A Elements of functional analysis A.1 Separating hyperplane theorem Let v Rn and γ R be given and consider the set H = x Rn : v, x = γ . For x 2 H we have2 { 2 h i } 2 v, x (γ/ v 2)v =0, − k k 2 so H = v? +(γ/ v )v. The⌦ complement of↵ H consists of the two sets x : v, x <γ and xk: kv, x >γ on the two “sides” of the hyperplane. { h iThe following} { theoremh i says} that for two disjoint, convex sets, one compact and one closed, there exists two “parallel” hyperplanes such that the sets lie strictly one di↵erent sides of those hyperplanes. The assumption that one of the sets is compact can not be dropped (see Exercise 1) Theorem A.1.1 (Separating hyperplane theorem). Let K and C be dis- joint, convex subsets of Rn, K compact and C closed. There exist v Rn and 2 γ1,γ2 R such that 2 v, x <γ <γ < v, y h i 1 2 h i for all x K and y C. 2 2 Proof. Consider the function f : K R defined by f(x) = inf x y : y C , i.e. f(x) is the distance of x to C.! The function f is continuous{k (check)− k and2 } since K is compact, there exists x0 K such that f attains its minimum at x0. Let y C be such that x y 2 f(x ). By the parallelogram law we have n 2 k 0 − nk! 0 y y 2 y x y x 2 n − m = n − 0 m − 0 2 2 − 2 y + y 2 = 1 y x 2 + 1 y x 2 n m x . 2 k n − 0k 2 k m − 0k − 2 − 0 34 Elements of functional analysis By convexity (yn + ym)/2 C, so that (yn + ym)/2 x0 f(x0). Hence, we have 2 k − k y y 2 n − m 1 y x 2 + 1 y x 2 f 2(x ). 2 2 k n − 0k 2 k m − 0k − 0 The right-hand side of this display converges to 0 as n, m . So the yn !1n form a Cauchy sequence and hence they converge to some y0 R . Since C 2 is closed, y0 C. Let v = y0 x0. Since K and C are disjoint, v = 0. It follows that 02< v 2 = v, y −x = v, y v, x . It remains to show6 that k k h 0 − 0i h 0ih 0i v, x v, x0 and v, y0 v, y for all x K and y C. h ihTake y iC. Sinceh Cihis convex,i the line2 segment y2+ λ(y y ), λ [0, 1], 2 0 − 0 2 belongs to C. Since y0 minimizes the distance to x0, we have y x y x + λ(y y ) k 0 − 0kk 0 − 0 − 0 k for every λ. By squaring this we find that 0 2λ y x ,y y + λ2 y y 2. h 0 − 0 − 0i k − 0k Dividing by λ and then letting λ 0 gives v, y y0 0, as desired. A similar argument shows that! v, x h v, x− forix K. h ih 0i 2 The polar C0 of a set C Rn is defined as ✓ 0 n C = y R : x, y 1 for all x C . { 2 h i 2 } Note that in the special case that C is closed under multiplication with positive scalars, we have C0 = y Rn : x, y 0 for all x C (check). For a given 0 { 2 h i 2 } x, the set Cx = z : x, z 0 is the set of all vectors that lie on the same { h i } 0 side of x? as x. The polar is in this case the intersection of all the sets Cx for x C. − 2 To illustrate the bipolar theorem geometrically, consider a V -shaped set: C the union of two rays emanating from the origin. Then one readily sees that the polar of the polar of C precisely equals the convex hull of C. The general result is as follows. Theorem A.1.2 (Bipolar theorem). Let C Rn contain 0. Then the bipolar C00 =(C0)0 equals the closed convex hull✓ of C. Proof. It is clear that C00 is a closed, convex set containing C, so the closed convex hull A of C is a subset of C00. Suppose that the converse inclusion does 00 not hold. Then there exists a point x0 C that is not in A. By the separating 2 n hyperplane theorem there then exists a vector v R and γ1,γ2 R such that 2 2 x0,v >γ1 >γ2 > y, v for all y A. Since 0 C A we have γ1 > 0. h i h i 2 n 2 ✓ Dividing by γ1 shows there exists a vector v R such that x0,v > 1 > y, v for all y A. The second inequality implies that2 v C0, andh theni the firsth onei implies that2 x C00, which gives a contradiction.2 0 62 A.2 Topological vector spaces 