Monotone operators and applications

Project Abstract

Monotone operators are fundamental objects in optimization theory and variational analysis. They play a crucial role in solving a wide range of problems in mathematics, engineering, economics, and physics. The study of monotone operators dates back to the work of Minty and Rockafellar in the 1960s and has since then developed into a rich area of research with numerous applications. This research project focuses on the theory of monotone operators and its applications in various fields. Monotone operators are operators that preserve an order structure, meaning that they satisfy a certain monotonicity property with respect to a partial order. This property allows for the generalization of many important concepts such as convexity, subdifferentiability, and optimality conditions. One of the key aspects of this research is the study of the fundamental properties of monotone operators, such as the existence of fixed points, resolvents, and the relationship between monotonicity and continuity. Understanding these properties is essential for developing efficient algorithms for solving optimization problems involving monotone operators. The applications of monotone operators are vast and diverse. In convex optimization, monotone operators are used to model a wide range of problems, including variational inequalities, convex feasibility problems, and equilibrium problems. They also arise in the study of partial differential equations, where monotone operators are used to establish the existence and uniqueness of solutions. Moreover, monotone operators have found applications in machine learning, image processing, and signal processing. In these fields, monotone operators are used to design efficient algorithms for solving inverse problems, denoising images, and learning complex patterns from data. Another important application of monotone operators is in game theory, where they are used to model strategic interactions between rational agents. Monotone operators play a central role in the study of Nash equilibria, stable matching problems, and competitive equilibrium in markets. Overall, the study of monotone operators and their applications is a rich and active area of research with broad significance in mathematics and its applications. This research project aims to contribute to the understanding of monotone operators and develop new algorithms and applications that leverage the power of monotonicity for solving complex optimization problems across various disciplines.

Project Overview

<p> </p><p>Preliminaries<br>The aim of this chapter is to provide some basic results pertaining<br>to geometric properties of normed linear spaces and convex functions.<br>Some of these results, which can be easily found in textbooks are given<br>without proofs or with a sketch of proof only.<br>1.1 Geometry of Banach Spaces<br>Throughout this chapter X denotes a real norm space and X denotes<br>its corresponding dual. We shall denote by the pairing hx; xi the value<br>of the function x 2 X at x 2 X. The norm in X is denoted by k k,<br>while the norm in X is denoted by k k. If there is no danger of<br>confusion we omit the asterisk from the notation kk and denote both<br>norm in X and X by the symbol k k.