A minimax theorem for linear operators
ABSTRACT. The aim of this note is to prove the following minimax theorem which generalizes a result by B. Ricceri : let E be a infinite-dimensional Banach space not containing `1, F be a Banach space,X be a convex subset of E whose interior is non- empty for the weak topology on bounded sets,SandT be linear and continuous operators fromEtoF,':F !Rbe a continuous convex coercive map,J ⇢Ra compact interval and :J !R a convex continuous function. Assume moreover thatS⇥T has a closed range in F ⇥F and that S is not compact. Then
sup
x2X
inf2J '(T x Sx) + ( ) = inf
2J sup
x2X
'(T x Sx) + ( )
In particular, if 'is the norm ofF and = 0, we get sup
x2X
inf2JkT x Sxk= inf
2J sup
x2XkT x Sxk
For any two normed spaces E andF we will denote byL(E, F) the space of continuous linear mappings from E to F, itself normed by kTk= supkxk61kT xk.
It was shown by E. Asplund and V. Pt´ak in [AP] that for two normed spacesE and F of dimension>2 and any twoA and B in L(E, F) the following inequality holds
kxk61sup inf
2RkAx+ Bxk6 inf
2RkA+ Bk= inf
2R sup
kxk61kAx+ Bxk
and that the equality in the above relation is attained for every pair A, B in L(E, F) if and only if both E and F are inner product spaces. In the sequel the unit ball of E is replaced by a convex set whose interior is non-empty for the weak topology (E, E⇤) or more generally the ‘weak topology on bounded sets’ and we are interested in proving such a minimax equality.
For a Banach spaceE, we will denote by (E, E⇤) the finest locally convex topology on Ewhich agrees with the weak topology (E, E⇤) on the bounded subsets ofE. This topology is finer than (E, E⇤) and coarser than the norm topology : so it is compatible with the duality betweeen E and E⇤. Converging sequences for (E, E⇤) are all weakly converging sequences, and a subsetU of E is open for (E, E⇤) if and only if every weakly converging sequence inE whose weak limit is in U has all but finitely many terms inside U. It can be noticed that a closed convex subset C ofE is a neighborhood of 0 for (E, E⇤) i↵ its polar set C0 is norm-compact inE⇤.
The following theorem has been proved by B. Ricceri in [R] (Theorem 3).
Theorem 1. Let E be a infinite-dimensional reflexive Banach space,T :E !E a non-zero linear compact operator, ': E !R a convex continuous and coercive functional, J ⇢ R a
compact interval with 0 2 J, : J ! R be a continuous convex function. Then, for each r >'(0), one has
sup
x2X
inf2J('(T x x) + ( )) =r+ (0) whereX ={x2E :'(T x)6r}.
Our main theorem extends this latter statement : indeed supx2X'(T x x) = +1for all 2J \ {0} and so inf 2Jsupx2X'(T x x) + ( ) is clearly equal to r+ (0).
In order to prove it, we need some preliminary results.
Lemma 2. Let E be a separable Banach space not containing `1, N be a closed linear subspace of E, q : E ! E/N be the quotient mapping and (yn) be a sequence in the quotient spaceE/N which converges weakly to 0. Then there is a subsequence (ynk) and a sequence(zk) weakly converging to 0 in E such that ynk =qzk.
The weakly converging sequence (yn) is bounded. So there is M 2 R+ such that supnkynk < M, and for all n we can find xn 2 q 1(yn) such that kxnk < M. Let ✓ an accumulation point of (xn
M) in the compact setBE⇤⇤ equipped with the topology (E⇤⇤, E⇤).
By Rosenthal’s and Bourgain-Fremlin-Talagrand’s theorems [BFT],✓is a Baire-1 function on the compact spaceBE⇤ equipped with (E⇤, E), and there is a subsequence (xnk) extracted from (xn) which converges to M.✓ for (E⇤⇤, E⇤). Then ynk = q⇤⇤(xnk) ! q⇤⇤(M.✓) for (E⇤⇤, E⇤), hence q⇤⇤(M.✓) = 0 and since q⇤⇤ is the projection of E⇤⇤ onto E⇤⇤/N⇤⇤ it follows that✓ 2N⇤⇤. SinceN ⇢E cannot contain`1,✓ is a Baire-1 function on the compact space BN⇤ equipped with (N⇤, N) and there is a sequence (wk) in BN such that wk ! ✓ for (N⇤⇤, N⇤) which agrees on BN⇤⇤ with (E⇤⇤, E⇤). It follows that h⇠, wki ! h✓,⇠i for all ⇠ 2 E⇤ and that zk = xnk M.wk converges to 0 for (E⇤⇤, E⇤) what means that q(zk) =q(xnk) M.q(wk) =q(xnk) =ynk and zk !0 weakly. ⌅ Theorem 3. Let E be a separable Banach space not containing `1, V be a Banach space and T 2 L(E, V). If for every sequence (vn) in E weakly converging to 0, the sequence (T vn) is relatively compact in V then the operator T is compact.
