on singular trudingerâmoser type inequalities for unbounded domains and their best exponents

Potential Anal (2013) 38:1091–1101DOI 10.1007/s11118-012-9308-7

On Singular Trudinger–Moser Type Inequalities forUnbounded Domains and Their Best Exponents

Manassés de Souza · João Marcos do Ó

Received: 6 January 2011 / Accepted: 6 August 2012 / Published online: 9 September 2012© Springer Science+Business Media B.V. 2012

Abstract Generalizations of the Trudinger–Moser inequality to Sobolev spaces withsingular weights are considered for any smooth domain � ⊂ R

N . Furthermore, weshow that the resulting inequalities are sharp obtaining the best exponents.

Keywords Trudinger–Moser inequality · Unbounded domain · Critical growth

Mathematics Subject Classifications (2010) 46E35 · 26D10 · 35B33

1 Introduction

Let � be a smooth domain in RN, N ≥ 2 and 1 ≤ p < ∞. Consider the Sobolev

space

W1,p(�) = {u ∈ Lp(�) : ∂xi u ∈ Lp(�), i = 1, . . . , N}with the Sobolev norm

‖u‖1,p :={∫

�

(|∇u|p + |u|p) dx}1/p

. (1.1)

We recall that W1,p(�) is a Banach space. Let C∞c (�) the space of smooth functions

with compact support contained in �. We define W1,p0 (�) as the closure of C∞

c (�)

Research partially supported by the National Institute of Science and Technology ofMathematics INCT-Mat, CAPES and CNPq grants 307400/2009-3, 620108/2008-8,142002/2006-2 and MCT/CNPq/MEC/CAPES—grant 552758/2011-6.

M. de Souza · J. M. do Ó (B)Departamento de Matemática, Universidade Federal da Paraíba,58051-900 João Pessoa, PB, Brazile-mail: [email protected]

M. de Souzae-mail: [email protected]

1092 M. de Souza, J.M. do Ó

in W1,p(�). In the case 1 ≤ p < N, the classical Sobolev imbedding theorem [2]asserts that

W1,p0 (�) ↪→ Lr(�) for 1 ≤ r ≤ p∗ = Np

N − p. (1.2)

When � is a bounded domain, using the Dirichlet norm ‖u‖ = (∫�

|∇u|p dx)1/p

(equivalent to Eq. 1.1 in W1,p0 (�)), we have that Eq. 1.2 is equivalent to

supu∈W1,p

0 (�) : ‖u‖≤1

‖u‖rr < +∞, for 1 ≤ r ≤ p∗, (1.3)

where ‖u‖r = (∫�

|u|r dx)1/r. We note that the supremum in Eq. 1.3 is infinite forr > p∗. The maximal growth |u|p∗

is called “critical” Sobolev growth. In the casep = N, we have p∗ = ∞ and every polynomial growth is admitted, so we mightexpect u ∈ L∞(�), provided that u ∈ W1,N

0 (�), but one knows by easy examples thatW1,N

0 (�) �↪→ L∞(�) (note that u(x) = log log (1 + 1/|x|) belongs to W1,N0 (B(0, 1)),

but not to L∞(B(0, 1))), where B(0, 1) ⊂ RN denotes the unitary open ball centered

at the origin. For � a smooth domain in RN with finite measure, Trudinger [15]

proved that W1,N0 (�) may be continuously imbedded in the Orlicz space Lφ(�) with

defining “N-function” φ(t) = eα|t|N/(N−1) − 1, that is, for all u in the unit ball ‖u‖ ≤ 1, inW1,N

0 (�) there exists some α (independent of u) such that the integral∫�

eα|u|N/(N−1)

dxis bounded. Hempel–Morris–Trudinger [12] proved that this result is optimal in thesense that space Lφ(�) above cannot be replaced by any smaller Orlicz space. J.Moser [13] later found the best exponent α and proved the following sharp result:for all α > 0 and u ∈ W1,N

0 (�) it holds eα|u|N/(N−1) ∈ L1(�) and there exists C(N) > 0such that

supu∈W1,N

0 (�) : ‖u‖≤1

∫�

eα|u|N/(N−1)

dx ≤ C(N)|�| for α ≤ αN, (1.4)

where αN = Nω1/(N−1)

N−1 and ωN−1 is the surface area of the unit sphere in RN .

