BY VARIATIONAL PROPERTIES
JEAN SAINT RAYMOND
Abstract. LetX be a reflexive Banach space and f :X !R[ {+1}be weakly l.s.c. We prove that if for every'in a convex dense subsetY of the dual spaceX⇤the set of points where the functionf 'attains its minimum is convex thenf is convex.
It is a common observation that if f is a non-convex continuous real function on the real line R which satisfies lim|x|!1f(x)
|x| = +1 then there is an affine functionhf such that the functionf hvanishes at several points, and even on a non-convex set.
So it is a natural question to know whether this fact could characterize convex functions on an infinite-dimensional Banach spaceX.
It is shown in [5] that iff :X !R[ {+1}
is a weakly l.s.c. function and iff 'attains its minimum on a convex non-empty subset of X for all ' in the dual space X⇤ then f is convex. It is worth noting that the hypothesis thatf 'attains its minimum for every'in X⇤ guarantees that the sublevels of f are all weakly compact inX.
Another result in the same direction, with a weaker assumption on the set of those'inX⇤ for which f 'attains its minimum, can be found in [4] : ifX is a reflexive Banach space and f : X ! R a coercive continuous and weakly l.s.c. function such thatf is bounded on every bounded subset ofX and thatf ' has a unique global minimum for all ' in a convex dense subsetY ofX⇤, thenf is convex.
In this paper we mix these two kinds of hypotheses : as in [5] we assume that f is weakly l.s.c. and thatf ' attains its minimum for every' 2X⇤ in order to ensure the weak compactness of all sublevels of f (one can notice that if X is reflexive this condition is automatically satisfied for every coercive f) ; and as in [4] we shall assume something on the shape of the setM' where f 'attains its minimum (except that it is non-empty) only for 'in a convex dense subset Y of X⇤.
2010Mathematics Subject Classification. 52A40 , 46A25, 46B10 .
Key words and phrases. convex functions, reflexive Banach spaces, weak compactness.
1
Recall that ifX is a Banach space a functionf :X !R[ {+1}is said to be coercive if limkxk!1f(x)
kxk = +1.
Definition 1. Let X be a Banach space. A function f : X ! R[ {+1} will be said to be quasi-coercive if for every continuous linear functional ' on X the functionf 'is bounded from below.
It is clear that a coercive function is quasi-coercive and well-known that the converse is false in every infinite-dimensional Banach space. A proof of the following lemma can be found in [5].
Lemma 2. Let X be a Banach space and f : X ! R[ {+1} be bounded from below. Then the following three are equivalent
- f is quasi-coercive.
- 8'2X⇤ lim infkxk!1f(x) '(x) kxk >0.
- 8'2X⇤ limkxk!1f(x) '(x) = +1.
The following result generalizes Theorem 4 in [4].
Theorem 3. Let X be a reflexive space and f : X ! R be quasi-coercive and weakly l.s.c. If Y is a convex dense subset ofX⇤ and iff 'attains its minimum at a unique point for every'2Y, thenf is convex.
For the proof of this theorem we need several lemmas.
Lemma 4. If Y is a convex dense subset of the Banach spaceE," >0,R 0and (a1, a2, . . . , an)a finite family of points inE, then there exist a point! inE and a finite-dimensional linear subspaceV ofE such thatk!k< ",V\B(0, R)⇢Y !, d(aj, V)< "forj = 1,2, . . . , nand d(x, V0)< " for allx2V \B(0, R), where V0 denotes the linear span of(a1, a2, . . . , an).
Denote by m the dimension of V0, and choosem+ 1 points (b1, b2, . . . , bm+1) affinely independent in V0 such that 0 be an interior point of the convex hull of (b1, b2, . . . , bm+1). We can for example take for (b1, b2, . . . , bm) a basis of V0 and bm+1 = Pm
1 bj. Increasing R if necessary we can assume that supjkajk R.
and leta0= 0.
