0 < (D) < +1 and f : D ! lR is an arbitrary summable function. Then the function F() := R. fx2D:f(x) g(f(x) ? )d ( 2 lR) is continuous, non-negative, non-increasing, ...
ESSENTIAL SUPREMUM AND SUPREMUM OF SUMMABLE FUNCTIONS H. X. Phu Institute of Mathematics P.O.Box 631 Bo Ho 10000{Hanoi, Vietnam
A. Ho mann Institut fur Mathematik, TU Ilmenau PF 327 D{98694 Ilmenau, Germany
Key words: Essential supremum, supremum, global maximum, summable function, convex function. AMS(MOS) subject classi cations: 26 A 51, 46 A 35, 90 C 30, 90 C 48.
ABSTRACT Let D lRN , 0 < (D) < +1 and Rf : D ! lR is an arbitrary summable function. Then the function F ( ) := fx2D f x g(f (x) ? ) d ( 2 lR) is : ( )
continuous, non-negative, non-increasing, convex, and has almost everywhere the derivative F 0 ( ) = ?[f ]. Further on, it holds ess sup f = supf 2 lR : F ( ) > 0g, where ess sup f denotes the essential supremum of f . These properties can be used for computing esssup f . As example, two algorithms are stated. If the function f is dense, or lower semicontinuous, or if ?f is robust, then sup f = ess sup f . In this case, the algorithms mentioned can be applied for determining the supremum of f , i.e., also the global maximum of f if it exists. 1
1. INTRODUCTION
During investigating the optimal control of a hydroelectric power plant ([10]) we met an interesting phenomenon: the water volume in (almost) all natural lakes is convex with respect to the height of the water surface. More precisely, if the height of water surface in a lake is denoted by h and the corresponding water volume is V (h), then the function V is convex. The explanation is simple: the higher the height of water the greater the water surface, and therefore, the faster the increasing of the volume. I.e., V 0 (h) is non-decreasing. Hence, V is convex. It was for us a pleasant surprise that this nice property holds true although the structure of the lake can be very wild. We thought, if this ideal property can appear in the complicated nature, why not in our theoretical world? What is similar to this? The answer is as follows. Let f : D lRN ! lR be an arbitrary summable function, and 0 < (D) < +1, then the function F de ned by
F ( ) :=
Z
fx2D:f (x) g
(f (x) ? ) d;
is convex (Theorem 2.1). It is really similar, because ?f can be regarded as a function describing the lake bottom, as the height of water surface, and F ( ) as the corresponding water volume. This is the basic result of the present paper. Why do we tell this story here? In the present paper, we shall use the mentioned property for determining the essential supremum of summable functions. It is of theoretical interest, because the essential supremum is needed for the norm k k1 in the Lebesgue space L1. On the other hand, it can take a part in optimization. For instance, if the maximum of f on D does not exist, since D is not closed or f is not upper semicontinuous, so that neither the Weierstrass theorem nor the concept of -quasisolution ([5], [13]) can be applied, the supremum can be determined instead of maximum. If f is not completely de ned for each point, as in Lebesgue spaces Lp, then only essential supremum come into consideration. Even if the existence of maximum is guaranteed and the information on f is complete, the maximum is sometimes nonsense, from practical point of view, because it re ects only an individual case and is not representative for a general situation. This is again a case for essential supremum. In Section 2, some related properties will be proved which are used for nding essential suprema. It is well-known that global optimization plays an important role in practice. There are many methods for nding global optima, but many deterministic approaches were developed for special classes of functions with favorable structure, for instance, maximizing of convex, or Lipschitz-continuous functions, or of functions representable as di erence of two convex functions... (compare [8]). For more complicated functions, one often used probabilistic methods ([11]). Although our rst interest consists in the essential supremum, Section 3 will point out some cases where the supremum is equal to the essential supremum, i.e., when the result of Section 2 can be used for nding the global maximum of some worse functions having no favorable structure, if it turns out to exist. 2
Similar considerations can be found in [2, 9] which describes the \mean value{level set method" for nding global minima of robust functions. One can say that the Newton method in Algorithm 2 (Section 2) is another foundation of the integral method in [2, 9] because they have the same iteration. Their distinction consists in the approach, the choice of initial point, and in the stopping rule.