<br>As usual We shall use the symbol ! and * to indicate strong and<br>weak convergence in X and X respectively. We shall also use w-lim<br>to indicate the weak-star convergence in X. The space X endowed<br>with the weak-star topology is denoted by Xw<br>1.1.1 Uniformly Convex Spaces<br>Denition 1.1. Let X be a normed linear space. Then X is said to<br>be uniformly convex if for any ” 2 (0; 2] there exist a = (“) &gt; 0 such<br>that for each x; y 2 X with kxk 1, kyk 1, and kx ô€€€ yk “, we<br>have k1<br>2 (x + y)k 1 ô€€€ .<br>Theorem 1.2. Let X be a uniformly convex space. Then for any<br>d &gt; 0; ” &gt; 0 and x; y 2 X with kxk d, kyk d, and kx ô€€€ yk “,<br>there exist a = ( ”<br>d ) &gt; 0 such that k1<br>2 (x + y)k (1 ô€€€ )d.<br>7<br>Proof. Let x; y 2 X, set k1 = x<br>d ; k2 = y<br>d and ” = ”<br>d . Then obviously we<br>see that ” &gt; 0. Moreover , kk1k 1; kk2k 1 and kk1 ô€€€ k2k ”<br>d = “.<br>Now, by the uniform convexity of X, we have for some ( ”<br>d ) &gt; 0,</p><p>1<br>2<br>(k1 + k2)</p><p>1 ô€€€ (“);<br>that is,<br>1<br>2d<br>(x + y)</p><p>1 ô€€€ “();<br>which implies,<br>1<br>2<br>(x + y)</p><p>[1 ô€€€ (<br>”<br>d<br>)]d:<br>Hence we have the result.<br>Proposition 1.3. Let X be a uniformly convex space, let 2 (0; 1)<br>and ” &gt; 0, then for any d &gt; 0, x; y 2 X such that kxk d, kyk d,<br>and kx ô€€€ yk ” there exist (“) &gt; 0 independent of x and y such that<br>kx + (1 ô€€€ )yk [1 ô€€€ 2(“) minf; 1 ô€€€ g]d:<br>Proof. Without loss of generality we shall assume that 2 (0; 1<br>2 ], we<br>also observe that<br>kx+(1ô€€€)yk = k(x + y)+(1ô€€€2)yk 2k<br>1<br>2<br>(x+y)k+(1ô€€€2)kyk<br>Thus from the uniform convexity of X we have for some (“) &gt; 0<br>kx + (1 ô€€€ )yk 2<br>1<br>2<br>(x + y)</p><p>+ (1 ô€€€ 2)kyk<br>2(1 ô€€€ (“))d + (1 ô€€€ 2)d<br>= (1 ô€€€ 2(“))d<br>[1 ô€€€ 2(“)minf; 1 ô€€€ g]d:<br>Which completes the proof.<br>8<br>1.1.2 Strictly Convex Spaces<br>Denition 1.4. A normed linear space X is said to be strictly convex<br>if for all x; y 2 X x 6= y, kxk = kyk = 1, we have<br>kx + (1 ô€€€ )yk &lt; 1 for all 2 (0; 1):<br>Theorem 1.5. Every uniformly convex space is strictly convex.<br>Proof. Suppose X is uniformly convex, since x 6= y, set ” = kxô€€€yk &gt;<br>0 and d = 1. Then in view of proposition (1:3) we see that for each<br>2 (0; 1); kx + (1 ô€€€ )yk &lt; 1, which gives the desired result.<br>We now give some examples to illustrate uniformly and strictly convex<br>spaces.<br>Example 1. Every inner product space H is uniformly convex. In<br>particular Rn with the euclidean norm is uniformly convex.<br>To see this we shall apply parallelogram law which is valid in any inner<br>product space. That is for all x; y;2 H, we have<br>kx + yk2 + kx ô€€€ yk2 = 2(kxk2 + kyk2)<br>Now let ” 2 (0; 2] be given, let x; y 2 H such that kxk 1; kyk 1,<br>and kx ô€€€ yk “, then from the above identity we have</p><p>1<br>2<br>(x + y)</p><p>2</p><p>1<br>4</p><p>2(2) ô€€€ kx ô€€€ yk2<br>= 1 ô€€€<br>1<br>2<br>(x ô€€€ y)</p><p>2<br>1 ô€€€<br>1<br>4<br>“2<br>So that<br>1<br>2<br>(x + y)</p><p>r<br>1 ô€€€<br>1<br>4<br>“2<br>To complete the proof we choose =<br>q<br>1 ô€€€ 1<br>4″2 &gt; 0.