38.3 Is there something like this for the fractional order spaces? Yes there is. However, in
order to prove it, it is convenient to use an equivalent norm for H m + s
which does
not depend explicitly on the Fourier transform. The following theorem is similar to one in
[? ] . It describes the norm in
H m + s in terms which are free of the Fourier transform.
This is also called an intrinsic norm
[? ] .
Theorem 38.3.1 Let s ∈
and let m be a nonnegative integer. Then an equivalent
norm for H m + s is
∑ ∫ ∫
|||u|||2 ≡ ||u||2 n + |D αu(x)− D αu(y)|2|x − y |−n− 2sdxdy.
m+s m,2,ℝ |α|=m
Also if
≤ m, there are constants, m and M such that
∫ ∫ ∫
m (s) |F u(z)|2||zβ||2 |z|2s dz ≤ ||D βu(x)− D βu(y)||2|x − y|− n−2sdxdy
∫
≤ M (s) |F u(z)|2 ||zβ||2|z|2sdz (38.3.10)
(38.3.10)
Proof: Let u ∈ S which is dense in H m + s
. The Fourier transform of the
function,
y → D α u − D α u equals
Now by Fubini’s theorem and Plancherel’s theorem along with the above, taking
=
m,
∫ ∫
|D αu (x) − D αu (y )|2|x − y|−n−2sdxdy
∫ ∫
= |Dαu (y+ t)− D αu(y)|2 |t|− n−2sdtdy
∫ ∫
−n−2s α α 2
= |t| |D u(y+ t)− D u(y)| dydt
∫ −n−2s∫ |( it⋅z ) α |2
= |t| | e − 1 FD u(z)| dzdt
∫ ( ∫ |( )| )
= |FD αu(z)|2 |t|−n−2s| eit⋅z − 1 |2dt dz. (38.3.11)
Consider the inside integral, the one taken with respect to
t .
(∫ − n− 2s |( )|2 )
G (z) ≡ |t| |eit⋅z − 1 |dt .
The essential thing to notice about this function of z is that it is a positive real number
whenever z ≠ 0 . This is because for small
, the integrand is dominated by
C − n +2 . Changing to polar coordinates, you see that
∫
|t|−n−2s||(eit⋅z − 1)||2dt < ∞
[|t|≤1]
Next, for
> 1
, the integrand is bounded by 4
− n − 2 s , and changing to polar
coordinates shows
∫ ∫
|t|−n−2s||(eit⋅z − 1)||2dt ≤ 4 |t|−n−2sdt < ∞.
[|t|>1] [|t|>1]
Now for α > 0,
∫
G (αz ) = |t|−n−2s||(eit⋅αz − 1)||2dt
∫
−n−2s||(iαt⋅z )||2
= |t| e − 1 dt
∫ ||r||−n−2s|(ir⋅z )|2 1
= |α| | e − 1 | αn-dr
∫ |( )|2
= α2s |r|−n−2s| eir⋅z − 1 | dr = α2sG (z).
Also
G is continuous and strictly positive. Letting
0 < m (s) = min {G(w ) : |w| = 1}
and
M (s) = max {G (w ) : |w| = 1},
it follows from this, and letting α =
, w ≡ z ∕ , that
( 2s 2s)
G (z) ∈ m (s)|z| ,M (s) |z| .
More can be said but this will suffice. Also observe that for s ∈
and
b > 0
,
(1 + b)s ≤ 1+ bs,21−s(1+ b)s ≥ 1 + bs.
In what follows, C
will denote a constant which depends on the indicated
quantities which may be different on different lines of the argument. Then from
38.3.11 ,
∫ ∫
|D αu (x) − D αu (y )|2|x − y|−n−2sdxdy
∫
≤ M (s) |FD αu(z)|2 |z|2s dz
∫
= M (s) |Fu (z)|2|zα |2|z|2sdz.
No reference was made to
=
m and so this establishes the top half of
38.3.10 .
Therefore,
∫ ∫
|||u|||2 ≡ ||u||2 n + ∑ |D αu(x)− D αu(y)|2 |x − y|−n− 2sdxdy
m+s m,2,ℝ |α|=m
∫ ( )m ∫ ∑
≤ C 1+ |z|2 |F u(z)|2 dz + M (s) |Fu (z)|2 |zα|2|z|2s dz
|α|=m
Recall that
( n ) m
∑ z2α1⋅⋅⋅z2αn≤ (1 + ∑ z2) ≤ C(n,m ) ∑ z2α1 ⋅⋅⋅z2αn. (38.3.12)
|α|≤m 1 n j=1 j |α|≤m 1 n
(38.3.12)
Therefore, where C
is the largest of the multinomial coefficients obtained in the
expansion,
( n ) m
(1 + ∑ z2) .
j=1 j
Therefore,
|||u|||2
∫ (m+s )m ∫ ∑
≤ C 1+ |z|2 |F u(z)|2dz +M (s) |F u(z)|2 |zα|2|z|2sdz
|α|=m
∫ ( 2)m+s 2 ∫ 2( 2)m 2s
≤ C 1+ |z| |F u(z)| dz + M (s) |F u(z)| 1 + |z| |z| dz
∫ ( )m+s
≤ C 1+ |z|2 |F u(z)|2dz = C||u ||Hm+s(ℝn).
It remains to show the other inequality. From 38.3.11 ,
∫ ∫
|D αu (x) − D αu (y )|2|x − y|−n−2sdxdy
∫
≥ m (s) |FD αu(z)|2|z|2sdz
∫
2 α 2 2s
= m (s) |Fu (z)| |z | |z| dz.