2. ESSENTIAL SUPREMUM Consider a measurable function f : D ! lR de ned on a measurable set D of lRN with respect to the Lebesgue measure , where 0 < (D) < +1. De nition 2.1: B is said to be an essential upper bound for f i f (x) B for almost all x 2 D, i.e., fx 2 D : f (x) > Bg = 0: The least essential upper bound is called essential supremum of f . We write ess sup f := inf fB : B is an essential upper bound for f g:
(2:1)
Analogously, the essential in mum (ess inf f ) is the largest essential lower bound of f . This de nition is similar to the one in the excellent book of E. Zeidler [12] which contains a compact survey on measurable sets and functions, on Lebesgue integral and Lebesgue spaces. Throughout this paper, assume ess sup f < +1, for example, f 2 L1(D). For the sake of shortness, we use the following abbreviations [f t] := fx 2 D : f (x) tg; [f > t] := fx 2 D : f (x) > tg; [f = t] := fx 2 D : f (x) = tg; [t f < t0 ] := fx 2 D : t f (x) < t0 g; [t < f < t0 ] := fx 2 D : t < f (x) < t0 g:
Lemma 2.1: The function m( ) := [f ] is non-negative and non-increasing. Let := supf 2 lR : m( ) > 0g; (2:2) then esssup f = . Proof: We only need to prove the second assertion, because the rst one is obvious. Clearly, B if [f > B] = 0 and [f ] > 0. Therefore, due to (2.1){(2.2), ess sup f . If ess sup f > , then ess sup f > := 21 (ess sup f + ) > . Due to de nition, this implies both [f > ] > 0 and [f ] = 0, which con ict to each other. Hence, the assertion of the lemma must be true. Basing on the above lemma, we can determine ess sup f by the following bisection algorithm. 3
Algorithm 1: Choose a stopping parameter > 0, an initial step size s > 0, and an initial values satisfying [f ] > 0. 0
0
0
(i) Begin: k = 0 ( k = 0) and s = s0 . (ii) Test: if [f k + s] > 0 then k+1 = k + s and go to (iv).
(iii) Test: if s < then go to (v), else reduce the step size, for instance, by s = 2s and go to (ii). (iv) k = k + 1 and go to (ii). (v) k is an approximate value of ess sup f with 0 esssup f ? k < . STOP. This procedure can nish for the function f having a nite essential supremum. The described algorithm takes advantage of the monotonicity property of m to determine the search direction. But it has two disadvantages: { The step size can only be chosen arbitrarily, independently on the current position k and the structure of the function f ; { If k reaches the true value , this algorithm cannot recognize it and goes on, and at the end, an interval containing the true value is delivered instead of the exact result. Here, a sucient condition is missing what could be taken as stopping rule. If [f k + s] = 0, then it is very possible that k + s > , therefore, [f k + s] = 0 cannot ensure that k + s = . On the other hand, [f k ] > 0 does not mean that it should step forward, since f is able to have a discontinuity at k , and [f ] = 0 for all > k , i.e., k is already equal to . Example 2.1: For illustration, consider now the function f : [0; 1] ! lR
n f (x) = 12 ifif xx 2= [01.; 1[ Obviously, ess sup f = supf : m( ) > 0g = = 1 and n 1 m( ) = 10 ifotherwise. After starting at some point 0 lying far away from the true value 1 ( 0 0g; where F ( ) :=
Z
f ]
[
(f (x) ? ) d
(2:3)
instead of (2.2) (f is supposed to be summable and the integral above is Lebesgue integral). The motivation for this is given by the following. Theorem 2.1: The function F : lR ! lR is Lipschitz-continuous, non-negative, nonincreasing, convex, and has almost everywhere the derivative F 0 ( ) = ?[f ]. Further on, it holds ess sup f = = ~. Proof: F is obviously non-negative. Let > 0, then
F ( ) ? F ( + ) = =
Z
f ]
Z
[
(f (x) ? ) d ?
