An Improved Leray-Trudinger Inequality

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by  Arka Mallick

AN IMPROVED LERAY-TRUDINGER INEQUALITY
ARKA MALLICK † AND CYRIL TINTAREV⋄
Abstract. Inthisarticle,wehavederivedthefollowingLeray-Trudingertypeinequality
on a bounded domain Ω in Rn containing theorigin.
n
16 u∈W01,n(Ωs)u,Ipn[u,Ω,R]≤1ZΩecn E|2βu((x|Rx)||)!n−1dx<+∞ , for some cn >0 dependingonly on n.
0
2 Here β = n2, In[u,Ω,R] := RΩ|∇u|ndx − (cid:0)n−n1(cid:1)nRΩ |x|n|Eu1n|n(|Rx|)dx, R ≥ xsu∈pΩ|x| and
r E (t) := log(e), E (t) := log(eE (t)) for t ∈ (0,1]. This improves an earlier result by
a 1 t 2 1
Psaradakis and Spector. Also we have proved that for any c > 0 in the place of c the
M n
above inequality is false if we take β< 1.
n

1
2
MSC2010 Classification: 46E35, 26D15,35J92

]
P Keywords: Hardy inequality, Leray potential, Borderline Sobolev embedding
A
. 1. Introduction
h
t
a In this article we intend to discuss some Leray-Trudinger type inequalities. These type
m of inequalities are closely related to different types of Trudinger-Moser inequalities and
[ Hardy inequalities. So let us quickly recall some relevant results about these inequalities.

2
v Let Ω be any bounded domain in Rn. Then the celebrated Sobolev embedding theorem
4 asserts that for p < n
9
np
1 W1,p(Ω)⊂ Lq(Ω) for any q satisfying 1 ≤ q ≤ p∗ = .
3 0 n−p
0
. This immediately leaves us with the following question. What happens when p = n? Note
1
thatherep∗ isformally +∞.Asexpected, inthiscaseonecan provethatW1,n(Ω)⊂ Lq(Ω)
0 0
6 for any q satisfying, 1 ≤ q < +∞. However, for q = +∞, easy example shows that one can
1 not get the above embedding. This raises the the following interesting question. What is
:
v the maximal growth function f(t) for which the following is true:
i
X
1,n
u∈ W (Ω) implies f(u)dx < ∞? (1.1)
r 0
a ZΩ
By Sobolev embedding, (1.1) is true when f is a polynomial, but in fact it is allowed to
have exponential growth, namely there exists some constant c(n) > 0 depending only on
n such that
n
sup ec(n)|u|n−1dx < ∞. (1.2)
u∈W01,n(Ω),||∇u||Ln(Ω)≤1ZΩ
† TIFR Centre for Applicable Mathematics, Post Bag No. 6503 Sharadanagar,
Yelahanka NewTown, Bangalore 560065.
Email:[email protected]
⋄Email:[email protected].
1
2 AN IMPROVED LERAY-TRUDINGER INEQUALITY
This result is most often attributed to Trudinger [30], although it was obtained earlier
by Yudovich [33], Peetre [24] and Pohozaev [26]. The proof in this paper still follows the
methodof[30]basedongettingsomeexpedientupperboundontheLq normofuthatyields
(1.2). Later Hempel, Morris and Trudinger (see [17]) together proved that the function
n
f(t) = etn−1 is in fact the function with maximal growth. In 1971, Moser proved a sharp
versionof (1.2)in[20]. Infact,heprovedthatforanycsatisfying,0 < c≤ α(n),(1.2)holds
1
truebutfailsforanyc> α(n),whereα(n) = nwn−1 andw isasusualsurfaceareaofthe
n−1 n−1
unitballinRn.Thisleadustoaveryrichliterature. See[2],[4],[7],[10],[13],[18],[23],[28],[29]
and references therein for more information.
Now let us discuss some results concerning Hardy inequalities. Let B be the unit ball
n
in Rn. Then the Hardy inequality says
|u|2
H (u) := |∇u|2dx− dx ≥ 0 , ∀u∈H1(B ). (1.3)
n (1−|x|2)2 0 n
ZBn ZBn
When n ≥ 2 one can prove that (see [11]) there exists a positive constant C, depending
only on n such that
H (u) ≥ C u2dx, ∀u∈H1(B ). (1.4)
n 0 n
ZBn
Later Maz’ya proved in [22] that for n > 2 there exists a constant C >0, depending only
n
on n such that for any u∈ H1(B ),
0 n
n−2
2n n
Hn(u) ≥ Cn |u(x)|n−2 . (1.5)
(cid:18)ZBn (cid:19)
This inequality is called Hardy-Sobolev-Maz’ya inequality. This inequality raises the fol-
lowing question. If n =2 wheather one can derive the Trudinger-Moser type of inequality
or not. In [32] Wang and Ye proved that indeed in this case, one can derive such an
inequality. Their result is the following.