Then there exists⌘ >0 such thatV0\B(0, ⌘)⇢conv(b1, b2, . . . , bm+1), and we defineb0j= R+"
⌘ bj. It is clear thatV0\B(0, R+")⇢conv(b01, b02, . . . , b0m+1). Since Y is dense, we can findb00j 2Y such that b0j b00j < "/2.
If 0 i n, the point ai belongs to conv(b01, b02, . . . , b0m+1), and there are
j 0 with sum 1 such that ai = P
j jb0j. Then a00i = P
j jb00j belongs to conv(b001, b002, . . . , b00m+1) ⇢ Y, and a00j aj P
j j b00j b0j "/2. Define
! = a000, then V as the linear span of the (b00j !)jm+1. We have k!k < "/2 andd(aj, !+V)"/2, henced(aj, V)< ".
The faces of the simplex S0 of vertices (b01, b02, . . . , b0m+1) are in the boundary of this simplex, hence disjoint from the open ball B(0, R+"). In particular, if a2conv(b001, b002, . . . , b00m), we havea=P
jmµjb00j withµj 0 andP
jµj= 1, thus d(a,P
jmµjb0j)< "/2 and since P
jmµjb0j2/B(0, R+"), we haveka !k> R.
More generally, ifa ! belongs to some face of the simplexS00of vertices b00j !, we have kak > R; it follows that V \B(0, R) is a convex subset of V, hence is connected, contains 0 and is disjoint from the boundary of the simplex S00, thus that
V \B(0, R)⇢S00= conv(b001, b002, . . . , b00m+1) !⇢Y ! .
Finally, ifx2V \B(0, R), we havex+!2conv(b001, b002, . . . , b00m+1), and there are non-negative real numbers µj with sum 1 such that x+! = P
jµjb00j ; we then have x+! P
jµjb0j P
jµj b00j b0j "/2, whence d(x+!, V0)"/2 and d(x, V0)< "since allb0j’s belong to the spaceV0. ⇤ Recall that a multivalued mappingGdefined on a topological spaceTwith (non- empty) values in a locally convex linear space E is said to be usco if it is upper semi-continuous and takes compact values. It is said to be cusco if moreover its values are convex. It is an easy and well-known result that Gis usco if its graph {(x, y)2T ⇥E :y2G(x)}is compact in T⇥E.
The following avatar of Brouwers’s theorem is well-known : its proof can be found in [1] (Th´eor`eme 1, p. 523).
Lemma 5. Let E be a finite-dimensional normed space and G a cusco mapping defined on the ballB(0, R) of E to the dual spaceE⇤. Ifh⇠, xi>0 for each xin the sphere S(0, R) and all ⇠ 2 G(x), then there is a point xˆ 2B(0, R) such that 02G(ˆx).
Lemma 6. Let X be a Banach space, f : X ! R[ {+1} be a quasi-coercive function and K a norm-compact subset of X⇤. Assume f(0) <+1. Then there exist > 0 and M 2 R+ such that kck M holds for every ' 2 X⇤ satisfying d(', K)< and every csuch that f(c)'(c) +f(0).
It follows from lemma 2 that, for all⇠2K, there are"⇠ >0 andr⇠ 0 such that, for allx2X,f(x) h⇠, xi f(0)+"⇠.kxk r⇠holds. Then the open ballsB(⇠, "⇠/3) coverKand there exists a finite subsetJ ofK such thatK⇢S
⇠2JB(⇠, "⇠/3).
Then put = min⇠2J "⇠
3 and M = max⇠2J 3r⇠
"⇠ . If d(', K) < there exists
⇠02K such that k' ⇠0k< and⇠2J such that k⇠0 ⇠k< "⇠
3. Thus we have k' ⇠k +"⇠
3 <2"⇠
3 . If moreoverf(c)'(c) +f(0), 2"⇠
3 kck k' ⇠k.kck h', ci h⇠, ci f(c) f(0) h⇠, ci "⇠kck r⇠
what shows that "⇠
3 kck r⇠, hence thatkck 3r⇠
"⇠ M. ⇤
Corollary 7. Let X be a Banach space, f : X ! R[ {+1}be a proper quasi- coercive function andK a norm-compact subset ofX⇤. Then there exist >0and M 2 R+ such that kck M holds for every '2 X⇤ satisfying d(', K)< and every cwhere f 'attains its global minimum onX.