It should be first noticed that a Banach space which contains an isomorphic copy of `1 cannot be reflexive. So the present theorem applies in particular to reflexive spaces. One can also remark that the hypothesis that E does not have a subspace isomorphic to `1 cannot be removed : indeed by Schur property, if E =V =`1 and T is the identity mapping of E, the operator T is not compact but every weakly converging sequence (xn) in E converges in norm, and that implies that (T xn) is relatively compact in V.
Let (vn) be a sequence in the unit ball BE of E. We want to show that there exists a subsequence (vnk) such that (T vnk) converges in V. Let ✓ an accumulation point of (vn) in the compact set BE⇤⇤ equipped with the topology (E⇤⇤, E⇤). Again by Rosenthal’s and Bourgain-Fremlin-Talagrand’s theorems, ✓ is a Baire-1 function on the compact space BE⇤ equipped with (E⇤, E), and there is a subsequence (vnk) extracted from (vn) which converges to ✓ for (E⇤⇤, E⇤). If the set{T vnk :k2N} is relatively compact in V, there is a subsequence which converges inV.
If not there is some "> 0 and a sequence extracted from (vnk), still denoted by (vnk), such thatkT vnk T vn`k>"fork 6=`. Consider then the sequencewk=vnk+1 vnk which satisfies kT wkk>" for all k. Nevertheless we have for allx⇤ 2E⇤ :
hx⇤, wki=hx⇤, vnk+1i hx⇤, vnki ! h✓, x⇤i h✓, x⇤i= 0
which shows that the sequence (wk) converges weakly to 0 in E. Then the sequence (T wk) should be relatively compact in V and converge weakly to 0, hence converge in norm to 0.
And this is in contradiction with kT wkk= T vnk+1 T vnk >" for all k.
Moreover since the compactness of T follows from the compactness of TE0 for all separable subspaceE0 ofE, the conclusion of Theorem 3 holds even if E is not assumed to
be separable. ⌅
Lemma 4. Let E be a infinite-dimensional Banach space not containing `1 and X be a convex subset of E whose interior for (E, E⇤) is non-empty. Then there exists a Banach space V, a compact linear mapping ⇡ :E !V with dense range and a convex open subset Y of V such that ⇡ 1(Y)⇢X ⇢⇡ 1(Y).
It is clear that if ⇡ :E !V is compact, it is continuous from (E, ) to (V,k.k), hence that the set⇡ 1(Y) has necessarily a non-empty (E, E⇤)-interior as soon as the interior of Y in V itself is non-empty.
Let X0 be the interior of X for (E, E⇤) and a 2 X0. Put W = (X0 a)\(a X0) which is a symmetric convex subset ofE. ThenW is open for (E, E⇤) and contains 0. Since (E, E⇤) is coarser than the norm-topology, W is absorbing and the Minkowski functional pW : x 7! inf{r > 0 : r 1x 2 W} is a semi-norm on E. Denote by V the separated completion of (E, pW) and ⇡ the canonical mapping from E to V. By definition ⇡(E) is dense in the Banach spaceV, and⇡(W) =⇡(E)\B(0,1). We will show that ⇡(X0) is open in ⇡(E) and that if Y denotes the interior of ⇡(X0) = ⇡(X), we have X0 = ⇡ 1(Y). Then since X0 6=; we have ⇡ 1(Y) =X0 ⇢X ⇢X0 =⇡ 1(Y).
Indeed let b2⇡(X0) andu 2X0 such that ⇡u=b. The set{t 2R:u+t(a u)2X0} is open and contains 0. Hence it contains also " for some">0 and we letv=u "(a u).
The homothety with center v 2 X0 and ratio ⌘ = "
1 +" < 1 transforms X0 into itself and in particular a into u, hence a+W ⇢ X0 into u+⌘ ·W. Thus ⇡(X0) ⇡(E)\B(b,⌘) and ⇡(X0) is a neighborhood of b in ⇡(E). Finally if Y is the interior of the convex set
⇡(X0) then Y \⇡(E) is a convex open subset of ⇡(E) contained in ⇡(E) \ ⇡(X0) and containing ⇡(X0) : indeed if u 2 ⇡(X0) we have shown the existence of a ball B(u, r) such that⇡(E)\B(u, r)⇢⇡(X0), hence that ⇡(E)\B(u, r)⇢B(u, r)⇢⇡(X0) and u2Y.
It remains to show the compactness of⇡. For this by theorem 3 it is enough to show that whenever (wn) is a sequence which converges weakly to 0 inE the sequence (⇡wn) converges to 0 inV, hence show thatpW(wn)!0. LetR >0 ; the sequences (a+R.wn) and (a R.wn) converge both weakly toawhich is an interior point ofX0 for (E, E⇤). So there is an integer N such thata±R.wn2X0 for alln>N. This means thatR.wn 2(X0 a)\(a X0) =W, hence thatk⇡wnk=pW(wn)6 1
R if n>N and that ⇡wn !0 in the normed spaceV. ⌅ Lemma 5. LetE be an infinite-dimensional Banach space not containing`1,F be a Banach space,J ⇢R a compact interval,S andT in L(E, F). Assume that S⇥T :E !F ⇥F is one-to-one and that(S⇥T)(E) is closed in F ⇥F. If there is no 2J such that T S is compact then there exists a sequence(wn)in E converging weakly to 0 and" >0such that kT wn .Swnk>" for all 2J and alln2N.
Notice first that since S ⇥T is one-to-one and (S ⇥T)(E) closed in the space F ⇥F equipped with the norm (x, y) 7! kxk+kyk, there exists by the open mapping Theorem some >0 such that kSxk+kT xk> .kxk for all x2E.
Assume that for all sequence (wn) weakly converging to 0 in the unit ball BE of E we had limninf 2JkT wn .Swnk = 0. Then for every sequence (wn) weakly converging to 0 in BE it would exist a sequence ( n) in J such that kT wn n.Swnk ! 0 hence an accumulation point 2J of ( n) and since
kT wn .Swnk6kT wn n.Swnk+| n |.kSwnk6kT wn n.Swnk+| n |.kSk we would have a subsequence of (wn) and a fixed 2J such that kT wn .Swnk !0.
Then two cases could occur : — either this is the same for all sequences (wn) converging to 0 for the weak topology but no in norm, inBE — or there are two sequences (wn) and (xn) in BE both converging to 0 weakly but not in norm, and 6= µ in J such thatT wn .Swn !0 and T xn µ.Sxn!0.
We show that in the first case the operatorT Sis compact. Indeed for every sequence (wn) weakly converging to 0 we have T wn Swn ! 0 and the conclusion follows from Theorem 3.
In the second one take subsequences of (wn) and (xn) satisfying infnkwnk > 0 and infnkxnk > 0 and choose for all n some x⇤n 2 F⇤ of norm 1 such that hx⇤n, Sxni = kSxnk. Since (wp) converges weakly to 0 we have limp!1hx⇤n, Swpi = limp!1hS⇤x⇤n, wpi = 0. So we can replace (wn) by some subsequence still denoted by (wn) such that |hx⇤n, Swni|2 n and for every⇣ 2R we get
kSxn ⇣.Swnk hx⇤n, Sxn ⇣.Swni kSxnk |⇣|.|hx⇤n, Swni| kSxnk |⇣|.2 n Put zn = 1
2(wn +xn) ; since (zn) is a sequence in BE weakly converging to 0 there must exist⌫ 2J such that T zn ⌫.Szn !0. So we have :
T wn .Swn !0 ; T xn µ.Sxn !0
2(T zn ⌫.Szn) =T wn+T xn ⌫.S(xn+zn)!0
hence .Swn+µ.Sxn ⌫(Swn+Sxn) = ( ⌫).Swn+ (µ ⌫).Sxn !0. If µ= ⌫ we get ( µ).Swn!0, hence kSwnk !0 and kT wnk kT wn Swnk+| |.kSwnk !0. Then kwnk 1.(kSwnk+kT wnk)!0, a contradiction. And else, with ⇣ = ⌫
µ ⌫, kSxnk kS(xn ⇣wn)k+|⇣|.2 n
1
|µ ⌫|k( ⌫).Swn+ (µ ⌫).Sxnk+|⇣|.2 n!0
andkT xnk kT xn µSxnk+|µ|.kSxnk !0. So kxnk 1.(kSxnk+kT xnk)!0, what is again a contradiction. This second case cannot occur and the proof is complete. ⌅ Lemma 6. Let E be an infinite-dimensional Banach space not containing `1, F and V Banach spaces, 0 6= 0 be a real number, S 2 L(E, F) , H 2 L(E, F) be a compact operator andT = 0.S+H. Assume thatS⇥T is a one-to-one operator with closed range, Y is a non-empty convex open subset of V, ⇡ 2 L(E, V) a compact operator with dense range andX =⇡ 1(Y). Then for 6= 0, supx2XkT x .Sxk= +1.
Notice first that S ⇥ H is also one-to-one with closed range, since the mapping (u, v)7!(u, v 0u) is an isomorphism from F ⇥F onto itself.
As above there is a > 0 such that kSxk + kHxk > .kxk for all x 2 E. Let y0 2 Y \ ⇡(E), x0 2 ⇡ 1(y0) and r > 0 such that B(y0, r) ⇢ Y. One can find in the unit sphere SE of E a vector w0 and a sequence (wn)n>1 such that for all n > 0 d(wn+1,span(w0, w1, . . . , wn)) > 2
3. As in the proof of Theorem 3 we can extract from (wn) a sequence (wnk) which converges for (E⇤⇤, E⇤) to some w⇤⇤ 2 E⇤⇤. Then for k > 1, zk = wnk wnk 1 satisfies kzkk > 2
3 and (zk) converges weakly to 0 in E. We deduce that kHzkk ! 0 and that k⇡zkk ! 0, since H and ⇡ are compact. Then put xk =x0+ r
2 k+kHzkk+k⇡zkk ·zk. It follows that
k⇡xk y0k=k⇡xk ⇡x0k= r.k⇡zkk
2 k+kHzkk+k⇡zkk < r , hence that⇡xk 2B(y0, r)⇢Y. It means that xk2X and that
kT xk Sxkk=k( 0 ).Sxk+Hxkk
>| 0|.(kSxkk+kHkk) (1 +| 0|).kHxkk
> .| 0|.kxkk (1 +| 0|).kHxkk , and at the same time
kxkk> r.kzkk
2 k+kHzkk+k⇡zkk kx0k> 2r
3 · 1
2 k+kHzkk+k⇡zkk kx0k !+1 andkHxkk6kHx0k+ r.kHzkk
2 k+kHzkk+k⇡zkk 6kHx0k+r.
