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Fact Sheet Functional Analysis Fact Sheet Functional Analysis Literature: Hackbusch, W.: Theorie und Numerik elliptischer Differentialgleichungen. Teubner, 1986. Knabner, P., Angermann, L.: Numerik partieller Differentialgleichungen. Springer, 2000. Triebel, H.: H¨ohere Analysis. Harri Deutsch, 1980. Dobrowolski, M.: Angewandte Funktionalanalysis, Springer, 2010. 1. Banach- and Hilbert spaces Let V be a real vector space. Normed space: A norm is a mapping k · k : V ! [0; 1), such that: kuk = 0 , u = 0; (definiteness) kαuk = jαj · kuk; α 2 R; u 2 V; (positive scalability) ku + vk ≤ kuk + kvk; u; v 2 V: (triangle inequality) The pairing (V; k · k) is called a normed space. Seminorm: In contrast to a norm there may be elements u 6= 0 such that kuk = 0. It still holds kuk = 0 if u = 0. Comparison of two norms: Two norms k · k1, k · k2 are called equivalent if there is a constant C such that: −1 C kuk1 ≤ kuk2 ≤ Ckuk1; u 2 V: If only one of these inequalities can be fulfilled, e.g. kuk2 ≤ Ckuk1; u 2 V; the norm k · k1 is called stronger than the norm k · k2. k · k2 is called weaker than k · k1. Topology: In every normed space a canonical topology can be defined. A subset U ⊂ V is called open if for every u 2 U there exists a " > 0 such that B"(u) = fv 2 V : ku − vk < "g ⊂ U: Convergence: A sequence vn converges to v w.r.t. the norm k · k if lim kvn − vk = 0: n!1 1 A sequence vn ⊂ V is called Cauchy sequence, if supfkvn − vmk : n; m ≥ kg ! 0 for k ! 1. A normed space is called complete if every Cauchy sequence converges to an element v 2 V .A Banach space is a complete normed space. Bilinear form: A bilinear form is a mapping (·; ·): V × V ! R which is linear in each argument: a(u; v1 + λv2) = a(u; v1) + λa(u; v2); u; v1; v2 2 V; λ 2 R; a(u1 + λu2; v) = a(u1; v) + λa(u2; v); u1; u2; v 2 V; λ 2 R: The bilinear form (·; ·) is called • symmetric, if (u; v) = (v; u); u; v 2 V , • positive, if (v; v) ≥ 0; v 2 V , • definite, if (u; u) = 0 , u = 0. Hilbert space: A symmetric, positive definite bilinear form is called inner product.A inner product induces a norm on V by kuk := p(u; u). If the bilinear form is not definite, it only induces a seminorm. A space which is endowed with a inner product and complete with respect to the induced norm is called Hilbert space. Important properties of a (semi-)norm which is induced by a inner product: a) Cauchy{Schwarz-inequality: j(u; v)j ≤ kuk kvk; u; v 2 V; b) parallelogram equation: ku − vk2 + ku + vk2 = 2(kuk2 + kvk2) hold. c) Any norm fulfilling the parallelogram equation is induced by a inner product: 1 (u; v) = (ku + vk2 − ku − vk2): 4 2. Sobolev spaces Lipschitz domain: The domain Ω is said to have a Lipschitz boundary if for some N 2 N there are open sets U1;U2;:::;UN such that 2 SN a) @Ω ⊂ i=1 Ui b) For each i = 1;:::;N the part of the boundary @Ω \ Ui can be represented as a graph of a function which is Lipschitz continuous. A domain Ω with a Lipschitz boundary is called Lipschitz domain. Example: The following domain is an example of a domain without a Lipschitz boundary Now, let Ω ⊂ Rd be a bounded domain with Lipschitz boundary Γ = @Ω. 1 Lloc(Ω) is the linear space containing all functions which are Lebesgue integrable on every compact subset of Ω. L2(Ω) is the linear space containing all functions which are Lebesgue square integrable on Ω. Endowed with the inner product Z (u; v)0 = uv Ω and the induced norm p kvk0 = (v; v)0; it is a Hilbert space. 