f< +]
[
Z
f +]
[
(f (x) ? ? ) d
(f (x) ? ) d + [f + ]
(2:4)
0;
hence, F is non-increasing. Because
Z
f< +]
[
(f (x) ? ) d [ f < + ];
it yields
F ( ) ? F ( + ) ([ f < + ] + [f + ]) = [f ]:
(2:5)
Consequently, 0 F ( ) ? F ( + ) (D) < +1, i.e., F is Lipschitz-continuous. Similar to (2.4), we have
F ( ? ) ? F ( ) =
Z
?f< ]
[
(f (x) ? + ) d + [f ]
(?)(?[f ]) for all > 0. On the other hand, (2.5) yields
F ( + ) ? F ( ) (?[f ]) for all > 0. Hence, F is convex and ?[f ] 2 @F ( ) for all 2 lR. Since a F is almost everywhere di erentiable, it follows that F 0 ( ) = ?[f ] a.e. Due to Lemma 2.1, it Rremains to prove = ~. In fact, ~ because > implies [f ] = 0, and by this, [f ] (f (x) ? ) d = 0, i.e., ~. Assume the contrary that 5
R > ~, then [f ^] > 0 and [f ^] (f (x) ? ^) d = 0, where ^ = ( + ~)=2. This yields [f > ^] = 0, and further on, [f (^ + )=2] = 0. Consequently, due to de nition, (^ + )=2 , which con icts with ~ < ^ < . Hence, = ~ must hold true. Actually, F is strictly decreasing when it is positive. Due to de nition, = ~ yields immediately [f ] > 0 if F ( ) > 0; F ( ) = 0 if [f ] = 0: Basing on Theorem 2.1, we can state a Newton algorithm for nding esssup f . Assume that one initial value 0 satisfying [f 0] > 0 has been found. Now, begin the forward search by setting k = 0, k = 0. The next point to be checked after k is given by Proposition 2.1: Suppose F ( k ) > 0. Then k < k+1 := k + [Ff ( k ) ] ess sup f: k The sequence ( k ) generated above converges to ess sup f = . If F ( k+1) = 0 then k+1 = ess sup f . Proof: k+1 is just the point where the tangential of the function F at k meets axis of abscissas. Because F ( k ) > 0, and F is convex and meets the axis of abscissas at = ~ = ess sup f the rst time, k < k+1 ess sup f holds true. The sequence ( k ) is monotonous and converges to some ^ < 1. Then ^ . From the continuity of F we get F (^ ) = 0. Hence ^ , i.e., ^ = . If F ( k+1) = 0, then k+1 = ess sup f , that implies k+1 = esssup f . Hence, if [f k ] > 0, then the next approximate value k+1 de ned in Proposition 2.1 never passes over the true value ess sup f , and it holds either k+1 = esssup f or F ( k+1) > 0. Here, the step size keeps changing, corresponding to the current position. Ideally, the algorithm stops after nite steps at the true value = ~ = esssup f if and only if there exists ^ < such that the function F is quasilinear (ane) on [^ ; ]. This happens if [f ] = [f = ess sup f ] for all 2]^ ; [, i.e., f (x) ^ for almost all x 2 D n [f = ess sup f ]. But such a case is really seldom. In general, in nitely many steps must be repeated to arrive at the true value of the essential supremum. Of course, it must stop earlier. Because it is insucient to stop somewhere without knowing how far the current position is still away from the exact solution, we need an additional test: If F ( k + ) = 0 ( is the stopping parameter), then stop there. Obviously, due to Proposition 2.1, F ( k + ) = 0 holds true only if k+1 ? k = [Ff ( k ) ] : k Therefore, we only need to do this test if the latter inequality is ful lled. 6
Algorithm 2: (Newton method) Suppose ess sup f < +1, the stopping parameter be chosen, and one initial value satisfying [f ] < +1 and F ( ) > 0 be found. 0
0
0