Thus limkkT xk Sxkk= +1, hence supx2XkT x .Sxk= +1. ⌅ Corollary 7. LetE be an infinite-dimensional Banach space not containing`1,F a Banach space, 0 6= 0 be a real number, S 2 L(E, F) , H 2 L(E, F) be a compact operator, T = 0.S +H and ': F !R be a convex continuous and coercive function. Assume that S ⇥T is one-to-one with closed range, that Y is a non-empty convex open subset of the normed spaceV, ⇡ 2L(E, V) is a compact operator with dense range and X =⇡ 1(Y).
Then for 6= 0, supx2X'(T x .Sx) = +1.
Since ' is coercive, for eachM 2R+, there is R >0 such that '(u)< M =) kuk< R for allu 2F. Following lemma 6, there exists x 2X such that kT x .Sxk>R; then we
have for this vectorx : '(T x .Sx)>M. ⌅
Lemma 8. Let E be an infinite-dimensional Banach space not containing `1, F a Banach space, S 2 L(E, F), J ⇢ R be a compact interval, T 2 L(E, F) such that T .S be compact for none 2J. Assume thatS⇥T is a one-to-one operator with closed range, that Y is a non-empty open subset of the normed space V, ⇡ 2L(E, V) is a compact operator with dense range andX =⇡ 1(Y).
Then supx2Xinf 2JkT x .Sxk= +1.
Let y0 2 Y \ ⇡(E), x0 2 ⇡ 1(y0) and r > 0 such that the closed ball ˜B(y0, r) is contained in Y. It follows from Lemma 5 that one can find in BE a sequence (wn)
converging weakly to 0 and " > 0 such that for all n 2 N and all 2 J the inequality kT wn .Swnk > " holds. Then the sequence (⇡wn) converges to 0 in V and we put for n>1 : xn =x0+ r
2 n+k⇡wnk ·wn. We then have
k⇡xn y0k=k⇡xn ⇡x0k= r.k⇡wnk
2 n+k⇡wnk < r hence⇡xn2B(y0, r)⇢Y, it is xn2X, and for 2J :
kT xn .Sxnk> r
2 n+k⇡wnk ·k(T .S)wnk kT x0 Sx0k
> r."
2 n+k⇡wnk kT x0 Sx0k , hence inf 2JkT xn .Sxnk> r."
2 n+k⇡wnk kT x0 Sx0k and sup
x2X
inf2JkT x .Sxk>sup
n
r."
2 n+k⇡wnk kT x0 Sx0k= +1 ,
which completes the proof of the lemma. ⌅
Corollary 9. LetE be an infinite-dimensional Banach space not containing`1,F a Banach space, S 2 L(E, F), J ⇢ R be a compact interval, T 2 L(E, F) such that T .S be compact for none 2J and':F !R a convex continuous and coercive function. Assume thatS⇥T is a one-to-one operator with closed range,Y is a non-empty convex open subset of the normed spaceV, ⇡ 2L(E, V)a compact operator with dense range and X =⇡ 1(Y).
Then supx2Xinf 2J'(T x .Sx) = +1.
Since ' is coercive, for eachM 2R+, there is R >0 such that '(u)< M =) kuk< R for all u 2 F. Following lemma 8, there exists x 2 X such that kT x .xk > R for all 2J; then we have for this vector x : inf 2J'(T x .Sx)>M. ⌅ Lemma 10. Let ' be a convex continuous and coercive function on the Banach space E.
Then the dual function '⇤ has a proper domain D which is a convex neighborhood of 0 in E⇤. And if Z is any dense subset ofE⇤, we have for allx2E :'(x) = sup⇠2Zh⇠, xi '⇤(⇠).
Since '⇤ is convex and the proper domain of '⇤ is D={⇠ :'⇤(⇠)<+1}=[
n
{⇠:'⇤(⇠)6n}
it is clear thatD is convex. Since' is coercive the set {x:'(x)61 +'(0)} is bounded in E, and there exists R >0 such that '(x)<1 +'(0) =) kxk< R. By convexity we deduce that'(x)>'(0) + kxk
R ifkxk>R hence that '(x) h⇠, xi is bounded from below outside of the ball of radiusR if⇠ 2E⇤ and k⇠k< 1
R.
The convex continuous function x 7! '(x) h⇠, xi is necessarily bounded from below on the bounded convex complete set ˜B(0, R) ; it follows that D B(0, 1
R) hence that the
interior Do of D is non-empty. The convex l.s.c. function '⇤ is finite on D. Hence, if Lm denotes the closed subset of Do defined by {⇠ 2 Do :'⇤(⇠) 6m} we have Do = S
m2NLm and some Lm has non-empty interior by Baire’s Category Theorem. So '⇤ is bounded from above on a neighborhood of some point of Do , hence is continuous on Do. Since h⇠, xi 6 '(x) +'⇤(⇠) for all x 2 E and all ⇠ 2 E⇤, we have '(x) >sup⇠2Z[h⇠, xi '⇤(⇠)]
for any Z ⇢E⇤.