<br>Example 2. Rn with kk1 is not strictly convex. To see this we choose<br>the canonical bases e1; e2 in Rn and take = 1<br>2 . Clearly ke1k = ke2k =<br>1; e1 6= e2 and<br>1<br>2<br>e1 +<br>1<br>2<br>e2</p><p>=<br>1<br>2<br>ke1 + e2k = 1:<br>Thus we have Rn with k k1 is not strictly convex.<br>Example 3. The space C[a; b] of all real valued continuous func-<br>tions on the compact interval [a; b] endowed with the “sup norm” is<br>9<br>not strictly convex. To see this we choose two functions such that<br>f(t) := 1 for all t 2 C[a; b]; g(t) :=<br>b ô€€€ t<br>b ô€€€ a<br>for all t 2 C[a; b]:<br>Take ” = 1<br>2 . Clearly, f; g 2 C[a; b], kfk = kgk = 1 and kfô€€€gk = 1 &gt; “.<br>But k1<br>2 (x + y)k = 1. Thus, C[a; b] is not strictly convex.<br>Theorem 1.6. Let X be a re exive Banach space with norm k k.<br>Then there exist an equivalent norm k k0 such that X is strictly con-<br>vex in this norm and X is strictly convex in the dual norm k k0<br>.<br>1.1.3 Duality Mappings.<br>Denition 1.7. Dene a map J : X ô€€€! 2X by<br>Jx :=<br>n<br>x 2 X : hx; xi = kxkkxk; kxk = kxk<br>o<br>:<br>By Hahn Banach theorem for each x 2 X; x 6= 0, there exist y 2 X<br>such that kyk = 1, and hx; yi = kxk. So if we set x = kxky, then<br>we see that for each x 2 X 9 x 2 X such that kxk = kxk and<br>hx; xi = kxk2. So we see that for each x 2 X; Jx 6= ;. The mapping<br>J : X ô€€€! 2X is called the duality mapping of the space X. In general<br>J is multivalued.<br>Remark. More generally, given an increasing continuous function ‘ :<br>[0; +1) ! [0; +1) such that ‘(0) = 0 and lim<br>+1<br>‘ = +1, one denes<br>the duality map J’ corresponding to the (normalization) function ‘,<br>by<br>J’x :=<br>n<br>x 2 X : hx; xi = kxkkxk; kxk = ‘(kxk)<br>o<br>:<br>Proposition 1.8. Let H be a real Hilbert space and identify H with<br>H, then<br>Jx = fxg for all x 2 H;<br>i.e The duality map J is the identity map.<br>Proof. Let a 2 H. Dene<br>‘a(x) = ha; xi 8x 2 H:<br>10<br>Then ‘a 2 H, k’ak = kak and ‘a(a) = kak2: Therefore ‘a 2 J(a)and<br>since ‘a is identied with a, via Riesz representation theorem, we can<br>write a 2 J(a). Conversely, if y 2 Ja then ha; yi = kakkyk and<br>kak = kyk so that<br>kaô€€€yk2 = ha ô€€€ y; aô€€€yi = kak2 +kyk2 ô€€€2ha; yi = 2kak2 ô€€€2kyk2 = 0:<br>So we have y = a. Therefore Ja = fag.<br>Proposition 1.9. Let X be a real Banach and J be the duality mapping<br>on X, then<br>J(x) = J(x) 8 2 R 8 x 2 X:<br>Proof. Let y 2 J(x) and 2 R. For = 0 the result follows trivially.<br>suppose 6= 0, then we have<br>hx; yi = 2hx; yi = kxkkyk;we also have kxk = kyk:<br>Thus we have y 2 J(x), which implies J(x) J(x). From the<br>preceding inclusion we also obtained that 1<br>J(x) J(x) which implies<br>J(x) J(x). Therefore J(x) = J(x) 8 2 R, 8x 2 X.<br>Denition 1.10. Let f : X ô€€€! Y be a map. Then f is said to<br>be demi-continuous if it is norm to weak-star continuous, i.e f is continuous<br>from X (endowed with the strong topology) to Y (endowed<br>with the weak-star topology).