The Trudinger-Moser inequality
Bernhard Ruf, Milano
June 23, 2010
Bernhard Ruf, Milano
The Trudinger-Moser inequality
The classical Sobolev embeddings
Let Ω ⊂ RN be a smooth and bounded domain.
Sobolev space:
W01,2 (Ω) : closure of C0∞ (Ω) with respect to the norm
Z
1/2
|∇u|2
kuk1,2 :=
Ω
Sobolev embedding theorem: for N > 2
W01,2 (Ω) ⊂ Lq (Ω)
if
1≤q≤
Bernhard Ruf, Milano
2N
,
N −2
The Trudinger-Moser inequality
The Trudinger-Moser inequality concerns the borderline cases
N=2
Sobolev exponent:
2N
∼ +∞ (formally)
N −2
Question: W01,2 ⊂ L∞ ?
No !
Example: Ω = B1 (0) , u(x) = log(1 − log |x|), then
Z
Z 1
2
d
|∇u|2 dx = 2π
log(1
−
log
r
)
rdr
dr
0
Ω
Z 1
1
−1 2
= 2π
rdr
1 − log r r
0
Z 1
1
1
dr < ∞
= 2π
2
0 (1 − log r ) r
but
log(1 − log |x|) 6∈ L∞
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Look for:
Maximal growth function g (t) such that
u ∈ W01,2 (Ω)
implies
Z
g (u)dx < ∞
Ω
S. Pohozaev (1965) and N. Trudinger (1967): maximal growth is
of exponential type.
More precisely, let N = 2, then
u∈
W01,2 (Ω)
Z
=⇒
Bernhard Ruf, Milano
2
e |u| dx < ∞ ,
Ω
The Trudinger-Moser inequality
2
• The growth function g (t) = e t is optimal:
for any higher growth exist functions for which the corresponding
integral becomes infinite.
• Proofs of Pohozaev and Trudinger use same idea, namely
- develop the exponential into a power series:
- this reduces the problem to show that a series of Lm -norms
(m ∈ N) remains finite.
- by controlling the embedding constants of W01,2 ⊂ Lm one
obtains the result.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Sharp result by J. Moser (1979):
Theorem:
(
Z
(1)
sup
k∇uk2 ≤1
e
α |u|2
dx
Ω
≤ c |Ω| , if α ≤ 4π
= ∞ , if α > 4π
Inequality (1) is called the Trudinger-Moser inequality.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
,
Proof:
J. Moser uses symmetrization:
to function u associate u ∗ : radially symmetric function
such that sublevel-sets of u ∗ are balls with the same area as the
corresponding sublevel-sets of u,
i.e. |{x ∈ R2 : u ∗ (x) < d}| = |{x ∈ Ω : u(x) < d}|,
where |A| denotes the Lebesgue measure of the set A.
Then u ∗ is a positive and non-increasing function defined on BR (0)
with |BR | = |Ω|.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Bernhard Ruf, Milano
The Trudinger-Moser inequality
By construction, one has the following property:
Let f ∈ C (R), then
Z
∗
Z
f (u ) dx =
f (u) dx ;
BR
Ω
furthermore, one has the well-known
Pólya-Szegö inequality
Z
∗ 2
Z
|∇u | dx ≤
BR
Bernhard Ruf, Milano
|∇u|2 dx
Ω
The Trudinger-Moser inequality
From this we clearly deduce that
Z
2
sup
e α|u(x)| dx ≤ sup
k∇u ∗ k2 ≤1
k∇uk2 ≤1 Ω
Z
e α|u
∗ (|x|)|2
dx ,
BR
hence: sufficient to consider the
radial case
Next, we perform a change of variables: set
1
r = |x| = Re −t/2 and w (t) = √ u ∗ (r ) .
4π
Note that
w : [0, +∞) → R , w (0) = 0 , w 0 (+∞) = 0
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Bernhard Ruf, Milano
The Trudinger-Moser inequality
One checks that
Z
∗ 2
Z
∞
|∇u | dx =
BR
|w 0 (t)|2 dt
0
and
Z
e
α|u ∗ |2
Z
dx = |BR |
BR
∞
α
e 4π |w (t)|
2 −t
dt
0
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Thus, we have reduced the problem to
Z
(2)
R∞
0
|BR |
sup
|w 0 |2
∞
α
e 4π |w (t)|
2 −t
dt .
