:
[email protected] 