<br>Proposition 1.11. Let X be a real norm space and J be the dual-<br>ity mapping on X. Then the following are true.<br>a) For each x 2 X, Jx is a closed, convex subset of B[0; kxk] in X.<br>Where B[0; kxk] = fy 2 X : kyk kxk:g<br>b) If X is strictly convex, then for each x 2 X, Jx is single valued.<br>Moreover the mapping J is demi-continuous, i.e J is continuous as<br>a mapping from X with the strong topology to X with the weak-star<br>topology.<br>c) If X is uniformly convex, then for each x 2 X, Jx is single valued<br>and the mapping x 7ô€€€! Jx is uniformly continuous on bounded subsets<br>of X.<br>Proof. (a) Obviously we have Jx B[0; kxk]. Let fyn<br>gn1 Jx<br>such that yn<br>! y, for each n 1 we have hx; yn<br>i = kxkkyn<br>k and<br>kxk = kyn<br>k. Letting n ! +1 we see that hx; yi = kxkkyk and<br>11<br>kxk = kyk. Hence we have y 2 Jx, which implies that Jx is closed.<br>For convexity, let x; y 2 Jx and 2 (0; 1), then<br>hx; x + (1 ô€€€ )yi = hx; xi + (1 ô€€€ )hx; yi<br>= kxkkxk + (1 ô€€€ )kyk = kxk2<br>We also have from the triangular inequality that kx + (1 ô€€€ )yk<br>kxk, also,<br>kxk2 = hx; x + (1 ô€€€ )yi<br>kxkkx + (1 ô€€€ )yk;<br>which implies that kxk kx + (1 ô€€€ )yk. Hence we have<br>kxk = kx + (1 ô€€€ )yk; which shows that x + (1 ô€€€ )y 2 Jx:<br>(b) Assume X is strictly convex, and suppose that there exit x; y 2<br>Jx such that x 6= y, then kxk = kyk = kxk and by the strict<br>convexity of X we have that for any 2 (0; 1), kx + (1 ô€€€ )yk &lt;<br>kxk. In particular taking = 1<br>2 , we have k1<br>2 (x + y)k &lt; kxk, which<br>contradicts the fact that k1<br>2 (x + y)k = kxk. (Since Jx is convex)<br>Let fxngn1 X such that xn ! x. using the fact that kJxnk = kxnk,<br>i.e fJxngn1 is bounded and the fact that the unit ball is w-compact<br>in X (Banach Alaoglo Theorem) we see that there exist a limit point<br>y of fJxngn1. Now let fJxnkgk1 X such that wô€€€limJxnk = y,<br>then we have lim<br>k!1<br>hxnk ; Jxnki = hx; yi. We also have that<br>lim<br>k!1<br>hxnk ; Jxnki = lim<br>k!1<br>kxnkk2 = kxk2:<br>So we get hx; yi = kxk2, which implies kxk kyk. To get the<br>reverse inequality we use the fact that w ô€€€ limJxnk = y implies<br>kyk lim inf kJxnkk = lim inf kxnkk = kxk: Thus we have kxk = kyk<br>and hx; yi = kxk2. i.e y = Jx. Therefore J is demicontinuous.<br>(c) Since a uniformly convex space is also strictly, then by part (b)<br>above we see that J is single valued.<br>Assume J is not uniformly continuous on bounded subsets of X, then<br>there exist a constant M &gt; 0, 0 &gt; 0, and subsequences fung; fvng<br>X such that<br>kunk M; kvnk M; n 1;<br>kun ô€€€ vnk ! 0 as n ! 1;<br>kJun ô€€€ Jvnk 0; n 1: (1.1)<br>12<br>Let &gt; 0 such that kunk ; kvnk ; for n 1: Such exist, for<br>if there exist a subsequence funkg X such that unk ! 0 as n ! +1,<br>then we see that vnk ! 0. From the denition of duality map we<br>obtained that Junk ! 0 and Jvnk ! 0, and this contradicts (1:1).<br>Now set<br>xn =<br>un<br>kunk<br>; yn =<br>vn<br>kvnk<br>un; vn 6= 0: Then we have,<br>kxn ô€€€ ynk =<br>1<br>kunkkvnk<br>kunkvnk ô€€€ kunkvnk</p><p>1<br>2 (kvnkkun ô€€€ vnk + kkvnk ô€€€ kunkkkvnk)</p><p>2M<br>2<br>kun ô€€€ vnk ! 