Introduction to Sobolev Spaces

Home , Sobolev space

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.

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. | |≤ ∈

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. →∞ →

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 operator 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 − ∈ −

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 . →∞ 1.2. Weak derivatives ......

First we introduceageneralization of the integration by parts formula known for smooth functions u,v :[a,b] R: → b b b u′v dx uv uv′ dx. = |a − a a

6 introduction to sobolev spaces

We start with acharacterization of geometric properties of a domain Rd. ⊂ Definition 1.11. The domain Rd issaid to haveaLipschitz continuous boundary, iffor ⊂ every point of the boundary Ŵ ∂ thereexists a local coordinate system in which the bound- = ary correspondstosomehypersurface with the domain lying on one sideof that surface and ∂ can locally be representedbythegraph of aLipschitz continuous mapping. Theorem 1.4 (Gauss theorem). Let Rd be a bounded domain with Lipschitz continuous ⊂ boundary Ŵ and outer normal n. Then, for w C1()d we have ∈ d ∂w divwdx i dx w ndγ . = ∂x = · i 1 i Ŵ = Let j 1,2,...,d beafixed index. The formulaof integration by parts follows by setting ∈{ } wi uv for i j and wi 0 for i j,and using the product rule. = = = = Lemma 1.5 (Integration by parts). For u,v C1() we have ∈ ∂u ∂v v dx uvnj dγ u dx, j 1,...,d. ∂x = − ∂x = j Ŵ j Nowwe have thetoolstogeneralize theconceptofdifferentiability. Definition 1.12. Thesupportof a function f : R is definedby → suppf x : f (x) 0 . = { ∈ = } Thesetof all functions that are differentiableof any orderwith compact support in will be denotedbyC0∞(). For (0,1), the function x x(1 x) does not belong to C () because its support = → − 0∞ is [0,1] (0,1). Indeed, functions in C ()arezero intheneighbourhood of the boundary ⊂ 0∞ ∂. Asaconsequence, a function f C () can beextended to a smoothfunction defined on ∈ 0∞ Rd by setting f 0outsideof . = Definition 1.13. We define

L1 (): f : R : f L1(A) for all compact A . loc ={ → ∈ ⊂ } Forboundeddomains , we have L2() L1() L1 (). ⊂ ⊂ loc Example 1.3. Consider (0,1). Themapping x 1/x belongs to L1 () but not to L1(). = → loc Themapping x 1/√x belongs to L1() but not to L2(). → Deﬁnition 1.14. Let α (α1,...,αd) beamulti-index, that is, αi are non-negative integersand = 1 α 1 α α1 α . A function, f L () has a weakderivative v D f L () iffor any | |= +···+ d ∈ loc = ∈ loc ϕ C () ∈ 0∞ α α v ϕ dx ( 1)| | fD ϕ dx. = −

1.3. sobolev spaces 7

The notion of a weakderivative generalizes theclassicalpartialderivatives. Indeed,apartial derivative intheclassical sense satisﬁes theabove condition by partial integration. Example 1.4. Let f :( 1, 1) R with f (x) 2 √ x . Then, thegeneralizedderivative − + → = − | | Df L1 ( 1, 1) isgiven by ∈ loc − + sgnx Df (x) x ( 1,0) (0, 1). =−2√ x ∈ − ∪ + | | Proof. Applying integration by parts separately inthetwosubintervals, wegetfor ϕ C ∈ 0∞ ( 1, 1) − + 1 sgnx 0 1 1 1 + ϕ dx ϕ dx + ϕ dx − 1 2√ x = 1 2√ x − 0 2√ x − | | − | | | | 0 0 2 x ϕ x 1 2 x ϕ′ dx = − | | | =− − 1 − | | − 1 1 + 2 x ϕ + 2 x ϕ′ dx + − | | |x 0 − − | | = 0 1 + 2 x ϕ′ dx, =− 1 − | | − where we used thecontinuity of ϕ at x 0and ϕ( 1) 0. = ± = 1.3. Sobolev spaces ......