(i) Begin: k = 0 ( k = 0). (ii) sk = [Ff ( k ) ] . k (iii) Test: if sk > then k+1 = k + sk and k = k + 1 and go to (v). (iv) Test: if F ( k + ) = 0 then go to (vi), else k+1 = k + and k = k + 1 and go to (ii). (v) Test: if F ( k ) > 0 go to (ii). (vi) ess sup f = k for coming from (v), or k is an approximate value with 0 ess sup f ? k if coming from (iv). STOP. Actually, if k+1 = k + sk , then we have the same iteration as in the mean value { level set method in [2, 9]. But by considering the convex function F , we can easily come to other stopping conditions and further interesting aspects. One of them is that, under the assumption 0 < ess inf f , this algorithm always leads after one iteration to the same point of the domain of function values, independently from the fact how far the initial 0 lies under the essential in mum of the function f . Precisely, we have Proposition 2.2: Let 1 be determined as in Proposition 2.1. Then inf f ) 2 [ess inf f; ess sup f ] 1 = ess inf f + F (ess (D) holds true for all 0 < ess inf f . Proof: Let 0 < ess inf f . Due to de nition, for all 2 [ 0; ess inf f ], we have [f ] = [f 0] = (D), and with this
F ( ) ? F ( 0 ) = ?( ? 0) (D) (recall 0 < (D) < +1). Therefore, inf f ) : 1 = 0 + F(( D0)) = essinf f + F (ess (D) Further on, 1 ess inf f follows from F (ess inf f ) 0 and Proposition 2.1 implies 1 ess sup f . This proposition is useful for nding an initial 0 for Algorithm 2. Choose now an arbitrary 0 2 lR. If F ( 0 ) > 0, then 0 is suitable for Algorithm 2. If it does not hold, then choose another value which is much smaller than the previous one. Do not worry that it could be possibly too far away, because the algorithm shall lead at once, after one 7
iteration, to one unique point of the domain of function values. This is quite di erent to the situation in Algorithm 1 where its running strongly depends on the initial value 0. We now demonstrate the principal behavior of Algorithm 2 by means of some academic examples. Example 2.2: Consider the function 8 sin x if x 2 [?4; 0[ > < if x = 0 f (x) := > 10 (2:6) x if x 2 ]0; 3[ : 5 if x 2 [3; 4]: In this example, we restrict ourselves by the case D = [?4; 3]. Let us begin in 0 = ?20 by choosing = 0:01. Algorithm 2 delivers the following result. k k F ( k ) [f k ] sk F ( k + ) 0 {.2000E+02 .1445E+03 .7000E+01 .2064E+02 1 .6429E+00 .3039E+01 .3468E+01 .8763E+00 2 .1519E+01 .1096E+01 .1481E+01 .7404E+00 3 .2260E+01 .2741E+00 .7404E+00 .3702E+00 4 .2630E+01 .6853E{01 .3702E+00 .1851E+00 5 .2815E+01 .1713E{01 .1851E+00 .9255E{01 6 .2907E+01 .4283E{02 .9255E{01 .4628E{01 7 .2954E+01 .1071E{02 .4628E{01 .2314E{01 8 .2977E+01 .2677E{03 .2314E{01 .1157E{01 9 .2988E+01 .6692E{04 .1157E{01 .5785E{02 .1231E{05 10 .2998E+01 .1231E{05 .1569E-02 .7845E{03 .0000E+00 STOP Because the algorithm comes from (iv) to STOP, 10 = 2:998 is an approximate value for ess sup f , and 2:998 ess sup f 0 + = 3:008. In fact, the exact solution is ess sup f = 3. Example 2.3: We now deal with the same function f de ned in (2.6), but with a larger domain D = [?4; 4]. By choosing the same initial 0 = ?20 and = 0:01, Algorithm 2 delivers the following. k k F ( k ) [f k ] sk F ( k + ) 0 {.2000E+02 .1695E+03 .8000E+01 .2119E+02 1 .1187E+01 .5455E+01 .2813E+01 .1940E+01 2 .3127E+01 .1873E+01 .1000E+01 .1873E+01 3 .5000E+01 .0000E+00 .1000E+01 .0000E+00 STOP 8
