Fact Sheet Functional Analysis

Fact Sheet Functional Analysis Literature: Hackbusch, W.: Theorie und Numerik elliptischer Differentialgleichungen. Teubner, 986. Knabner, P., Angermann, L.: Numerik partieller Differentialgleichungen. Springer, 2. Triebel, H.: Höhere Analysis. Harri Deutsch, 98. Dobrowolski, M.: Angewandte Funktionalanalysis, Springer, 2.. Banach- and Hilbert spaces Let V be a real vector space. Normed space: A norm is a mapping : V [, ), such that: u = u =, (definiteness) αu = α u, α R, u V, (positive scalability) u + v u + v, u, v V. (triangle inequality) The pairing (V, ) is called a normed space. Seminorm: In contrast to a norm there may be elements u such that u =. It still holds u = if u =. Comparison of two norms: Two norms, 2 are called equivalent if there is a constant C such that: C u u 2 C u, u V. If only one of these inequalities can be fulfilled, e.g. u 2 C u, u V, the norm is called stronger than the norm 2. 2 is called weaker than. Topology: In every normed space a canonical topology can be defined. A subset U V is called open if for every u U there exists a ε > such that B ε (u) = {v V : u v < ε} U. Convergence: A sequence v n converges to v w.r.t. the norm if lim v n v =. n

A sequence v n V is called Cauchy sequence, if sup{ v n v m : n, m k} for k. A normed space is called complete if every Cauchy sequence converges to an element v 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, v + λv 2 ) = a(u, v ) + λa(u, v 2 ), u, v, v 2 V, λ R, a(u + λu 2, v) = a(u, v) + λa(u 2, v), u, u 2, v V, λ R. The bilinear form (, ) is called symmetric, if (u, v) = (v, u), u, v V, positive, if (v, v), v V, definite, if (u, u) = u =. Hilbert space: A symmetric, positive definite bilinear form is called inner product. A inner product induces a norm on V by u := (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: (u, v) u v, u, v V, b) parallelogram equation: u v 2 + u + v 2 = 2( u 2 + v 2 ) hold. c) Any norm fulfilling the parallelogram equation is induced by a inner product: (u, v) = 4 ( u + v 2 u v 2 ). 2. Sobolev spaces Lipschitz domain: The domain is said to have a Lipschitz boundary if for some N N there are open sets U, U 2,..., U N such that 2

a) N i= U i b) For each i =,..., N the part of the boundary U i can be represented as a graph of a function which is Lipschitz continuous. A domain with a Lipschitz boundary is called Lipschitz domain. The following domain is an example of a domain without a Lipschitz boundary Now, let R d be a bounded domain with Lipschitz boundary Γ =. L loc () is the linear space containing all functions which are Lebesgue integrable on every compact subset of. L 2 () is the linear space containing all functions which are Lebesgue square integrable on. Endowed with the inner product (u, v) = uv and the induced norm it is a Hilbert space. v = (v, v), Weak derivative: Consider a function u L loc (). We say that v L loc () is the α th -weak derivative of u, if vϕ = ( ) α ud α ϕ, ϕ C (). We use the notation D α α ϕ ϕ = x α,... x αn n for a multi-index α = (α... α n ), where α = α i N. C () is the linear space of all infinitely differentiable functions with a compact support in. We denote D α u := v. Sobolev-space: For k N, set H k () := {v L 2 () : D α v L 2 (), α k}. It holds: H () = L 2 (). Consider = B(, 2 ), s R and u(x) = x s. 3

For s = we have u H () and the first weak derivative is given by {,.5 < x < Du(x) =., < x <.5.5.5.45.4 s = ; alpha = s = ; alpha =.35.5.3.25.2.5.5..5 2 3 4 5 6 7 8 9.5.5.4.3.2...2.3.4.5 For s = 2 we have u / H () and the first weak derivative is given by Du(x) = x.5 / L 2 ()..8 45.7 s =.5; alpha = 4 s =.5 ; alpha =.6 35.5 3 25.4 2.3 5.2. 5 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 The linear space H k () is a Hilbert space endowed with the inner product (u, v) k = (D α u, D α v), and the induced norm Examples: α k v k = (v, v) k. L 2 orh -norm : u 2 = u2, H -seminorm : u 2 = u 2, H -norm : u 2 = u 2 + u 2, H -seminorm : u 2 = u 2, H k -norm : u 2 k = u 2 k + D α u 2. α =k 4

Important dense embedding: The set C () H k () is dense in H k (). Thus H k () is the completion of the linear space {v C () : v k < } in L 2 () with respect to the norm k. Zero-boundary conditions: The linear space H k () is defined as the completion of the space C () in L 2 () with respect to the norm k. Poincaré-Friedrichs-inequality: There exists a constant C such that v k C v k, v H k (). Hence in H k () the norm k and the seminorm k are equivalent. Further norm-equivalences: ( v 2 v 2 + v where Γ D Γ has a positive measure. ) 2 (, v 2 v 2 + Γ D v ) 2, 3. Trace-, Extensions- and Embedding theorems, Dual spaces Sobolev-Slobodeckij-space: For s >, s N, where s = k + t, k N, < t <, define H s () = {v H k () : I t (D α v) <, α = k}, (w(x) w(y)) 2 where I t (w) = dxdy. x y d+2t Consider = (, ) and v(x) = {, x >, x. Question: For which s is v H s () fulfilled? We already know v H (), v / H (). 5