No reference was made to
=
m and so this establishes the bottom half of
38.3.10 .
Therefore, from
38.3.12 ,
2
|||u|||m+s
∫ ( 2)m 2 ∫ 2 ∑ α 2 2s
≥ C 1 + |z| |Fu (z)| dz + m (s) |Fu (z)| |z | |z| dz
∫ ( ) ∫ ( |α|=m)
≥ C 1 + |z|2 m|Fu (z)|2dz + C |Fu (z)|2 1+ |z|2 m |z|2sdz
∫
( 2)m ( 2s) 2
= C 1 + |z| 1 + |z| |F u(z)|dz
∫ ( 2)m ( 2)s 2
≥ C 1 + |z| 1 + |z| |F u(z)|dz
∫ ( )m+s
= C 1 + |z|2 |Fu (z)|2dz = ||u||Hm+s(ℝn).
This proves the theorem.
With the above intrinsic norm, it becomes possible to prove the following version of
Theorem 38.2.7 .
Lemma 38.3.2 Let h : ℝ n → ℝ n be one to one and onto. Also suppose that D α h and
D α
exist and are Lipschitz continuous if ≤ m for m a positive integer.
Then
∗ m+s n m+s n
h : H (ℝ ) → H (ℝ )
is continuous, linear, one to one, and has an inverse with the same properties, the inverse
being
∗ .
Proof: Let u ∈ S . From Theorem 38.2.7 and the equivalence of the norms in
W m, 2
and
H m ,
∫ ∫ ∑
||h∗u||2Hm (ℝn) + |α|=m |Dαh ∗u(x)− D αh∗u(y)|2 |x − y|−n− 2sdxdy
2 ∫ ∫ ∑ 2 −n−2s
≤ C ||u||Hm (ℝn) + |α|=m |Dαh ∗u (x)− Dαh ∗u(y)||x − y | dxdy
2 ∫ ∫ ∑ ||∑ ( )
= C ||u||Hm (ℝn) + |α|=m | |β(α)|≤m h∗ D β(α)u gβ(α)(x)
( ) |2 −n−2s
− h∗ D β(α)u gβ(α)(y )| |x− y| dxdy
2 ∫ ∫ ∑ ∑ || ∗( β(α) )
≤ C ||u||Hm (ℝn) + C |α|=m |β(α)|≤m h D u gβ(α)(x)
− h∗(D β(α)u)g (y )||2|x− y|−n−2sdxdy
β(α)
(38.3.13)
(38.3.13)
A single term in the last sum corresponding to a given α is then of the form,
∫ ∫ | ( ) ( ) |2
|h ∗ Dβu gβ(x)− h∗ D βu gβ(y)| |x − y|− n−2sdxdy (38.3.14)
(38.3.14)
[∫ ∫
≤ ||h ∗(Dβu) (x)g (x )− h∗(D βu)(y)g (x)||2 |x − y|− n− 2s dxdy +
β β
∫ ∫ | ∗( β ) ∗ ( β ) |2 −n−2s ]
|h D u (y)gβ(x)− h D u (y)gβ (y )| |x− y| dxdy
[ ∫ ∫ | ( ) ( ) |2
≤ C (h) |h∗ Dβu (x) − h ∗ Dβu (y )| |x− y|−n−2sdxdy +
∫ ∫ |∗ ( β ) |2 2 −n−2s ]
|h D u (y)| |gβ (x )− gβ(y)||x− y| dxdy .
Changing variables, and then using the names of the old variables to simplify the
notation,
[ ( )∫ ∫ |( ) ( ) |2
≤ C h,h −1 | Dβu (x )− Dβu (y)| |x − y|−n−2sdxdy +
∫ ∫ | ( ) |2 ]
|h∗ Dβu (y)| |gβ (x )− gβ(y)|2|x− y|−n−2sdxdy .
By 38.3.10 ,
∫ 2 | |2 2s
≤ C (h) |F (u)(z)| |zβ| |z| dz
∫ ∫ | ( ) |
+ |h∗ D βu (y)|2 |gβ(x) − gβ (y )|2|x− y|−n−2sdxdy.
In the second term, let
t =
x − y . Then this term is of the form
∫ | ( ) |2∫ 2 −n−2s
|h∗ D βu (y)| |gβ(y + t) − gβ (y )| |t| dtdy (38.3.15)
∫ | ( ) |
≤ C |h ∗ Dβu (y )|2dy ≤ C||u||2Hm(ℝn). (38.3.16)
because the inside integral equals a constant which depends on the Lipschitz constants
and bounds of the function,
g β and these things depend only on
h . The reason this
integral is finite is that for
≤ 1
,
2 −n−2s 2 −n−2s
|gβ(y + t) − gβ (y)| |t| ≤ K |t| |t|
and using polar coordinates, you see
∫
2 −n−2s
[|t|≤1]|gβ(y +t)− gβ (y )| |t| dt < ∞.
Now for
> 1
, the integrand in
38.3.15 is dominated by 4
− n − 2 s and using polar
coordinates, this yields
∫ 2 −n−2s ∫ −n−2s
[|t|>1]|gβ(y + t) − gβ (y)| |t| dt ≤ 4 [|t|>1]|t| dt < ∞.
It follows 38.3.14 is dominated by an expression of the form
∫
C (h) |F (u)(z)|2||zβ ||2|z|2sdz + C||u||2m n
H (ℝ )