0 as n ! +1<br>We also have 2 kJxn + Jynk hxn; Jxn + Jyni which together with<br>hxn; Jxn + Jyni = kxnk2 + kynk2 + hxn ô€€€ yn; Jyni<br>= 2 + hxn ô€€€ yn; Jyni 2 ô€€€ kxn ô€€€ ynk<br>implies<br>lim<br>n!1<br>kJxn + Jynk = 2 i.e lim<br>n!1<br>k<br>1<br>2<br>(Jxn + Jyn)k = 1: (1.2)<br>Now suppose there exist “0 &gt; 0 and a subsequence fxnkg; fynkg X<br>such that kJxnk ô€€€ Jynkk “0, for n 1. Observing that kJxnkk =<br>kJynkk = 1 and using the uniform convexity of X we see that there<br>exist (“0) &gt; 0 such that k1<br>2 (Jxnk+Jynk)k 1ô€€€(“0) which contradicts<br>(1.2). Therefore we have lim<br>n!1<br>kJxn ô€€€ Jynk = 0, which implies<br>kJun ô€€€ Jvnk = kJ(kunkxn) ô€€€ J(kvnkyn)k<br>= kkunkJxn ô€€€ kvnkJynk<br>kunkkJxn ô€€€ Jynk + kvnkkkunk ô€€€ kvnkk<br>MkJxn ô€€€ Jynk + kun ô€€€ vnk ! 0 as n ! +1:<br>This contradicts (1.1). Hence we have the result.<br>1.1.4 Duality maps of Lp Spaces (p &gt; 1)<br>Proposition 1.12. The duality map on Lp([0; 1]), p &gt; 1 is given by<br>J(0) = f0g and for f 6= 0<br>J(f) = ffg<br>13<br>where f 2 (Lp) = Lq is dened by<br>f(g) =<br>Z 1<br>0<br>f(t)g(t)dt 8g 2 Lp<br>and<br>f =<br>jfjpô€€€1signf<br>kfkpô€€€2<br>p<br>Observe that when p 2, this f has the following expression:<br>f =<br>fjfjpô€€€2<br>jjfjjpô€€€2<br>p<br>:<br>Proof. Now set Af = ffg. By denition of the duality map<br>J(f) = f 2 (Lp) : (f) = kfkkk; kfk = kkg:<br>Let 2 J(f), since 2 (Lp), then by Riez representation theorem<br>there exist a unique f 2 Lq, 1<br>p + 1<br>q = 1 p; q &gt; 1 such that<br>(f) = hf; fi =<br>Z 1<br>0<br>f(t)f(t) dt; kk = kfk:<br>Setting = f , we have<br>f(f) = hf; fi =<br>Z 1<br>0<br>f(t)f(t) dt; kfk = kfk:<br>So we have<br>kfkkfk =<br>f (f) =<br>Z 1<br>0<br>f(t)f(t) dt; k<br>fk = kfk = kfk:<br>which implies<br>Z 1<br>0<br>f(t)f(t) dt = kfk2; kfkq = kfkp: (1.5)<br>We now show that f(t) := jf(t)jpô€€€1signf(t)<br>kfkpô€€€2<br>p<br>satises (1.5). But we have<br>Z 1<br>0<br>jf(t)jq dt<br>1<br>q<br>=<br>Z 1<br>0<br>jf(t)jq(pô€€€1)<br>kfk(pô€€€2)q dt<br>1<br>q<br>=<br>1<br>kfk(pô€€€2)<br>Z 1<br>0<br>jf(t)jq dt<br>1<br>q<br>=<br>kfk<br>p<br>q<br>kfkpô€€€2 =<br>kfkpô€€€1<br>kfkpô€€€2 = kfkp:<br>14<br>Therefore kfkq = kfkp. Also<br>Z 1<br>0<br>f(t)f(t) dt =<br>Z 1<br>0<br>jf(t)jpô€€€1signf(t)<br>kfkpô€€€2<br>p<br>f(t) dt<br>=<br>1<br>kfkpô€€€2<br>Z 1<br>0<br>jf(t)jp dt<br>=<br>kfkp<br>kfkpô€€€2 = kfkpkfkp = kfkpkfkq = kfkkfk:<br>Thus J(f) Af . On the other hand for arbitrary h 2 Lp([0; 1]),<br>f(h) =<br>Z 1<br>0<br>f(t)h(t) dt; kfk = kfk:<br>In particular<br>f(f) =<br>Z 1<br>0<br>g(t)f(t) dt<br>=<br>Z 1<br>0<br>f(t)<br>jf(t)jpô€€€1signf(t)<br>kfkpô€€€2<br>p<br>dt<br>=<br>1<br>kfkpô€€€2<br>Z 1<br>0<br>jf(t)jp dt = kfk2<br>p:<br>So we have f(f) = kfkpkfkp and kfkp = kfkq so that kfk = kfkp:<br>Thus Af J(f). Therefore Af = J(f).<br>1.2 Convex Functions and Subdierentials<br>In this section we present the basic properties of convex functions and<br>subdierentials as we shall use them in the next chapter.