Deﬁnition 1.15. We denote by W m,p(), 1 p , m 0, thesetof all functions, f Lp() ≤ ≤∞ ≥ ∈ withweakderivatives Dαf Lp()up to theorder α m. ThesetsW m,p()arecalledSobolev ∈ | |≤ spaces. Lemma 1.6. The Sobolev space W m,p() equipped with the norm 1/p α p f m,p : D f (x) dx if 1 p < W () = ⎛ | | ⎞ ≤ ∞ α m | |≤ ⎝ ⎠ and

α f W m,p() : max ess supx D f (x) if p = α m ∈ | | =∞ | |≤ is a Banach space. In particular, the Sobolev space Hm(): W m,2() is a Hilbert space with = respect to the inner product

α α α α m (u,v) m : (D u,D v) D uD v dx for all u,v H (). H ( ) = = ∈ α m α m | |≤ | |≤

8 introduction to sobolev spaces

Proof. Here, wegive thearguments for 1 p < . For the case p , we refer to Adams ≤ ∞ =∞ (1975). ThesetsW m,p()are reallinear spaces (belowweshow that the sum (f g)of two + elements f ,g W m,p() isanelement of W m,p()). Next we havetoshow that themapping ∈ f f m,p isanorm, that is, thethree conditions in Definition 1.1 hold. If f 0, then → W () = f m,p 0 by definition of m,p . Further,since 0 f m,p f p wecon- W () = ·W () = W () ≥ L () clude f 0 inthe sense of Lp(), that is, functions which areequal up to a set of measurezero = are identified. Thus, thefirst condition in Definition 1.1 holds. Thesecond condition follows directly from the definition

1/p 1/p α p p α p λf m,p D λf (x) dx λ D f (x) dx W () = ⎛ | | ⎞ = ⎛ | | | | ⎞ α m α m | |≤ | |≤ ⎝ ⎠ 1/p ⎝ ⎠ α p λ D f (x) dx λ f m,p . =| |⎛ | | ⎞ =| | W () α m | |≤ ⎝ ⎠ Let f ,g W m,p()and α m. TheMinkowski’s inequality, Lemma 1.1, implies that ∈ | |≤ Dα(f g) Dαf Dαg Lp()and + = + ∈ α α α D (f g) p D f p D g p . + L () ≤ L () + L () Using theMinkowski’s inequalityfor sums, weget

p α p α α p f g D (f g) D f p D g p + W m,p() = + Lp() ≤ L () + L () α m α m | |≤ | |≤ 1/p 1/p p Dαf p Dαg p ≤ ⎡⎛ Lp()⎞ + ⎛ Lp()⎞ ⎤ α m α m ⎢ | |≤ | |≤ ⎥ ⎣⎝ ⎠ p ⎝ ⎠ ⎦ f m,p g m,p ≤ W () + W () from which thethird condition follows. It remains to prove thecompleteness of thenormed m,p m,p α space W (). Let (fn)n N beaCauchy sequence in W (). Then, (D fn)n N isaCauchy ∈ ∈ sequence in Lp() for any multi-index α with α m.Since Lp() isaBanach space, there is α p α α p | |≤ α α 0 f L () with D fn f in L (). It remains to show that f D f where f f . We show ∈ → = = this for α (1,0,...,0). For any ϕ C () we have = ∈ 0∞ α α α α α α (f ϕ fD ϕ)dx f D fn ϕ dx (D fnϕ fnD ϕ)dx + ≤ | − || | + + α f fn D ϕ dx. + | − || | α Thesecond termonthe right hand sideequalszerosince D fn isthe weakderivative of fn. The α α p 1 convergence of fn to f and D fn to f in L (), respectively, imply theconvergence in L (), thus theﬁrst and third termonthe right hand sidetend to zero for n . The left hand side →∞

1.3. sobolev spaces 9

is non-negative and independent of n, consequently weconcludethat it iszero, which means f α isthe weakderivative of f . The properties for the innerproduct in Hm(), see Definition 1.9, followfrom its definition 2 m,2 and theobservation that (v,v) m v for all v W (). H ( ) = W m,2() ∈ niti n m,p Defi o 1.16. We denote by W0 ()theclosureof C0∞() inthenorm W m,p(). m m · Analogously, H0 () istheclosureof C0∞() in H ().

