1. Introduction

2. Background

Let f,g be Orlicz functions. If for every c>0, , we write g

3. Tools for the generalized Trudinger type inequality

b) for every cube Q, ,

satisfies and

4. The imbeddings theorems

.

(4.5)

.

(4.7) ,

) respectively, we may assume that diam(Q0)=1, ( )=1/2 and ||--u||Ln+s( , )=1. We must thus show that fs(|u|/t) is bounded by a constant for some t>0 uniformly bounded from above. Let where . By lemma 2.1 fs' and fs define the same Orlicz space. By the triangle inequality and the convexity of fs' we have the pointwise estimate fs'(|u(x)-uQ0, |/t)<fs'(|u(x)-uQ, |/t)+fs'(|uQ0, uQ, |/t), where x belongs to Q. We call the first type of term "good" and the second kind "bad" and denote by G and B, respectively, their total integrals.
Expanding fs' as a power series we have conclude that
(4.8) .
Assuming s=0 we may use the estimate from lemma 4.2(c) with p=n in (4.8) by the remark following that lemma. Noting that Qn/ (Q) m(Q)/ (Q) 1/cw, and that ||--u|| 1, we arrive at the inequality:
,
so any G is bounded for any t>a1/n'.
If s>0, we instead apply the estimate (4.5) from Lemma 4.2(a) (again invoking the local equivalence of and the Lebesgue measure):
.
Inserting this into (4.8) noting that ||--u||n+s=1 and interchanging the order of summation (positive summands), we see that
,
where the local balance condition was used for the last inequality. Hence the good terms are bounded by assumption (4.7) in the case s>0 case also.
We still have to estimate the bad terms, i.e.
(4.9) .
Using the triangle inequality, we estimate
(4.10),
where {Qi} is the chain of Whitney cubes from the global balance condition (we implicitly assume, without loss of generality, that k-1 is even). Now we derive the estimate:
.
The second inequlity follows from the fact that 9Qi+1 contains Qi and the local doubling condition, the third from estimate (4.6) with q=1 and p=n+s. Note that the quantity before the gradient's norm is 1/(n+s)'(9Qi+1).
Inserting this estimate in (4.10) and applying the Hölder inequality for sums with the exponent n+s followed by the global balance condition deduce:
,
Inserting this into (4.9) we get
,
where the last inequality follows by assumption (4.7) and the second to last is valid when t>(C0/ )1/(n+s)'. Hence also the bad terms have been controlled for t bounded by max{2a1/n', 2(C0/ )1/(n+s)') where "a"(from lemma 4.2(c) depends only on n and C0 (from (4.6) through some intermediate steps) only on n and s. Hence the Orlicz norm has a finite bound, which is what we needed to show since we assumed ||--u||=1.
Note that in the proof of this theorem we separated the estimation of "good" and "bad" terms by means of the triangle inequality. It is far from clear that nothing is lost in this approximation. Of course the whole proof revolves around this estimate, in that we basically assume our measures to be locally equivalent to the Lebesgue measure, use the old results for this case and bound the non-local difference by means of a chain. The sharpness properties of this method of proof is wanting of further investiagtion.

Recall that r:={x s< | (x)<r}, the interior boundary of thickness r of .
Lemma 4.3 Let be QHBC(with constant c ) and sDC1( ). If C1>C(c ,n) then there exist C2, >0 such that
.
Proof: [h].

Lemma 4.4 If is QHBC then any finite (i.e. ( )< ) sDC1( ) satisfies the sub-summability condition (4.7) of theorem 3.
Proof: Assume, without loss of generality, that diam( )=1 and ( )=1 and define Li:= 2-i+1\ 2-i. By definition, Q 2i if Q has center zQ in Li and moreover the Q is contained in 2-i+2. The number of such Whitney cubes is bounded by 2in since ( )=1 and hence by the Hölder inequality for sums with exponent (1 ) we have
(4.11) ,
where we have used Lemma 4.3 to bound the second term for the second inequality. Hence, for any satisfying 0< < /( + ), we derive the sub-summability condition by summing (4.11) over i.