The inequality 1.4 is optimal, in the sense that for any growth eα|u|N/(N−1)

with α >

αN the correspondent supremum is +∞.Estimate 1.4 is now referred as Trudinger–Moser inequality and plays an important

role in partial differential equations. For more details we refer the reader to thereview article on this subject by D. G. de Figueiredo et al. [7] and references therein.

Adimurthi–Sandeep in [3] proved a singular Trudinger–Moser inequality:∫

�

eα|u|N/(N−1)

|x|a dx < ∞, for all u ∈ W1,N0 (�) and α > 0,

where � is a smooth bounded domain in RN containing the origin and a ∈ [0, N).

Moreover, there exists C(N, a) > 0 such that

supu∈W1,N

0 (�) : ‖u‖≤1

∫�

eα|u|N/(N−1)

|x|a dx ≤ C(N, a)|�|, iff α/αN + a/N ≤ 1. (1.5)

Note that the supremums in Eqs. 1.4 and 1.5 become infinite for domains � whichdoes not have finite measure, and therefore the Trudinger–Moser inequality is notavailable for this class of domains in the classical formulation. Related inequalities

Singular Trudinger–Moser Type Inequalities 1093

for � = R2 have been proposed by Cao in [6] and by J. M. do Ó in [10] and R. Panda

in [14], for the general case � = RN (N ≥ 2), where it was proved that for all α > 0

and u ∈ W1,N(RN), it holds �N(α|u|N/(N−1)) ∈ L1(RN), where

�N(t) = et −N−2∑j=0

1

j! t j.

Moreover, it was proved that if α < αN and ‖u‖2 ≤ M, there exists C(N, M, α) > 0such that

supu∈W1,N

0 (�) : ‖u‖≤1

∫RN

�N(α|u|N/(N−1)) dx ≤ C(N, M, α). (1.6)

Adachi–Tanaka [1] proved that for any α ∈ (0, αN) there exists a constant C(N, α) >

0 such that∫

RN�N

(α

( |u|‖u‖

)N/(N−1))

dx ≤ C(N, α)‖u‖N

N

‖u‖Nfor all u ∈ W1,N(RN) \ {0}.

(1.7)In a recent paper Yuxiang–Ruf [16] proved that if the Dirichlet norm is replaced bystandard Sobolev norm, then

supu∈W1,N

0 (�) : ‖u‖1,N≤1

∫�

�N(α|u|N/(N−1)) dx = dN(α), (1.8)

where

dN(α) ={

< ∞ if α ≤ αN

+∞ if α > αN,

independently of the domain �.The main purpose of this paper is two-fold: Motivated by the results in [1, 3, 4, 6, 8–

11, 13–16], our first aim is to investigate Trudinger–Moser type inequalities for thespace of functions W1,N

0 (�) endowed with the Dirichlet norm as well as in the casethat W1,N

0 (�) is equipped with Sobolev norm with a singular weight for any domain� ⊂ R

N containing the origin. Moreover, we obtain the best exponent for this classof inequalities.

Next, for easy reference, we state our main results on Trudinger–Moser inequalityfor any smooth domain � ⊂ R

N (not necessarily bounded) contained the origin forthe space of functions W1,N

0 (�) endowed with the Dirichlet norm.

Theorem 1 Let a ∈ [0, N) and 0 < α < αN(N − a)/N, then there exists C(N, α, a) >

0 (independent of �) such that

∫�

�N

(α

(|u|‖u‖

)N/(N−1))

|x|a dx ≤ C(N, α, a)

‖u‖N

∫�

|u|N

|x|a dx for u ∈ W1,N0 (�) \ {0}.

(1.9)

Remark 1 The Theorem 1 improve the main result in [1], since in the nonsingularcase, that is, if a = 0, we have exactly the result stated in Eq. 1.7. Moreover,


Theorem 1 improve the main results in [3, 6, 10, 14], because we consider the singularcase and here � could be any smooth domain in R

N , not necessarily bounded.

Next, we prove that α = αN(N − a)/N is the best exponent for Eq. 1.9. Further-more, we show that differently from the classical Trudinger–Moser inequality (seeEq. 1.4), the exponent α = αN(N − a)/N, which corresponds to α = αN in the casea = 0, is excluded in Eq. 1.9. We show that if α ≥ αN(N − a)/N we can construct asequence (uk) ⊂ W1,N

0 (�) such that ‖uk‖ = 1 and for which an inequality similar toEq. 1.9 does not hold.