Since the algorithm comes from (v) to STOP, ess sup = 3 = 5:0, and it is really the right result. In general, we often get an approximation for esssup f even coming from (v), because k only can be approximately computed. We see here an interesting phenomenon that Algorithm 2 comes in Example 2.3 faster to STOP as in Example 2.2, although the function considered is almost the same. One possibly things that the reverse must happen, because the initial 0 = ?20 is in Example 2.3 farer away from the destination ess sup f = 3 than from the one in Example 2.2 (ess sup f = 5). There are two reasons for this paradox. First, while enlarging the domain from [?4; 3] to [?4:4], the integral denoted by F , which corresponds the volume, increases much more than the measure [f ], that corresponds the area of the ground. Therefore, the increasing rate sk in Example 2.3 is greater than the one in Example 2.2. Second, in Example 2.3, because the function f has an upward jump at the level f = 3 and remains constant in the set [f = 5] = [f = ess sup f ] = [3; 4] at the level f = ess sup f = 5, the function F is ane in the interval [3; 5]. So the algorithm runs at once to the destination 3 = esssup f = 5, once coming into the domain [3; 5] (at 2 = 3:127). Hence, there are no re ned iterations around the level F = 0, as it happens in Example 2.2. Such a case has been discussed after Proposition 2.1. In the previous examples, we intend to choose an initial value lying far away from the true one, in order to show that the realization of Algorithm 2 is really independent from the choice of 0, as Proposition 2.2 says. In fact, 1 already belongs the the interval [ess inf f; ess sup f ]. R Generally, it is not easy to determine [f ] (f (x) ? ) d and [f ] or their quotient. But this trouble cannot be saved once seeking or ensuring the essential supremum. Fortunately, one can use positive experiences made for this purpose as stated in [2] and [9]. In principle, Monte Carlo methods ([6]) can be used for approximating the expressions just montioned. Another approach without calculating [f ] is as follows. Here we need two initial values 0 and 1. The rst one can be got as in the remark after Proposition 2.2, and the second one can be found as in Proposition 2.1. After having at least two approximate values, the further ones are determined corresponding to Proposition 2.3: Suppose k?1 < k , F ( k?1) > 0, and F ( k ) > 0. Then k < k+1 := F ( Fk )( k?) 1??FF( ( k?)1 ) k ess sup f: k k?1 The sequence ( k ) generated above converges to ess sup f = . If F ( k+1) = 0 then k+1 = ess sup f . Proof: Draw a line going through two points ( k?1 ; F ( k?1)) and ( k ; F ( k )). Since F is convex and meets the axis of abscissas at ess sup f , this line cuts the axis of abscissas at 9
one point being not greater than ess sup f . Denote it by k+1, then k < k+1 ess sup f and F ( k ) = k+1 ? k ; F ( ) ? a k?1
which implies
k?1
k+1
k+1 = F ( Fk )( k?) 1??FF( ( k?)1 ) k : k?1
k
The proof of convergence is similar to the one in Proposition 2.1. Because F ( k+1) = 0 means k+1 ess sup f , it yields k+1 = ess sup f . In such a way, one point more can be calculated, which is either the exact solution we are looking for, or which is a better approximation that does not exceeds the true solution. To determine k+1 (k 1), we only have to estimate F ( k ), because F ( k?1 ) is known since the latter step. Analogously as in Algorithm 2, if (and only if)
k+1 ? k = F ( Fk )( k?) 1??FF( ( k?)1 ) k ? k < ; k?1
k
then we check whether F ( k + ) = 0. In this case we stop there and take k as an approximate value, otherwise reset k+1 = k + and proceed to nd k+2, and so on.
