Unformatted text preview: ), m → R, then D � u → H m−|�| (Rn )
(10.3) D � : H m (Rn ) � H m−|�| (Rn ) is continuous.
Proof. First it is enough to show that each Dj deﬁnes a continuous
(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 =
πj · u and
|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 >
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)
implies that if uj � u in H m (Rn ), m > n + k , then uj � u� in C0 (Rn )
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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.
- Fall '13