Since f is proper there exists some a 2 X such that f(a) < +1. Applying Lemma 6 to the functionf1:x7!f(x+a) we getM1and . Ifd(', K)< and if f 'attains its minimum atcthe functionf1 ':x7!f(x+a) '(x+a) +'(a) attaints its minimum atc a. So we havef1(c a) '(c a)f1(0) '(0) =f1(0).
Thus it follows thatkc ak M1 hence thatkck M=M1+kak. ⇤
Lemma 8. Let X be a Banach space and f :X !R[ {+1}be a quasi-coercive weakly l.s.c. function. Assume that the proper domain of f is dense in X and contains 0. Then, for every finite-dimensional linear subspace V of X⇤ and every
>0, there existR >0and >0such thatf⇤(') .Rfor all '2X⇤ satisfying both d(', V)< andR k'k R+ .
The convex l.s.c. functionf⇤ :'7!supx2Xh', xi f(x) is finite at every point ofX⇤, hence norm-continuous on X⇤ and a fortiori onV. Let 0 = + 1. The set C={'2V :f⇤(')f⇤(0) + 1 + 0k'k}is a closed neighborhood of 0 inV. IfC is unbounded inV, there exist for all integerka'k2V such thatk'kk= 1 and a
k k such that k.'k 2C. Passing to a subsequence one can assume that ('k) converges to some '2 V. Then k'k = 1. Moreover, the function ⌧k defined on R+ by⌧k(t) =f⇤(t'k) f⇤(0) 1 0t is convex, negative at 0 and at k, hence at ` k. In particular if `is any integer, we have `.'k 2C for allk `. Thus
`.'2C for all`2N. Then for all` kwe havef⇤(`'k)1 +f⇤(0) + 0`, hence f⇤(`')1 +f⇤(0) + 0`. For everyx2Df and every`2N:
`'(x) f(x)f⇤(`')1 +f⇤(0) + 0.`
whence'(x) 0+1 +f⇤(0) +f(x)
` , and finally'(x) 0. And this contradicts the hypothesis that the non-empty open set{x:h', xi> 0}meetsDf.
It follows thatC is compact, and there exists R1 such thatC ⇢V \B(0, R1).
Then, ifk'k> R1, we have' /2C and 1 +f⇤(0) + 0R1< f⇤(R1
k'k') =f⇤(R1
k'k'+ (1 R1
k'k).0)
R1
k'kf⇤(') + (1 R1
k'k)f⇤(0) R1
k'kf⇤(') +f⇤(0) hencef⇤(') 1 + 0R1
R1 .k'k 0k'kwhenever'2V.
Applying lemma 6 to the compact setK=V \B(0, R1+ 1), we findM and 0 such that ifd(', K)< 0 and ifc2X is a point such thatf(c)f(0) +'(c) , we havekck M. Put = min(1
4, 0, 1
M+ 0) andR=R1+ 2 .
Then if 2X⇤ withR k k R+ andd( , V)< , there exists'2V such thatk' k< ; we havek'k> R1andk'k R+ 2 R1+ 4 R1+ 1, hencef⇤(') 0R1 and'2K.
Moreover there exists a sequence (cn) inX such that c0= 0 andf(cn) '(cn) decreases to infxf(x) '(x). Thus f⇤(') = supn '(cn) f(cn) and we have f(cn) '(cn)f(c0) '(c0) =f(0) hencekcnk M. Then
f⇤( ) lim inf
n (cn) f(cn) lim inf
n '(cn) f(cn) sup
n kcnk.k' k f⇤(') M " 0R1 M " 0R (M+ 0)" ( 0 1)R= R ,
and that completes the proof of the lemma. ⇤
Proof of Theorem 3.
In fact we will deduce theorem 3 from the more general following statement.