Algorithm 3: (Secant method) Suppose ess sup f < +1, the stopping parameter be chosen, and two initial values 0 and 1 satisfying 0 < 1, F ( 0 ) > 0, and F ( 1 ) > 0 be found. (i) Begin: k = 1 ( k?1 = 0 and k = 1). (ii) k = F ( Fk )( k?) 1??FF( ( k?)1 ) k and sk = k+1 ? k . k k?1 (iii) Test: if sk k = k + 1 and go to (v). (iv) Test: if F ( k + ) = 0 then go to (vi), else reset k+1 = k + , k = k + 1, and go to (ii). (v) Test: if F ( k ) > 0 go to (ii). (vi) ess sup f = k for coming from (v), or k is an approximate value with 0 ess sup f ? k for coming from (iv). STOP.
Example 2.4: Consider again the function f de ned in (2.6) and restricted by D = [?4; 3]. Applying Algorithm 3, by choosing = ?20, = 0:6429, and = 0:01, we 0
10
1
get
k k F ( k ) sk F ( k + ) 0 {.2000E+02 .1445E+03 1 .6429E+00 .3039E+01 .4435E+00 2 .1086E+01 .1831E+01 .6722E+00 3 .1759E+01 .7706E+00 .4885E+00 4 .2247E+01 .2835E+00 .2843E+00 5 .2531E+01 .1098E+00 .1798E+00 6 .2711E+01 .4173E{01 .1102E+00 7 .2821E+01 .1597E{01 .6831E{01 8 .2890E+01 .6096E{02 .4216E{01 9 .2932E+01 .2329E{02 .2607E{01 10 .2958E+01 .8895E{03 .1611E{01 11 .2974E+01 .3398E{03 .9957E{02 .1291E{03 12 .2984E+01 .1291E{03 .6128E{02 .1841E{04 13 .2994E+01 .1841E{04 .1664E{02 .0000E+00 STOP Because the algorithm comes from (iv) to STOP, 13 = 2:994 is an approximate value for ess sup f with 2:994 ess sup f 13 + = 3:0094. We see that the Algorithm 3 needs more steps to come to an end.
3. SUPREMUM
The idea of nding essential (global) optima instead of global optima is rather old ([11]). Already nearly thirty years ago, Chichinadze (Ref. 3) proposed to evaluate the function ( ) = fx 2 D : f (x) g=(D) which has ess sup f as its rst root. This method approximating the level sets has been followed by several papers, for instance, [1] and [4]. The main diculty of this approach consists in evaluating the function . But our aim is not to discuss its realization and its eciency. We only want to investigate when the essential supremum of the function f is equal to its supremum, i.e., when ess sup f = sup f holds. If this happens and if f attains its global maximum, then the result in Section 2 can be applied for seeking global maxima. De nition 3.1: ([2]) A set S lRN is called robust if cl S = cl int S . A function f : D ! lR is said to be robust if [f < ] = fx : f (x) < g is robust for any real . Example 3.1: ([2]) Let
0 f1(x) = 10 ifif xx < 0
n 0 if x 0 and f2 (x) = 1 if x > 0:
Then f1 and f2 are robust, but f1 + f2 is not robust. 11
Proposition 3.1: If ?f is robust then esssup f = sup f .
