solutions:, and it cannot be the case that a supersolution is always greater than or equal to a subsolution.

Chapter 4 Comparison The basic problem to be considered here is the question when one can say that a supersolution is always greater than or equal to a subsolution of a problem, where one in most cases assume that this inequality holds on the boundary. It will then immediately lead to uniqueness for the Dirichlet problem (where the boundary value are specified). One cannot hope to establish the comparison principle in complete generality because if Å Ê is an open set and is a continuously differentiable function in Ê which vanishes on Å but not in all of Å, then the equation Ù ¾ ¼ ܵ ¾ ¼ with Ù ¼ on Å, then this equation has two solutions:, and it cannot be the case that a supersolution is always greater than or equal to a subsolution. 1. The simplest cases First we consider the case where we have a bounded open set and first we state an intermediate result that does not involve the equation at all. Proposition 4.1. Assume that and that (i) Å Ê is open, bounded, and nonempty; (ii) Ù ¾ ÍË Åµ is such that ÙÔ Ü¾Å Ù Üµ ; (iii) Ú ¾ ÄË Åµ is such that Ò Ü¾Å Ú Üµ ; (iv) ÙÔ Ü¾Å Ù Üµ Ú Üµµ ÙÔ Ý¾Å Ù Ýµ Ú Ýµµ; Then there is a point Ü ¾ Å such that Ù Ü µ Ú Ù µ ÙÔ Ü¾Å Ù Üµ Ú Üµµ and and for each there are points Ü Ý ¾ Å, Ô ¾ Ê, ¾ Ë µ such that ÐÑ Ü ÐÑ Ý Ü, ÐÑ Ù Üµ Ù Ü µ, ÐÑ Ú Ýµ 25

26 4. Comparison 11.11.2005 Ú Ü µ, Ü Ù Üµ Ô µ ¾  ¾ ÅÙ, Ý Ú Ýµ Ô µ ¾  ¾ ÅÚ,, ÐÑ Ü Ý Ô Ü Ý ¾ µµ ¼. Proof of Proposition 4.1. Let Ü Ýµ ٠ܵ Ú Ýµ ¾ «Ü ݾ Ü Ý ¾ Å Since is upper semicontinuous and Å Å is compact the maximum of is achieved at some point Ü«Ý«µ. If Þ is a cluster point of Ü«Ý«µ then it follows from the assumption ÙÔ Ü¾Å Ù Üµ Ú Üµµ ÙÔ Ý¾Å Ù Ýµ Ú Ýµµ and from Lemma 4.6 that Þ Ü Ü µ for some Ü ¾ Å. It also follows that Ù Ü µ Ú Ü µ ÙÔ Ü¾Å Ù Üµ Ú Üµµ and this implies, since Ù is upper and Ú is lower semicontinuous that ÐÑ Ù Üµ Ù Ü µ and ÐÑ Ú Ýµ Ú Ü µ. By taking subsequences we find Ü and Ý such that ÐÑ Ü ÐÑ Ý Ü. The remaining claims follow from Theorem 2.9 by taking Ù Ù, Ù ¾ Ú, ³ Ü Ýµ ¾ «Ü ݾ, «, so that ܳ Ü Ýµ Ý ³ Ü Ýµ «Ü ݵ Á Á and ¾ ³ Ü Ýµ «. Á Á Now we get a first comparison result (where Ø Ñܼ Ø): Theorem 4.2. Assume that and that (i) Å Ê is open, bounded, and nonempty; (ii) ¾ Å Ê Ê Ë µ ʵ is nonincreasing in its fourth argument; (iii) There is a function ¾ Å Å Ê Ê Êµ such that Ü Ü Ö ¼µ ¼ and Ü Ö Ô µ Ý Ö Ô µ Ü Ý Ö Ü ÝÔ Ü Ý ¾ µ for all Ü Ý ¾ Å, Ö ¾ Ê, Ô ¾ Ê and ¾ Ë µ; (iv) There is a function ¾ Å Ê ¼ µ ʵ such such that Ü Ö Ô µ Ü Ô µ Ü Ö µ ¼ for all Ü ¾ Å, Ö, Ô ¾ Ê, and ¾ Ë µ. (v) Ù ¾ ÍË Åµ is a subsolution of ¼ in Å and ÙÔ Ü¾Å Ù Üµ ; (vi) Ú ¾ ÄË Åµ is a subsolution of ¼ in Å and Ò Ü¾Å Ú Üµ ; Then ٠ܵ Ú Üµ ÙÔ Ý¾Å Ù Ýµ Ú Ýµ Ü ¾ Å Proof. If the claim does not hold, then the assumptions of Proposition 4.1 hold. Since Ù is a subsolution, Ú is a supersolution and is continuous we see that Ü Ù Üµ Ô µ ¼ and Ý Ú Üµ Ô µ ¼

2. Parabolic equations 27 Now it follows from the assumptions that ¼ Ü Ù Üµ Ô µ Ý Ú Üµ Ô µ Ü Ù Üµ Ô µ Ý Ú Üµ Ô µ Ü Ù Üµ Ô µ Ü Ú Üµ Ô µ Ü Ú Üµ Ô µ Ý Ú Üµ Ô µ Ü Ù Üµ Ú Ýµµ Ü Ý Ú Üµ Ü ÝÔ Ü Ý ¾ µ Ü Ù Ü µ Ú Ü µ Ü Ü Ú Ü µ ¼µ ¼ and we have a contradiction. 2. Parabolic equations Just as in the case of elliptic equations there are very many different variants of the comparison results. We start by presenting a basic one. Theorem 4.3. Assume that and that (i) Å Ê is open, bounded, and nonempty and ¼ Ì ; (ii) ¾ ¼ Ì µ Å Ê Ê Ë µ ʵ is nondecreasing in its third and nonincreasing in its fifth argument; (iii) There is a function ¾ ¼ Ì µ Å Å Ê ¼ µ ʵ such that Ø Ü Ü Ö ¼µ ¼ and Ø Ü Ö Ô µ Ø Ý Ö Ô µ Ø Ü Ý Ö Ü ÝÔ Ü Ý ¾ µ for all Ø ¾ ¼ Ì µ, Ü Ý ¾ Å, Ö ¾ Ê, Ô ¾ Ê and ¾ Ë µ; (iv) Ù ¾ ÍË ¼ Ì µ ŵ is a subsolution of Ù Ø Ø Ü Ù Ü Ù ¾ ÜÙµ ¼ in ¼ Ì µ Å and ÙÔ Ø¾ ¼ µü¾å Ù Ø Üµ for all Ì ; (v) Ú ¾ ÄË ¼ Ì µ ŵ is a supersolution of Ú Ø Ø Ü Ú Ü Ú ¾ ÜÚµ ¼ in ¼ Ì µ Å and Ò Ø¾ ¼ µü¾å Ú Ø Üµ for all Ì ; Then Ù Ø Üµ Ú Ø Üµ ÑÜ ÙÔ ¾ ¼Ì µý¾å ٠ݵ Ú Ýµµ ÙÔ Ù ¼ ݵ Ú ¼ ݵµ for every Ø ¾ ¼ Ì µ and Ü ¾ Å. In particular, if Ù Ø Ýµ Ú Ø Ýµ when Ø ¾ ¼ Ì µ and Ý ¾ ŵ or when Ø ¼ and Ý ¾ Å, then Ù Ø Üµ Ú Ø Üµ for all Ø ¾ ¼ Ì µ and Ü ¾ Å. ݾÅ