Since Do is open,Do\Z is dense inDo. By continuity of '⇤ onDo we necessarily have '(x)> sup
⇠2Do\Z
[h⇠, xi '⇤(⇠)] = sup
⇠2Do
[h⇠, xi '⇤(⇠)] .
If↵<'(x), the point (x,↵) does not belong to the closed convex setG={(u, t) :t>'(u)}. Then it follows from Hahn-Banach’s theorem that exists⇠ 2E⇤ such that
h⇠, xi ↵> sup
(u,s)2G
[h⇠, ui s] = sup
u [h⇠, ui '(u)] ='⇤(⇠)
what implies ⇠ 2 D and h⇠, xi '⇤(⇠) > ↵, whence '(x) = sup⇠2D[h⇠, xi '⇤(⇠)]. For
⇠ 2 D, we have t⇠ 2 Do for all t 2 [0,1[ since 0 2 Do. The function t 7! '⇤(t⇠) is convex and l.s.c. on [0,1] , hence continuous, and it follows that
h⇠, xi '⇤(⇠) = lim
t!1,t<1ht⇠, xi '⇤(t⇠)6 sup
⇠2Do
[h⇠, xi '⇤(⇠)] , hence that
'(x) = sup
⇠2D
[h⇠, xi '⇤(⇠)]6 sup
⇠2Do
[h⇠, xi '⇤(⇠)]6sup
⇠2Z
[h⇠, xi '⇤(⇠)]6'(x) . ⌅ Lemma 11. Let E be an infinite-dimensional Banach space , V be a normed space and K : E !V a compact linear operator. Then there exists a -compact set Z0 ⇢E⇤ and for all⇠ 2/ Z0 a sequence (wn) in E such that h⇠, wni= 1 for all n and kKwnk !0.
Since K is compact, the operator K⇤ : V⇤ ! E⇤ is compact too and for all m 2 N the set Tm = K⇤(mBV⇤) is a compact subset of the space E⇤. Thus Z0 = S
mTm is - compact and if⇠2/ Z0 there cannot exist any continuous linear functional ˆ⌘ on V such that
⇠=K⇤(ˆ⌘).
If it existed > 0 such that {x : h⇠, xi = 1} be disjoint from {x : kKxk 6 } the set {h⇠, xi : kKxk 6 } would be convex and symmetric hence an interval centered at 0 and non containing 1. So it would exist some r 6 1 such that |h⇠, xi| 6 r
·kKxk for all x 2 E and it would exist a linear functional ⌘ on K(E) ⇢ V whose norm would be at most r
such that ⇠ = ⌘ K : indeed if y 2 K(E) satisfies y = Kx = Kx0, we have
|h⇠, x x0i|6 r
kKx Kx0k= 0 ; hence⌘(y) =h⇠, xidepends only ony =Kxand satisfies
|⌘(y)|6 r
kKxk= r kyk.
By Hahn-Banach’s theorem it would exist ˆ⌘ 2 V⇤ extending ⌘ and we would have
⇠= ˆ⌘ K =K⇤(ˆ⌘), a contradiction with the choice ofZ0. We conclude that if ⇠2/ Z0 we can find for all n2N some wn such that kKwnk62 n and h⇠, wni= 1. ⌅
Lemma 12. Let E be an infinite-dimensional Banach space, F be a normed space, S : E ! F be a continuous non-compact linear operator, H 2 L(E, F) be a compact operator,Y be a non-empty convex open subset of the normed space V and ⇡ :E !V be a compact operator with dense range,X0 =⇡ 1(Y), J ⇢ R be a compact interval, 0 2J and :J ! R be a convex continuous function. Then there exists a dense subset Z of F⇤ such that
sup
x2X0
⇣
0h⇠, Sxi+h⇠, Hxi ⇤(h⇠, Sxi)⌘
> ( 0) + sup
x2X0
h⇠, Hxi
for all ⇠ 2Z.
Let K be the operator x7!(Hx,⇡x) fromE to the product spaceW =F⇥V normed by k(y, u)k =kyk+kuk. SinceH and ⇡ are compact, K is compact too and by lemma 11 one can find an -compact set Z0 ⇢ E⇤ such that for ⇠0 2/ Z0 there exists a sequence (wn) in E such that kKwnk=kHwnk+k⇡wnk !0 and that h⇠, wni= 1.
Let (Tm) be a sequence of compact subsets of E⇤ such that Z0 =S
mTm. If the closed set S⇤ 1(Tm) had non-empty interior in F⇤ there would be some ball B(⇠, r) in F⇤ such that S⇤(B(⇠, r)) ⇢ Tm and S⇤ would be a compact operator which in turn would imply thatS is compact. So S⇤ 1(Tm) is nowhere dense and M =S
mS⇤ 1(Tm) is meager in F⇤. Therefore Z = F⇤ \M is dense in F⇤. Moreover, if ⇠ 2 Z, then ⇠0 = S⇤⇠ 2/ Z0 and there exists a sequence (wn)2E such that Kwn!0 and h⇠, Swni=h⇠0, wni= 1.