Corollary 1 Let satisfy the QHBC and C0-slice condition. Further let w sA1( ) interDC0( ) and v sRH ( ) interDC1( ) for C1>c(C ). Then ( ,w,v) ss-T if and only if (v,w) is globally balanced and w is bounded away from zero when s=0.
Proof: Follows from Theorems 2 and 3, proposition 1 and Lemma 4.4.

5. Sharpness results

We no longer implicitly assume the measures to be locally doubling, however, still denotes a bounded domain in Euclidian n-space.
In trying to prove sharpness results of the imbeddings established in chapter 4, it important to realize that some of those results are not sharp at all! The easiest way to see this is to consider why we started investigating Trudinger type inequlities in the first place, viz. because of the gap between L and Lp for a finite p. However in chapter 4 no precausions are taken to ensure that the L norm is infinite, and in fact for some measures it might not be.
Owing to theorem 1, we may conclude that if is locally doubling and is not too complicated (in particular it might be QHBC) then the Trudinger inequality implies the local balance condition which implies that for every >0, Qn+s+ / (Q) lim0 as Q approaches the boundary. Therefore some growth assumtion on the mearsures seems appropriate, although the assumptions contained in what follows are quite a bit more specific than those of the positive results of the previous section. Another obvious difference to the previous theorems is that in proving sharpness resluts, we can restrict our attention to a neighbourhood of one arbitrary boundary point, since if an inequality is not valid there it can impossibly be valid for the whole domain.

Proposition 2 (sharpness) Let R>0, z be a fixed boundary point and {xi} be an sequence in satisfying:
(a) | xi-z| <2 (xi)=:4ri,
(b) (2jBi) (2jri)n+s, for all j 0, where Bi:=B(xi,ri) and
(c) (Bi)>riR.
Then s in the Trudinger inequality is sharp, i.e. (2.1) becomes false if we replace s by any satisfying s .
Note: Condition (b) is equivalent to (2iB) 1 for every i 0. Note that (b) constrains the boundary behavior of at z, whereas (c) is more like a local condition, never actually reaching the boundary.
Proof: Assume x0 is in the center cube Q0 and (x0)=1. Further assume ri<1/6. Fix m>0 and denote r:=rm, B:=Bm and x:=xm. Let M be the smallest integer such that 2M+2 includes x0. From this definition and (a) it follows that
.
Hence as (x) lim0, so we may assume that this ratio is between 1/2 and 2.
Set Ai:=(2iB)\(2i-1B) for every i M. Define a radial function u(y):=f(|x-y|), linear on each Ai, zero outside 2MB, constant on B and continuous. Moreover fix the slopes so that f(2i-1)-f(2i)=M-1/(n+s). Hence |--u|=21-i r-1 M-1/(n+s) for points in Ai and

because Ai is contained in 2iB by assumption (b). Summing over i we conclude that ||--u||n+s is bounded by one.
B0 lies completely outside 2MB, because if they intersected, 2M+1 would contain B0 contrary to definition. Hence u(x)=0 on B0 and u(x)=M1 1/(n+s)= M1/(n+s)' on B. Thus
,
where Q0 is contained in B0. Fixing t>0 greater than the psi-Luxemburg-norm of u-uQ0, , we derive, assuming a psi-Trudinger inequality can be applied (which it of course can, after the usual smoothening argument) :
(5.1) ,
provided C0>t(2R+1)ln(2). The constant C0 is the reciprocal of the c in the definition of , above (lemma 2.1), and may hence be chosen arbitrarily. For points xi converging to z, M and r lim0, contradicting (5.1) and thus showing that fs is sharp (if it's applicable, in the first place).

