1.3 Hilbert space methods 25 1.3 Hilbert space methods One of the most often used theorems of functional analysis is the Riesz rep- resentation theorem. Theorem 1.30 (Riesz representation theorem). If x∗ is a continuous linear functional on a Hilbert space H, then there is exactly one element y 2 H such that < x∗; x >= (x; y): (1.32) 1.3.1 To identify or not to identify{the Gelfand triple Riesz theorem shows that there is a canonical isometry between a Hilbert space H and its dual H∗. It is therefore natural to identify H and H∗ and is done so in most applications. There are, however, situations when it cannot be done. Assume that H is a Hilbert space equipped with a scalar product (·; ·)H and that V ⊂ H is a subspace of H which is a Hilbert space in its own right, endowed with a scalar product (·; ·)V . Assume that V is densely and continuously embedded in H that is V = H and kxkH ≤ ckxkV , x 2 V , for some constant c. There is a canonical map T : H∗ ! V ∗ which is given by restriction to V of any h∗ 2 H∗: ∗ ∗ < T h ; v >V ∗×V =< h ; v >H∗×H ; v 2 V: We easily see that ∗ ∗ kT h kV ∗ ≤ Ckh kH∗ : Indeed ∗ ∗ ∗ kT h kV ∗ = sup j < T h ; v >V ∗×V j = sup j < h ; v >H∗×H j kvkV ≤1 kvkV ≤1 ∗ ∗ ≤ kh kH∗ sup kvkH ≤ ckh kH∗ : kvkV ≤1 ∗ ∗ Further, T is injective. For, if T h1 = T h2, then ∗ ∗ ∗ ∗ 0 =< T h1 − T h2; v >V ∗×V =< h1 − h2; v >H∗×H for all v 2 V and the statement follows from density of V in H. Finally, the image of TH∗ is dense in V ∗. Indeed, let v 2 V ∗∗ be such that < v; T h∗ >= 0 for all h∗ 2 H∗. Then, by reflexivity, ∗ ∗ ∗ ∗ ∗ 0 =< v; T h >V ∗∗×V ∗ =< T h ; v >V ∗×V =< h ; v >H∗×H ; h 2 H implies v = 0. Now, if we identify H∗ with H by the Riesz theorem and using T as the canonical embedding from H∗ into V ∗, one writes 26 1 Basic Facts from Functional Analysis and Banach Lattices V ⊂ H ' H∗ ⊂ V ∗ and the injections are dense and continuous. In such a case we say that H is the pivot space. Note that the scalar product in H coincides with the duality pairing < ·; · >V ∗×V : (f; g)H =< f; g >V ∗×V ; f 2 H; g 2 V: Remembering now that V is a Hilbert space with scalar product (·; ·)V we see that identifying also V with V ∗ would lead to an absurd { we would have V = H = H∗ = V ∗. Thus, we cannot identify simultaneously both pairs. In such situations it is common to identify the pivot space H with its dual H∗ bur to leave V and V ∗ as separate spaces with duality pairing being an extension of the scalar product in H. An instructive example is H = L2([0; 1]; dx) (real) with scalar product 1 Z (u; v) = u(x)v(x)dx 0 and V = L2([0; 1]; wdx) with scalar product 1 Z (u; v) = u(x)v(x)w(x)dx; 0 where w is a nonnegative bounded measurable function. Then it is useful to ∗ −1 identify V = L2([0; 1]; w dx) and 1 1 Z Z p g(x) < f; g >V ∗×V = f(x)g(x)dx ≤ f(x) w(x) dx ≤ kfkV kgkV ∗ : pw(x) 0 0 1.3.2 The Radon-Nikodym theorem Let µ and ν be finite nonnegative measures on the same σ-algebra in Ω. We say that ν is absolutely continuous with respect to µ if every set that has µ-measure 0 also has ν measure 0. Theorem 1.31. If ν is absolutely continuous with respect to µ then there is an integrable function g such that Z ν(E) = gdµ, (1.33) E for any µ-measurable set E ⊂ Ω. 1.3 Hilbert space methods 27 Proof. Assume for simplicity that µ(Ω); ν(Ω) < 1. Let H = L2(Ω; dµ + dν) on the field of reals. Schwarz inequality shows that if f 2 H, then f 2 L1(dµ+ dν). The linear functional Z < x∗; f >= fdµ Ω To be completed 1.3.3 Projection on a convex set Corollary 1.32. Let K be a closed convex subset of a Hilbert space H. For any x 2 H there is a unique y 2 K such that kx − yk = inf kx − zk: (1.34) z2K Moreover, y 2 K is a unique solution to the variational