0
dt≤1
for w ∈ C 1 we have
Z t
1/2
Z
t 0
1/2
0
2
≤ t 1/2 ,
|w (t)| =
w (s) ds ≤ t
|w (t)| dt
The case α < 4π:
0
0
by assumption, i.e. |w (t)|2 ≤ t
Inserting this in (2) we find
Z ∞
Z ∞
α
α
2
e 4π t−t dt
e 4π |w (t)| −t dt ≤
0
0
Z ∞
α
=
e ( 4π −1)t dt =
0
Bernhard Ruf, Milano
1
α <∞
1 − 4π
The Trudinger-Moser inequality
The case α = 4π: more delicate, see below.
The exponent α = 4π is optimal:
Suppose that α > 4π; consider the
Moser-sequence
wn (t) :=
Z
Then clearly
∞
|wn0 (t)|2


√t
 √n , n ≤ t
Z
0
∞
α
e 4π |wn |
n
1 2
√
= 1, and
n
dt =
0
Z
, 0≤t≤n
n
0
2 −t
Z
∞
dt ≥
α
α
e 4π n−t dt = e ( 4π −1)n → ∞
n
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Bernhard Ruf, Milano
The Trudinger-Moser inequality
The case α = 4π: J. Moser shows that
if vn (t), t ∈ (0, +∞) is sequence with
Z +∞
|vn |2 dt = 1 and kvn k∞ → +∞
0
then
vn is ”close to” wn , and for α = 4π we have
Z ∞
Z n 2
Z ∞
t
|wn |2 −t
−t
e
dt =
e n dt +
e n−t dt
0
0
n
Second integral is equal 1.
First integral is nice exercise:
Bernhard Ruf, Milano
The Trudinger-Moser inequality
n
Z
Exercise:
e
t2
−t
n
dt → 2
0
t
n
Let s =
Z
; then
n
e
t2
−t
n
Z
1
dt =
0
n(s 2 −s)
e
1/2
Z
e n(s
nds = 2
0
2 −s)
nds
0
For every ε > 0:
s 2 − s ≤ (−1 + ε)s , for 0 ≤ s ≤ ε
So,
Z
1/2
e
0
n(s 2 −s)
Z
ε
e
nds =
n(s 2 −s)
0
Bernhard Ruf, Milano
Z
nds +
1/2
e n(s
2 −s)
ε
The Trudinger-Moser inequality
nds
First integral:
Z ε
Z
n(s 2 −s)
e
nds ≤
0
ε
ε
n
n
−
e −n 1−ε
n(1 − ε) n(1 − ε)
1
=
+ on (1)
1−ε
e n(−1+ε)s nds =
0
Second integral:
Z
1/2
e
n(s 2 −s)
Z
nds ≤
ε
1/2
e n(ε−1)ε nds = on (1)
ε
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Supremum attained ?
(
Z
(3)
sup
k∇uk2 ≤1
e
α |u|2
dx
Ω
attained , if α < 4π
? ? ,
if α = 4π
,
For α < 4π: compactness:
for (un ) ∈ W01 (Ω) with kun k ≤ 1 ⇒
Z
2
e un dx →
Ω
Z
Ω
(for subsequence);
see e.g. P.L. Lions: concentration-compactness
Bernhard Ruf, Milano
2
e u dx
The Trudinger-Moser inequality
For α = 4π: loss of compactness: indeed, for Ω = B1 (0)
Moser sequence wn (t) ↔ un (r ) , (r = e −t/2 ) , where

log 1
1
1  √logr n , if n ≤ r ≤ 1
√
un (r ) =
2π  √log n , if 0 ≤ r ≤ 1
n
Then: kun k = 1 , un * 0 in W01,2 ,
and un (r ) → 0 pointwise, for a.e. r ∈ (0, r ),
but
Z
e
4πun2 (r )
Z
rdr 6→
B1 (0)
Bernhard Ruf, Milano
e 0 rdr
B1 (0)
The Trudinger-Moser inequality
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Comparison with Sobolev-embedding:
W01,2 (Ω) ⊂ Lq (Ω) , Ω ⊂ RN bounded
Best embedding constant
SNq
(4)
•
if 1 ≤ q < 2∗ =
Z
=
2N
N−2 :
sup
|u|q dx
k∇uk2 =1 Ω
compactness (Rellich-Kondrachov):
supremum attained
•
if q = 2∗ : loss of compactness:
supremum not attained, if Ω 6= RN (in particular, on bounded
domains)
Bernhard Ruf, Milano
The Trudinger-Moser inequality
By contrast, in the situation of Trudinger-Moser inequality:
surprising result by L. Carleson and A. Chang.