Theorem 2 Let a ∈ [0, N) and α ≥ αN(N − a)/N, then there exists a sequence (uk) ⊂W1,N

0 (�) such that ‖uk‖ = 1 and

(∫�

|uk|N

|x|a dx)−1 ∫

�

�N(α|uk|N/(N−1))

|x|a dx → +∞, as k → +∞. (1.10)

Following some ideas of [6, 8, 10, 11], next we show that replacing the Dirichletnorm by the standard Sobolev norm on W1,N

0 (�) yields a bound independent of �,and for this case, the best exponent α = αN(N − a)/N is included.

Theorem 3 (Sobolev Norm) Let a ∈ [0, N) and 0 < α ≤ αN(N − a)/N, then thereexists a constant C = C(N, α, a) > 0 (independent of �) such that

supu∈W1,N

0 (�) : ‖u‖1,N≤1

∫�

�N(α|u|N/(N−1))

|x|a dx ≤ C(N, α, a). (1.11)

Theorem 4 Let a ∈ [0, N) and 0 < α ≤ αN(N − a)/N, then the inequality 1.11 issharp, more precisely, for any α > αN(N − a)/N the supremum in Eq. 1.11 is +∞.

Remark 2 We call attention to the fact that in Theorem 3, we consider the singularand nonsingular case, consequently our result is an improvement of the inequality1.8 proved by Yuxiang–Ruf in [16]. It is worth mentioning that our theoremscomplement some results contained in [4, 8, 9, 11].

Remark 3 We focus our analysis of inequalities 1.9 and 1.11 on the singular case, thatis, when � ⊂ R

N is a smooth domain containing the origin, since the nonsingular casewas already studied in the papers [1, 6, 10, 14, 16].

2 Proof of Theorem 1

We note that it suffices to show that for any 0 < α < αN(N − a)/N, there existsC(N, α, a) > 0 (independent of �) such that

∫�

�N(α|u|N/(N−1)

)|x|a dx ≤ C(N, α, a)

∫�

|u|N

|x|a dx (2.1)

for all u ∈ W1,N0 (�) such that ‖u‖ = 1.


It is clear that

∫�

�N(α|u|N/(N−1)

)|x|a dx ≤

∫RN

�N(α|u|N/(N−1)

)|x|a dx,

since any function u ∈ W1,N0 (�) can be extended by zero outside of �, obtaining a

function in W1,N(RN). Hence, it suffices to show the desired inequality for functionsu ∈ W1,N(RN). Moreover, we can make use of symmetrization, to show the desiredinequality for functions u(x) = u(|x|), which are non-negative, radially symmetric anddecreasing.

Thus assuming that u is radially symmetric we can write

∫RN

|u(x)|N dx = ωN−1

∫ +∞

0|u(r)|NrN−1 dr, (2.2)

∫RN

|∇u(x)|N dx = ωN−1

∫ +∞

0|u′(r)|NrN−1 dr, (2.3)

∫RN

�N(α|u(x)|N/(N−1)

)|x|a dx = ωN−1

∫ +∞

0�N

(α|u(r)|N/(N−1)

)rN−a−1 dr. (2.4)

Now let γ := 1 − a/N, then 0 < γ ≤ 1 and α/γ < αN . Setting v(r) :=γ (N−1)/Nu(r1/γ ). We have

∫ +∞

0|v′(r)|NrN−1 dr =

∫ +∞

0|u′(r)|NrN−1 dr

and ‖u‖ = ‖v‖. Moreover,

∫ +∞

0|v(r)|NrN−1dr = γ N

∫ +∞

0|u(r)|NrN−a−1 dr.

Thus,

∫RN

|v|N dx = γ N∫

RN

|u|N

|x|a dx. (2.5)

Now notice that

∫RN

�N(α|u|N/(N−1)

)|x|a dx = 1

γ

∫RN

�N

(α

γ|v|N/(N−1)

)dx. (2.6)

Indeed,

∫RN

�N

(α

γ|v|N/(N−1)

)dx = ωN−1

∫ +∞

0�N

(α

γ|v(r)|N/(N−1)

)rN−1 dr

= ωN−1

∫ +∞

0�N

(α|u(r1/γ )|N/(N−1)

)rN−1 dr.