There exists an constant C > 0 such that
0
4πu2
eH2(u)dx ≤ C0 < ∞, ∀u∈ H01(B2)\{0}. (1.6)
ZB2
One may wonder if this kind of Moser-Tridunger inequality holds for bounded convex
domainwithsmoothboundary. Forthatletusrecallthecorrespondingversionofboundary
Hardy inequality. Let Ω be a bounded, convex domain in R2 with smooth boundary, then
(see [11])
1 u2
H (u) := |∇u|2dx− dx > 0, ∀u∈ H1(Ω)\{0}. (1.7)
d 4 d(x,∂Ω)2 0
ZΩ ZΩ
One can easily check that for Ω = B , (1.6) is true if we replace H (u) by H (u). In fact,
2 2 d
in a very recent paper by Lu and Yang ([19]), it has been established that the above type
of inequality is true for any bounded and convex domain. In this context let us mention
thefollowing improved version of Moser-Trudinger inequality on unitdiskon R2 whichwas
proved in [6, 21].
e4πu2 −1
sup dx < ∞. (1.8)
(1−|x|2)2
u∈H01(B2),||∇u||L2(B2)≤1ZB2
Now one may ask the following question. What will be the scenario if we replace the
distant function from boundary by simply the distant function from the origin in (1.7)?
AN IMPROVED LERAY-TRUDINGER INEQUALITY 3
To give an answer of this question, let us recall the corresponding Hardy inequality and
the literature associated with it.
Let Ω be a domain domain in Rn containing the origin. Then the classical Hardy’s
inequality asserts that for n ≥ 3,
n−2 2 |u|2
I [u] := |∇u|2dx− dx ≥ 0 ,∀u∈ H1(Ω). (1.9)
Ω 2 |x|2 0
ZΩ (cid:18) (cid:19) ZΩ
Here n−2 2 isthebestconstantandneverachieved. Hencethereisascopeofimprovement
2
of (1.9). The first improvement was done by Brezis and Vazquez (see [12]). Their result is
(cid:0) (cid:1)
the following.
If Lebesgue measure of Ω is finite then, for any 1 ≤ q < 2∗ there exists a positive constant
C depending only on n , q and Ω, such that
1
(IΩ[u])12 ≥ C |u|q q ,∀u∈ H01(Ω). (1.10)
(cid:18)ZΩ (cid:19)
Here as usual 2∗ = 2n is the critical Sobolev constant.
n−2
In the past few years a lot of efforts have been made to improve (1.10) and generalize it
to Lp Hardy-Sobolev inequality. We refer [1], [3], [5], [15], [31] and references therein for
further details. Observing the relevance of (1.10) with Sobolev’s inequality one may won-
der what would be the case when n = 2 or in analogy with Lp Hardy-Sobolev inequality
what would be the case when p = n. For that let us recall Hardy’s inequality for p = n
(see [1],[5], [8] and [9]).
Let Ω be a bounded domain in Rn (n ≥ 2) containing the origin and R := sup|x|. Then
Ω
x∈Ω
1,n
for any u∈ W (Ω) and R ≥ R
0 Ω
n−1 n |u|n
I [u,Ω,R] := |∇u|ndx− dx ≥ 0, (1.11)
n
ZΩ (cid:18) n (cid:19) ZΩ |x|nE1n(|Rx|)
where E (t):= log(e) for t ∈ (0,1] and n−1 n is the best possible constant which is never
1 t n
achieved.