I s (v) = =... + }{{} = 2 = 2 = s = (v(x) v(y)) 2 dx dy x y +2s + dx dy x y +2s 2s (x y) 2s dy ( dy y 2s ) dy. y 2s Thus I s (v) < for s < 2 and v Hs () for all s < 2. Trace theorem: There is a unique linear mapping Remarks: +... }{{} tr : H () H /2 (Γ) such that tr (u) = u Γ, for all u H () C( ). For u H (), there exists a sequence u n C() such that u n u, tr (u n ) tr (u). The mapping tr is surjective (onto H /2 (Γ)), but not injective. For s > /2 we have tr : H s () H s /2 (Γ). H () = {v H () : tr v = }. Consider = { (x, y) R 2 < x < < y < x 4} Γ = (, ) {}. is not a Lipschitz domain. For u : R, it holds u / L 2 (Γ), but u H (), as u 2 (x, y)dx dy = Γ u(x, y) = x, = lim ɛ ɛ x 2 dx dx = lim x2 ɛ = [ ] =, x ɛ 6

and Du(x, y) 2 dx dy = = x 4 x 2 2 dy dx x 2 2 x 4 dx = dx = <. We see, that the trace theorem is not valid! Extension theorem: For any u H /2 (Γ) there is a v H () such that Remarks: The extension operator is not unique. tr (v) = u and v, C u /2,Γ For s > /2 it maps u H s /2 (Γ) v H s (). Sobolev lemma: H s (R d ) C k (R d ), H s () C k (), s > k + d/2, s > k + d/2. Compact embedding: A Banach space V is said to be compactly embedded in a Banach space W (V K W ) if every sequence v n V, v n V contains a subsequence which is convergent in W. It holds: H s () K H t (), s, t R, s > t, H k () K H l (), k, l N, k > l. Dual space: Given a real vector space V, we call V the dual space, i.e. the space of all continuous, linear mappings from V onto R. If V is a normed space, the dual norm φ V = sup v V φ(v) v V defines a norm on V. The dual space of H s () is denoted by H s (). Riesz representation theorem: Every Hilbert space is isometrically (i.e. norms are preserved) isomorphic to its own dual space. 7

Sobolev spaces for second order, elliptic boundary value problems usually can be H () as the unlimited solution space, H /2 (Γ) for the Dirichlet boundary conditions and H /2 (Γ) for Neumann boundary conditions. 4. Lax-Milgram lemma Banach fixed point theorem: Let V be a Banach space and T : V V a contraction mapping, i.e. T v T w L v w, v, w V, where L [, ) is a constant. Then the map T admits a unique fixed-point. Continuity: A bilinear form a(, ) is called continuous if there is a constant M such that: a(u, v) M u v. A linear functional F : V R is called continuous if there is a constant C such that: F (v) C u. V -Ellipticity: A bilinear form a(, ) is called V -elliptic if there is a constant α > such that: a(v, v) α v 2. The bilinear form a(u, v) := u v dx is symmetric and continuous on H () H (). It is H elliptic as the Poincaré Friedrichs inequality a(v, v) = v 2 α v 2 in H() holds. However, the bilinear form is not H elliptic as a(, ) =. Let be a bounded domain; The bilinear form a(u, v) = u v dx + ( x 2 + ) u v dx is continuous on H () H () and H elliptic as we know: a(v, v) = v 2 + ( x 2 + ) v 2 dx v 2 + v 2 = v 2 }{{}. Consider = (, ) and a(u, v) = x 2 u v dx. 8

We would like to know, whether a(u, v) is L 2 elliptic. We are looking for some α > such that a(u, u) α u 2. Defining + nx if x n u n (x) = nx if < x n else. we see u n 2 = 2 a(u n, u n ) = 2 n We conclude, that there is no α > such that a(u, u) α u 2 ( nx) 2 dx = 2 3 n, x 2 ( nx) 2 dx = 5 n. 3 u L 2 (). If there was one, α 2 would be necessary so α 5 n 3 3 n n2 must hold for all n. This is not possible for any α > which does not depend on n. Hence a(u, v) cannot be L 2 elliptic. Lax-Milgram lemma: Given a Hilbert space (V, (, )), a continuous, V -elliptic bilinear form a(, ), and some continuous, linear functional F : V R. There is one and only one u V such that a(u, v) = F (v), v V. () Proof: For any u V, Au(v) = a(u, v), v V, defines a continuous, linear functional Au V. The continuity constant of Au is the continuity constant M of a(, ). Using the Riesz representation theorem, for any φ V there is a unique τφ V such that φ(v) = (τφ, v), v V. As τ : V V is an isomorphism, we can reformulate the variational problem () equivalently: There is a unique u V such that τau = τf. (2) To solve (2), we desire a contraction mapping T, T v = v ρ(τav τf ) with an unknown parameter ρ. If Banach fixed-point theorem can successfully be applied, we know that there is a unique u V such that u = T u = u ρ(τau τf ). This is sufficient for the statement. Thus, we need to show the existence of a ρ such that Banach fixed-point theorem can be applied. Let v, v 2 V, consider v = v v 2. T v T v 2 2 = v ρ(τav) 2 = v 2 2ρ(τAv, v) + ρ 2 τav 2 (τ, A linear) = v 2 2ρa(v, v) + ρ 2 a(v, τav) (Definition of τ, A) v 2 2ρα v 2 + ρ 2 M v τav ( 2ρα + ρ 2 M 2 ) v 2 = L 2 v v 2 2, (a continuous, elliptic) (A continuous, τ isometric) where we have defined L 2 = 2ρα + ρ 2 M 2. For ρ (, 2α/M 2 ), we know L <. 9