Another approach to these sorts of problems is to use trace spaces. This allows the
consideration of fractional order Sobolev spaces. In so far as the subject of Sobolev spaces
is concerned, I will present this material in a manner which is essentially independent of
the previous material on interpolation spaces.
As in the case of interpolation spaces, suppose A_{0} and A_{1} are two Banach spaces
which are continuously embedded in some topological vector space, X.
Definition 43.1.1Define a norm on A_{0} + A_{1}as follows.
, the measure in this case being usual Lebesgue measure.
Then
∫ t ′ ∫ t ν ′ −ν
f (t)− f (s) = s f (τ)dτ = s τ f (τ) τ dτ.
For
1
p
+
-1
p′
= 1, νp^{′} =
( 1)
θ − p
p^{′}< 1 because θ < 1 =
1-
p′
+
1
p
. Therefore,
||f (t) − f (s)||A0+A1
∫ t
≤ ||f′(τ)||A0+A1 dτ
∫st ∫ t
≤ ||f′(τ)|| dτ = ||τνf′(τ )|| τ−νdτ
s A1 s A1
( ∫ t )1∕p(∫ t ′ )1∕p′
≤ ||τνf′(τ )||pA1 dτ τ−νp dτ
s ( ′ ′) s
-t1−-νp- s1−νp-
≤ ||f ||W 1 − νp′ − 1− νp′ (43.1.3)
1−νp′
≤ ||f || t-----.
W 1− νp′
which converges to 0 as t → 0. This shows that lim_{t→0+}f
(t)
exists in A_{0} + A_{1}.
Clearly Z is a subspace. Let f_{n}→ f in W and suppose f_{n}∈ Z. Then since f ∈ W,43.1.3 implies f is continuous. Using 43.1.3 and replacing f with f_{n}− f_{m} and then
taking a limit as s → 0,
||f (t)− f (t)|| ≤ ||f − f || C t1− νp′
n m A0+A1 n m W ν
Taking a subsequence, it can be assumed f_{n}
(t)
converges to f
(t)
a.e. But the above
inequality shows that f_{n}
(t)
is a Cauchy sequence in C
([0,β];A0 + A1)
for all β < ∞.
Therefore, f_{n}
(t)
→ f
(t)
for all t. Also,
||fn(t)||A0+A1 ≤ Cν ||fn ||W t1−νp′ ≤ Kt1−νp′
for some K depending on max
{||f || : n ≥ 1}
n
and so
||f (t)||A +A ≤ Kt1−νp′
0 1
which implies f
(0)
= 0. Thus Z is closed.
Consider the last claim. For a generic t^{θ}g ∈ L^{p}
provided δ is small enough due to continuity of translation in L^{p}. Thus changing variables
in 43.1.4, letting τ = ln
(t)
and g_{δ}
(t)
≡
^g
_{δ}
(ln(t))
, it follows g_{δ}∈ C^{∞}
(0,∞; A)
and this
integral equals
( ∫ ∞ θp p dt)1∕p
0 t ||gδ(t)− g(t)|| t
This result applied to f and f^{′} with A = A_{0} and then A = A_{1} shows the last claim. This
proves the lemma.
Definition 43.1.5Let W be a Banach space and let Z be a closed subspace. Thenthe quotient space,denoted by W∕Z consists of the set of equivalence classes
[x ]
where the equivalence relation is defined by x ∽ y means x − y ∈ Z. Then W∕Z isa vector space if the operations are defined by α
[x]
≡
[αx]
and
[x]
+
[y]
≡
[x + y]
and these vector space operations are well defined. The norm on the quotient spaceis defined as
||[x]||
≡ inf
{||x + z|| : z ∈ Z}
.
The verification of the algebraic claims made in the above definition is left to the
reader. It is routine. What is not as routine is the following lemma. However, it is similar
to some topics in the presentation of the K method of interpolation.