(cid:0) (cid:1)
Inequality of type (1.6) is generally expected to hold when the functional H is replaced
2
by a similar nonnegative functional with a general potential,
H (u) = |∇u|2dx− V(x)|u|2dx (1.12)
V
ZΩ ZΩ
as long as H does not possess a null sequence, that is, a sequence u ∈ C∞(Ω), such that
V k c
H (u ) → 0, while for some bounded open set B, u dx =1 (such sequence is known to
V k B k
converge to a positive solution, often called generalized ground state). If a null-sequence
R
exists, the corresponding analog of (1.6) is immediately false. On the other hand, it is
shown in [29] for Ω = B and for radial potential V, that if H does not admit a null-
2 V
sequence, then the analog of (1.6) remains true, as long as conditions of the ground state
alternative ([25])aresatisfied. Intheradialcase thisamounts tothecondition (2.3) of [29]
(we bring attention to the reader that we correct here a misprint in the publishedversion):
1
∃α> 0 : limr2(log )2+αV(r)→ 0. (1.13)
r→0 r
Inthecase whenH hastheformI ,potential V correspondstotheborderlinecaseα = 0,
V 2
does not satisfy condition (1.13), and, remarkably, as it was first observed by Psaradakis
4 AN IMPROVED LERAY-TRUDINGER INEQUALITY
and Spector in [27], the analog of (1.6) is false, and, more generally, there does not exist
any positive constant c depending only on n for which the following is true:
n
sup ec|u|n−1dx < ∞.
u∈W01,n(Ω),In[u,Ω,RΩ]≤1ZΩ
However, introducing a logarithmic factor, in the same paper they proved the following
Hardy-Trudinger type inequality.
Let Ω be a bounded domain in Rn; n ≥ 2, containing the origin. For any ǫ > 0 there
exist positive constants A depending only on n, ǫ and B depending only on n such that
n,ǫ n
n
n−1
eAn,ǫE1ǫ(cid:18)|uR||xΩ|(cid:19) dx ≤ Bnvol(Ω) for all u ∈ W01,n(Ω) satisfying In[u,Ω,RΩ]≤ 1.
ZΩ
(1.14)
Let B denote the unit ball in Rn centered at the origin and consider the following space
of radial functions
W := {u∈ C1 (B): u is radially symmetric and I [u,B,1] < ∞}.
0,rad n
Then one can easily derive using Lemma 5 of [14] that there exists an positive constant C
n
depending only on n such that
n
sup eCn|Eu2|n(|−x|1)dx < ∞, (1.15)
u∈W,In[u,B,1]≤1ZB
where E (t) := log(eE (t)) for t ∈ (0,1].
2 1
This simple observation motivated us to investigate wheather (1.14) is true if we replace
E by E in the power of exponential and we got the following result.
1 2
Theorem 1.1. Let 0 ∈ Ω be a bounded domain in Rn, (n ≥ 2). Then for any β ≥ 2/n,
and R ≥ R , there exists constants A and B depending only on n such that, for any
Ω n n
0 < c< A
n
n
n−1
c |u(x)|
e E2β(|Rx|)! dx ≤ Bnvol(Ω), ∀u∈ W01,n(Ω) satisfying, In[u,Ω,R] ≤ 1. (1.16)
ZΩ
In other words,
n
n−1
c |u(x)|
β |x|
sup e E2(R)! dx < +∞ , for 0< c < An. (1.17)
u∈W01,n(Ω),In[u,Ω,R]≤1ZΩ
Moreover, the above supremum is +∞ if β < 1 for any c > 0.
n
Remark 1.2. The situation is not clear when β satisfies 1 ≤ β < 2. However, for n = 2,
n n
when β satisfies 1 ≤ β < 2, (1.16) is true if we consider Ω to the unit ball in Rn and
n n
u∈ W defined above. This fact follows from (1.15) .
AN IMPROVED LERAY-TRUDINGER INEQUALITY 5
2. Preliminaries
In this section we are going to recall some lemmas, propositions and theorems that we
have used in the subsequent sections. But before that let us fix the notations for the rest
of this article.
Ω is a bounded domain in Rn and R = sup|x|. We define E (t) := log(e) , E (t) :=
Ω 1 t 2
x∈Ω
log(eE (t)) for t ∈(0,1] and C (n):= 2n−1−1.
1 1
The first lemma is a very basic one. It is basically a representation formula for smooth
functions in terms of its derivative. The proof of this lemma can be found in [16] (Lemma
7.14).
Lemma 2.1. Let Ω be any open set in Rn, (n ≥2) and u∈ C1(Ω). Then ,
c
1 (x−y).∇u(y)
u(x) = dy,
nw |x−y|n
n ZΩ
where, w is the volume of unit ball in Rn.
n
The proof of following result can be found in [27] (Proposition 2.6). Easy to see here
that we will get an equality instead of the inequality for n = 2.