Theorem:
If Ω ⊂ R2 is the ball B1 (0), then
Z
2
sup
e 4π |u| dx
k∇uk2 ≤1
Ω
is achieved
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Idea of the proof (in the case N = 2):
By symmetrization, can consider radially symmetric functions.
a) assume that the supremum is not attained:
then any maximizing sequence (un ) must concentrate at origin
b) determine the limit value of the integral
Z
2
lim
e 4πun dx = (1 + e) |B1 |
n→∞ B (0)
1
along any concentrating maximizing sequence (un )
Thus, this level represents a non-compactness level.
c) construct explicit function w with
Z
2
e 4πw dx > (1 + e) |B1 | :
Contradiction !
B1 (0)
Bernhard Ruf, Milano
The Trudinger-Moser inequality
M. Struwe (1984): result holds for small perturbations of B1 (0)
M. Flucher (1992): result holds for any bounded Ω ∈ R2
K.C. Lin (1996) generalized result to any dimension
In (de Figueiredo-do O’-R., 2002): explicit concentrating sequence
was constructed, along which the integral (1) converges to the
non-compactness level found by Carleson-Chang.
Furthermore, it was shown (by an asymptotic analysis) that along
this sequence the integral (1) converges from above to this value.
This gives a new proof of the result of Carleson-Chang, and it also
shows that that there is a strong analogy with the famous result of
Brezis-Nirenberg (1983) concerning perturbations of the Sobolev
case.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Unbounded domains:
From Moser’s inequality:
Z
2
e 4π |u| dx
sup
k∇uk2 ≤1
≤ c |Ω|
Ω
one sees:
inequalities valid only for bounded domains
For unbounded domains:
Replace Dirichlet norm
Z
|∇u|2 dx
Ω
by the full W 1,2 -norm
kuk21
Z
=
(|∇u|2 + |u|2 ) dx
Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Theorem: (R., JFA, 2004)
Let Ω be any domain in R2 , and let
Z
1/2
(|∇u|2 + |u|2 ) dx
kuk1 =
Ω
Then exists constant d (independent of Ω) such that
Z 2
(5)
sup
e 4πu − 1 dx ≤ d
kuk1 ≤1 Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Proof:
Sufficient to consider the case Ω = R2 .
By symmetrization: can reduce to radial case, i.e. we may assume
that
u is radial and non-increasing.
Divide integral into two parts:
Z
Z
Z 4πu 2
4πu 2
e
− 1 dx =
e
− 1 dx +
R2
|x|≤r0
|x|≥r0
= J1 + J2
(r0 to be determined)
Bernhard Ruf, Milano
The Trudinger-Moser inequality
2
e 4πu − 1 dx
Write second integral as a series:
Z
(6)
J2 =
e
4πu 2
∞ Z
X
− 1 dx =
|x|≥r0
k=1
(4π)k |u|2k
dx .
k!
|x|≥r0
Using ”Radial Lemma” (H. Berestycki - P.L. Lions, W. Strauss)
one estimates
1
1
|u(r )| ≤ √ kukL2 , for all r > 0 ;
r
π
from this one obtains easily that the series in (6) converges for r0
sufficiently large, and hence
Z
2
J2 =
e 4πu − 1 dx ≤ c(r0 )
|x|≥r0
Bernhard Ruf, Milano
The Trudinger-Moser inequality
First integral: write
u(r ) = u(r ) − u(r0 ) + u(r0 ) =: v (r ) + u(r0 )
and estimate, using again the Radial Lemma:
u 2 (r ) = v 2 (r ) + 2v (r )u(r0 ) + u 2 (r0 )
1
1
kuk2L2 + 1 + 2 kuk2L2
πr02
πr0
1
≤ v 2 (r ) 1 + 2 kuk2L2 + d(r0 )
πr0
≤ v 2 (r ) + v 2 (r )
and hence:
1
u(r ) ≤ v (r ) 1 + 2 kuk2L2
πr0
1/2
Bernhard Ruf, Milano
+ d 1/2 (r0 ) =: w (r ) + d 1/2 (r0 ) ;
The Trudinger-Moser inequality
Hence
Z
2
e 4πu dx ≤ c
Br0
Z
2
e 4πw dx ≤ d
Br0
by Moser’s inequality, since w ∈ H01 (Br0 ) with
Z
Z
1
|∇w |2 dx =
|∇v |2 dx 1 + 2 kuk2L2
πr0
Br0
Br0
Z
1
|∇u|2 dx
= 1 + 2 kuk2L2
πr0
Br0
1
2
≤ 1 + 2 kukL2 1 − kuk2L2 ≤ 1 ,
πr0
provided that πr02 ≥ 1.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Again, one can ask the question:
is the supremum attained ?