Theorem 4 (a necessary condition) Let satisfy a c0-weak slice property with center x0. Suppose that (B(x,t)) tn+s and (Q) QR for every x and Q sW( ) and R>0, t 0. If ( , , ) satisfies an s-Trudinger inequality then is QHBC.
Note: The condition on can again be succinctly stated using : for every t>0 and every ball B, (tB)>1.
Proof: As usual, we need only consider the QHBC condition for points far from x0 in the quasi-hyperbolic metric, so we assume that mx 2. We will denote the fixed mx by m and further assume that c0>2. Akin to the proof of theorem 2, we define
,
where the infimum is taken over all rectifiable curves through x0 and y. Using the same intermediate steps as in the proof of theorem 2 we have, by the condition on :
(5.2) ,
Condition (d) of the definition of the weak slice property implies ui(x)=0 for every x in B0 and condition (b) implies ui(x) m 1/(n+s) for x in B (B0 and B are defined in section 3.1.3, above). Define u:=Sui. It follows from (5.2) that and u(x) m(n+s 1)/(n+s) for x belongsB. Hence, using the s-Trudinger inequality (after the usual smoothening argument) we conclude:
,
for some bounded t. Solving for t, we derive
,
and so, by condition (c) of the weak-slice property,
(5.3) .
Recalling that B is defined in terms of its centers distance to the boundary of (it is a Whitney type object), we conclude by that (B) R(x), where x is a point in B, by the assumption on . Then (5.3) clearly implies the QHBC condition.

Corollary 2 is QHBC iff it has a weak slice property and ( , t , s) ss-T.
Proof: If is QHBC, lemma 3.2 implies that it has a weak slice property. From the inequality, where we first estimate (Q) 1 and then use the definition of QHBC, formula (2.6):
(5.6)
we deduce that ( t, s) is globally balanced since the constant c1 depends on the arbitrary constant in the balance condition and can hence be made smaller than t, so that the right hand side of (5.6) dominates n+t(x). Also t satisfies the sub-summability condition (4.7) by lemma 4.4. Hence the corollary follows by theorems 3 and 4.

Let us define measures , absolutely continuous with respect to the Lebesgue measure satisfying
(5.5a) (Q) g( Q) Qn+s and
(5.5b) (Q) f( Q)
on the unit sphere at the vicinity of some boundary point. The sense of (5.5) is of course to be able to control the behavior of the balance condition directly, since (Q) g( Q)-1/(n+s-1). To see that the conditions (5.5) can be satisfied consider
(5.6) ,
where z is a boundary-point and f and g are functions belonging to D(0, ):
(5.7) .
Then if {Qi} is a chain of Whitney cubes converging to z, all but a finite number have the desired property. With the help of these functions we derive the following

Proposition 3 (sharpness) Let be the unit sphere and z a boundary point. For the measures in (5.6) define ai:=f(2-i) and bi:=g(2-i). Assume: (i) , (ii) and (iii) where c0>1 and
.
Then ( , , ) does not support an s-Trudinger inequality .
Note: Because of (5.5b) i is the sum in the global balance condition for the chain consisting of the i first cubes on a path from the center to the boundary point z. Assumption (i) tells us that does not satisfy the sub-summability condition (4.7) even for this chain. Hence (ii) is like the global balance condition for this chain, except that the inequality is reversed. Since (i) and (ii) imply that i with i, condition (iii) is merely a technical assumption concerning the speed of divergence.
Proof: Fix an integer M>1 and define and
.
For i M, define , where , as usual, the infimum is over all rectifiable curves in passing through y and the origin (=center of ). Continuing in the usual fashion we define u:=Sui and deduce ||--ui||n+s ci and ||--u||n+s 1. Hence by considering only regions Sj for 1 j i we deduce, since the width of Sj is at least 2-j,
(5.8)
for every y belongsTi. By condition (iii) for i-1>M/2 i-1>c M with c<1. Hence (5.8) implies u(y)>c 1-1/(n+s)(i-1)=c 1/(n+s)'(i-1) with c>1. Hence
,
where the last inequality follows from assumption (ii). As {ai} if l1 but not l p for any p, 0<p<1, we conclude that the left hand side is not bounded, an so, after the usual smoothening argument, we conclude that ( , , ) does not support a s-Trudinger inequality.

