The above presentation is very abstract, involving the trace of a function in
W (A0,A1,p,θ)
and a norm which was the infimum of norms of functions in W which have trace equal to
the given function. It is very useful to have a description of the norm in these fractional
order spaces which is defined in terms of the function itself rather than functions which
have the given function as trace. This leads to something called an intrinsic norm. I am
following Adams [?].
The following interesting lemma is called Young’s inequality. It holds more generally
than stated.
Lemma 44.3.1Let g = f ∗ h where f ∈ L^{1}
(ℝ )
,h ∈ L^{p}
(ℝ )
, and f,h are all Borelmeasurable, p ≥ 1. Then g ∈ L^{p}
(ℝ)
and
||g|| p ≤ ||f|| 1 ||h||p
L (ℝ) L (ℝ) L (ℝ)
Proof: First of all it is good to show g is well defined. Using Minkowski’s
inequality
The following is a very interesting inequality of Hardy Littlewood and Pólya.
Lemma 44.3.2Let f be a real valued function defined a.e. on [0,∞) and letα ∈
(− ∞, 1)
and
∫
1 t
g(t) = t 0 f (ξ)dξ (44.3.6)
(44.3.6)
For 1 ≤ p < ∞
∫ ∫
∞ tαp|g(t)|p dt ≤---1--- ∞ tαp|f (t)|p dt (44.3.7)
0 t (1− α)p 0 t
(44.3.7)
Proof: First it can be assumed the right side of 44.3.7 is finite since otherwise there is
nothing to show. Changing the variables letting t = e^{τ}, the above inequality takes the
form
∫ ∫
∞ τpα τ p ---1--- ∞ τpα τ p
−∞ e |g (e)| dτ ≤ (1 − α)p −∞ e |f (e )| dτ
Now from the definition of g it follows
∫ τ
τ −τ e
g(e ) = e −∞ f (ξ)dξ
∫ τ
= e−τ f (eσ)eσdσ
−∞
(∫ 0 )p∫ ∞
≤ e(α−1)udu epσα|f (eσ)|p dσ
− ∞ − ∞
( 1 )p ∫ ∞ pσα σ p
= 1−-α- −∞ e |f (e )|dσ
which was to be shown. This proves the lemma.
Next consider the case where G
(t)
,t > 0 is a continuous semigroup on A_{1} and
A_{0}≡ D
(Λ)
where Λ is the generator of this semigroup. Recall that from Proposition
17.14.5 on Page 1743 Λ is a closed densely defined operator and so A_{0} is a Banach space
if the norm is given by
||u||A ≡ ||u||A + ||Λu ||A
0 1 1
Also assume
||G (t)||
is uniformly bounded for t ∈ [0,∞). I have in mind the case where
A_{1} = L^{p}
n
(ℝ )
and G
(t)
u
(x)
= u
(x+ tei)
but it is notationally easier to discuss this in
the general case. First here is a simple lemma.
.Then there exists a unique solution to the initial value problem
y′ − Λy = g,y(0) = y0 ∈ D (Λ)
and it is given by
∫ t
y(t) = G(t)y0 + 0 G (t− s)g (s)ds. (44.3.8)
(44.3.8)
This solution is continuous having continuous derivative and has values in D
(Λ)
.
Proof: First I show the following claim.
Claim:∫_{0}^{t}G
(t− s)
g
(s)
ds ∈ D
(Λ)
and
(∫ t ) ∫ t
Λ G (t − s)g (s)ds = G (t)g (0) − g(t)+ G (t− s)g′(s)ds
0 0
Proof of the claim:
1 ( ∫ t ∫ t )
h- G (h) G(t− s)g(s)ds− G (t− s)g(s)ds
0 0
1( ∫ t ∫ t )
= -- G(t− s+ h)g (s)ds − G(t− s)g(s)ds
h 0 0
1( ∫ t−h ∫ t )
= -- G(t− s)g(s+ h) ds − G(t− s)g(s)ds
h −h 0
1∫ 0 ∫ t−h g(s+ h)− g (s)
= h- G (t− s)g (s+ h)ds+ G (t− s)------h-------
−∫ht 0
− 1 G (t− s) g(s) ds
h t−h
Using the estimate in Theorem 17.14.3 on Page 1738 and the dominated convergence
theorem the limit as h → 0 of the above equals
∫
t ′
G (t)g(0)− g(t)+ 0 G (t− s)g (s)ds
which proves the claim.
Since y_{0}∈ D
(Λ)
,
G (h)y0 − y0
G (t)Λy0 = G (t)hli→m0 -----h-----
G(t+ h) − G (t)
= lhim→0 ------h-------y0
G(h)G (t)y − G (t)y
= lim ----------0-------0- (44.3.9)
h→0 h
Since this limit exists, the last limit in the above exists and equals