Theorem: (R., 2004)
For Ω = BR (0), the ball of radius R, and for Ω = R2 the
supremum in (5) is attained
Proof: As in proof of Carleson and Chang:
determine the limit of
Z
2
e 4πu − 1 dx
BR (0)
along a concentrating maximizing sequence.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Indeed, for any concentrating maximizing sequence (un ):
Z
lim
n→∞ B (0)
R
where
2
e 4πun − 1
dx = πe 1−D(R)
D(R) = 2K0 (R) 2RK1 (R) −
1
I0 (R)
;
here K0 , K1 , I0 are modified Bessel functions, i.e. solutions of
(for k = 0, 1)
−x 2 u 00 (x) − xu 0 (x) + (x 2 + k 2 )u(x) = 0
Bernhard Ruf, Milano
The Trudinger-Moser inequality
One shows that
D(R) ∼ −2 log R , for R small
D(R) ∼
1
Re R
, for R large
and hence one obtains for Ω = R2 that for concentrating
maximizing sequences
Z 2
e 4πun − 1 = πe
lim
n→∞ R2
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Finally, construct explicit concentrating maximizing sequence
such that the convergence is from above, i.e. there exists n0 such
that for all n ≥ n0
Z
2
e 4πun − 1 > πe 1−D(R)
BR (0)
and hence
Z
sup
2
e 4πu − 1 > πe 1−D(R)
k∇uk22 +kuk22 ≤1 BR (0)
From this it follows immediately that the supremum in (5) is
attained.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Generalized Trudinger-Moser inequalities
Numerous generalizations, extensions and applications of the
Trudinger-Moser (TM) inequality have been given in recent years:
TM-type inequalities involving higher order derivatives:
D.R. Adams, Annals of Math., (1988)
Extensions of the TM-inequality to manifolds:
P. Cherrier (1981), L. Fontana (1993), Y. Li (2001,2005), Y. Yang
(2006).
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Recently, Adimurthi - O. Druet (2004) have given an
improved TM-inequality with remainder term
they proved:
Theorem: Let Ω ⊂ R2 be a bounded domain, and set
Z
2
2
Cα = sup
e 4πu (1+α)kuk dx
k∇uk2 ≤1 Ω
Let λ1 denote the first eigenvalue of the Laplacian in W01,2 (Ω).
Then
Cα < ∞ , if α < λ1
Cα = +∞ , if α ≥ λ1
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Sharpness: obtained by suitably modified Moser-type sequences
Convergence: inspired by the following concentration-compactness
result by P.-L. Lions (1985):
Theorem: Let (u )>0 be a sequence of functions in W01,2 (Ω) with
k∇u k = 1 such that u * u0 weakly in W01,2 (Ω).
For any p < 1/(1 − ku0 k22 ) holds:
Z
lim sup
→0
2
e 4π p u dx < +∞
Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Result gives more precise information than Trudinger-Moser
inequality (1) in the case when u * u0 in H01 (Ω) with u0 6= 0.
Adimurthi-Druet extend the result of Lions, giving extra
information even when u * 0 in W01,2 (Ω).
They obtain this extra information by doing a careful blow-up
analysis of sequences of solutions to approximate elliptic equations
with near critical Trudinger-Moser growth.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
We also mention that recently TM-inequalities with other
boundary data, and Trudinger-Moser trace inequalities have been
obtained, see A. Cianchi
Finally, we mention TM-type inequalities in other function spaces,
in particular in Orlicz spaces, Zygmund spaces, Lorentz spaces,
Besov spaces etc., see e.g. A. Cianchi, N. Fusco - P.-L. Lions - C.
Sbordone, A. Alvino - V. Ferone - G. Trombetti, D.E. Edmunds P. Gurka - B. Opic, S. Hencl, H. Brezis - S. Wainger.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
In particular, we recall here some recent results for embeddings of
Lorentz-Sobolev spaces into Orlicz spaces and the related
TM-inequalities.