Then if x0 2 X0, y0 = ⇡(x0) 2 Y and t 2 R, xn = x0 + (t h⇠0, x0i).wn satisfies hS⇤⇠, xni = h⇠0, xni = t, ⇡xn =y0 + (t hS⇤⇠, x0i).⇡wn !y0. Because Y is open in V we have ⇡xn 2Y for nlarge enough, hence xn 2X0. Moreover
h⇠, Hxni=h⇠, Hx0i+ (t hS⇤⇠, x0i).h⇠, Hwni ! h⇠, Hx0i , from what it follows that
sup
x2X0
⇣
0h⇠, Sxi+h⇠, Hxi ⇤(h⇠, Sxi)⌘
>lim sup
n
⇣
0h⇠, Sxni+h⇠, Hxni ⇤(h⇠, Sxni)⌘
= lim sup
n
⇣
0h⇠0, xni+h⇠, Hxni ⇤(h⇠0, xni)⌘
= 0t+h⇠, Hx0i ⇤(t) , and since this holds for all t2R and all x0 2X0 we get
sup
x2X0
⇣
0h⇠, Sxi+h⇠, Hxi ⇤(h⇠, Sxi)⌘
>sup
t 0t ⇤(t) + sup
x2X0
h⇠, Hxi
= ( 0) + sup
x2X0
h⇠, Hxi ,
what is the wanted inequality. ⌅
Lemma 13. LetE andF be infinite-dimensional Banach spaces,S :E !F be a continuous non-compact linear operator,⇡ be a compact linear mapping with dense range from E to a normed space V, Y be a convex subset of V with non-empty interior, H 2 L(E, F) be a compact operator,J ⇢Rbe a compact interval, 0 2J,':F !R be a convex continuous
and coercive function and : J ! R be a continuous convex function. Assume S ⇥T is one-to-one with closed range. Then
sup
⇡(x)2Y
inf2J' ( 0 ).Sx+Hx + ( ) > ( 0) + sup
⇡(x)2Y
'(Hx) .
Put ( , x) =' ( 0 ).Sx+Hx . It follows from lemma 10 that, for a dense subset Z of F⇤, ( , x) = sup⇠2Zh⇠,( 0 ).Sx+Hxi '⇤(⇠), hence
inf2J ( , x) + ( ) = inf
2Jsup
⇠2Z
⇣h⇠,( 0 ).Sx+Hxi '⇤(⇠) + ( )⌘
>sup
⇠2Z
inf2J
⇣h⇠,( 0 ).Sx+Hxi '⇤(⇠) + ( )⌘
>sup
⇠2Z
⇣
0h⇠, Sxi+h⇠, Hxi '⇤(⇠) + inf
2J h⇠, Sxi+ ( ) ⌘
>sup
⇠2Z
⇣
0h⇠, Sxi+h⇠, Hxi '⇤(⇠) ⇤(h⇠, Sxi)⌘ ,
and after Lemma 12, sup
⇡(x)2Y
inf2J ( , x) + ( ) > sup
⇡(x)2Y
sup
⇠2Z
⇣
0h⇠, Sxi+h⇠, Hxi '⇤(⇠) ⇤(h⇠, Sxi)⌘
>sup
⇠2Z
⇣ '⇤(⇠) + sup
⇡(x)2Y 0h⇠, Sxi+h⇠, Hxi ⇤(h⇠, Sxi) ⌘
>sup
⇠2Z
'⇤(⇠) + ( 0) + sup
⇡(x)2Yh⇠, Hxi
> ( 0) + sup
⇡(x)2Y
sup
⇠2Zh⇠, Hxi '⇤(⇠)
= ( 0) + sup
⇡(x)2Y
'(Hx)
what is the wanted inequality. ⌅
Theorem 14. LetE be an infinite-dimensional Banach space not containing`1,F a Banach space,S, T 2L(E, F),J ⇢Rbe a compact interval,':F !Rbe a convex continuous and coercive function, X be a convex subset of E whose interior is non-empty for the topology (E, E⇤), J ⇢ R be a compact interval and : J ! R be a convex continuous function.
AssumeS⇥T has a closed range andS is not compact. Then the following holds : sup
x2X
inf2J '(T x Sx) + ( ) = inf
2J sup
x2X
'(T x Sx) + ( ) .
Lemma 15. We can reduce the problem to the case where S⇥T :x7!(Sx, T x)is one-to- one from E to F ⇥F.