Proof: Assume the contrary that esssup f < sup f . Then [f > ] 6= ; and [f > ; 6= [?f < ? ], which con icts with the robustness of ?f . De nition 3.2: A function f : D ! lR is said to be dense if
] = 0 for = 21 (ess sup f + sup f ). Therefore cl int[?f < ? ] = [y ? < f < y + ] > 0 for all > 0 and y 2 f (D):
Example 3.2: The functions
n 1 if x 2 D is rational f1 (x) := 0 if x 2 D is irrational
and f2 (x) := 1 ? f1 (x) are not dense, because
[1 ? < f1 < 1 + ] = 0 and [? < f2 < ] = 0 for all 2]0; 1[: Let
f3 (x) := f1 (x) sin x; f4 (x) := f2 (x) sin x: Then f3 is not dense for every domain D, because [y ? < f3 < y + ] = 0 for all y 2 f3 (D) n f0g and 0 < < jyj. But if D is an interval containing a real number of the form k, k 2 ZZ, then the function f4 is dense, since [y ? < f4 < y + ] > 0 for all y 2 f4 (D) and > 0. On the other hand, if D = [a; b] ]k; (k + 1)[ for a certain k 2 ZZ, then [? < f4 < ] = 0 for all 0 < < minfj sin aj; j sin bjg, although 0 2 f4 (D). Hence, f4 is not dense in the latter case. We just see that even a totally discontinuous function, such as f4 , can be dense. On the other hand, continuity of a function is not enough for its density. For example, let D be some subset of lRN , N > 1, with (D) = 0 and let f be an arbitrary continuous function on it, then 0 [y ? < f < y + ] (D) = 0, for all y 2 f (D). It is not sucient if we only assume (D) > 0. A counter example is D = D1 [ D2, where cl D1 \ cl D2 = ;, (D1 ) = 0, (D2 ) > 0, and f (x) = 1 for x 2 D1, f (x) = 2 for x 2 D2 . Obviously, this function is continuous on D, but is not dense. What we additionally need is as follows. Proposition 3.2: Let D be a non-empty robust set. Then each continuous function f : D ! lR is dense. This is only a special case of the following. Proposition 3.3: Suppose D = Sj2J Dj and each Dj is a non-empty robust set. Further on, suppose that the function f : D ! lR is continuous on each Dj . Then it is dense. 12
Proof: Let y 2 f (D), then y = f (x ) for some x 2 Dj . Since f is continuous on Dj , for all > 0 there exists an open neighborhood U of x such that f (U \ Dj ) ]y ? ; y + [, i.e., x 2 U \ Dj [y ? < f < y + ]. Therefore, ; 6= Dj cl int Dj implies that the open subset U \ int Dj is not empty, and hence, [y ? < f < y + ] > 0. We now are in a position to formulate a sucient condition for the identity of the supremum and the essential supremum. Proposition 3.4: The supremum of a dense function is equal to its essential supremum. Proof: Assume the contrary that f : D ! lR is dense but ess sup f < sup f . Then there exists x 2 D with ess sup f < f (x ) sup f and [f (x ) ? ; f (x ) + ] > 0, where 0 < = 21 (f (x ) ? esssup f )) < f (x ) ? ess sup f , in contradiction to the de nition of essential supremum. A further sucient condition is Proposition 3.5: Suppose that D is a non-empty robust set and that f : D ! lR is lower semicontinuous, i.e.,
f (x) lim inf f (t); for all x 2 D: t!x
Then the supremum of f is equal to its essential supremum. Proof: Assume the contrary that esssup f < sup f . Then there exists x 2 D with ess sup f < f (x ) sup f . Because f is lower semicontinuous, the set L = fx 2 D : f (x) > 21 (ess sup f + f (x ))g is relatively open. Since L contains x , it is not empty. Hence, ; 6= D cl int D implies (L) > 0. Due to de nition, ess sup f 21 (ess sup f + f (x )); which con icts with ess sup f < f (x ). Recall that the Weierstrass theorem needs the upper semicontinuity for ensuring the existence of maximum, but it can not guarantee the equality of maximum and essential supremum. For functions on the real line, Proposition 3.5 can be generalized as follows. Proposition 3.6: Suppose that D is a non-empty open interval of lR, and that the function f : D ! lR satis es
