In this section, several significant theorems are presented. Unless indicated otherwise, the
measure will be Lebesgue measure. First here is a lemma.
Lemma 32.2.1Suppose g ∈ L^{1}
([a,b];X )
where X is a Banach space. Then if∫_{a}^{b}g
(t)
ϕ
(t)
dt = 0 for all ϕ ∈ C_{c}^{∞}
(a,b)
, then g
(t)
= 0 a.e.
Proof: Let S be a measurable subset of
(a,b)
and let K ⊆ S ⊆ V ⊆
(a,b)
where K
is compact, V is open and m
(V ∖K )
< ε. Let K ≺ h ≺ V as in the proof of
the Riesz representation theorem for positive linear functionals. Enlarging K
slightly and convolving with a mollifier, it can be assumed h ∈ C_{c}^{∞}
(a,b)
. Then
|∫ | |∫ |
|| bX (t) g(t)dt|| = || b (X (t)− h(t))g (t)dt||
||a S || || a S ||
∫ b
≤ |XS (t)− h(t)|||g (t)||dt
∫a
≤ ||g(t)||dt.
V∖K
Now let K_{n}⊆ S ⊆ V_{n} with m
(Vn ∖ Kn)
< 2^{−n}. Then from the above,
| |
||∫ b || ∫ b
||a XS (t)g (t)dt|| ≤ a XVn∖Kn (t)||g (t)||dt
and the integrand of the last integral converges to 0 a.e. as n →∞ because
∑_{n}m
(Vn ∖Kn )
< ∞. By the dominated convergence theorem, this last integral
converges to 0. Therefore, whenever S ⊆
(a,b)
,
∫ b
XS (t)g(t)dt = 0.
a
Since the endpoints have measure zero, it also follows that for any measurable S, the
above equation holds.
Now g ∈ L^{1}
([a,b];X )
and so it is measurable. Therefore, g
([a,b])
is separable. Let D
be a countable dense subset and let E denote the set of linear combinations of the form
∑_{i}a_{i}d_{i} where a_{i} is a rational point of F and d_{i}∈ D. Thus E is countable. Denote by Y
the closure of E in X. Thus Y is a separable closed subspace of X which contains all the
values of g.
Now let S_{n}≡ g^{−1}
(B (yn,||yn||∕2))
where E =
{yn}
_{n=1}^{∞}. Therefore,
∪_{n}S_{n} = g^{−1}
(X ∖{0})
. This follows because if x ∈ Y and x≠0, then in B
( )
x, ||x4||
there
is a point of E,y_{n}. Therefore,
||y ||
n
>
3
4
||x||
and so
||yn||
2
>
3||x||
8
>
||x||
4
so
x ∈ B
(y ,||y ||∕2)
n n
. It follows that if each S_{n} has measure zero, then g
(t)
= 0 for a.e. t.
Suppose then that for some n, the set, S_{n} has positive measure. Then from what was
shown above,
|| ∫ || || ∫ ||
||y || = ||||--1--- g(t)dt− y ||||= ||||--1--- g(t)− y dt||||
n ||m (Sn) Sn n|| ||m (Sn ) Sn n ||
1 ∫ 1 ∫
≤ m-(Sn-) ||g (t)− yn||dt ≤ m-(Sn) ||yn||∕2dt = ||yn||∕2
Sn Sn
and so y_{n} = 0 which implies S_{n} = ∅, a contradiction to m
(Sn )
> 0. This contradiction
shows each S_{n} has measure zero and so as just explained, g
(t)
= 0 a.e. ■
Definition 32.2.2For f ∈ L^{1}
(a,b;X )
, define an extension, fdefined on
[2a − b,2b− a] = [a− (b− a),b+ (b− a)]
as follows.
(
-- { f (t) if t ∈ [a,b]
f (t) ≡ f (2a − t) if t ∈ [2a − b,a]
( f (2b− t) if t ∈ [b,2b− a]
Definition 32.2.3Also if f ∈ L^{p}
(a,b;X )
and h > 0, define for t ∈
[a,b]
,f_{h}
(t)
≡f
(t− h)
for all h < b − a. Thus the map f → f_{h}is continuous and linear onL^{p}
(a,b;X )
. It is continuous because
∫ b ∫ a+h ∫ b−h
||fh(t)||pdt = ||f (2a − t+ h)||p dt+ ||f (t)||p dt
a ∫a ∫ a
a+h p b− h p p
= a ||f (t)|| dt+ a ||f (t)|| dt ≤ 2||f ||p.
The following lemma is on continuity of translation in L^{p}
(a,b;X)
.
Lemma 32.2.4Let fbe as defined in Definition 32.2.2. Then for f ∈ L^{p}
(a,b;X)
forp ∈ [1,∞),
∫
lim b||||f-(t − δ) − f (t)||||p dt = 0.
δ→0 a X
Proof: Regarding the measure space as
(a,b)
with Lebesgue measure, by regularity
of the measure, there exists g ∈ C_{c}
The other thing to notice about this is the following lemma. It follows immediately
from the definition.
Lemma 32.2.7Suppose f,f^{′}∈ L^{1}
(a,b;X )
. Then if
[c,d]
⊆
[a,b]
, it follows that
( )
f |[c,d]
^{′} = f^{′}|_{[c,d]
}. This notation means the restriction to
[c,d]
.
Recall that in the case of scalar valued functions, if you had both f and its weak
derivative, f^{′} in L^{1}
(a,b)
, then you were able to conclude that f is almost everywhere
equal to a continuous function, still denoted by f and
∫ t
f (t) = f (a) + f′(s)ds.
a
In particular, you can define f
(a)
to be the initial value of this continuous function. It
turns out that an identical theorem holds in this case. To begin with here is the same sort
of lemma which was used earlier for the case of scalar valued functions. It says that if
f^{′} = 0 where the derivative is taken in the sense of X valued distributions, then f equals
a constant.