28 4. Comparison 11.11.2005 Proof of Theorem 4.3. Let Ì and ÑÜ ÙÔ ¾ ¼Ì µý¾å ٠ݵ Ú Ýµµ ÙÔ Ý¾Å Ù ¼ ݵ Ú ¼ ݵµ. If we can show that Ù Ø Üµ Ú Ø Üµ for all Ø ¾ ¼ µ, Ü ¾ Å, and ¼ then we are done. Suppose that this Ø is not the case. Let Ø Ü Ýµ Ù Ø Üµ Ú Ø Ýµ Ø ¾ «Ü ݾ Ø ¾ ¼ µ Ü Ý ¾ Å By assumption we know that this function is bounded from above and since Å is bounded the maximum value is achieved at some point Ø «Ü«Ý«µ ¾ ¼ µ Å Å. We may clearly choose a convergent subsequence and since we by Lemma 4.6 know that «Ü«Ý«¾ ¼ as «we find some point Ø Ü µ such that ÐÑ Ø «Ø and ÐÑ Ü«ÐÑ Ý«Ü and ÐÑ Ø «Ø. Our assumption that Ù Ø Üµ Ú Ø Üµ does not hold implies that Ø Ø ¾ ¼ µ and Ü ¾ Å By Theorem 2.16 (and the fact that È ¾ ¾ Ú Ø Üµ È there are Ø «Ü«µ Ù Ø «Ü«µ ««Ü«Ý«µµ «µ ¾ È ¾ ¼ µ ÅÙ Ø «Ý«µ Ù Ø «Ý«µ ««Ü«Ý«µµ «µ ¾ È ¾ ¼ µ ÅÚ Ú Ø Üµ) we know that such that (4.1) ««Ø «µ ¾ and «Á ¼ ¼ Á Á «Á Á Thus we have ««and ««and provided Ø «¼ and Ü«Ý«¾ Å (which we have seen to be the case for sufficiently large «) we have and «Ø «Ü«Ù Ø «Ü«µ «Ü«Ý«µ «µ ¼ «Ø «Ý«Ú Ø «Ý«µ «Ü«Ý«µ «µ ¼ By subtracting these equations from each other, using (4.1), the monotinicity assumptions on together with the facts that Ù Ø «Ü«µ Ú Ø «Ý«µ and ««

3. Some other results 29 we conclude that ¼ Ø «µ Ø «Ü«Ù Ø ¾ «Ü«µ «Ü«Ý«µ «µ Ø «Ý«Ú Ø «Ý«µ «Ü«Ý«µ «µ Ø «µ Ø «Ü«Ù Ø ¾ «Ü«µ «Ü«Ý«µ «µ Ø «Ý«Ù Ø «Ü«µ «Ü«Ý«µ «µ Ø «Ü«Ý«Ù Ø «Ü«µ «Ü«Ý«¾ Ü«Ý«¾ «µµ If we now take ««and let we get a contradiction because ÐÑ ««Ü«Ý«¾ ¼ and ««. 3. Some other results First we prove a version of the strong maximum principle. Theorem 4.4. Assume that and that (i) Å Ê is open and nonempty; (ii) Å Ê Ê Ë µ is nonincreasing in its fourth variable; (iii) For every ¼ there are ¼ and Æ ¼ such that Ü Ô ÔÁ Ô Ô Å Ô ¼ or Ü Ô ÔÁ Ô Ô Å Ô ¼ for all Ü ¾ Å, ¼, (or ¼ ), and ¼ Ô Æ. (iv) Ù ¾ ÍË Åµ (or ÄË Åµ) is a subsolution of Ü Ù Ù ¾ Ùµ ¼ (or a supersolution of Ü Ù Ù ¾ Ùµ ¼µ in Å and Ù achieves a local maximum (resp. minimum) at the point Ü ¼ ¾ Å with Ù Ü ¼ µ ¼ (Ù Ü ¼ µ ¼ ). Then Ù is constant in a neighbourhood of Ü ¼. Proof. We choose Ö ¼ so small that Ü ¼ Öµ Å and ٠ܵ Ù Ü ¼ µ for all Ü ¾ Ü ¼ Öµ. If Ù is not a constant in Ü ¼ Öµ, then there is, of course, a point Þ ¼ ¾ Ü ¼ Öµ such that Ù Þ ¼ µ Ù Ü ¼ µ but we claim that Þ ¼ can be chosen so that ٠ܵ Ù Ü ¼ µ when Ü Þ ¼ Ê and Ù Ý ¼µ Ù Ü ¼ µ for some Ý ¼ ¾ Þ ¼ ʵ where Ê Ö Þ ¼ Ü ¼. To see that this is the case we observe that since Ù is upper semi-continuous it follows that the set Ü ¾ Š٠ܵ Ù Ü ¼ µ is open so we may certainly choose Ê Ö Þ ¼ Ü ¼ such that ٠ܵ Ù Ü ¼ µ when Ü Þ ¼ Ê. If there is no Ý ¼ ¾ Þ ¼ ʵ such that Ù Ý ¼ µ Ù Ü ¼ µ we can replace Þ ¼ by Þ ¼ Ê Ü¼ Þ¼ Ü ¼ Þ ¼ µ and repeat this procedure until we find such a

30 4. Comparison 11.11.2005 point Ý ¼. Using once more the fact that Ù is upper semicontinuous we see that there is a number ¼ such that ٠ܵ Ù Ü ¼ µ when Ü Þ ¼ Ê. Furthermore ¾ we may choose Ù Ü ¼ µ ¼. Next define the function by where «¼. We have and ܵ Ù Ü ¼ µ e «Ê¾ e «Ü Þ ¼ ¾ ܵ ¾«e «Ü Þ ¼ ¾ Ü Þ ¼ µ ¾ ܵ «e «Ü Þ ¼ ¾ ¾Á «Ü Þ ¼ µ Å Ü Þ ¼ µ If now Ü Þ ¼ Ê and Ô Üµ then we have ¾ ¾ ܵ ¾Ô Ê Á «Ê Ô Ô Å Ô Thus we take ¾ in (iii) and if we choose «so large that Ô Æ, then we Ê Ê conclude from (ii) and (iii) that, provided ٠ܵ ¼, Ü Ù Üµ ܵ ¾ ܵµ ¼ Ü ¾ Þ ¼ ʵ Ò Þ ¼ Ê ¾ µ Now ܵ Ù Ü ¼ µ ٠ܵ for all Ü with Ü Þ ¼ Ê and by choosing «large enough we have ܵ Ù Ü ¼ µ e «Ê¾ e «Ê¾ Ù Ü ¼ µ ٠ܵ for all Ü with Ü Þ ¼ Ê. But we assumed that there is a point ¾ Ý ¼ with Ê ¾ Ý ¼ Þ ¼ Ê such that Ù Ý ¼ µ Ù Ü ¼ µ so that Ý ¼ µ Ù Ü ¼ µ Ù Ý ¼ µ. Thus we see that the maximum of the function Ù in Ü Ê ¾ Ý ¼ Þ ¼ Ê is achieved at some interior point Ü and since the maximum is positive we have Ù Ü µ Ü µ ¼. But since Ù is a subsolution and Ü Ù Ü µ Ü µ ¾ Ü µµ ¼ we get a contradiction. This implies that Ù is a constant in Ü ¼ Öµ. Theorem 4.5. Assume that and that (i) Å Ê is open, bounded, and nonempty; (ii) Ê Ò ¼ Ë µ Ê is nonincreasing in its second variable and Lipschitz-continuous on compact subsets of Ê Ò ¼ Ë µ; (iii) For every ¼ there are ¼ and Æ ¼ such that Ô ÔÁ Ô Ô Å Ô ¼ and Ô ÔÁ Ô Ô Å Ô ¼ for all Ô ¾ Ê with ¼ Ô Æ;

3. Some other results 31 (iv) For each compact set Ã Ê Ò ¼ Ë µ there is a positive constant à such that Ô Å «Ô Å Ôµ Ô Å Ô Å Ôµ à «µ for all Ô Å µ ¾ à and «; (v) Ù ¾ ÍË Åµ is a subsolution of and Ú ¾ ÄË µ is a supersolution of the equation ¼ in Å, ÙÔ Ü¾Å Ù Üµ and Ò Ü¾Å Ú Üµ ; Then ٠ܵ Ú Üµ ÙÔ Ý¾Å Ù Ýµ Ú Ýµ Ü ¾ Å Proof. The fact that the equation