f (x) maxftlim inf f (t); tlim inf f (t)g for all x 2 D: !x;tx 13
Then the supremum of f is equal to its essential supremum. Proof: Assume the contrary that ess sup f < sup f . Then there exists x 2 D such that ess sup f < f (x ). Due to de nition, [f 12 (ess sup f + f (x ))] = 0. Therefore, there exist two sequences (t?n ) and (t+n ) converging to x from below, or from above (respectively) such that limn!+1 f (t?n ) and limn!+1 f (t+n ) are less than 21 (ess sup f +f (x )). This yields 1 (ess sup f + f (x )) < f (x ); maxftlim inf f ( t ) ; lim inf f ( t ) g !x;tx 2 which con icts with the assumption of the proposition. Typical examples for Proposition 3.6 are functions without external saltus, where a function f : D ! lR is said to have an external saltus at x if f (x 0) exist and f (x) does not lie between f (x ? 0) and f (x + 0) ([7], p. 234). I.e., f is a function without external saltus if f (x) 2 [minff (x ? 0); f (x + 0)g; maxff (x ? 0); f (x + 0)g] for all x 2 D. Note that a semicontinuous function can have external saltus. Proposition 3.7: Let D be convex and int D 6= ;. Suppose that there exists one ( xed) 2 ]0; 1[ such that
f ( x1 + (1 ? )x2 ) f (x1 ) + (1 ? ) f (x2 ) holds for all x1 ; x2 2 D. Then the supremum of f is equal to its essential supremum. Proof: Assume the contrary that ess sup f < sup f . Then there exists x 2 D such that ess sup f < f (x ). By the assumption,
f ( x + (1 ? )x) f (x ) + (1 ? ) f (x) holds for all x 2 D. Let ) ? ess sup f ) and M = [esssup f ? f ess sup f ]: = (f (x2(1 ? ) Then, for x 2 M ,
f ( x + (1 ? )x) f (x ) + (1 ? ) (ess sup f ? ) ) ? ess sup f ) ( f ( x = f (x ) + (1 ? ) ess sup f ? 2(1 ? ) = ess sup f + 2 (f (x ) ? ess sup f ); i.e.,
x + (1 ? )M M := [f ess sup f + 2 (f (x ) ? esssup f )]: 14
Therefore, (M ) > 0 implies (M ) > 0. This con icts with the de nition of the essential supremum, because ess sup f + 2 (f (x ) ? ess sup f ) > ess sup f . Hence, the essential supremum cannot be less than its supremum.
4. CONCLUDING REMARKS
In the previous sections we only discussed the essential supremum and the supremum or global maximum values, but not the maximizer or the corresponding values of the variable. Actually, during determining F ( ) or [f ], some elements of the level set [f ] must be found at the same time. So the algorithms stated in Section 2 can deliver some maximizer or "-maximizer, too. The e ectivity of the algorithms described here mainly depends on the fact how the integral F and the measure or their quotient are determined. As mentioned above, one can use the computational experiences of the integral method in [2, 9] for this purpose. It is hopeful that that the convexity structure of F can deliver us some additional valuable advantages. But the computational realization and the modi cation of our algorithms are not the subject of the present paper. They will be considered in another one. Here, we only like to show the convexity and some other useful properties of the auxiliary function F and to mention some applications of them.
ACKNOWLEDGMENTS
Supported by the Deutsche Forschungsgemeinschaft. The rst author would like to express his sincere gratitude to Prof. Dr. E. Zeidler and Prof. Dr. H. G. Bock for inviting him to the Universities of Leipzig and Heidelberg, where parts of this work were done.
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