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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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 Fall '13
 Melrose
 Derivative

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