Proposition 2.2. Suppose, u∈ C∞(Ω\{0}), andu≥ 0.Then J [v,Ω,R] ≤ C (n)I [u,Ω,R].
c n 1 n
HereC (n) =2n−1−1, v(x) = E−n−n1(|x|)u(x)∀x ∈ Ω,J [v,Ω,R] = |∇v(x)|nEn−1(|x|)dx
1 1 R n Ω 1 R
and R ≥ R .
Ω
R
The following theorem is due to Barbatis, Filippas and Tertikas(See [8] Theorem B and
Proposition 3.2 ). We are going to state a much simplified version of their result, which is
about an improved version of Ln Hardy’s inequality.
Theorem 2.3. Let Ω be a bounded domain in Rn. Then for any R ≥ R and all u ∈
Ω
1,n
W (Ω) there holds
0
n−1 n |u|n 1 n−1 n−1 |u|n
|∇u(x)|ndx− dx ≥ dx.
ZΩ (cid:18) n (cid:19) ZΩ |x|nE1n(|Rx|) 2 (cid:18) n (cid:19) ZΩ |x|nE1n(|Rx|)E22(|Rx|)
(2.1)
Moreover, if Ω contains the origin and for R ≥ R there exists a positive constant B > 0
Ω
for which the following holds true
n−1 n |u|n |u|n
|∇u(x)|ndx− dx ≥ B dx, ∀u∈ W1,n(Ω)
ZΩ (cid:18) n (cid:19) ZΩ |x|nE1n(|Rx|) ZΩ |x|nE1n(|Rx|)E2γ(|Rx|) 0
and for some γ ∈ R, then
• γ ≥ 2 ;
• and if γ = 2 then B ≤ 1 n−1 n−1.
2 n
This makes 1 n−1 n−1 the best (cid:0)const(cid:1)ant of inequality (2.1).
2 n
For the res(cid:0)t of t(cid:1)his article we will denote |∇u(x)|ndx − n−1 n |u|n dx as
Ω n Ω |x|nEn(|x|)
1 R
In[u,Ω,R]. R (cid:0) (cid:1) R
6 AN IMPROVED LERAY-TRUDINGER INEQUALITY
3. Estimate of Lq norm
The proposition that we are going to prove in this section will help us to prove the first
partof our main theorem. Ourplan is to follow Trudinger’sapproach of proving Trudinger
inequality. To execute this, we need an expedient upper bound of Lq norm of u/E2/n. In
2
the next proposition we will derive such an upper bound.
Proposition 3.1. Let u ∈W1,n(Ω), then for any R ≥ R and q > n, we have the following
0 Ω
estimate,
q 1/q
1−1+1
ZΩ(cid:12)(cid:12)Eun2((x|x)|)(cid:12)(cid:12) dx ≤ Cn(cid:20)1+ q(nn−1)(cid:21) n q (vol(Ω))q1 (In[u,Ω,R])1/n. (3.1)
(cid:12) 2 R (cid:12)
 (cid:12) (cid:12) 
(cid:12) (cid:12) n−1
Where Cn(cid:12) = 11 (cid:12)(C1(n))n1 +2n1 n−n1 n n+n1
nwnn (cid:20) (cid:21)
(cid:16) (cid:17)
Proof. Firstofallnotethat,itisenoughtoprovethepropositionforpositivevaluedsmooth
functions. So, let u ∈ C∞(Ω\{0}), u ≥ 0 and define, v(x) = E−n−n1(|x|)u(x), ∀x ∈ Ω.