In the following, for f W m,p(ω), m 0, 1 p ,thenorm will beshortly denotedby ∈ ≥ ≤ ≤∞ f m,p,ω : f m,p and aseminorm is introducedby = W (ω) 1/p α p f m,p,ω : D f (x) dx if 1 p < | | = ⎛ | | ⎞ ≤ ∞ ω α m | |= ⎝ ⎠ and

α f m,p,ω : max ess supx D f (x) if p . | | = α m ∈ | | =∞ | |= In case, ω and/or p 2 weomit and p, respectively. Thus, = =

f m f m,2,, f m,p f m,p,, f m f m,2,, f m,p f m,p,. = = | | =| | | | =| |

Lemma 1.7 (Poincaré inequality). There is a positive constant cP 2a such that ≤ 1,p v 0,p cP v 1,p for all v W (). ≤ | | ∈ 0 Here, ‘a’ denotes the diameter of in an arbitrary but ﬁxed direction.

Proof. Suppose the Poincaré inequality istrue forfunctions from C0∞(). Then, for any 1,p 1,p v W ()there isasequence vn C () with vn v in W (). We have ∈ 0 ∈ 0∞ → 0

v 0,p v vn 0,p vn 0,p v vn 0,p cP vn 1,p ≤ − + ≤ − + | | (1 cP) v vn 1,p cP v 1,p. ≤ + − + | |

Since v vn 1,p tendstozero for n ,the Poincaré inequality istrue forfunctions from 1,p − →∞ W0 (). In order to prove the Poincaré inequalityforfunctions v C () weextend v : R by ∈ 0∞ → setting v(x) 0 for all x Rd . Let [ a, a] Rd 1,then = ∈ \ ⊂ − + × − x1 ∂v v(x) (s,x2,...,xd)ds = a ∂x1 −

10 introduction to sobolev spaces

and Hölder’s inequalitywith 1/p 1/q 1 implies + = / x1 p 1 p x1 1/q ∂v q v(x) (s,x2,...,xd) ds 1 ds , | | ≤ a ∂x1 a − − a p p p/q + ∂v v(x) (2a) (s,x2,... ,xd) ds, | | ≤ a ∂x1 − a a p + p 1 p/q + ∂v v(x) dx1 (2a) + (s,x2,...,xd) ds, | | ≤ ∂x a a 1 − − ∂v p v(x) p dx (2a)p (x) dx. Rd | | ≤ Rd ∂x 1 Using v(x) 0 for x Rd the inequalityfollows. = ∈ \ The Poincaré inequality allowstoshow that theseminorm m,p isequivalenttothenorm m,p m,p α 1,p |·| m,p on W (). Let v W (), then D v W () for α m 1. We apply · 0 ∈ 0 ∈ 0 | |≤ − successively the inequality α α D v 0,p cP D v 1,p for α 0,1,...,m 1 | | ≤ | | | |= − to estimate the lowerderivatives by the highest derivative and obtain

1/p α p v m,p v m,p D v C v m,p, | | ≤ = ⎛ | |0,p⎞ ≤ | | α m | |≤ ⎝ ⎠ that is, theequivalence of theseminormtothenorm. An important toolwill bethe Sobolev embedding theorems which state that functions from m,p m 1,p W () belong to W − ∗ for some p∗ > p. Moreover, it can beshownthat for suitable numbers m, p functions from W m,p() belong to classical spaces of continuous and continuously differentiable functions, respectively. Deﬁnition 1.17. We introduce thespace of Hölder continuous functions with Hölder exponent β (0,1] ∈

0,β 0 u(x) u(y) C (): u C (): sup | − β | < = ∈ x,y ,x y x y ∞ ∈ = | − | equippedwith thenorm u(x) u(y) v C0,β () sup u(x) sup | − β |. = x | |+x,y ,x y x y ∈ ∈ = | − | Hölder continuous functions with aHölder exponent β 1arealso called Lipschitz continu- = ous. Exercise 1.1. Consider f :[ 1, 1] R given by f (x) √ x .Prove that f C0,1/2([ 1, 1]) − + → = | | ∈ − + and that there isnoβ (1/2, 1] with f C0,β ([ 1, 1]). ∈ + ∈ − +