inequality (x − y; z − y) ≤ 0 (1.35) for any z 2 K. Proof. Let d = inf kx − zk. We can assume z2 = K and so d > 0. Consider z2K f(z) = kx − zk, z 2 K and consider a minimizing sequence (zn)n2N, zn 2 K such that d ≤ f(zn) ≤ d + 1=n. By the definition of f,(zn)n2N is bounded and thus it contains a weakly convergent subsequence, say (ζn)n2N. Since K is closed and convex, by Corollary 1.24, ζn * y 2 K. Further we have 1 j(h; x−y)j = lim j(h; x−ζn)j ≤ khk lim inf kx−ζnk ≤ khk lim inf d+ = khkd n!1 n!1 n!1 n for any h 2 H and thus, taking supremum over khk ≤ 1, we get f(y) ≤ d which gives existence of a minimizer. To prove equivalence of (1.35) and (1.34) assume first that y 2 K satisfies (1.34) and let z 2 K. Then, from convexity, v = (1 − t)y + tz 2 K for t 2 [0; 1] and thus kx − yk ≤ kx − ((1 − t)y + tz)k = k(x − y) − t(z − y)k and thus kx − yk2 ≤ kx − yk2 − 2t(x − y; z − y) + t2kz − yk2: Hence tkz − yk2 ≥ 2(x − y; z − y) for any t 2 (0; 1] and thus, passing with t ! 0, (x − y; z − y) ≤ 0. Conversely, assume (1.35) is satisfied and consider 28 1 Basic Facts from Functional Analysis and Banach Lattices kx − yk2 − kx − zk2 = (x − y; x − y) − (x − z; x − z) = 2(x; z) − 2(x; y) + 2(y; y) − 2(y; z) + 2(y; z) − (y; y) = 2(x − y; z − y) − (y − z; y − z) ≤ 0 hence kx − yk ≤ kx − zk for any z 2 K. For uniqueness, let y1; y2 satisfy (x − y1; z − y1) ≤ 0; (x − y2; z − y2) ≤ 0; z 2 H: Choosing z = y2 in the first inequality and z = y1 in the second and adding 2 them, we get jy1 − y2j ≤ 0 which implies y1 = y2. We call the operator assigning to any x 2 K the element y 2 K satisfying (1.34) the projection onto K and denote it by Pk. Proposition 1.33. Let K be a nonempty closed and convex set. Then PK is non expansive mapping. Proof. Let yi = PK xi, i = 1; 2: We have (x1 − y1; z − y1) ≤ 0; (x2 − y2; z − y2) ≤ 0; z 2 H so choosing, as before, z = y2 in the first and z = y1 in the second inequality and adding them together we obtain 2 ky1 − y2k ≤ (x1 − x2; y1 − y2); hence kPK x1 − PK x2k ≤ kx1 − x2k. 1.3.4 Theorems of Stampacchia and Lax-Milgram 1.3.5 Motivation Consider the Dirichlet problem for the Laplace equation in Ω ⊂ Rn −∆u = f in Ω; (1.36) uj@Ω = 0: (1.37) Assume that there is a solution u 2 C2(Ω) \ C(Ω¯). If we multiply (1.37) by a 1 test function φ 2 C0 (Ω) and integrate by parts, then we obtain the problem Z Z ru · rφdx = fφdx. (1.38) Ω Ω 1.3 Hilbert space methods 29 Conversely, if u satisfies (1.38), then it is a distributional solution to (1.37). Moreover, if we consider the minimization problem for 1 Z Z J(u) = jruj2dx − fudx 2 Ω Ω 2 over K = fu 2 C (Ω); uj@Ω = 0g and if u is a solution to this problem then 1 for any > 0 and C0 (Ω) we have J(u + φ) ≥ J(u); then we also obtain (1.38). The question is how to obtain the solution. In a similar way, we consider the obstacle problem, to minimize J over 2 K = fu 2 C (Ω); uj@Ω = 0; u ≥ gg over some continuous function g satisfying gj@Ω < 0. Note that K is convex. Again, if u 2 K is a solution then for any > 0 and φ 2 K we obtain that u + (φ − u) = (1 − )u + φ is in K and therefore J(u + (φ − u)) ≥ J(u): Here, we obtain only Z Z ru · r(φ − u)dx ≥ f(φ − u)dx: (1.39) Ω Ω for any φ 2 K. For twice differentiable u we obtain Z Z ∆u(φ − u)dx ≥ f(φ − u)dx Ω Ω 1 and choosing φ = u + , 2 C0 (Ω) we get −∆u ≥ f almost everywhere on Ω. As u is continuous, the set N = fx 2 Ω; u(x) > 1 g(x)g is open. Thus, taking 2 C0 (N), we see that for sufficiently small > 0, u ± φ 2 K. Then, on N −∆u = f Summarizing, for regular solutions the minimizer satisfies −∆u ≥ f u ≥ g (∆u + f)(u − g) = 0 on Ω. 30 1 Basic Facts from Functional Analysis and Banach Lattices Hilbert space theory We begin with the following definition.