c 1 R
|x| |x|
Then using Lemma 2.1 and the fact E ( ),E ( ) ≥ 1, ∀x∈ Ω, we have for x ∈ Ω,
1 R 2 R
1−1 |x|
u(x) v(x)E n( )
= 1 R
(cid:12) (cid:12)
2/n |x| 2/n
(cid:12)E ( )(cid:12) (cid:12) |x| (cid:12)
(cid:12) 2 R (cid:12) (cid:12) E2(R) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:16) (x(cid:17)−y)(cid:12)(cid:12)(cid:12).∇ v(y) E11−n1(|Ry|)
= (cid:12)(cid:12) 1 E2(|Ry|) 2/n!dy(cid:12)(cid:12)
(cid:12)(cid:12)(cid:12)nwn ZΩ |x−y(cid:16)|n (cid:17) (cid:12)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12) 1 |∇v(y)|En−n1(|y|) 1 (cid:12)(cid:12)v(y) E1−n1(|y|)
≤ (cid:12) 1 R dy+ (cid:12) ∇ 1 R dy
nwn ZΩ |x−y|n−1 nwn ZΩ |x−y|n−1 (cid:12)(cid:12) (E2(|Ry|))2/n(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12)  (cid:12)(3.2)
(cid:12) (cid:12)
(cid:12) (cid:12)
Now,
∇ E11−n1 (|y|) = −E1−n1(|Ry|)(1− n1)|yy|2E22/n(|Ry|)+ |yy|2n2E2n2−1(|Ry|)E1−n1(|Ry|)
(cid:12)(cid:12) (E2)2/n R (cid:12)(cid:12) (cid:12)(cid:12) En4(|y|) (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) 2 R (cid:12)
(cid:12)   (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) y (n−1)E (|y|)− 2 (cid:12)
(cid:12) (cid:12) = (cid:12)− n 2 R n (cid:12)
(cid:12)(cid:12) |y|2En1(|y|) En2+1(|y|) (cid:12)(cid:12)
(cid:12) 1 R 2 R (cid:12)
(cid:12) n+1 (cid:12)
(cid:12) (cid:12)
≤ (cid:12) 1 n 2 (Since, E2(cid:12)≥ 1, on Ω)
|y| |y|
|y|En( )En( )
1 R 2 R
So, (3.2) implies,
1−1 |y|
u(x) 1 |∇v(y)|E n( )
≤ 1 R dy
(cid:12)(cid:12)En2(|x|)(cid:12)(cid:12) nwn ZΩ |x−y|n−1
(cid:12) 2 R (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
AN IMPROVED LERAY-TRUDINGER INEQUALITY 7
n+1 v(y)
+ dy.
n2wn ZΩ |y|En1(|y|)En2(|y|)|x−y|n−1
1 R 2 R
Define,
1−1 |y|
|∇v(y)|E n( )
K(x):= 1 R dy
|x−y|n−1
ZΩ
v(y)
M(x) := dy.
1 2
ZΩ |y|En(|y|)En(|y|)|x−y|n−1
1 R 2 R
So, we have,
u 1 n+1
≤ ||K|| + ||M|| (3.3)
(cid:12)(cid:12)(cid:12)(cid:12)E22/n(cid:12)(cid:12)(cid:12)(cid:12)Lq(Ω) nwn (cid:20) Lq(Ω) n Lq(Ω)(cid:21)
(cid:12)(cid:12) (cid:12)(cid:12)
Now, let q > n. Defi(cid:12)(cid:12)(cid:12)(cid:12)ne r b(cid:12)(cid:12)y(cid:12)(cid:12) 1/n+1/r = 1+1/q. Then clearly, 1 < r < n . For x ∈ Ω, let
n−1
us define
1
h (x) := dy
r
|x−y|(n−1)r
ZΩ
LetR˜ besuchthatvol(Ω) = w R˜n.Forx ∈ Ω,letusdenoteB(x,R˜)tobetheballofradius
n
R˜ and centre at x in Rn. Then as vol(Ω) = vol(B(x,R˜)), we have vol Ω∩B(x,R˜)c =
vol Ωc∩B(x,R˜) . Now, (cid:16) (cid:17)
(cid:16) (cid:17)
1 1
h (x) = dy+ dy
r
ZΩ∩B(x,R˜)c |x−y|(n−1)r ZΩ∩B(x,R˜) |x−y|(n−1)r
1 1
≤ vol Ω∩B(x,R˜)c + dy
R˜(n−1)r (cid:16) (cid:17) ZB(x,R˜)∩Ω |x−y|(n−1)r
vol Ωc∩B(x,R˜)
1
= + dy
(cid:16) R˜(n−1)r (cid:17) ZB(x,R˜)∩Ω |x−y|(n−1)r
1 1
= dy+ dy
ZΩc∩B(x,R˜) R˜(n−1)r ZΩ∩B(x,R˜) |x−y|(n−1)r
1 1
≤ dy+ dy
ZΩc∩B(x,R˜) |x−y|(n−1)r ZΩ∩B(x,R˜) |x−y|(n−1)r
1
= dy
ZB(x,R˜) |x−y|(n−1)r
nw R˜n−(n−1)r
n
= .