does not involve Ù implies that if Ù is a subsolution (or a supersolution) then Ù is a subsolution (supersolution) as well for any constant. If the claim of the theorem does not hold we may therefore assume that (4.2) ÙÔ Ù Üµ Ú Üµµ ¼ and ÙÔ Ù Ýµ Ú Ýµµ ¼ ܾŠݾŠSince Å is compact it follows from the second inequality that there is a number Æ ¼ such that the set Ü Ýµ Ü Ý ¾ Š٠ܵ Ú Ýµ ¼ Ü Ý ¾Æ is a compact subset of Å Å. We define Let Ò Ú Üµ and ÙÔ Ù Üµ ܾŠܾŠص ÙÔ Ù Üµ Ú Ýµµ ÜݾÅÜ ÝØ Ø ¼ We claim that we have ص ¼µ for Ø ¼. Suppose that this is not the case. Since Å is bounded, (4.2) holds and Ù Ú is upper semi-continuous there is a point Ü ¼ ¾ Å is such that Ù Ü ¼ µ Ú Ü ¼ µ ÙÔ Ü¾Å Ù Üµ Ú Üµµ ¼µ. since Ø ¼ µ ¼ for some Ø ¼ ¼ it follows that Ü ¼ is a local maximum for Ù and a local minumum for Ú so that by Theorem 4.4 we know that Ù and Ú are constants in a neighbourhood of Ü ¼. But this implies that the set where ٠ܵ Ú Üµ achieves its maximum is open and since it is, by the semi-continuity assumptions, closed, we conclude that this set is the whole of Å which is a contradiction in view of (4.2) and we conclude in particular that Ƶ ¼µ ¼. Since is upper semi-continuous and nondecreasing we have ÐÑ Ø¼ ص ¼µ so there exists a number ¼ so that µ ¼µ Ƶ ¼µµ

32 4. Comparison 11.11.2005 à where The compact set Ã Ê Ò ¼ Ë µ is defined by Ò Æµ ¼µ µ Ô Å µ Ô É Å É Ó Æ Æ ¾ ɾ We choose É Æµ ¼µ ¾Æ ³ ص Ø e ¾Ò¾ ÒØ where Ò is a positive integer chosen so that Ò Ò Ò ¾ e Ò¾ Òe Ò¾ ¾ µ Æ ÑÒ Æµ ¼µ Ò Ä Ã ÙÔ Ô Å Ã ÔÅ µ¾ã where Ä Ã is the Lipschitz-constant of in Ã. Furthermore, let and (4.3) ³ Ù Ùµ Ú Úµ Ö Õ Æ µ ³ ¼ ÖµÕ ³ ¼ ÖµÆ ³ ¼¼ ÖµÕ Å Õ Then we see that Ù is a subsolution of ¼ and Ú a supersolution of ¼ and also that Ò Ú Üµ and ÙÔ Ù Üµ ܾŠܾŠIf we define ص ÙÔ ÜݾÅÜ ÝØ Ù Üµ Ú Ýµµ for Ø ¼ then we conclude that and Ƶ ¼µ Ƶ Òe Ò¾ ¼µ Ƶ ¼µ µ ¼µ µ ¼µ Òe Ò¾ because ¼ ص µòe Ò¾ Òe Ò¾ Ò ¾ µòe ¼µµ Ƶ Òe ¾Ò ¾ Ò Òe Ò¾ when ص and Ò. Ƶ ¼µµ Ƶ ¼µµ

