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Unformatted text preview: 1 Lecture 13 Theorem 1 Let Ω be a domain with Lipschitz boundary Γ . Then for all u ∈ C 1 ( ¯ Ω) → C (Γ) , Tu = u  Γ satisfies k Tu k L 2 (Γ) ≤ C k u k H 1 (Ω) Proof is long, so we will omit. Theorem 2 (Trace Theorem) Let Ω be a bounded domain in R n with Lipschitz boundary Γ . Then (i) there exists a unique bounded linear operator T : H 1 (Ω) → L 2 (Γ) (i.e. k Tu k L 2 (Γ) ≤ C k u k H 1 (Ω) ) with the property that if u ∈ C 1 ( ¯ Ω) , then Tu = u  Γ in the conventional sense. (ii) The range of T is dense in L 2 (Γ) . More generally, if u ∈ H m (Ω) and T j u ≡ ∂ j u ∂ν , ≤ j ≤ m 1 , then T j : H m (Ω) → H m j 1 / 2 (Γ) is a continuous, linear surjection. Now if u ∈ H 1 (Ω), we can unambiguously define boundary values of u on Γ. u = u means u = u a.e. on Γ. Lastly, recall the thoerem from last time: Any differential operator of order m is a bounded linear mapping from H s into H s m for s > m . Then for example (as in EIT) if u ∈ H 1 (Ω), then Tu = u  Γ ∈ L 2 (Γ). In fact, since j = 0, Tu ∈ H 1 1 / 2 (Γ) = H 1 / 2 (Γ) and ∂u ∂ν  Γ ∈ H 1 / 2 (Ω)....
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This note was uploaded on 03/14/2011 for the course MATH 676 taught by Professor Staff during the Fall '08 term at Colorado State.
 Fall '08
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