n−(n−1)r
Using 1/n+1/r = 1+1/q we get
||hr||L1r∞(Ω) ≤ wn1−n1+q1 1+ q(nn−1) 1−n1+q1 R˜nq
(cid:18) (cid:19)
1−1+1
≤ wn1−n1 1+ q(n−1) n q (vol(Ω))q1 (Since, vol(Ω) = wnR˜n). (3.4)
n
(cid:18) (cid:19)
8 AN IMPROVED LERAY-TRUDINGER INEQUALITY
Now let us break the integrand of K(x) as
|∇v(y)|E1−n1(|y|) |∇v(y)|nEn−1(|y|) q1 1 |y| n1−q1
1 R = 1 R |∇v(y)|nEn−1( ) .
|x−y|n−1 |x−y|(n−1)r ! |x−y|(n−1)(1−rq) (cid:18) 1 R (cid:19)
Note that, 1+n−1+ 1 −1 = 1. So, by applying H¨older’s inequality with exponent q, n
q n n q n−1
and 1 we get,
1−1
n q
1
K(x)≤ ZΩ |∇v|x(y−)|nyE|(n1n−−11)(r|Ry|)dy!q (hr(x))1−n1 (cid:18)ZΩ|∇v(y)|nE1n−1(|Ry|)dy(cid:19)n1−q1 .
Integrating,
||K|| ≤ ||h ||1−n1 |∇v|nEn−1(|y|)dy n1−1q |∇v(y)|nE1n−1(|Ry|)dydx 1/q.
Lq(Ω) r L∞(Ω)(cid:18)ZΩ 1 R (cid:19) ZΩZΩ |x−y|(n−1)r !
So, by Tonelli’s theorem, we have,
1−1 1/q
||K|| ≤ ||h ||1−n1 |∇v|nEn−1(|y|)dy n q |∇v(y)|nEn−1(|y|)h (y)dy .
Lq(Ω) r L∞(Ω) 1 R 1 R r
(cid:18)ZΩ (cid:19) (cid:18)ZΩ (cid:19)
Now we use Proposition 2.2 to get,
1 1 1
||K||Lq(Ω) ≤ ≤ (C1(n))n ||hr||Lr∞(Ω)(In[u,Ω,R])n . (3.5)
Similarly writing integrand of M(x) as
1
|v(y)| 1 |v(y)|n q
=
|y|E1n1(|Ry|)E2n2(|Ry|)|x−y|n−1 |y|nE1(|Ry|)E22(|Ry|)|x−y|(n−1)r!
1−1
1 |v(y)|n n q
|x−y|(n−1)(1−rq) |y|nE1(|Ry|)E12(|Ry|)!
and applying H¨older’s inequality with the same exponent as in the case of K(x) and
Tonelli’s theorem we get,
1
1 vn(y) n
||M|| ≤ ||h ||r dy
Lq(Ω) r L∞(Ω) ZΩ |y|nE1(|Ry|)E22(|Ry|) !
1
1 un(y) n
= ||h ||r dy .
r L∞(Ω) ZΩ |y|nE1n(|Ry|)E22(|Ry|) !
−n−1 |x|
In the last inequality we have put v(x) = E n ( )u(x). Now we use Theorem 2.3 of
1 R
previous section to get,
un(y) dy n1 ≤2n1 n n−n1 In[u,Ω,R] n1.
(cid:16)ZΩ |y|nE1n(|Ry|)E22(|Ry|) (cid:17) n−1
(cid:0) (cid:1) (cid:0) (cid:1)
So,
n−1
1 1 n n 1
||M||Lq(Ω) ≤ ||hr||Lr∞(Ω) 2n n−1 (In[u,Ω,R])n . (3.6)
(cid:18) (cid:19)
AN IMPROVED LERAY-TRUDINGER INEQUALITY 9
Using (3.4) ,(3.5) and (3.6) we have from (3.2),
n−1 1−1+1
u 1 1 n n n+1 1 q(n−1) n q
≤ Cn(n)+2n 1+
(cid:12)(cid:12) 2 (cid:12)(cid:12) 1 n−1 n 1 n
(cid:12)(cid:12)(cid:12)(cid:12)E2n(cid:12)(cid:12)(cid:12)(cid:12)Lq(Ω) " (cid:18) (cid:19) #nwnn (cid:20) (cid:21)
(cid:12)(cid:12) (cid:12)(cid:12)
(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) (vol(Ω))1q (In[u,Ω,R])n1 .