3. Some other results 33 Next we construct a function as follows: ¼ Ø ¼ Ƶ ¼µµ ¾Ø Ø µ Æ Øµ ¼ Ø Æµ ¼µ Ø Ø Æ ¾ ¾Æ Ƶ ¼µ µ Ø Æµ¾ Ø Ø Æ ¾ ¾Æ Æ ¾ We see that we have a function ¾ ¾ Ê Êµ such that ص µ ¼ when Ø ¾Æ, Ƶ Ƶ ¼µ Ƶ ¼µµ, and ص ص ¼µ Ƶ ¼µµ when ¼ Ø. Thus the maximum of the the function is achieved at a point Ø ¾ ¾Æµ and Ƶ ¼µ ¾Æ ¼ ¼¼ Ø µ ¼ Ø µ ¾ µ Æ ¾ Ƶ ¼µ ¾Æ ¾ µ Æ Now we proceed to a standard argument for viscosity solutions and define Ü Ýµ ٠ܵ Ú Ýµ Ü Ýµ Ü Ý ¾ Å It follows from our choice of Æ, the definition of, the fact that Ø ¾Æ, and from the semi-continuity assumptions that the maximum of is achieved at some point Ü Ý µ ¾ Å Å with Ü Ý Ø. Define Õ ¼ Ü Ý µ Ü Ý ¼¼ Ü Ý µ Æ Ü Ý ¾ Ü Ý µ ¼ Ü Ý µ Á Ü Ý ¼ Ü Ý µ Ü Ý Ü Ý µ Å Ü Ý µ By Theorem 2.9 we have Ü Ù Ü µ Õ µ ¾  ¾ ÅÙ Â ¾ ÅÙ such that ¼ Á ¼ Á and Ý Ú Ý µ Õ µ ¾ Æ Æ Æ Æ Æ Æ Æ Æ where is e.g. Æ. From this we conclude that and that Æ. Using the facts that Ù and Ú are, respectively, sub and supersolutions we have Ù Ü µ Õ µ ¼ and Ú Ý µ Õ µ ¼ Since is nonincreasing in its third argument this implies in particular that Ù Ü µ Õ µ Ú Ý µ Õ µ

34 4. Comparison 11.11.2005 Now it remains to show that this is a contradiction, using the fact that Ù Ü µ Ù Ý µ We have in other words to show that the function Ö Ö Õ µ is strictly increasing on the interval µ. Taking into account definition (4.3), the facts that ¾ ³¼ ص and ³ ¼¼ ص when Ø ¾ µ, Õ É, Æ and Æ ¾ É we have ¾ µ Æ ¾ Ö Õ µ Õ µ ³ ¼ ÖµÕ ³ ¼ Öµ ³ ¼¼ ÖµÕ Å Õ ³ ¼ µõ ³ ¼ µ ³ ¼¼ ÖµÕ Å Õ ³ ¼ µõ ³ ¼ µ ³ ¼¼ ÖµÕ Å Õ ³ ¼ µõ ³ ¼ µ ³ ¼¼ µõ Å Õ Ä Ã ³ ¼ Öµ ³ ¼ µ Õ ³ ¼¼ µ ³ ¼¼ Öµµ à Òe ¾Ò¾ e Ò e ÒÖ µ Ä Ã Õ µ Ã Ò Ö Thus it follows from our choice of Ò that the function Ö Ö Õ µ is strictly increasing and the proof is thus complete. 4. Some auxiliary results Lemma 4.6. Assume that Ê, Û Ê is upper and ¼ µ lower semicontinuous. Suppose furthermore that and that Æ Þ ¾ Þµ ¼ ÙÔ Þ¾ Û Þµ Þµ Let Å «ÙÔ Þ¾ Û Þµ «Þµµ for «and assume that Þ «¾ is such that Then (4.4) ÐÑ Å «Û Þ«µ «Þ«µµµ ¼ «ÐÑ ««Þ «µ ¼ Moreover, if Þ is a cluster point of Þ«as «, then Þ ¾ Æ and Û Þµ Û Þ µ for all Þ ¾ Æ. We also need the following variant. Lemma 4.7. Assume that Ê, Û Ê is upper and µ Ñ ¼ µ nondecreasing in its first Ñ arguments and lower semicontinuous with respect to the last one. Suppose furthermore that Æ «Þ ¾ «Þµ ¼