We first recall the definition of Lorentz spaces:
Lorentz spaces Lp,q are ”scales of interpolation spaces” between
the Lebesgues spaces Lp :
obtained via spherically decreasing rearrangement:
for a measurable function u : Ω → R let u ∗ (s) denote its
decreasing rearrangement.
Then the function u belongs to the Lorentz space Lp,q (Ω) if
Z
kukp,q =
∞
∗
[u (t) t
0
Bernhard Ruf, Milano
dt
]
t
1/p q
1/q
< +∞ .
The Trudinger-Moser inequality
We refer to Adams, ”Sobolev spaces”, for the precise definitions
recall here only that, for Ω ⊂ RN of finite measure,
bigskip
Lp,p = Lp , Lp,q1 ⊂ Lp,q2 ,
if q1 < q2 ,
Lr ⊂ Lp,q ⊂ Ls ,
if 1 < s < p < r , for all 1 ≤ q ≤ ∞
We denote the norm in Lp,q by kukp,q . The following Hölder
inequality holds:
Z
q
p
(7) fg dx ≤ kf kp,q kg kp0 ,q0 , where p 0 =
, q0 =
p−1
q−1
Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
First, we recall that the standard Sobolev embeddings can be
sharpened by the use of Lorentz spaces, see e.g. [?]. Denoting
W01 Lp,q (Ω) = cl u ∈ C0∞ (Ω) : k∇ukp,q < ∞ , Ω ⊂ RN bounded
one has the following
Sobolev-Lorentz embedding: Suppose that 1 ≤ p < N; then
W01 Lp,q ⊂ Lp
∗ ,q
, where p ∗ =
pN
,
N −p
and hence in particular, since p < p ∗
W01,p = W01 Lp,p ⊂ Lp
Bernhard Ruf, Milano
∗ ,p
⊂ Lp∗ ,p∗ = Lp∗ .
6
=
The Trudinger-Moser inequality
For the limiting case p = N, once has the following important
refinement of the Trudinger embedding, see Brezis-Wainger [?] and
A. Alvino, V. Ferone and G. Trombetti [?]:
Theorem: If
q0
q
u ∈ W01 L2,q (Ω) , then e |u| ∈ L1 (Ω) , where q 0 = q−1
, and the
following corresponding Moser-type inequality holds:
Brezis-Wainger inequality: There exist numbers βq > 0 such
that
(8)
(
Z
q
≤ C (N, q)| Ω| ,
for β ≤ βq
q−1
sup
e β|u(x)|
dx
= +∞ ,
for β > βq
{k∇ukN,q ≤1} Ω
The Trudinger-Moser inequality corresponds to the case
W01,N (Ω) = W01 LN,N (Ω). It is remarkable that in (8) the exponent
depends only on the second index q of the Lorentz space, and is
independent of N.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Inequalities (1) and (8) are sharp with respect to the coefficients
α, resp. β, in the exponents.
In fact, considering for simplicity the inequality (1) in the case
N = 2, one notes that if α = α2 = 4π, then any unbounded lower
order perturbation f (s) in the exponent (i.e. f (s) with
lim|s|→∞ f (s) = +∞ and lims→∞ f s(s)
2 = 0) will yield
Z
2
sup
e 4π|u(x)| +f (u(x)) dx = +∞ .
k∇uk2 ≤1
Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
In [?] the TM-inequality (1) and the more general Brezis-Wainger
inequality (8) were generalized with regard to such lower order
perturbations. More precisely, concerning inequality (1) (with
N = 2) it was asked: in the limiting case α = α2 = 4π, and given
an unbounded lower order perturbation function f (s), can we
characterize a largest space Λ(g ) of Lorentz type such that
Z
2
(9)
sup
e 4π|u(x)| +f (u(x)) dx < +∞ .
k∇ukΛ(g ) ≤1
Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Thus, we reverse the question:
given an unbounded perturbation of the Trudinger-Moser
nonlinearity,
what integrability condition on the gradient of u in order to have a
bounded integral of u with respect to this perturbed
TM-nonlinearity ?
This is a subtle question: note that if we replace in (1) the
condition k∇uk2 ≤ 1 by k∇uk2 ≤ 1 − δ, for an arbitrary δ > 0,
R
1
2
then sup{k∇uk2 ≤1−δ} Ω e 4π( 1−δ |u(x)|) dx ≤ c, and hence for any
subquadratic perturbation f (u) we get
Z
2
sup
e 4π|u(x)| +f (u(x)) dx ≤ c .