Theorem 9. LetX be a Banach space andf :X !R[ {+1}be a quasi-coercive weakly l.s.c. function. Assume that the proper domain off is dense inX and that
f ' attains its minimum for all '2 X⇤. If Y is a convex dense subset of X⇤ and if the set M' = {x 2X : f ' attains its minimum atx} is convex for all '2Y, thenf is convex.
Since f ' attains its minimum for all ' 2 X⇤ it follows from [5] that the sublevelsSf(↵) ={x2X:f(x)↵}off are weakly compact inX.
Assume that the function f is weakly l.s.c. and quasi-coercive but not convex.
There exist pointsa, b1, b2 inX and 2]0,1[ such that a= b1+ (1 )b2 and f(b1) + (1 )f(b2)< f(a).
By hypothesis the functionf attains its minimumµat some pointb.
We replace f by f1 : x 7! f(x+b) µ, a, b1 and b2 by a b, b1 b and b2 b, andY byY1=Y , which is convex and dense in X⇤. And the proper domain of f1, Df1 = Df b, is dense in X : the function f now satisfies the hypotheses of the theorem with infxf(x) = f(0) = 0, b1+ (1 )b2 = a and
=f(a) f(b1) (1 )f(b2)>0.
Since f is weakly l.s.c., there exists a weak neighborhood W of a such that f(x)> f(a)
3 onW. Thus one can find⇠1, ⇠2, . . . ⇠n inX⇤ and" >0 such that sup
inh⇠i, x ai< "=)x2W
Denote then by V0 the linear subspace of X⇤ spanned by (⇠1, ⇠2, . . . , ⇠n). Using lemma 8 we chooseR 1 and 0"such thatf⇤(')> R.(kak+") for all'such thatR 0 k'k R+ 0 andd(', V0)< 0.
Since f is quasi-coercive, the sublevel S = Sf(f(a)) is bounded in X and we denote by D the diameter of S. Then if k⇠i0 ⇠ik < "
2D, if f(x) f(a) and if h⇠i, x ai ", we get kx ak D, hence h⇠i0, x ai "
2. It follows that if supinh⇠0i, x ai< "
2, we have
- eitherkx ak D andx2W hencef(x)> f(a) 3, - orkx ak> D andx /2S hencef(x)> f(a),
thusf(x)> f(a)
3 in both cases.
FixK0=V0\B(0, R+"), < 0"andM 1 such thatkc0k M whenever f 'attains its minimum atc0 andd(', K0)< .
Applying lemma 4 one finds ! 2 X⇤ and a finite-dimensional subspace V of X⇤ such that k!k< "0
M1, k!k<
3D, V \B(0, R) ⇢Y ! and d(', K0) < if '2V \B(0, R), and also points⇠0i2V such thatk⇠i ⇠i0k< "
2D. The points⇠i0 spanV. Denote by⇡the restriction toX of the canonical projection of X⇤⇤ onto V⇤ : for x 2 X and ' 2 V, h⇡(x), 'i = '(x). Then, since ⇠i0 2 V, we see that if x2 X and if ⇡(x) = ⇡(a), we have h⇠i0, x ai = 0 for all i, and in particular f(x)> f(a)
3. Moreover if'2V andx2X, we haveh⇡(x), 'i='(x). Observe that if'2V andk'k Rthen!+'2Y.
If'2V,k'k=Randf (!+') attains its minimum atc, we then have (!+')(c) =f(c) +f⇤(!+')> f(c) +R.(kak+") R(kak+") =k'k.kak+"
sincek!k< "0, hence'(c) '(a)>k'k.kak+" '(a) M.k!k 0.
Consider the multivalued functionG:V \B(0, R)◆X defined by x2G(') =M!+' () f⇤(!+') = (!+')(x) f(x)
Ifx2G(') we havekxk Mhencef(x) (!+')(x)f(0) (!+')(0) = 0 and f(x)(!+')(x) k!+'k.kxk ↵:= (k'k+k!k).M
So the values ofG are contained in the weakly compact subsetSf(↵) of X. And the graph ofGis compact in (V \B(0, R))⇥X, ifY ⇢X⇤ is equipped with the norm topology andX with the weak topology ; indeed ⇢(V \B(0, R))⇥Sf(↵) and
(', c)2 () c2G(') () 8x2X '(c) f(c) '(x) f(x) thus G is usco from Y equipped with the norm topology into X with the weak topology. And since G takes convex values by hypothesis this shows that it is a cusco mapping.