1 1 Weak derivative: Consider a function u 2 Lloc(Ω). We say that v 2 Lloc(Ω) is the αth-weak derivative of u, if Z Z jαj α 1 v' = (−1) uD '; ' 2 C0 (Ω): Ω Ω We use the notation @jαj' Dα' = ; α1 αn @x1 : : : @xn P 1 for a multi-index α = (α1 : : : αn), where jαj = αi 2 N0. C0 (Ω) is the linear space of all infinitely differentiable functions with a compact support in Ω. We denote Dαu := v. Sobolev-space: For k 2 N0, set Hk(Ω) := fv 2 L2(Ω) : Dαv 2 L2(Ω); jαj ≤ kg: It holds: H0(Ω) = L2(Ω): Example: 1 s Consider Ω = B(0; 2 ), s 2 R and u(x) = jxj . 3 • For s = 1 we have u 2 H1(Ω) and the first weak derivative is given by ( −1; −0:5 < x < 0 Du(x) = : 1; 0 < x < 0:5 0.5 1.5 0.45 s = 1 ; alpha = 1 s = 1 ; alpha = 0 1 0.4 0.35 0.5 0.3 0.25 0 0.2 −0.5 0.15 0.1 −1 0.05 0 −1.5 0 100 200 300 400 500 600 700 800 900 1000 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 1 1 • For s = 2 we have u2 = H (Ω) and the first weak derivative is given by Du(x) = jxj−0:5 2= L2(Ω): 0.8 45 0.7 s = 0.5; alpha = 0 40 s = 0.5 ; alpha = 1 35 0.6 30 0.5 25 0.4 20 0.3 15 0.2 10 0.1 5 0 0 0 100 200 300 400 500 600 700 800 900 1000 0 100 200 300 400 500 600 700 800 900 1000 The linear space Hk(Ω) is a Hilbert space endowed with the inner product X α α (u; v)k = (D u; D v)0; jα|≤k and the induced norm p kvkk = (v; v)k: Examples: 2 0 2 R 2 L orH -norm : kuk0 = Ω u ; 0 2 2 H -seminorm : juj0 = kuk0; 1 2 2 2 H -norm : kuk1 = kuk0 + kruk0; 1 2 2 H -seminorm : juj1 = kruk0; k 2 2 P α 2 H -norm : kukk = kukk−1 + kD uk0: jαj=k 4 Important dense embedding: The set C1(Ω)\Hk(Ω) is dense in Hk(Ω). Thus Hk(Ω) 1 2 is the completion of the linear space fv 2 C (Ω) : kvkk < 1g in L (Ω) with respect to the norm k · kk. k Zero-boundary conditions: The linear space H0 (Ω) is defined as the completion of the 1 2 space C0 (Ω) in L (Ω) with respect to the norm k · kk. Poincar´e-Friedrichs-inequality: There exists a constant C such that k kvkk−1 ≤ Cjvjk; v 2 H0 (Ω): k Hence in H0 (Ω) the norm k · kk and the seminorm j · jk are equivalent. Further norm-equivalences: Z 2 Z 2 2 2 2 2 kvk1 ∼ jvj1 + v ; kvk1 ∼ jvj1 + v ; Ω ΓD where ΓD ⊂ Γ has a positive measure. 3. Trace-, Extensions- and Embedding theorems, Dual spaces Sobolev-Slobodeckij-space: For s > 0, s 62 N0, where s = k + t, k 2 N0, 0 < t < 1, define s k α H (Ω) = fv 2 H (Ω) : It(D v) < 1; jαj = kg; Z Z (w(x) − w(y))2 where I (w) = dxdy: t kx − ykd+2t Ω Ω Example: 1; x > 0 Consider Ω = (−1; 1) and v(x) = 0; x ≤ 0: Question: For which s ≥ 0 is v 2 Hs(Ω) fulfilled? We already know v 2 H0(Ω); v2 = H1(Ω). 5 Z 1 Z 1 (v(x) − v(y))2 Is(v) = 1+2s dx dy −1 −1 jx − yj Z 0 Z 0 Z 0 Z 1 Z 1 Z 0 Z 1 Z 1 = ::: + ··· + ··· + ::: −1 −1 −1 0 0 −1 0 0 | {z } | {z } =0 =0 Z 0 Z 1 1 = 2 1+2s dx dy −1 0 jx − yj Z 0 1 −1 1 = 2 2s j0 dy −1 2s (x − y) 1 Z 0 1 Z 0 1 = − 2s dy − 2s dy : s −1 j1 − yj −1 jyj 1 s 1 Thus Is(v) < 1 for s < 2 and v 2 H (Ω) for all 0 ≤ s < 2 . Trace theorem: There is a unique linear mapping tr : H1(Ω) ! H1=2(Γ) 1 ¯ such that tr (u) = ujΓ; for all u 2 H (Ω) \ C(Ω): Remarks: 1 • For u 2 H (Ω), there exists a sequence un ⊂ C(Ω) such that un ! u; tr (un) ! tr (u). • The mapping tr is surjective (onto H1=2(Γ)), but not injective. • For s > 1=2 we have tr : Hs(Ω) ! Hs−1=2(Γ). 1 1 • H0 (Ω) = fv 2 H (Ω) : tr v = 0g. Example: Consider 2 4 Ω = (x; y) 2 R j 0 < x < 1 0 < y < x Γ = (0; 1) × f0g: Ω is not a Lipschitz domain. For u :Ω ! R, 1 u(x; y) = ; x it holds u2 = L2(Γ), but u 2 H1(Ω), as Z Z 1 2 1 u (x; y)dx dy = 2 dx Γ 0 jxj Z 1 1 1 1 = lim dx = lim − = 1; !0 2 !0 x x 6 and Z Z 1 Z x4 2 1 2 jDu(x; y)j dx dy = j 2 j dy dx Ω 0 0 x Z 1 Z 1 1 2 4 = j 2 j x dx = 1 dx = 1 < 1: 0 x 0 We see, that the trace theorem is not valid! Extension theorem: For any u 2 H1=2(Γ) there is a v 2 H1(Ω) such that tr (v) = u and kvk1;Ω ≤ Ckuk1=2;Γ Remarks: • The extension operator is not unique.
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