35 A.2 Topological vector spaces A vector space X is called a topological vector space if it is endowed with a topology which is such that every point of X is a closed set and the addition and scalar multiplication operations are continuous. It is easy to see that translation by a fixed vector and multiplication by a nonzero scalar are homeomorphisms of a topological vector space. This implies in particular that the topology is translation-invariant, meaning that a set E X is open if and only if each of its translates x + E is open. ✓ Topological vector spaces have nice separation properties. Combined with the fact that points are closed sets, the next theorem implies for instance that they are always Hausdor↵. Theorem A.2.1. Suppose that K and C are disjoint subsets of a topological vector space X, K compact and C closed. Then there exits a neighborhood V of 0 such that K + V and C + V are disjoint. Proof. The continuity of addition implies that for every neighborhood W of 0 there exist neighborhoods V and V of 0 such that V + V W (check). Now 1 2 1 2 ✓ put U = V1 V2 ( V1) ( V2). Then U is symmetric (i.e. U = U) and U + U W .\ Applying\ − the\ same− procedure to the neighborhood U we− see that for every✓ neighborhood W of 0 there exists a symmetric neighborhood U such that U + U + U W (etc.). Pick an x ✓K.ThenX C is an open neighborhood of x. By translation invariance and the2 preceding paragraph\ there exists a symmetric neighborhood Vx of 0 such that x + Vx + Vx + Vx does not intersect C. By the symmetry of Vx this implies that x + Vx + Vx and C + Vx are disjoint (check). Since K is compact, it is covered by finitely many sets x1 + Vx1 ,...,xn + Vxn . Put V = V V . Then x1 \···\ xn K + V (x + V + V ) (x + V + V ) ✓ i xi ✓ i xi xi [ [ and none of the terms in the last union intersects C + V . The following lemma implies that if V is a neighborhood of 0 in a topo- logical vector space X, then for every x X it holds that x rV if r is large enough. A set V with this property is called2 absorbing. 2 Lemma A.2.2. Suppose V is a neighborhood of 0 in a topological vector space X and rn is a sequence of positive numbers tending to infinity. Then rnV = X. [ Proof. Fix x X. Then since V is open in X and λ λx from R to X is 2 7! continuous, λ : λx V is open in R. The set contains 0, and hence it contains { 2 } 1/rn for n large enough. This completes the proof. 36 Elements of functional analysis For an arbitrary absorbing subset A (for instance a neighborhood of 0) of a topological vector space X we define the Minkowsky functional µA : X [0, ) by ! 1 µ (x) = inf t>0:x/t A . A { 2 } Note that µA is indeed finite-valued, since A is absorbing. The following lemma collects properties that we need later. Lemma A.2.3. Let A be a convex, absorbing subset of a topological vector space X and let µA be its Minkowsky functional. (i) µ (x + y) µ (x)+µ (y) for all x, y X. A A A 2 (ii) µ (tx)=tµ (x) for all x X and t 0. A A 2 ≥ Proof. For x, y X and ">0, consider t = µA(x)+", s = µA(y)+".Then by definition of µ2 , x/t A and y/s A. Hence, the convex combination A 2 2 x + y t x s y = + s + t s + t t s + t s belongs to A as well. This proves (i). The proof of (ii) is easy. For the proof of the following characterization of continuous linear func- tionals we need the notion of a balanced neighborhood. A set B X is said to ✓ be balanced if ↵B B for every scalar ↵ R with ↵ 1. ✓ 2 | | Lemma A.2.4. Every neighborhood of 0 contains a balanced neighborhood of 0. Proof. Let U be a neighborhood of 0. Since scalar multiplication is continuous, there exists a δ>0 and a neighborhood V of 0 in X such that ↵V U whenever ✓ ↵ <δ.