Cambridge University Press 978-1-108-41570-5 — Finite Elements Sasikumaar Ganesan , Lutz Tobiska Excerpt More Information chapter 1 ........................................................................................................ INTRODUCTION TO SOBOLEV SPACES ........................................................................................................ In thischapterwe recall some basics on functional analysisandprovideabrief introduction to Sobolev spaces. For amore detailed and comprehensive study, we refer to Adams (1975). 1.1. Banach and hilbert spaces ................................................................................................................ Definition 1.1. Let X beareallinear space. Amapping : X R iscalled anormonX if · → x 0 x 0 for all x X, • = ⇔ = ∈ λx λ x for all x X, λ R, • =| | ∈ ∈ x y x y for all x,y X. • + ≤ + ∈ The pair (X, ) iscalled anormed space. · Setting y x inthe inequality and using theother two properties weget x 0 for all =− ≥ x X. ∈ Definition 1.2. Asequence (xn)n N iscalled Cauchy sequence, iffor all ε>0thereexists an ∈ index n0(ε)such that for all m,n > n0(ε) it holds xm xn <ε. − Definition 1.3. Asequence (xn)n N converges to x X iffor all ε>0there isanindex n0(ε) ∈ ∈ such that for all n > n0(ε) it holds xn x <ε. − Definition 1.4. Anormed space (X, ) iscalled complete if every Cauchy sequence in X · converges in X. Definition 1.5. Acomplete normed space iscalledBanach space. © in this web service Cambridge University Press www.cambridge.org Cambridge University Press 978-1-108-41570-5 — Finite Elements Sasikumaar Ganesan , Lutz Tobiska Excerpt More Information 2 introduction to sobolev spaces Example 1.1. Let Rd, d isthe dimension, beanopen andboundeddomain. We denote by ⊂ Lp(), 1 p < ,thesetof all measurable functions f : R forwhich ≤ ∞ → f (x) p dx < . | | ∞ Similarly,thesetof all measurable functions f : R satisfying → esssup f (x) : x < {| | ∈ } ∞ will be denotedbyL (). Then, Lp(), p [1, ] isareallinear space.Setting for 1 p < ∞ ∈ ∞ ≤ ∞ 1/p p f p : f (x) dx and f : esssup f (x) : x , L () = | | L∞( ) = {| | ∈ } respectively, Lp() becomesanormed space for any p [1, ] ifwe identify functions which ∈ ∞ areequal up toasetof measurezero. This identification is necessary because weonly have for a set M of measurezero ⊂ f (x) p dx 0 f (x) 0 for all x M. | | = ⇒ = ∈ \ Strictly speaking, theelements in Lp(), p [1, ]areclasses of equivalent functions (equal up ∈ ∞ to a set of measurezero). Then, thefirst condition in Definition 1.1 becomes true. Thesecond condition follows directly from the definition of p and thethird condition is just the ·L () inequality stated in Lemma 1.1. Lemma 1.1 (Minkowski’s inequality). For f ,g Lp(),p [1, ],wehave ∈ ∈ ∞ f g p f p g p . + L () ≤ L () + L () Theorem 1.1. Lp(),p [1, ], is a Banach space. ∈ ∞ Lemma 1.2 (Hölders’s inequality). For f Lp() and g Lq() with p,q [1, ], 1/p ∈ ∈ ∈ ∞ + 1/q 1,wehavefg L1() and = ∈ fg 1 f p g q . L () ≤ L () L ( ) Definition 1.6. Twonorms 1 and 2 onanormed space X arecalled equivalent if there · · exist positive constants C1, C2 such that C1 x 1 x 2 C2 x 1 for all x X. ≤ ≤ ∈ Note that topologicalproperties do not change when switching to an equivalent norm, for example,asequence isconvergent in(X, 1) iff it isconvergent in(X, 2). · · Definition 1.7. Let (X, X ) beanormed space. Amapping g : X R iscalledlinear if · → g(αx βy) αg(x) βg(y) for all α,β R, x,y X. + = + ∈ ∈ A linear mapping g : X R iscontinuous if there is a constant C such that → g(x) C x X for all x X. | |≤ ∈ © in this web service Cambridge University Press www.cambridge.org Cambridge University Press 978-1-108-41570-5 — Finite Elements Sasikumaar Ganesan , Lutz Tobiska Excerpt More Information 1.1. banach and hilbert spaces 3 Definition 1.8. We define the sum of twocontinuous linearfunctionals, g1 and g2,and the multiplication of acontinuous linearfunctional g with a real number α by (g1 g2)(x): g1(x) g2(x)and (αg)(x): αg(x), α R, x X. + = + = ∈ ∈ Then, it can beshownthat thesetof all continuous linearfunctionalscreate a reallinear space, the dual space X . In the following, for theelements g X we use the notation g,x : g(x). ∗ ∈ ∗ = Lemma 1.3. The set X of continuous linear functionals x g,x on X is a Banach space with ∗ → respect to the norm g,x g X∗ : sup | |. = 0 x X x X = ∈ Proof. Using the definition of the dual norm X , wesee · ∗ g X 0 g,x 0 for all x X g 0. ∗ = ⇔ = ∈ ⇔ = Further, we have λg,x λ g,x g,x λg X∗ sup | | sup | | λ sup | | λ g X∗ = 0 x X x X = 0 x X x X =| | 0 x X x X =| | = ∈ = ∈ = ∈ andfor the sum of two functionals it holds g1 g2,x g1,x g2,x g1 g2 X∗ sup | + | sup | + | + = 0 x X x X = 0 x X x X = ∈ = ∈ g1,x g2,x sup | | sup | | g1 X∗ g2 X∗ ≤ 0 x X x X + 0 x X x X = + = ∈ = ∈ It remains to show that (X∗, X ) iscomplete. Let (gn)n N beaCauchy sequence in X∗,that · ∗ ∈ is, for all ε>0there isanindex n0(ε)such that for all m,n > n0(ε) it holds gm gn X <ε. − ∗ Then, for afixed x X,thesequence ofreal numbers(gn,x )n N isaCauchy sequence in R ∈ ∈ due to gm,x gn,x gm gn,x gm gn X x X <ε x X | − | = | − | ≤ − ∗ for all m,n > n0(ε). Any Cauchy sequence ofreal numbers isconvergent in R,that means limn gn,x g(x) for all x X. The linearity of g follows from →∞ = ∈ g(αx βy) lim gn,αx βy lim α gn,x β gn,y n n + = →∞ + = →∞ + α lim gn,x β lim gn,y αg(x) βg(y). n n = →∞ + →∞ = + Concerning thecontinuity of g we recall that Cauchy sequences are bounded,thus gn,x gn X x X C x X for all x X. | | ≤ ∗ ≤ ∈ The limit n showsthat the linear mapping g : X R iscontinuous. →∞ → © in this web service Cambridge University Press www.cambridge.org Cambridge University Press 978-1-108-41570-5 — Finite Elements Sasikumaar Ganesan , Lutz Tobiska Excerpt More Information 4 introduction to sobolev spaces Definition 1.9. Let V bealinear space. Amapping ( , ):V V R iscalled an innerproduct · · × → on V if (u,v) (v,u) for all u,v V, • = ∈ (αu βv,w) α(u,w) β(v,w) for all u,v,w V, α,β R, • + = + ∈ ∈ (u,u) > 0 for u 0. • = AnormonV is inducedbysetting u : √(u,u). If (V, V ) iscomplete wecall V aHilbert = · space. Lemma 1.4 (Schwarz inequality). LetVbeaHilbertspace.Then, (u,v) u V v V for all u,v V. | |≤ ∈ Example 1.2. V L2()equippedwith the innerproduct = (f ,g): f (x)g(x)dx = isaHilbertspace. Theorem 1.2 (Riesz representation theorem). Let V be a Hilbert space. There is a unique linear mapping R : V V from the dual space V into the Hilbert space V such that for all g V and ∗ → ∗ ∈ ∗ v V ∈ (Rg,v) g,v and Rg V g V . = = ∗ Fixedpoint theorems areoften used to establish theunique solvability of an abstract oper- ator equation. Let us consider the following problem inaBanach space V with an operator P : V V: → Find u V such that u Pu. (1.1) ∈ = Solutions u V ofProblem (1.1) arecalled fixedpoints of themapping P : V V. ∈ → Definition 1.10. Themapping P : V V iscalled contractive or to beacontraction if there is → a positive constant ρ<1such that Pv1 Pv2 V ρ v1 v2 V for all v1,v2 V. − ≤ − ∈ Theorem 1.3 (Banach’sfixedpoint theorem). Let V be a Banach space and P : V Vbea → contraction. Then, we have the following statements: (1) There is a unique fixed point u V solving Problem (1.1). ∗ ∈ (2) The sequence un Pun 1,n 1, converges to the fixed point u∗ V for any initial guess = − ≥ ∈ u0 V. ∈ (3) It holds the error estimate ρn un u∗ V Pu0 u0 V for all n N. − ≤ 1 ρ − ∈ − © in this web service Cambridge University Press www.cambridge.org Cambridge University Press 978-1-108-41570-5 — Finite Elements Sasikumaar Ganesan , Lutz Tobiska Excerpt More Information 1.2. weak derivatives 5 Proof. We start by estimating the distance of ui V to ui 1 V ∈ − ∈ 2 ui ui 1 V Pui 1 Pui 2 V ρ ui 1 ui 2 V ρ ui 2 ui 3 V − − = − − − ≤ − − − ≤ − − − . ≤ i 1 ui ui 1 V ρ − Pu0 u0 V for all i N. − − ≤ − ∈ From that wegetfor m > n 1 ≥ m m um un V (ui ui 1) ui ui 1 V − = − − ≤ − − i n 1 V i n 1 =+ =+ m m n i 1 n 1 ρ − ρ − Pu0 u0 V ρ − Pu0 u0 V ≤ − ≤ 1 ρ − i n 1 =+ − ρn Pu0 u0 V 0 for n . ≤ 1 ρ − → →∞ − Therefore, for any ε>0, thereexists an index bound n0(ε)such that for all m > n n0(ε) ≥ ρn um un V Pu0 u0 V <ε, − ≤ 1 ρ − − consequently,(un) is a Cauchy sequence and convergestosomeu V. We show that u isa ∗ ∈ ∗ fixedpoint of P. Indeed, u∗ Pu∗ V u∗ un V Pun 1 Pu∗ V − ≤ − + − − u∗ un V ρ un 1 u∗ V 0 for n . ≤ − + − − → →∞ Next we prove that there is at most one fixedpoint. Assume that therearetwofixedpoints u u . Then, weconclude 1∗ = 2∗ u∗ u∗ V Pu∗ Pu∗ V ρ u∗ u∗ V (1 ρ) u∗ u∗ V 0. 1 − 2 = 1 − 2 ≤ 1 − 2 ⇒ − 1 − 2 ≤ Since ρ<1 wegetu u . Finally, we recall the estimate for m > n 1 1∗ = 2∗ ≥ ρn un u∗ V un um V um u∗ V Pu0 u0 V um u∗ V − ≤ − + − ≤ 1 ρ − + − − from which the last statement follows by m .