<br>1.2.1 Basic notions of Convex Analysis<br>Denition 1.13. Let C be a non empty subset of a real norm linear<br>space X. The set C is said to be convex if for each x; y 2 C and for<br>each t 2 (0; 1) we have tx + (1 ô€€€ t)y 2 C:<br>Denition 1.14. Let C be a non empty convex subset of X. Then the<br>convex hull of C denoted by coC is the intersection of all convex sets<br>containing C. (Equivalently, convex hull of C is the set of all convex<br>15<br>combinations of nite subsets of points of C).<br>Denition 1.15. Let C be a non empty convex subset of X. Let<br>f : C ô€€€! R [ f+1g. Then f is said to be convex if for each t 2 (0; 1)<br>and for all x; y 2 C we have<br>f(tx + (1 ô€€€ t)y) tf (x) + (1 ô€€€ t)f(y):<br>Moreover f is said to be proper if f is not identically +1 (i.e 9 x0 2 C<br>such that f(x0) 2 R)<br>Denition 1.16. Let f : C ô€€€! R [ f+1g be a map. The eective<br>domain of f is dened by<br>D(f) = fx 2 X : f(x) &lt; +1g:<br>The set<br>epi(f) = f(x; ) 2 X R : f(x) g<br>is called the epigraph of f, while<br>S = fx 2 X : f(x) g<br>is called the section of f.<br>Proposition 1.17. A mapping f : X ô€€€! R[f+1g is convex if and<br>only if its epigraph is convex.<br>Proposition 1.18. (Slope Inequality) Let I be an interval of R and<br>f : I ô€€€! R be a convex function. Assume r1 &lt; r2 &lt; r3 with ri 2 I for<br>i = 1; 2; 3: and f(r1); f(r2) are nite. Then<br>f(r2) ô€€€ f(r1)<br>r2 ô€€€ r1</p><p>f(r3) ô€€€ f(r1)<br>r3 ô€€€ r1</p><p>f(r3) ô€€€ f(r2)<br>r3 ô€€€ r2<br>:<br>Proposition 1.19. Suppose f : I ô€€€! R is convex and derivable on I.<br>Then f0 is increasing.<br>Proof. Let r &lt; t we show that f0(r) f0(t). Now<br>f0(r) = lim<br>s!r+<br>f(s) ô€€€ f(r)<br>s ô€€€ r</p><p>f(t) ô€€€ f(r)<br>t ô€€€ r<br>lim<br>s!rô€€€<br>f(s) ô€€€ f(t)<br>s ô€€€ t<br>= f0(t)<br>Denition 1.20. Let f : X ô€€€! R[ f+1g be a map. Let x0 2 D(f),<br>16<br>then f is lower semicontinuous at x0 if for each ” &gt; 0 there exist &gt; 0<br>such that f(x0) ô€€€ ” &lt; f(x) for all x 2 B(x0; ).<br>Proposition 1.21. Let f : X ô€€€! R[f+1g be a map. Let x0 2 D(f),<br>then f is lower semicontinuous at x0 if and only if<br>lim inf f(xn) f(x0)<br>for all fxng X such that xn ! x0.<br>Proposition 1.22. Let f : X ô€€€! R [ f+1g be a map. Then the<br>following are equivalent.<br>(a) f is lower semicontinuous,<br>(b) epi(f) is closed,<br>(c) S is closed for each 2 R.<br>Denition 1.23. Let f : X ô€€€! R [ f+1g be a map. Then f is<br>said to be coercive if<br>lim<br>kxk!1<br>f(x) = +1:<br>Proposition 1.24. Let f : X ô€€€! R [ f+1g be a map. Then f is<br>convex and l.s.c if and only if f is convex and weakly l.s.c.<br>Proof.<br>f is convex and l.s.c , epi(f) is convex and closed<br>, epi(f) is convex and weakly closed<br>, f is convex and weakly l.s.c.<br>Theorem 1.25. Assume X is re exive. Let f : X ô€€€! R [ f+1g<br>be proper, convex, coercive and l.s.c function on X. Then f has a<br>minimum on X. That is there exist x0 2 X such that<br>f(x0) = inf<br>x2X<br>f(x):<br>Proof. Let = inf<br>x2X<br>f(x). Since f is proper we see that &lt; +1.