If'2V \B(0, R) andc2G(') we necessarily have f(c) (!+')(c)min⇣
f(b1) (!+')(b1), f(b2) (!+')(b2)⌘
f(b1) + (1 )f(b2) (!+')( b1+ (1 )b2)
(f(a) ) (!+')(a) =f(a) (!+')(a) but if⇡(x) =⇡(a) we have'(x) ='(a) hence
f(x) (!+')(x) =f(x) '(a) !(x)> f(a)
3 (!+')(a) k!k.kx ak
> f(a) (!+')(a)
3 D.k!k f(a) (!+')(a) 2 3 and that shows that⇡(c)6=⇡(a).
The multivalued function G1 : ' 7! ⇡ G(') ⇡(a) takes convex compact values in V⇤ and is upper semi-continuous from V \B(0, R) to V⇤, since ⇡ is weakly continuous from X to V⇤ and it follows from what precedes that G1(') cannot contain 0 for any'2Y \V.
By the choice ofR, we havehz, 'i='(c) '(a)>0, for all'2 V such that k'k=Rand allz=⇡(c) ⇡(a)2G1('). We deduce then from lemma 5 that the cusco mappingG1 should contain 0 at some point ofV \B(0, R), in contradiction with what precedes. And this shows thatf must be convex. ⇤ Observe that most of the arguments of the above proof do not require any hy- pothesis of weak compactness. The only point where such an hypothesis is used (by the weak compactness of the sublevels off) is the proof thatGis cusco (and that so isG1). If we attempt to replace eachG1(') by its closure G01(') =G1(') inV⇤, we get a multivalued mapping with convex compact values but we cannot ensure that this new multivalued mapping is upper semi-continuous. And if we try to replace the graph 1={(', z) :z2G1(')}by its closure inB(0, R)⇥V⇤, we get the graph of a new u.s.c. multivalued mappingG001 with compact values avoiding 0 but could loose the convexity of its valuesG001(').
Theorem 10. LetX be a Banach space andf :X !R[{+1}be a proper weakly l.s.c. function. Assume thatf 'attains its minimum for all '2X⇤ and that the setM'={x2X :f 'attains its minimum atx}is convex for all'in a convex dense subset Y ofX⇤. Thenf is convex.
If the proper domain off is dense inX, the statement follows from theorem 9.
In the other case, it follows from [5] that the sublevels Sf(↵) off are all weakly compact inX. IfKn denotes the sublevelSf(2n) and ifmn = sup{kxk:x2Kn}, the setK =P 2 n
1 +mn
Kn is weakly compact in X. It follows then from [2] that there exists a subspaceZ ofX containingK and a normk|.|konZ such thatZ be a reflexive Banach space fork|.|kand thatK be bounded inZ. The unit ball ofZ is compact for the weak topology (Z, Z⇤) and a fortiori for (Z, X⇤).
SinceKn⇢2n(1 +mn)K⇢Z, we haveDf =S
nKn ⇢Z. And replacingX by the closure ¯Z ofZ inX, we can also assumeZ to be dense inX.
For any positive integerq, define fq(x) = inf
y2Xf(y) +qk|x y|k2 Ifa2X satisfiesf(a)<+1, we havea2Df ⇢Z and
fq(x)f(a) +qk|x a|k2<+1
wheneverx2Z; this shows that the functionfq is everywhere finite onZ, hence Dfq Z and Dfq is dense in X. In fact, if f(y) <+1 and kx yk<+1, we havey2Zandx y2Z, thusx2Z, whenceDfq =Z. One sees also thatfqf. By hypothesisf is bounded from below, and we defineµ= infx2Xf(x).
