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# Since dj u j u and j m cm m1 m we conclude that

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Unformatted text preview: ), m → R, then D � u → H m−|�| (Rn ) and (10.3) D � : H m (Rn ) � H m−|�| (Rn ) is continuous. Proof. First it is enough to show that each Dj deﬁnes a continuous linear map (10.4) Dj : H m (Rn ) � H m−1 (Rn ) � j since then (10.3) follows by composition. � If m → R then u → H m (Rn ) means u → ⊂π ≤−m L2 (Rn ). Since Dj u = ˆ πj · u and ˆ, |πj | ⊂π ≤−m ∀ Cm ⊂π ≤−m+1 � m we conclude that Dj u → H m−1 (Rn ) and ∈Dj u∈H m−1 ∀ Cm ∈u∈H m . � Applying this result we see Corollary 10.3. If k → N0 and m > (10.5) n 2 + k then k H m (Rn ) � C0 (Rn ) . 0 Proof. If |�| ∀ k , then D � u → H m−k (Rn ) � C0 (Rn ). Thus the ‘weak derivatives’ D � u are continuous. Still we have to check that this means that u is itself k times continuously diﬀerentiable. In fact this again follows from the density of S (Rn ) in H m (Rn ). The continuity in (10.3) k implies that if uj � u in H m (Rn ), m > n + k , then uj � u� in C0 (Rn ) 2 k (u...
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## This note was uploaded on 01/22/2014 for the course MATH 18.155 taught by Professor Melrose during the Fall '13 term at MIT.

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