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By Olaf Steinbach

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Bu, q = g, q Let u ¯ ∈ (ker B)⊥ be a second solution satisfying for all q ∈ Π. Bu ¯, q = g, q Then, B(u − u ¯), q = 0 for all q ∈ Π. Obviously, u − u ¯ ∈ (ker B)⊥ . 19) we then conclude B(u − u ¯), q 0 ≤ cS u − u ¯ X ≤ sup = 0 q Π 0=q∈Π and therefore uniqueness, u = u ¯. 19) for the solution u this gives cS u X ≤ sup 0=q∈Π Bu, q = q Π sup 0=q∈Π g, q ≤ g q Π Π′ . 4 Operator Equations with Constraints In many applications we have to solve an operator equation Au = f where the solution u has to satisfy an additional constraint Bu = g.

However, for a Lipschitz domain Ω ⊂ Rd the norms v are only equivalent for |s| ≤ 1. 23. 18) holds also for appropriate Sobolev spaces H s (Γ ). 24. Let Γ = ∂Ω the boundary of a C k−1,1 –domain Ω ⊂ Rd and let f : H k+1 (Γ ) → R be a bounded linear functional satisfying |f (v)| ≤ cf v for all v ∈ H k+1 (Γ ). H k+1 (Γ ) If f (q) = 0 is satisfied for all q ∈ Pk (Γ ) then we also have for all v ∈ H k+1 (Γ ). 1 Let u(x), x ∈ (0, 1), be a continuously differentiable function satisfying u(0) = u(1) = 0.

6) Proof. For u, v ∈ X we choose an arbitrary t ∈ R. Then we have 1 A(u + tv), u + tv − f, u + tv 2 1 = F (u) + t [ Au, v − f, v ] + t2 Av, v . 4) we then obtain 1 F (u) ≤ F (u) + t2 Av, v = F (u + tv) 2 for all v ∈ X and t ∈ R. 6). 6). Then, as a necessary condition, d F (u + tv)|t=0 = 0 for all v ∈ X. dt From this we obtain Au, v = f, v for all v ∈ X and therefore the equivalence of both the variational and the minimization problem. 3) we now consider a fixed point iteration. For this we need to formulate the following Riesz representation theorem.

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