The epigraphsEf off andEq of the functionnq:y7!qk|y|k2 are both weakly closed and the function : (y, t),(z, s) 7!(y+z, s+t) is weakly perfect from Ef⇥Eq intoX⇥R: indeed ifH is weakly compact inX⇥R, there existsh2R such thatH ⇢X⇥] 1, h]. Then if (y, t)2Ef, (z, s)2Eqand (y+z, s+t)2H, we must have µ+s s+t h, thus s h µ and k|z|k r = h µ
q
1/2. It follows that (z, s) belongs to the weakly compact setL={z:k|z|k r} ⇥[0, h µ]
and (y, t) belongs to the weakly compact set H L. Finally sinceEf andEq are weakly closed, H1 = Ef \(H L) and H2 = Eq\L are weakly compact and (Ef⇥Eq)\ 1(K) is closed in the compact setH1⇥H2.
Since the mapping |Ef⇥Eq is perfect, Ef+Eq is weakly closed in X⇥R and it is the epigraph offq. If'2X⇤ the minimum of fq 'is the minimum of the function (x, u)7!u '(x) onEf+Eq: thus it is attained at some point (x, u) if and only if (x, u) = (y, t) + (z, s), witht '(y) = inf{u '(w) : (w, u)2Ef}and s '(z) = inf{u '(w) : (w, u)2Eq}.
SinceZ is reflexive,
inf{u '(w) :w2Eq}= inf{q.k|w|k2 '(w) :w2W}
is attained at some point ofZ for any'2Z⇤, and a fortiori for any'2X⇤ ; and since the fonction nq is convex the set N' where the function nq ' attains its minimum is convex.
By hypothesis inf{u '(w) : (w, u)2Ef}= infw2Xf(w) '(w) and the set M'where this minimum is attained is convex for all'. It follows that for all'2Y the set where infw2Xfq(w) '(w) is attained is the convex setN'+M' .
One deduces from theorem 9 that fq is convex on X for all integer q. If it is shown thatf = supqfq, then the proof of the convexity off will be complete. Let x2X and < f(x). Sincef is weakly l.s.c., one can find some weak neighborhood
W ofxsuch thatf(y)> whenevery2W, then >0 such thatW Bk|.|k(x, ).
Choose thenqsuch thatq 2> µ; we have
f(y) +q.k|x y|k2
( +qk|x y|k2> ify2W µ+q 2> ify /2W
thusfq(x) . This completes the proof thatf = supqfq is convex. ⇤ It is shown in [5] that f is strictly convex as soon as it satisfies the conditions of theorem 3 with Y = X⇤. Nevertheless we shall prove that on every reflexive infinite-dimensional Banach spaceX, there exists a weakly l.s.c. functionf, convex but not strictly convex, and a dense linear subspaceY ofX⇤such thatf 'attains its minimum at a unique point for every'2Y.
Lemma 11. LetX be a reflexive Banach space and↵2X⇤. There exists a coercive convex non strictly convex weakly l.s.c. functionf :X !Rsuch thatf 'attains its global minimum at a unique point for all'2X⇤ distinct from↵.
Without loss of generality one can assume thanks to [3] thatX is strictly convex.
LetK be a strictly convex bounded closed subset of X containing more than one point, e.g. the unit ball. Define onX the coercive functionf by :
f(x) =d(x, K)2+h↵, xi It is clear that the function
f0:x7!d(x, K) = sup'2X⇤,k'k=1'(x) µK(')
where µK(') = supy2K'(y), is convex non-negative and weakly l.s.c. Thus so is f02, andf is convex and weakly l.s.c. The functionf ↵is constant onK, which is not a singleton ; thus f is not strictly convex. It remains to show that, for all '2Y distinct from↵, the minimum off 'is attained at a unique point, it is the minimum off0 'is attained at a unique point whenever'6= 0.