Setting ρ = r1/γ and using that Nγ = N − a, we have∫ +∞

0�N

(α|u(r1/γ )|N/(N−1)

)rN−1 dr =

∫ +∞

0�N

(α|u(ρ)|N/(N−1)

)ρNγ−γ γργ−1dρ

= γ

∫ +∞

0�N

(α|u(ρ)|N/(N−1)

)ρN−a−1 dρ,

which implies that Eq. 2.6 holds.Since α/γ < αN , using inequality 1.7, there exist C(N, α/γ ) > 0 such that∫

RN�N

(α

γ|v|N/(N−1)

)dx ≤ C(N, α/γ )‖v‖N

N.

Therefore, using Eq. 2.5 there exist C(N, α, a) > 0 such that∫

RN

�N(α|u|N/(N−1)

)|x|a dx ≤ C(N, α, a)

∫RN

|u|N

|x|a dx,

which completes the proof. �


Proof First we note that since the inequality 1.9 is invariant by scale, we can assumewithout loss of generality that B(0, 1) ⊂ �. Indeed, let λ > 0 and uλ(x) = u(λx), thuswe have ‖uλ‖ = ‖u‖ and∫

�λ

�N(α|uλ|N/(N−1))

|x|a dx = λ(a−N)

∫�

�N(α|u|N/(N−1))

|x|a dx,

where �λ := {z ∈ RN : z = λx with x ∈ �}. Consequently, in order to prove

Theorem 2, it is enough to obtain a sequence (uk) ⊂ W1,N0 (B(0, 1)) satisfying Eq.

1.10. Moreover, using the fact that �N(t) = et − ∑N−2j=0 t j/j! is an increasing function,

it is enough to consider the case α = αN(N − a)/N. Indeed for α ≥ αN(N − a)/N,we have (∫

�

|uk|N

|x|a dx)−1 ∫

�

�N(α|uk|N/(N−1))

|x|a dx

≥(∫

�

|uk|N

|x|a dx)−1 ∫

�

�N( N−a

N αN|uk|N/(N−1))

|x|a dx.

Next, following an argument proposed by Moser in [13], we construct a sequence(uk) ⊂ W1,N

0 (B(0, 1)) satisfying Eq. 1.10. For each k ∈ N, we consider the functionwk : R → R given by

wk(t) =

⎧⎪⎨⎪⎩

0, for t ≤ 0,

k(N−1)/N tk

, for 0 ≤ t ≤ k,

k(N−1)/N, for k ≤ t.

(3.1)

For convenience we introduce the variable |x|N = e−t and we set

wk(t) = NN

N−1 ω1NN−1uk(|x|). (3.2)


Then uk : RN → R is radially symmetric, decreasing and non-negative function

compactly supported in B(0, 1). Moreover, we have

∫B(0,1)

|∇uk|N dx =∫ +∞

−∞|w′

k(t)|N dt = 1, (3.3)

∫B(0,1)

�N(α|uk|N/(N−1))

|x|a dx = ωN−1

N

∫ +∞

−∞�N

(α

αNwk(t)N/(N−1)

)e

(a−N)

N t dt, (3.4)

∫B(0,1)

|uk|N

|x|a dx = 1

NN

∫ +∞

−∞|wk(t)|Ne

(a−N)

N t dt. (3.5)

Thus, it is enough to show that

∫ +∞

−∞|wk(t)|Ne

(a−N)

N t dt → 0 as k → +∞ (3.6)

and

∫ +∞

−∞�N

(α

αNwk(t)N/(N−1)

)e

(a−N)

N t dt ≥ NN − a

≥ 1, (3.7)

for any k ∈ N sufficiently large.First we prove Eq. 3.6. Using that a ∈ [0, N), an easy computation yields

∫ +∞

−∞|wk(t)|Ne

(a−N)

N t dt ≤ 1

k

∫ +∞

0tNe

(a−N)

N t dt + NkN−1

N − ae

(a−N)

N k → 0 as k → +∞.