<br>Let fxng X such that f(xn) ! &lt; +1, then from the coercivity<br>condition of f we see that fxng is bounded. Since X is re exive, then<br>there exist x0 2 X and a subsequence fxnkg such that xnk * x0. In<br>17<br>view of proposition (1.24) f is weakly lower semi continuous. So we<br>have<br>f(x0) lim inf f(xnk) = lim<br>k!1<br>f(xnk) = :<br>Therefore<br>f(x0) = = inf<br>x2X<br>f(x):<br>Theorem 1.26. Let f be proper, convex, and lower semi-continuous<br>on X. Then f is bounded from below by an ane function. i.e There<br>exist x 2 X and a constant c 2 R such that<br>f(x) hx; xi + c for all x 2 X:<br>Proof. Let x0 2 D(f) and 2 R such that f(x0) &gt; . This is<br>possible since f is proper, i.e D(f) 6= ;: Clearly (x0; ) =2 X R, also<br>in view of proposition (1.22) epi(f) is closed and convex, so by Hahn<br>Banach theorem there exists a closed hyperplane<br>H = f(x; ) 2 X R : hx; x<br>0i + = g<br>that separates epi(f) and (x0; ) i.e<br>hx; x<br>0i + hx0; x<br>0i +<br>Considering the left hand side of the inequality only, it is easy to see<br>that &lt; 0, otherwise we arrived at contradiction. Therefore we have</p><p>+ hx;ô€€€<br>x0</p><p>i for all (x; ) 2 epi(f):<br>Since for each x in X (x; f(x)) 2 epi(f), we see that<br>f(x) hx; xi + c for all x 2 X; where x = ô€€€<br>x0</p><p>and c =</p><p>:<br>Denition 1.27. Let X be Banach space. Let f : X ô€€€! R [ f+1g<br>be a function. Let x 2 D(f) and v 2 X, then we say that f has a<br>directional derivative at x in the direction of v 6= 0 if the limit<br>lim<br>t!0+<br>f(x + tv) ô€€€ f(x)<br>t<br>exist:<br>We denote by f0(x; v) the directional derivative of f at x in the direction<br>of v, and we write<br>f<br>0<br>(x; v) = lim<br>t!0+<br>f(x + tv) ô€€€ f(x)<br>t<br>:<br>18<br>The function f : X ô€€€! R is said to be G^ateaux dierentiable at x 2 X<br>if for all v 2 X f0(x; v) exists in R and the function v 7! f0(x; v) is<br>linear and continuous. We denote by DGf(x) the G^ateaux dierential<br>of f at x and<br>hDGf(x); vi := f<br>0<br>(x; v) for all v 2 X:<br>1.2.2 Subdierential of a Convex function<br>Denition 1.28. Let f : X ô€€€! R [ f+1g be a proper and convex<br>function. Let x 2 D(f), then the subdierential @f(x) of f at x is the<br>set<br>@f(x) = fx 2 X : hy ô€€€ x; xi f(y) ô€€€ f(x) 8y 2 Xg:<br>We remarked that if x is not in D(f) then @f(x) = ;:<br>Proposition 1.29. Let X be proper and convex function which is<br>G^ateaux dierentiable at x 2 D(f) then<br>@f(x) = fDGf(x)g:<br>Proof. To see this we pick y 2 X. Convexity of f implies that<br>f(x + t(y ô€€€ x)) ô€€€ f(x)<br>t<br>f(y) ô€€€ f(x); 0 &lt; t &lt; 1:<br>Since f is G^ateaux dierentiable at x we obtained that<br>hy ô€€€ x;DGf(x)i = lim<br>t!0+<br>f(x + t(y ô€€€ x)) ô€€€ f(x)<br>t<br>f(y) ô€€€ f(x):<br>Thus DGf(x) 2 @f(x).<br>Conversely, let w 2 @f(x), then for any y 2 X and t &gt; 0<br>f(x + ty) ô€€€ f(x)<br>t<br>hy;wi:<br>Using the G^ateaux dierentiability of f at x we obtained that<br>hy;DGf(x)i hy;wi 8y 2 X<br>which implies DGf(x) = w 2 @f(x). Therefore @f(x) = fDGf(x)g:<br>Example 1. Dene a function f : X ô€€€! R [ f+1g by<br>f(x) =<br>1<br>2<br>kxk2 8x 2 X:<br>19<br>Then f is proper, convex and continuous. Moreover @f(x) = J(x) for<br>each x 2 X where J is the duality map on X.