Trace spaces are equivalent to interpolation spaces. In showing this, a more general sort
of trace space than that presented earlier will be used.
Definition 43.2.1Define for m a positive integer, V^{m} = V^{m}
(A0,A1,p,θ)
to be the setof functions, u such that
( )
t → tθu (t) ∈ Lp 0,∞, dt;A0 (43.2.15)
t
(43.2.15)
and
( dt )
t → tθ+m −1u(m)(t) ∈ Lp 0,∞, --;A1 . (43.2.16)
t
(43.2.16)
V^{m}is a Banach space with the norm
( )
||u|| ≡ max ||||tθu (t)|||| ,||||tθ+m−1u(m)(t)|||| .
Vm Lp(0,∞,dtt ;A0) || ||Lp(0,∞,dtt ;A1)
Thus V^{m} equals W in the case when m = 1. More generally, as in [?] different
exponents are used for the two L^{p} spaces, p_{0} in place of p for the space corresponding to
A_{0} and p_{1} in place of p for the space corresponding to A_{1}.
Definition 43.2.2Denote by T^{m}
(A0,A1, p,θ)
the set of all a ∈ A_{0} + A_{1}such that forsome u ∈ V^{m},
a = lt→im0+ u(t) ≡ trace (u ), (43.2.17)
(43.2.17)
the limit holding in A_{0} + A_{1}. For the norm
||a||Tm ≡ inf{||u||V m : trace(u) = a} . (43.2.18)
(43.2.18)
The case when m = 1 was discussed in Section 43.1. Note it is not known at
this point whether lim_{t→0+}u
(t)
even exists for every u ∈ V^{m}. Of course, if
m = 1 this was shown earlier but it has not been shown for m > 1. The following
theorem is absolutely amazing. Note the lack of dependence on m of the right
side!
Theorem 43.2.3The following hold.
Tm (A ,A ,p,θ) = (A ,A ) = (A ,A ) . (43.2.19)
0 1 0 1 θ,p,J 0 1 θ,p
(43.2.19)
Proof:It is enough to show the first equality because of Theorem 42.7.5 which
identifies
(A0,A1 )
_{θ,p,J} and
(A0,A1)
_{θ,p}. Let a ∈ T^{m}. Then there exists u ∈ V^{m} such
that
a = lim u(t) in A0 +A1.
t→0+
The first task is to modify this u
(t)
to get a better one which is more usable in order to
show a ∈
(A0, A1)
_{θ,p,J}. Remember, it is required to find w
(t)
∈ A_{0}∩ A_{1} for all
t ∈
(0,∞ )
and a = ∫_{0}^{∞}w
(t)
dt
t
, a representation which is not known at this time. To get
such a thing, let
ϕ ∈ C∞c (0,∞ ),spt(ϕ) ⊆ [α,β] (43.2.20)
(43.2.20)
with ϕ ≥ 0 and
∫ ∞
ϕ (t) dt = 1. (43.2.21)
0 t
(43.2.21)
Then define
∫ ( ) ∫ ( )
^u(t) ≡ ∞ϕ t- u(τ) dτ-= ∞ ϕ(s)u t ds. (43.2.22)
0 τ τ 0 s s
(43.2.22)
Claim: lim_{t→0+}
^u
(t)
= a and lim_{t→∞}
^u
^{(k)
}
(t)
= 0 in A_{0} + A_{1} for all k ≤ m.
Proof of the claim:From 43.2.22 and 43.2.21 it follows that for
∫ ( )
-(−-1)m-- ∞ -tm-- (m) t
= (m − 1)! 0 ϕ(s)sm+1 u s ds ∈ A1
The last step may look very mysterious. If so, consider the case where m = 2.
( ( ))
-t ′′
ϕ(s) su s
( t ( t) ( t)) ′
= ϕ(s) − -u′ - + u -
(( s ) s ( )( s ) ( ) ( ))
= ϕ(s) − t u ′′ -t −-t + -tu′ t − t-u′ t
s s s2 s2 s s2 s
t2 ′′(t)
= ϕ(s)s3u s .
You can see the same pattern will take place for other values of m.
Now
(∫ ∞ ( −θ ( ( 1)) )p dt)1 ∕p
||a||θ,p,J ≤ t J t,v t t-
0
{∫ [( || ( )|| ) ( || ( ) || )]p }1∕p
≤ C ∞ t−θ ||||v 1 |||| + t1− θ||||v 1 |||| dt
p 0 || t ||A0 || t ||A1 t
( (∫ ( || ( )|| )p )1 ∕p
≤ C { ∞ t−θ ||||v 1 |||| dt
p( 0 || t ||A0 t