First remark that the functionf02 ':x7!d(x, K)2 '(x) attains its minimum at xif and only if the function (z, t)7!t '(z) attains at (x, g(x)) its minimum on the epigraph E ofg =f02. It is easily checked that E is the sum of the closed convex setsK0 =K⇥[0,+1[ andE0 ={(z, t) :t kzk2}because the mapping
“sum” is weakly perfect fromK0⇥E0 toX⇥R.
The set where the linear functional : (z, t)7! t '(z) attains its minimum on K0+E0 is the sum of the sets where this same linear functional attains its minimum onK0 and onE0 respectively. If attains its minimum onK0at (t, z), we necessarily havet= 0 and'(z) = supy2K'(y). SinceK is strictly convex and 'is non-zero, this minimum is attained at a unique pointzofK.
In the same way, if attains its minimum onE0 at the point (t, z), we must have t = kzk2 and '(z) = sup{'(y) : kyk kzk} ; thus the linear functional ' attains its maximum on the unit ball ofX at z
pt. We also have (s2t, sz)2E0 fors >0, hence
(st)2 s'(z) =kszk2 '(sz) t2 '(z) and (s 1)⇣
(s+ 1)t2 '(z)⌘
0, it is '(z) = k'k.kzk = 2t2 = 2kzk2, so k'k= 2kzk. And sinceX is strictly convex the pointuof the unit ball ofX where
'attains its maximumk'kis unique, and so is the pair (z, t) : indeedkzk=1 2k'k, z=kzk.uandt=1
4k'k2.
This completes the proof of uniqueness for the global minimum of f02 ' as '6= 0, hence the uniqueness for the global minimum off 'as '6=↵. ⇤ Theorem 12. LetXbe a reflexive infinite-dimensional Banach space. There exists a dense linear subspaceY of X⇤and a coercive weakly l.s.c. convex and non stricly convex functionf :X !Rsuch thatf 'attains its global minimum at a unique point for every'2Y.
Following the previous lemma, it is enough to find a dense linear subspaceY of X⇤distinct fromX⇤ and to choose↵2X⇤\Y.
If X is separable, so is X⇤ ; it is enough to take a dense sequence (yn⇤) in X⇤ and to defineY as the linear subspace spanned by the (yn⇤)’s.
In the general case, if (an) is a linearly independent sequence inX, the closed lin- ear subspaceV spanned by the (an)’s is reflexive separable and infinite-dimensional, and so isV⇤. The canonical projection⇡:X⇤!V⇤is onto and continuous, hence open. Then, if Z is a dense linear subspace of V⇤ distinct from V⇤, the linear subspaceY =⇡ 1(Z) is distinct fromX⇤ and dense : indeed ifU is a non-empty open set ofX⇤,⇡(U) is a non-empty open set ofV⇤and meetsZ. Thus there exists
y2U such that⇡(y)2Z, it isy2Y \U. ⇤
References
1. J.-P. Aubin,Mathematical methods of game and economic theory, North-Holland, 1979.
2. W.J. Davis, T. Figiel, W.B. Johnson, A. Pe lczy´nski,Factoring weakly compact operators J.
Functional Analysis17(1974), 311–327
3. J. Lindenstrauss,On nonseparable reflexive Banach spaces.Bull. Amer. Math. Soc.72(1966) 967–970
4. I.G. Tsar’kov,Approximate compactness and nonuniqueness in variational problems and their applications to di↵erential equations. 1468-4802Mat. Sb.202(2011), no. 6, 133–158; trans- lation in Sb. Math.202(2011), no. 5-6, 909–934.
5. J. Saint Raymond,Weak compactness and variational characterization of the convexity.to appear in Mediterranean Journal of Math. (2012), 13 p.
Analyse Fonctionnelle
Institut de Math´ematique de Jussieu Universit´e Pierre et Marie Curie (Paris 6) Boˆıte 186 - 4 place Jussieu
F- 75252 Paris Cedex 05 FRANCE
E-mail address:raymond@math.jussieu.fr
![[PDF] Cours Programmation JavaScript avec exemples pdf | Cours informatique](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACH5BAEAAAAALAAAAAABAAEAAAICRAEAOw==)