{k∇uk2 ≤1−δ} Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
The adequate class of Lorentz spaces for this problem are weighted
Lorentz spaces, which were proposed by G.G. Lorentz [?] already in
his original paper ”On the Theory of Spaces”. Weighted Lorentz
spaces are defined as follows: Let φ : Ω → R+ be a measurable
function, and let φ∗ (s) denote its decreasing rearrangement.
+
Furthermore,
R t let w (t) : R → R a nonnegative integrable function,
such that 0 w (s)ds < +∞ for all t > 0. The weighted Lorentz
space Λp (w ) is defined as follows: φ ∈ Λp (w ), 1 ≤ p < +∞, if
Z
(10)
kφkΛp (w ) =
+∞
∗
p
(φ (t)) w (t) dt
1/p
< +∞.
0
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Surprisingly, one can establish a precise relation between a weight
w (s) and the corresponding lower order perturbation function f (u)
to obtain sharp TM-type inequalities. To formulate the result, let
ϕ : R+ → R+ be a continuous function (the ”weight function”)
such that
(H1 )
lim ϕ(t) = 0
t→+∞
Z +∞
(H2 )
ϕ(t) = +∞
0
(H3 ) ϕ(t) is non increasing as t → +∞
Bernhard Ruf, Milano
The Trudinger-Moser inequality
One has the following optimal Moser type inequality:
Theorem: Let Ω ⊂ RN bounded, and let ϕ ∈ C(R+ ) satisfying
(H1 )–(H3 ). Let f (t) ∈ C 1 (R+ ):
N
Z
(11)
αN t N−1
f (t) =
0
1/(N−1)
where αN = NωN−1
. Then
Z
(12)
ϕ(s)
ds
1 + ϕ(s)
sup
e αN |u|
N
N−1 +f (u)
dx ≤ C |Ω| ,
{u∈C01 (Ω),k∇ukΛN ,ϕ ≤1} Ω
where
(13)
kv kN
ΛN,ϕ =
Z
+∞
N n
v ∗ (s)
1 + ϕ log
0
s oN−1
ds.
|Ω|
and C = C (kϕk∞ ): positive constant, depends only on kϕk∞ .
Furthermore, the result is sharp.
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Examples:
1) Let ϕ1 (s) = √
2
1
4π (s+1) − 1
, then f (s) = s , i.e.
Z
sup
k∇ukΛ2,ϕ ≤1
1
e 4πu
2 +u
dx ≤ C |Ω|
Ω
√
2) Let ϕ2 (s) =
π√p
s+4π s+p
, then f (s) = p log(1 + |u|) , i.e.
Z
sup
k∇ukΛ2,ϕ ≤1
2
2
(1 + |u|)p e 4πu dx ≤ C |Ω|
Ω
Bernhard Ruf, Milano
The Trudinger-Moser inequality
Proof: Relies on a generalization of Adams’ inequality (??)
Lemma: Let ϕ : R+ → R+ continuous, with (H1), (H2), and let
f (t) be defined by (11).
Let a(s, t) be a non-negative measurable function on R × [0, +∞)
such that
(14)
(15)
a(s, t) ≤ 1, for a.e. 0 < s < t
N−1
Z 0
Z +∞ N
N
a N−1 (s, t)
ds
=γ<∞
sup
+
1 + ϕ(s)
t>0
−∞
t
Then there exists a constant c0 = c0 (kϕk∞ , γ) such that for φ ≥ 0
with
Z +∞
(16)
φN (s) (1 + ϕ(s))N−1 ds ≤ 1
−∞
Bernhard Ruf, Milano
The Trudinger-Moser inequality
One has
Z
(17)
+∞
e −Ψ(t) dt ≤ c0 ,
0
where
(18)
Z
Ψ(t) = t−
+∞
a(s, t)φ(s) ds
−∞
N
N−1
+f
1
N−1
αNN
Z
+∞
a(s, t)φ(s) ds
−∞
Note that for ϕ(s) ≡ 0 we have f (t) ≡ 0, and hence Ψ(t) = F (t)
in Adams’ inequality.
Again, the Theorem follows directly from the above inequality.
Bernhard Ruf, Milano
The Trudinger-Moser inequality