<br>Indeed choose rst x 2 Jx. Then for any y 2 x we have<br>hy ô€€€ x; xi = hy; xi ô€€€ kxk2 kykkxk ô€€€ kxk2</p><p>1<br>2<br>kyk2 ô€€€<br>1<br>2<br>kxk2<br>= f(y) ô€€€ f(x):<br>Thus we have x 2 @f(x). Conversely, for x 2 @f(x) we have<br>hy ô€€€ x; xi f(y) ô€€€ f(x) 8y 2 X:<br>So considering x + ty, t 2 (0; 1) we get<br>hx; yi<br>1<br>2t<br>(kx + tyk2 ô€€€ kxk2) kxkkyk +<br>t<br>2<br>kyk:<br>As t ! 0+ we have hx; yi kxkkyk; which implies kxk kxk: Also<br>using the fact that x 2 @f(x) and considering x + tx 2 X we have<br>2thô€€€x; xi kx ô€€€ txk2 ô€€€ kxk2 = (t2 ô€€€ 2t)kxk2; t &gt; 0:<br>So we have (2 ô€€€ t)kxk2 2hx; xi: Now as t ! 0+ we obtained<br>kxk2 hx; xi kxkkxk which implies kxk kxk:<br>Therefore we have kxk = kxk and hx; xi = kxk2: Thus x 2 J(x).<br>Example 2. Let K be a closed, convex subset of X. Dene a map Ik<br>on X by<br>Ik(x) =<br>(<br>0 for x 2 K,<br>+1 for x =2 K.<br>It is easy to see that Ik is convex and lower semi-continuous (since K<br>is convex and closed). Futhermore for any x 2 K we get<br>@Ik(x) = fx 2 X : hy ô€€€ x; xi 0; 8y 2 Xg:<br>1.2.3 Jordan Von Neumann Theorem for the Existence of<br>Saddle point<br>We now state and prove Jordan von Neumann Theorem for the existence<br>of saddle point for an upper semi-continuous function dened on<br>a compact convex subset of a Banach space. But before that we state<br>20<br>Kakutani xed point theorem without proof.<br>Theorem 1.30. (Kakutani) Let K be a nonempty compact convex<br>subset of a Banach space and let<br>T : K ô€€€! 2K<br>be a mapping having a closed graph (i.e., T is upper semicontinuous)<br>and such that for every x 2 K, T(x) K is nonempty, closed and<br>convex. Then there exists at least one x 2 K such that x 2 T(x).<br>Theorem 1.31. (J.Von Neumann) Let X and Y be real Banach<br>spaces and let U X and V Y be compact convex subsets of X and<br>Y , respectively. Let H : U V ô€€€! R be a continuous, convex-concave<br>function (i.e H(u; v) is convex as a function of u and concave as a<br>function of v). Then there exists (u0; v0) 2 U V such that<br>H(u0; v) H(u0; v0) H(u; v0) 8 u 2 U and 8 v 2 V:<br>Such a point (u0; v0) is called the saddle point of the function H.<br>Proof. Dene the mappings T1 : U ô€€€! V , T2 : V ô€€€! U and<br>T : U V ô€€€! 2UV<br>respectively by<br>T1(u) =<br>n<br>v 2 V : H(u; v) H(u;w) 8 w 2 U<br>o<br>;<br>T2(v) =<br>n<br>u 2 U : H(u; v) H(w; v) 8 w 2 V<br>o<br>;<br>T(u; v) = T2(v) T1(u):<br>Since H is continuous we see that the graph of T;<br>Graph(T) =<br>nô€€€<br>(u; v); (x; y)</p><p>2 (U V )2 : (x; y) 2 T(u; v)<br>o<br>;<br>is closed. Also for each (u; v) 2 U V , T(u; v) is convex. To see this it<br>is enough to show that T1(u) and T2(v) are convex. Let u1; u2 2 T2(v)<br>and 2 (0; 1). Since H is convex as a function of u we see that<br>H(u1 + (1 ô€€€ )u2; v) H(u1; v) + (1 ô€€€ )H(u2; v)<br>H(w; v) + (1 ô€€€ )H(w; v)<br>= H(w; v) 8w 2 V:<br>21<br>which implies u1+(1ô€€€)u2 2 T2(v). So we have T2(v) is convex. Similarly<br>we have T1(u) is convex, and hence T(u; v) is convex. Therefore<br>by theorem (1:30) there exists (u0; v0) 2 U V such that<br>(u0; v0) 2 T(u0; v0) = T2(v0) T1(u0)<br>which implies<br>H(u0; v) H(u0; v0) H(u; v0) 8 u 2 U and 8 v 2 V:<br>The proof is complete.<br>22</p> <br><p></p>