This proves (3.1). (cid:3)
4. Proof of Theorem
1,n
Proof. Let u ∈ W (Ω) satisfying I [u,Ω,R] ≤ 1. Then applying Proposition 3.1 with
0 n
q = nk , k ∈ {n,n+1,.....} we get,
n−1
nk
|u(x)| n−1 nk
dx ≤ Cn−1 vol(Ω) (1+k)1+k.
n
2/n |x|
ZΩ E2 (R)!
Multiplying above inequality by ck and adding from n to m,(n ≤ k ≤ m) we get,
k!
n k
m 1 |u(x)| n−1 m n k (1+k)1+k
c dx ≤ c Cn−1 vol(Ω) .
n
ZΩkX=nk!  E22/n(|Rx|)!  kX=n(cid:16) (cid:17) k!
R.H.S converges as m → ∞ if c< 1 . Also by H¨older’s inequality
n
e Cnn−1
n s
n−1 |u(x)| n−1
S = c dx,
ZΩXs=0 E22/n(|Rx|)! 
is bounded by a constant depending only on n. This proves the first part of the result for
|x|
β = 2/n. For β > 2/n, the same A and B will work since E ( ) ≥ 1, ∀x ∈ Ω.
n n 2 R
Next we will prove the second part of our result i.e. for β < 1 the inequality (1.16) is
n
false. Let Ω = B , be the unit ball in Rn centered at the origin and β < 1. If possible let
1 n
c and c be two positive constants depending only on n such that
1 2
n
n−1
ec1 E|u2β|! dx < c2 ,∀u ∈ W01,n(B1) satisfying In[u,B1,R] ≤ 1, (4.1)
ZB1
where R ≥ 1 is a fixed real number. We choose 1 < θ < 2 such that 1 < nβ +θ < 2. In
the rest of the proof we will concentrate our focus on deriving the following inequality
|u|n(x)
1,n
dx ≤ cI [u,B ,R] ,∀u∈ W (B ), (4.2)
ZB1 |x|nE1n(|Rx|)E2nβ+θ(|Rx|) n 1 0 1
for some constant c> 0. ButTheorem 2.3 suggests that this can only happenif nβ+θ ≥ 2.
Hence we are through.
1,n
Let u∈ W (B ) be such that I [u,B ,R]≤ 1. Then
0 1 n 1
|u|n
dx
ZB1|x|nE1n(|Rx|)E2nβ+θ(|Rx|)
n−1 n−1 c n−1 |u|n 1
1
= dx
(cid:18) c1 (cid:19) ZB1"(cid:18)n−1(cid:19) E2nβ(|Rx|)#"|x|nE1n(|Rx|)E2θ(|Rx|)#
10 AN IMPROVED LERAY-TRUDINGER INEQUALITY
n
n−1
≤ nc−1 n−12n−2 ec1 E2β|(u||Rx|)! dx+PB1, (4.3)
(cid:18) 1 (cid:19) ZB1
where
1 n−1
1 1 n−1
P = 22(n−2) 1+ log 1+ dx.
B1 ZB1 |x|nE1n(|Rx|)E2θ(|Rx|)!  |x|nE1n(|Rx|)E2θ(|Rx|)! 
  
To derive the last inequality of (4.3) we have used the following inequality
ab≤ 2n−2 e(n−1)an−11 +2n−2(1+b) log(1+bn−11) n−1 ,∀a,b ≥ 0,
(cid:20) (cid:21)
(cid:16) (cid:17)
which in turn can be derived from the following version of Young’s inequality
ab≤ ea−a−1+(1+b)log(1+b)−b (See [27] for details).
Now using (4.1) we get from (4.3)
|u|n n−1 n−1
dx ≤ c 2n−2+P . (4.4)
ZB1 |x|nE1(|Rx|)E2nβ+θ(|Rx|) 2(cid:18) c1 (cid:19) B1
An easy calculation shows that P is finite (See [27], Theorem 1.1 for details). Hence for
B1
general u ∈ W1,n(B ), putting u˜ = u in (4.4) we get (4.2). This concludes the
0 1 1
(In[u,B1,R])n
theorem. (cid:3)
Acknowledgement
The authors would like to thank Prof. Sandeep and Prof Adimurthi for their helpful
suggestions.
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