Denote by N = kerS\kerT the kernel of S ⇥T and by q : E ! E/N the quotient mapping. It is clear thatS andT factor throughq :S = ˜S q andT = ˜T q; so for all 2R, T x Sx depends only on qx. So we can replace E by ˜E =E/N, S by ˜S, T by ˜T, and X by ˜X = q(X). Clearly if ˜S was compact, S would be compact too. Finally it is enough to
check that the convex subset ˜X of ˜E has non-empty interior for the topology ( ˜E,E˜⇤) : let abe an interior point of X for (E, E⇤) and ˜a=qa. We want to prove that ˜ais an interior point of ˜X for ( ˜E,E˜⇤). If not it would exist a sequence (yn) in ˜E \X˜ weakly converging to ˜a and we could find by lemma 2 a subsequence (ynk) and a sequence (zk) in E weakly converging to 0 such thatq(zk) =ynk a. Then (z˜ k+a) converges weakly toathus satisfies zk+a2X hence q(zk+a) =ynk 2X˜ fork large enough, a contradiction. ⌅ By lemma 15, we can and do assume that S⇥T is one-to-one. By lemma 4, we know that there exists a normed space V, a compact linear mapping ⇡ : E ! V and a convex open subsetY of V such thatV =⇡(E) and ⇡ 1(Y)⇢X ⇢⇡ 1(Y). It is then easy to see that the supremum overX is equal to the supremum on ⇡ 1(Y), and we will only prove the statement when X =⇡ 1(Y).
The inequality sup
x2X
inf2J '(T x Sx) + ( ) 6 inf
2J sup
x2X
'(T x Sx) + ( ) is standard. So we have only to prove the converse inequality.
Following Corollary 9, it is enough to consider the case where T = 0.S +H, 0 2 J andH 2L(E, F) is compact. Then by Corollary 7, we have
inf2J sup
x2X
'(T x Sx) + ( ) = sup
x2X
'(T x 0Sx) + ( 0) = sup
x2X
'(Hx) + ( 0) , and it follows from Lemma 13 that
sup
⇡(x)2Y
'(Hx) + ( 0)6 sup
⇡(x)2Y
inf2J' ( 0 ).Sx+Hx + ( )
= sup
⇡(x)2Y
inf2J'(T x Sx) + ( ),
and this completes the proof. ⌅
The hypothesis made on the operator S⇥T to have closed range and on S to not be compact could seem quite artificial. In fact without any hypothesis on S the statement of Theorem 14 becomes false, as shown by the following.
Example 16. There exist two continuous linear operators S and T from the Hilbert space E = `2 to a Banach space F, X a convex subset of E whose interior is non-empty for (E,E⇤), a compact interval J ⇢ R and a convex continuous and coercive function ':F !R such that
sup
x2X
inf2J '(T x Sx) < inf
2J sup
x2X
'(T x Sx)
We begin by exhibiting a concrete example of a pair A,B of operators between normed spaces which do not satisfy the minimax equality of Asplund and Pt´ak.
Define the two-dimensional normed spaces E and F as the linear space R2 equipped respectively with the normsk.k1 : (x, y)7!|x|+|y|andk.k1 : (x, y)7!max(|x|,|y|). Denote by A 2 L(E, F) the operator whose matrix in the canonical bases is
✓ 3 1 1 0
◆
and by I the identity mapping.
Lemma 17. We have : inf 2Rsupkzk
161kAz zk1 = inf 2RkA Ik= 3 2. It is easily noticed that the norm inL(E, F) of the operatorT having matrix
✓↵ ◆ is max(|↵|,| |,| |,| |). Indeed, denoting by (e1, e2) the canonical basis of E 'F⇤, we have ke1k=ke2k= 1 and |↵|=|he1, T e1i|6ke1k.kTk.ke1k hence|↵|6kTk; and similarly for
| |,| | and | |. Moreover, if z = (x, y),
kT zk1 = max(|↵x+ y|,| x+ y|)
6max(max(|↵|,| |).(|x|+|y|),max(| |,| |).(|x|+|y|))
= max(|↵|,| |,| |,| |).kzk1
ThuskA Ik= max(|3 |,1,1,| |). And since max(|3 |,1,1,| |)>max(|3 |,| |)> 1
2(|3 |+| |)> 1
2|3 + |= 3 2
and A 3
2I = max(3
2,1,1,3 2) = 3
2, we conclude that inf 2RkA Ik= 3
2. ⌅
Lemma 18. We have : supkzk
161inf 2RkAz zk1 6 5 4 < 3
2. More precisely, for eachz in the unit ball of E there is a ⇤ 2[0,2] such that kAz ⇤zk1 6 5
4. Let x2[0,1] and|y|61 x. Then for z = (x, y) we have
kAz zk1 = max(|(3 )x+y|,| x y|) = max(|(3 )x+y|,|x+ y|) 6max |3 |.x+ 1 x, x+| |.(1 x)
Then for ⇤ =x+ 12[0,2] we get
kAz ⇤zkF 6max (2 x).x+ 1 x, x+ (x+ 1)(1 x) = max(x+ 1 x2, x+ 1 x2)
=x+ 1 x2 6 5 4
Ifz = (x, y) satisfieskzk1 61, we have|x|61 and up to replacingz by z, which does not change kAz zk1, we can assume 0 6 x 6 1. It follows from the previous computation that there exists some ⇤ such that kAz ⇤zk1 6 5
4. Hence for anyz such that kzk1 61 we get inf 2RkAz zk1 6 5
4.