4. Some auxiliary results 35 and that Let Å «ÙÔ Þ¾ Û Þµ that ÐÑ «Þµ «Þ ÙÔ Þ¾ ¾ Ò Æ Û Þµ Þµ «Þµµ for «and assume that Þ «¾ is such ÐÑ «Å «Û Þ «µ «Þ «µµµ ¼ If Þ is a cluster point of Þ «as «, then Þ ¾ Æ and Û Þµ Û Þ µ for all Þ ¾ Æ and if ÐÑ Þ «Þ then ÐÑ «Þ «µ ¼. Moreover, if is compact, then ÐÑ ««Þ «µ ¼ Proof of Lemma 4.6. Since is nonnegative, Å «is a nonincreasing function on µ. On the other hand ÙÔ Þ¾Æ Û Þµ Å «and Æ is nonempty, so Å «is def bounded from below and Å ÐÑ «Å «exists (and is finite). Let «µ Å «Û Þ«µ «Þ«µµ. If now «, then Å «µ Þ«µ Û Þ«µ Þ«µµ «µ Þ«µ Now we take «so that we have ¾ «Å «¾ Û Þ«µ «Þ«µ Å ««µ ¾ Þ «µ Å ««µ µ «Þ«µ ¾ Å «¾ Å ««µ Since we know that ÐÑ «Å «exists and by assumption ÐÑ ««µ ¼ the right hand side of this inequality tends to zero and since is nonnegative we get the claim (4.4). Finally, assume that Þ«Þ ¾ where «. By (4.4) and the lower semicontiuity of we have ¼ ÐÑ Þ«µ Þ µ and since is nonnegative we have Þ ¾ Æ. Finally, since Û is upper semicontinuous and ÙÔ Þ¾Æ Û Þµ Å «we get Û Þ µ ÐÑ Û Þ«µ «Þ«µ ÐÑ Å ««µ Å ÙÔ Û Þµ Þ¾Æ Proof of Lemma 4.7. If «then we have «Þµ Þµ for all Þ ¾ and hence Å «Å as well. On the other hand, Æ is nonempty and ÙÔ Þ¾Æ Û Þµ Å «, so Å «is bounded from below and Å def ÐÑ «Å «exists (and is finite).

36 4. Comparison 11.11.2005 Assume next that Þ «Þ ¾ where «. If Þ ¾ Æ then there exists a vector «such that «Þ µ Û Þ µ Å ¾. Since Þ «Þµ is lower semicontinuous and Û is upper semicontinuous we see that for sufficiently large we have «Þ «µ Û Þ «µ Å. Since ««for sufficiently large values of and is nondecreasing in the first variables we see that «Þ «µ Û Þ «µ Å for all sufficiently large. But this contradicts the assumption that ¼ ÐÑ «Å «Û Þ «µ «Þ «µµ Å ÐÑ «Û Þ «µ «Þ «µµ. Since is nonnegative and Û is upper semicontinuous we have Û Þ «µ «Þ «µ Û Þ µ ÐÑ ÙÔ ÐÑ ÙÔ Û Þ «µ «Þ «µ Å «ÐÑ Å «Å ÙÔ Û Þµ Þ¾Æ Since Þ ¾ Æ and Û Þ µ ÐÑ ÙÔ Û Þ «µ this inequality implies in addition that ÐÑ «Þ «µ ¼. Finally, if we assume that is compact then every subsequence «µ has a subsequence for which ÐÑ Þ «Þ ¾ and the claim follows from the results already proven. 5. Comments The proofs of Theorems 4.4 and 4.5 are modifications of the proofs of these results in [1]