Lemma 32.2.8Suppose f ∈ L^{1}
(a,b;X )
and for all ϕ ∈ C_{c}^{∞}
(a,b)
,
∫ b
f (t)ϕ′(t)dt = 0.
a
Then there exists a constant, a ∈ X such that f
(t)
= a a.e.
Proof:Let ϕ_{0}∈ C_{c}^{∞}
(a,b)
,∫_{a}^{b}ϕ_{0}
(x)
dx = 1 and define for ϕ ∈ C_{c}^{∞}
(a,b)
∫ ( ∫ )
x b
ψϕ (x) ≡ a [ϕ (t)− a ϕ(y)dy ϕ0(t)]dt
Then ψ_{ϕ}∈ C_{c}^{∞}
(a,b)
and ψ_{ϕ}^{′} = ϕ −
(∫b )
a ϕ (y)dy
ϕ_{0}. Then
( ( ) )
∫ b ∫ b ′ ∫ b
f (t)(ϕ (t))dt = f (t) ψ ϕ(t)+ ϕ(y)dy ϕ0(t) dt
a a a
◜=0 by as◞su◟mption◝ ( )
∫ b ′ ∫ b ∫ b
= f (t)ψϕ(t)dt+ ϕ (y)dy f (t)ϕ0 (t)dt
(a ( a) ) a
∫ b ∫ b
= a a f (t)ϕ0(t)dt ϕ (y)dy .
where the derivative is taken inthe sense of X valued distributions. Then there exists a unique point of X, denoted byf
(a)
such that the following formula holds a.e. t.
∫ t
f (t) = f (a)+ f′(s)ds
a
Proof:
∫ ( ∫ ) ∫ ∫ ∫
b t ′ ′ b ′ b t ′ ′
a f (t)− a f (s)ds ϕ (t)dt = a f (t)ϕ (t)dt − a a f (s)ϕ (t)dsdt.
Now consider ∫_{a}^{b}∫_{a}^{t}f^{′}
(s)
ϕ^{′}
(t)
dsdt. Let Λ ∈ X^{′}. Then it is routine from approximating
f^{′} with simple functions to verify
(∫ ∫ ) ∫ ∫
Λ b tf′(s)ϕ′(t)dsdt = b tΛ(f′(s))ϕ ′(t)dsdt.
a a a a
Now the ordinary Fubini theorem can be applied to obtain
∫ b∫ b (∫ b∫ b )
= Λ(f′(s)) ϕ′(t)dtds = Λ f′(s)ϕ′(t)dtds .
a s a s
Since X^{′} separates the points of X, it follows
∫ ∫ ∫ ∫
b t ′ ′ b b ′ ′
a a f (s)ϕ (t)dsdt = a s f (s)ϕ (t)dtds.
Therefore,
( )
∫ b ∫ t ′ ′
a f (t)− a f (s)ds ϕ (t)dt
∫ b ∫ b∫ b
= f (t)ϕ′(t)dt− f′(s)ϕ′(t)dtds
∫a ∫a s ∫
b ′ b ′ b ′
= a f (t)ϕ (t)dt− a f (s) s ϕ (t)dtds
∫ b ∫ b
= f (t)ϕ′(t)dt+ f′(s)ϕ(s)ds = 0.
a a
Therefore, by Lemma 32.2.8, there exists a constant, denoted as f
(a)
such
that
∫ t
f (t)− f′(s) ds = f (a) ■
a
There is also a useful theorem about continuity of pointwise evaluation.
Corollary 32.2.10Let f,f^{′}∈ L^{1}
(a,b;X )
so that
∫ t ′
f (t) = f (0)+ 0 f (s)ds (32.2.8)
(32.2.8)
where in this formula, t → f
(t)
is the continuous representative of f. Then there exists aconstant C such that for each t ∈
( )
--1-- ( ′ )
∥f (t)∥X ≤ b − a + 1 ∥f∥L1(a,b;X) + ∥f ∥L1(a,b;X ) ■
Let X be the space of functions f ∈ L^{1}
(a,b;X )
such that their weak derivatives f^{′} are
also in L^{1}
(a,b;X )
. Then X is a Banach space with norm given by
∥f∥X ≡ ∥f∥L1(a,b;X) + ∥f′∥L1(a,b;X)
This is because the map f → f^{′} is a closed map. If f_{n}→ f in L^{1}
(a,b;X)
and f_{n}^{′}→ ξ in
L^{1}
(a,b;X )
, then for ϕ ∈ C_{c}^{∞}
(a,b)
,
∫ b ∫ b ∫ b ∫ b
ξϕdt = lim f′ϕdt = lim − fnϕ′dt = − fϕ′dt
a n→ ∞ a n n→∞ a a
showing that ξ = f^{′}. Thus if you have a Cauchy sequence in X,
{fn}
, then f_{n}→ f in
L^{1}
(a,b;X )
and f_{n}^{′}→ ξ in L^{1}
(a,b;X )
for some ξ. Hence f^{′} = ξ.
Then the above corollary says that pointwise evaluation is continuous as a map from
X to X. This is clearly a linear map. Also the formula obtained shows that in fact, this is
continuous into C
([a,b];X )
.
( )
∥f∥C([a,b];X ) = sup ∥f (t)∥X ≤ C ∥f∥L1(a,b;X) + ∥f′∥L1(a,b;X) = C ∥f∥X .
t∈ [a,b]