Example 5.1  There exists , satisfying the conditions of proposition 2 and further all the conditions of theorem 3 except the sub-summablility condition (4.7):



(5.9) .
Proof: Clearly (5.7) implies that and are both strongly doubling and Muckenhoupt and Reverse Hölder respectively. As (Q) is not sub-summable even over the single chain of cubes converging to z. As Q 2-i if Q intersects Si, (Q) g( Q)-1/(n+s-1)=log(3 Q-1) i for this Q. Moreover each Si contains only a bounded number of cubes intersecting this chain, so we have the following inequality:
(5.10) ,
where r is the distance of the final cube from the boundary; r 2-i for cubes in Si (see the picture on the following page). The second inequality follows from Euler's formula (the one defining 0.58...). From (5.9) we see that (Q) i-1log-2(i) for a cube intersecting Si. Hence by choosing c0<1/c1 (c0 is the constant in the definition of the balance condition, and may hence be chosen freely) in (5.10), we arrive at the global balance condition for cubes along the chain.
If s>0 the function g(t)tn+s has one minimum in the range [0,2], at 0. Thus we see that (Q1)>c(s) (Q2) when Q1 is the cube on the chain and Q2 is any other cube with Q1 Q2 and similarly (Q1)>c(s) (Q2). Hence the inequlity anlong the chain implies the global balance condition in the case s>0 for any chain.
If s=0, we fix an arbitrary point on the boundary, z'. We divide a chain of Whitney cubes converging to z' into two parts, one (which might be empty) in which |x-z| (x) and another in which |x-z|< (x). By the argument above the first part of the chain satisfies the global balance condition. However on the second part |x-z| |z'-z|, hence bounded from below, and so the global balance condition is trivially satisfied.

6. Conclusions

After a couple of theorems that were kept as general as possible, chapter 4 culminated in corollary 1, which states the tells us that the s-Trudinger inequlity is equivalent to the measures being globally balanced. To derive this reslut, however, the single, relatively weak, conditions from the preceeding theorems and lemmas were superimposed so that corrollary 1 in fact makes use of all the restrictions introduced in chapter 3.
One way to avoid this accumulation of constraints would be to use a more similar set throughout the chapter. In particular the approch taken here needs to assume that the measures are doubling in order to necessity of the geometrical conditions, whereas the A1 and RH classes are required for the sufficiency. Note that the slice-condition (or something similar) has to be assumed even with the Lebesgue measure to have the geometrical condition imply the inequlity [f].
Also, as already noted, the proof of theorem 3 involves an approximating technique that is not intrinsically sharp. It might advantageously be investigated on which domains this estimate is sharp, even for simple measures.
In chapter 5 it was shown that if the measures are sufficiently regular near a single boudary point, the the Trudinger inequlity is sharp (if it is satisfied). Furthermore, under similar conditions on the measures, it was shown that for a domain satisfying a weak slice property the inequlity implies a QHBC condition. Hence we derived corollary 2, which is an improvement of previous results [f] even when the measures involved are Lebesgue measures. Finally it was shown by example that the sub-summability condition (4.6) in theorem 3 is not superfluous.
It should be noted that the sharpness resluts of chapter 5 are more restrictive by far than the positive resluts of chapter 4; in chapter 5 it is necessary to assume that the measures approximate the "canonical" measures discussed in section 3.3. Hence also corollary 2 is a theorem only for the one dimensional subspace d = s(x)dm of the measure space.
Thus it is seen that there is a large discrepency between what can be proven and what is known to be sharp. As was mentioned in the introduction to chapter 5, it is vital to consider where Hölder(and other) type imbedings take over. This phase of the considerations where not indulged in at all, although they were sure performed implicitly.
As was promised in the introduction, it has now been indicated that there are still questions to settle also in the Euclidean setting. In particular it would be interesting to derive more necessary conditions for a Trudinger type inequlity, so far only the local balance condition was derived. On the other hand, the local balance condition suffice in a single proof.