And this achieves the proof that supkzk
161inf 2RkAz zk1 6 5
4. ⌅
Lemma 19. There exist two Banach spaces E2 andF2 which are isomorphic to the Hilbert space `2 and operators A2 and B2 in L(E2, F2) such that
sup
kzkE261 inf
2[0,2]kA2z B2zkF2 6 5 4 < 3
2 6 inf
2R sup
kzkE261kA2z B2zkF2 .
Take E2 =E⇥`2 with the norm : (z,⇠)7!
q
kzk21+k⇠k2,F2 =F⇥`2 with the norm : (z,⇠)7!
q
kzk21+k⇠k2. These spaces are clearly isomorphic to the Hilbert space`2. For (z,⇠) 2 E2 define A2(z,⇠) = (Az,⇠) and B2(z,⇠) = (z,⇠). Then A2 and B2 are clearly continuous linear operators fromE2 toF2. It is easily checked thatkA2 B2k> 3 for all real . Indeed 2
k(z,⇠)k6sup 1k(A2 B2)(z,⇠)k> sup
kzk161k(A2 B2).(z,0)k= sup
kzk161k(A I)zk1
=kA Ik> 3 2
Thus inf 2Rsupk(z,⇠)k61k(A2 B2)(z,⇠)k= inf 2RkA2 B2k> 3 2.
Conversely by lemma 18 and by homogeneity, if k(z,⇠)kE2 6 1 there is a ⇤ 2 [0,2]
such that kAz ⇤zk1 6 5
4 kzk1. Then
k(A2 ⇤B2)(z,⇠)k2 =kAz ⇤zk21+k⇠ ⇤⇠k2 6(5
4)2kzk21+ (1 ⇤)2k⇠k2 , and since|1 ⇤|616 5
4 we get k(A2 ⇤B2)(z,⇠)k2 6(5
4)2 kzk21+k⇠k2 = (5
4)2k(z,⇠)k2E2 6(5 4)2 , hence
k(z,⇠)k61sup inf
2[0,2]kA2(z,⇠) B2(z,⇠)kF2 6 5 4 < 3
2 6 inf
2R sup
k(z,⇠)k61k(A2 B2)(z,⇠)k ,
and this completes the proof. ⌅
SinceE2is isomorphic to`2, we can find a one-to-one compact operator⇡ :E =`2 !E2 with dense range, defineF =F2,T =A2 ⇡ andS =B2 ⇡ and'=k.kF2. ChooseJ = [0,2]
and Y as the unit ball of E2. Then define X = ⇡ 1(Y). So, for any 2 J, we have supy2Y k(A2 B2)yk= supy2Y\⇡(`2)k(A2 B2)yk, thus
sup
x2X
'(T x Sx) = sup
x2XkT x Sxk= sup
⇡(x)2Y kT x Sxk= sup
⇡(x)2Y kA2⇡x B2⇡xk
= sup
y2Y kA2y B2yk=kA2 B2k> 3 2 , whence inf 2Jsupx2X'(T x Sx)> 3
2.
Similarly, because the function ⌫ : y 7! inf 2Jk(A2 B2)ykF2 is Lipschitz hence supy2Y ⌫(y) = supy2Y\⇡(`2)⌫(y), and using lemma 19,
sup
x2X
⇣inf
2J'(T x Sx)⌘
= sup
x2X
⇣inf
2JkT x SxkF2
⌘= sup
x2X
⇣inf
2Jk(A2 B2)(⇡x)kF2
⌘
= sup
⇡(x)2Y
⇣inf
2Jk(A2 B2)(⇡x)kF2⌘
= sup
y2Y
⇣inf
2Jk(A2 B2)(y)kF2⌘
= sup
kyk61
⇣ inf
2[0,2]k(A2 B2)(y)kF2⌘ 6 5
4 ,
so supx2X⇣
inf 2J'(T x Sx)⌘
< inf 2Jsupx2X'(T x Sx), as announced. It can be noticed that in this example B2 is an isomorphism, so S is one-to-one. A fortiori S⇥T is one-to-one. But it is compact, and its range is not closed. ⌅ It would be interesting to know whether the interior of X for (E, E⇤) has to be non-empty for guaranteeing the validity of Theorem 14. Suppose E is a Banach space non containing`1,X ⇢E is a non-empty convex set and the conclusion of Theorem 14 holds for all F, S, T, J,', as in the hypotheses : should the interior of X for the topology (E, E⇤) be non-empty ?
References
[ AP ] E. Asplund and V. Pt´ak, A Minimax Inequality for Operators and a Related Numerical Range, Acta Math.126 ( 1971), 53—62.
[ BFT ] J. Bourgain, D.H. Fremlin, M Talagrand, Pointwise compact sets of Baire-measurable functions, Amer. J. Math. 100 (1978), no. 4, 845–886.
[ R ] B. Ricceri, A minimax Theorem in infinite-dimensional Topological Vector Spaces, Linear and Nonlinear Analysis 2, no. 1, (2016), 47–52
Jean SAINT RAYMOND
Sorbonne Universit´es, Universit´e Pierre et Marie Curie (Univ Paris 06), CNRS Institut de Math´ematique de Jussieu - Paris Rive Gauche
Boˆıte 186 - 4 place Jussieu F- 75252 Paris Cedex 05 FRANCE