Next we estimate Eq. 3.7. Setting

I =∫ +∞

−∞�N

(α

αNwk(t)N/(N−1)

)e

(a−N)

N t dt

and α = αN(N − a)/N, by the definition of wk(t) given in Eq. 3.1, we obtain

I =∫ k

0�N

(N − a

Nk

(tk

)N/(N−1))

e− (N−a)N t dt +

∫ +∞

k�N

(N − a

Nk)

e− (N−a)N t dt,

which together with the definitions of �N(t) implies

I =∫ k

0exp

[N − a

N

(k

(tk

)N/(N−1)

− t

)]dt

−N−2∑j=0

1

j!(

N − aN

) j

k− jN−1

∫ k

0t

NN−1 je

(a−N)

N t dt

+ NN − a

�N

(N − a

Nk)

e− (N−a)

N k.


Consequently, we get

I ≥∫ k

0e

(a−N)

N t dt −N−2∑j=0

1

j!(

N − aN

) j

k− jN−1

∫ k

0t

NN−1 je

(a−N)

N t dt

+ NN − a

⎛⎝e

N−aN k −

N−2∑j=0

1

j!(

N − aN

) j

k j

⎞⎠ e− (N−a)

N k.

Note that ∫ k

0e

(a−N)N t dt = −N

N − a

(e

(a−N)N k − 1

)→ N

N − aas k → +∞

and for j = 1, . . . , N − 2, it holds

1

j!(

N − aN

) j

k− jN−1

∫ k

0t

NN−1 je

(a−N)

N t dt → 0 as k → +∞.

Therefore, we obtain that I ≥ N/(N − a), which implies that Eq. 3.7 holds. Thiscompletes the proof of Theorem 2. �


Proof It is clear that

sup‖u‖1,N≤1

∫�

�N(α|u|N/(N−1))

|x|a dx ≤ sup‖u‖1,N≤1

∫RN

�N(α|u|N/(N−1))

|x|a dx,

since any function u ∈ W1,N0 (�) can extended by zero outside of �, obtaining a

function in W1,N(RN). Hence, it sufficient to show that

sup‖u‖1,N≤1

∫RN

�N(α|u|N/(N−1))

|x|a dx ≤ d. (4.1)

Moreover, by means of symmetrization, it suffices to show the desired inequality forfunctions which are non-negative, radially symmetric and decreasing.

In order to prove Eq. 4.1, we write∫

RN

�N(α|u|N/(N−1))

|x|a dx =∫

B(0,ρ0)

�N(α|u|N/(N−1))

|x|a dx

+∫

RN\B(0,ρ0)

�N(α|u|N/(N−1))

|x|a dx. (4.2)

where ρ0 > 0 will be chosen later. To estimate the first integral in Eq. 4.2, let v(ρ) =u(ρ) − u(ρ0) if 0 ≤ ρ ≤ ρ0. Notice that v ∈ W1,N

0 (B(0, ρ0)).Using the fact that the function g : (0,+∞) → R given by

g(t) = (t + 1)N/(N−1) − tN/(N−1) − 1

t1

N−1


is bounded, we have a positive constant A = A(N) such that

uN/(N−1)(ρ) ≤ vN/(N−1)(ρ) + Av1

N−1 (ρ)u(ρ0) + uN/(N−1)(ρ0).

By the Young inequality, we obtain

v1

N−1 u(ρ0) ≤ vN/(N−1)(ρ)uN(ρ0)

N+ N − 1

N≤ vN/(N−1)(ρ)uN(ρ0)

N+ 1. (4.3)

Using the Radial Lemma (see [5]), we have u(ρ0) ≤ 1ρ0

(N

ωN−1

)1/N ‖u‖N. Therefore,

uN(ρ0) ≤ N

ωN−1ρN0

‖u‖NN. (4.4)

From Eqs. 4.3 and 4.4, we obtain

v1

N−1 (ρ)u(ρ0) ≤ vN/(N−1)(ρ)‖u‖N

N

ωN−1ρN0

+ 1 (4.5)

and hence

uN/(N−1)(ρ) ≤ vN/(N−1)(ρ)

(1 + A

‖u‖NN

ωN−1ρN0

)+ A + uN/(N−1)(ρ0).