Lemma 43.1.6Let W be a Banach space and let Z be a closed subspace of W.Then W∕Z with the norm described above is a Banach space.
Proof: That W∕Z is a vector space is left to the reader. Why is
||⋅||
a norm? Suppose
α≠0. Then
||α[x]|| = ||[αx ]|| ≡ inf{||αx + z|| : z ∈ Z}
= inf{||αx + αz|| : z ∈ Z}
= |α|inf{||x+ z|| : z ∈ Z } = |α|||[x]||.
Now let
||[x]||
≥
||x+ z1||
− ε and let
||[y]||
≥
||y+ z2||
− ε where z_{i}∈ Z. Then
||[x ]+ [y]|| ≡ ||[x + y]|| ≤ ||x+ y + z + z ||
1 2
≤ ||x+ z1||+ ||y + z2|| ≤ ||[x]||+ ||[y]||+ 2ε.
Since ε is arbitrary, this shows the triangle inequality. Clearly,
||[x ]||
≥ 0. It remains to
show that the only way
||[x]||
= 0 is for x ∈ Z. Suppose then that
||[x]||
= 0. This means
there exist z_{n}∈ Z such that
||x+ zn||
→ 0. Therefore, −x is a limit of a sequence of
points of Z and since Z is closed, this requires −x ∈ Z. Hence x ∈ Z also because Z is a
subspace. This shows
||⋅||
is a norm on W∕Z. It remains to verify that W∕Z is a Banach
space.
Suppose
{[xn]}
is a Cauchy sequence in W∕Z and suppose
||[xn ]− [xn+1]||
<
2n1+1
.
Let x_{1}^{′} = x_{1}. If x_{n}^{′} has been chosen let x_{n+1}^{′} = x_{n+1} + z_{n+1} where z_{n+1}∈ Z be such
that
where the infimum is taken over all f ∈ W such that a = f
(0)
. Also, if a ∈ A_{0}∩A_{1}, thena ∈ T and
||a||T ≤ K ||a||1A−1θ||a||θA0 (43.1.7)
(43.1.7)
for some constant K. Also
A0 ∩A1 ⊆ T (A0, A1,p,θ) ⊆ A0 + A1 (43.1.8)
(43.1.8)
and the inclusion maps are continuous.
Proof: First suppose f
(0)
= a where f ∈ W. Then letting f_{λ}
(t)
≡ f
(λt)
, it follows
that f_{λ}
(0)
= a also and so
(||||θ |||| ||||θ ′|||| )
||a||T ≤ max t fλ Lp(0,∞,dtt ;A0), t (fλ) Lp(0,∞,dtt ;A1)
( − θ|||| θ |||| 1−θ||||θ ′|||| )
= max λ t f Lp(0,∞,dtt ;A0),λ t f Lp(0,∞,dtt ;A1)
≡ max (λ−θR,λ1−θS ).
Now choose λ = R∕S to obtain
|| ||1−θ || ||θ
||a||T ≤ R1− θSθ = ||tθf||Lp(0,∞,dt;A0)||tθf′||Lp(0,∞,dt;A1).
t t
Thus
{|| || || || }
||a||T ≤ inf ||tθf||1L−pθ(0,∞,dt;A0)||tθf′||θLp(0,∞,dt;A1) .
t t
Next choose f ∈ W such that f
(0)
= a and
||f ||
_{W}≈
||a||
_{T}. More precisely, pick
f ∈ W such that f
(0)
= a and
||a||
_{T}> −ε +
||f||
_{W}. Also let
R ≡ ||||tθf||||p dt ,S ≡ ||||tθf′|||| p dt .
L (0,∞,-t ;A0) L (0,∞,t ;A1)
Then as before,
|| || || ||
||tθfλ|| p dt = λ−θR, ||tθ(fλ)′||p dt = λ1− θS. (43.1.9)
L (0,∞,t ;A0) L (0,∞, t ;A1)