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1.3 Hilbert Space Methods 25 1.3 Hilbert Space Methods 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.
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