Using again the Radial Lemma, we have

uN/(N−1)(ρ) ≤ vN/(N−1)(ρ)

(1 + A

‖u‖NN

ωN−1ρN0

)+ A + 1

ρN/(N−1)

0

(N

ωN−1

)N−1

‖u‖N/(N−1)

N .

Thus, for ‖u‖1,N ≤ 1, we have

uN/(N−1)(ρ) ≤ vN/(N−1)(ρ)

(1 + A

‖u‖NN

ωN−1ρN0

)+ A + 1

ρN/(N−1)

0

(N

ωN−1

)1/(N−1)

= vN/(N−1)(ρ)

(1 + A

‖u‖NN

ωN−1ρN0

)+ d(ρ0).

Hence u(ρ) ≤ w(ρ) + d(ρ0)(N−1)/N , where

w(ρ) := v(ρ)

[1 + A

‖u‖NN

ωN−1ρN0

](N−1)/N

.

Next we are going to prove that ‖∇w‖N ≤ 1 if ‖u‖1,N ≤ 1. First we note that∫B(0,ρ0)

|∇v|N dx =∫

B(0,ρ0)

|∇u|N dx ≤ 1 − ‖u‖NN,

which implies

∫B(0,ρ0)

|∇w|N dx =(

1 + A‖u‖N

N

ωN−1ρN0

)N−1 ∫B(0,ρ0)

|∇v|N dx

≤(

1 + A‖u‖N

N

ωN−1ρN0

)N−1

(1 − ‖u‖NN).


Since

(1 + A

ωN−1ρN0

sN)N−1

(1 − s) ≤ 1 if s ∈ [0, 1] andA

ωN−1ρN0

≤ 1/(N − 1),

we obtain∫

B(0,ρ0)|∇w|N dx ≤ 1. Notice that

∫B(0,ρ0)

�N(α|u|N/(N−1))

|x|a dx ≤∫

B(0,ρ0)

e(α|u|N/(N−1))

|x|a dx

which together with u(ρ)N−1/N ≤ w(ρ)N/(N−1) + d(ρ0) implies

∫B(0,ρ0)

�N(α|u|N/(N−1))

|x|a dx ≤ eαd(ρ0)

∫B(0,ρ0)

e(α|w|N/(N−1))

|x|a dx.

But w ∈ W1,N0 (B(0, ρ0)) and

∫B(0,ρ0)

|∇w|Ndx ≤ 1, thus by inequality 1.5, there existd1 > 0 such that

∫B(0,ρ0)

�N(α|u|N/(N−1))

|x|a dx ≤ d1. (4.6)

For the second integral, choosing ρ0 sufficiently large, we have

∫RN\B(0,ρ0)

�N(α|u|N/(N−1))

|x|a dx ≤∫

RN\B(0,ρ0)

�N(α|u|N/(N−1)) dx. (4.7)

Hence, by Radial Lemma and using the expansion of the exponential function thereexist d2 > 0 such that

∫RN\B(0,ρ0)

�N(α|u|N/(N−1)) dx ≤ d2. (4.8)

Then the proof follows from Eqs. 4.6, 4.7 and 4.8. �


Next we will show that Eq. 1.11 does not hold if α/αN + a/N > 1. Consider the Moserfunction (cf. [13]):

Mn(x) = (ωN−1)−1/N

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

(log n)(N−1)/N if |x| ≤ 1/n

log ( 1|x| )

(log n)1/Nif 1/n ≤ |x| ≤ 1

0 if |x| ≥ 1.

Hence Mn(·) ∈ W1,N(RN), supp(Mn) = B1, ‖Mn‖ = 1 and Mn ⇀ 0 in W1,N0 (B(0, 1)).

Then by the compact embedding, we way assume∫

B1Mp

n dx → 0 for p ≥ 1, then


‖Mn‖1,N ≤ 1, for n large. Given any fixed α such that α/αN + a/N > 1, we take β ∈(αN(N − a)/N, α), then α(Mn/‖Mn‖1,N)N/(N−1) > βMN/(N−1)

n when n are sufficientlylarge. So, we get

limn→+∞

∫RN

�N(α|Mn/‖Mn‖1,N|N/(N−1))

|x|a dx ≥ limn→+∞

∫B1

(eβMN/(N−1)n − 1)

|x|a dx = +∞.

This completes the proof of result. �

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on singular trudingerâmoser type inequalities for unbounded domains and their best exponents

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