denotes the functions which are restrictions of Cbm
(ℝn)
toU.
It is clear this is a Banach space, the proof being a simple exercise in the use of the
fundamental theorem of calculus along with standard results about uniform
convergence.
Lemma 38.4.2Let u ∈S and let
n
2
+ m < t. Then there exists C independent of u suchthat
||u||Cmb (ℝn) ≤ C ||u||Ht(ℝn).
Proof: Using the fact that the Fourier transform maps S to S and the definition of
the Fourier transform,
α α
|D u(x)| ≤ C||F D u||L1(ℝn)
∫ α
= C |x ||Fu (x )|dx
∫ ( 2)|α|∕2
≤ C 1+ |x| |Fu(x)|dx
∫ ( )m ∕2( )−t∕2( )t∕2
≤ C 1+ |x|2 1+ |x |2 1+ |x|2 |Fu(x)|dx
( ∫ ) ( ∫ )
( 2)m −t 1∕2 ( 2)t t 1∕2
≤ C 1+ |x| dx 1+ |x| |F u(x)|
≤ C||u|| t n
H (ℝ )
because for the given values of t and m the first integral is finite. This follows from a use
of polar coordinates. Taking sup over all x ∈ ℝn and
|α|
≤ m, this proves the
lemma.
Corollary 38.4.3Let u ∈ Ht
(ℝn )
where t > m +
n2
. Then u is a.e. equal to a functionof Cbm
(ℝn )
still denoted by u. Furthermore, there exists a constant, C independent of usuch that
||u||m n ≤ C ||u|| t n .
Cb (ℝ ) H (ℝ )
Proof:This follows from the above lemma. Let
{uk}
be a sequence of functions of S
which converges to u in Ht and a.e. Then by the inequality of the above lemma, this
sequence is also Cauchy in Cbm
(ℝn)
and taking the limit,
||u||Cm (ℝn) = lim ||uk||Cm(ℝn) ≤ C lim ||uk||Ht(ℝn) = C ||u||Ht(ℝn).
b k→ ∞ b k→∞
What about open sets, U?
Corollary 38.4.4Let t > m +
n
2
and let U be an open set with u ∈ Ht
(U )
. Then u isa.e. equal to a function of Cm
(U)
still denoted by u. Furthermore, there exists aconstant, C independent of u such that