As in the case of scalar valued functions, two functions in L^{p}
(Ω;X )
are considered equal
if they are equal a.e. With this convention, and using the same arguments found in the
presentation of scalar valued functions it is clear that L^{p}
(Ω; X)
is a normed linear space
with the norm given by 19.5.33. In fact, L^{p}
(Ω;X )
is a Banach space. This is the main
contribution of the next theorem.
Lemma 19.5.2If x_{n}is a Cauchy sequence in L^{p}
(Ω; X)
satisfying
∑∞
||xn+1 − xn||p < ∞,
n=1
then there exists x ∈ L^{p}
(Ω;X )
such that x_{n}
(s)
→ x
(s)
a.e. and
||x− xn||p → 0.
Proof:Let
N∑
gN (s) ≡ ||xn+1(s)− xn(s)||X
n=1
Then by the triangle inequality,
( ∫ ) (∫ )
p 1∕p N∑ p 1∕p
ΩgN (s) dμ ≤ Ω ||xn+1(s)− xn(s)|| dμ
n∞=1
≤ ∑ ||xn+1 − xn|| < ∞.
n=1 p
Let
∑∞
g(s) = lim gN (s) = ||xn+1 (s)− xn (s)||X .
N→ ∞ n=1
By the monotone convergence theorem,
(∫ p )1∕p (∫ p )1∕p
Ω g(s) dμ = Nlim→∞ Ω gN (s) dμ < ∞.
Therefore, there exists a set of measure 0, E, such that for s
∕∈
E, g
(s)
< ∞. Hence, for
s
∕∈
E,
lim x (s)
N→ ∞ N+1
exists because
N∑
xN+1 (s) = xN+1 (s) − x1(s) +x1 (s) = (xn+1(s)− xn(s)) + x1(s) .
n=1
is complete. Also every Cauchy sequence has asubsequence which converges pointwise.
Proof:If {x_{n}} is Cauchy in L^{p}
(Ω;X )
, extract a subsequence {x_{nk}} satisfying
||||xnk+1 − xnk|||| ≤ 2−k
p
and apply Lemma 19.5.2. The pointwise convergence of this subsequence was
established in the proof of this lemma. This proves the theorem because if a
subsequence of a Cauchy sequence converges, then the Cauchy sequence must also
converge.
Observation 19.5.4If the measure space is Lebesgue measure then you havecontinuity of translation in L^{p}
(ℝn;X )
in the usual way. More generally, for μa Radon measure on Ω a locally compact Hausdorff space, C_{c}
(Ω;X )
is dense inL^{p}
(Ω; X)
. Here C_{c}
(Ω;X )
is the space of continuous X valued functions whichhave compact support in Ω. The proof of this little observation follows immediatelyfrom approximating with simple functions and then applying the appropriateconsiderations to the simple functions.
Clearly Fatou’s lemma and the monotone convergence theorem make no sense for
functions with values in a Banach space but the dominated convergence theorem holds in
this setting.
Theorem 19.5.5If x is strongly measurable and x_{n}
(s)
→ x
(s)
a.e. with
||xn(s)|| ≤ g(s) a.e.
where g ∈ L^{1}
(Ω )
, then x is Bochner integrable and
∫ ∫
x(s)dμ = lnim→∞ xn (s)dμ.
Ω Ω
Proof:
||xn(s)− x(s)||
≤ 2g
(s)
a.e. so by the usual dominated convergence
theorem,
∫
0 = lim ||xn (s) − x(s)||dμ.
n→ ∞ Ω
Also,
∫
||xn (s)− xm (s)||dμ
Ω
∫ ∫
≤ ||x (s) − x (s)||dμ+ ||x (s)− x (s)||dμ,
Ω n Ω m
and so {x_{n}} is a Cauchy sequence in L^{1}
(Ω;X)
. Therefore, by Theorem 19.5.3, there
exists y ∈ L^{1}
(Ω;X )
and a subsequence x_{n′} satisfying
1
xn′(s) → y(s) a.e. and in L (Ω; X).
But x
(s)
= lim_{n′→∞}x_{n′}
(s)
a.e. and so x
(s)
= y
(s)
a.e. Hence
∫ ∫
||x (s)||dμ = ||y(s)||dμ < ∞
Ω Ω
which shows that x is Bochner integrable. Finally, since the integral is linear,
||∫ ∫ || ||∫ ||
|||| x (s)dμ− x (s)dμ|||| = |||| (x(s)− x (s))dμ||||
|| Ω Ω n || || Ω n ||
∫
≤ Ω||xn (s)− x (s)||dμ,
and this last integral converges to 0. This proves the theorem.
One can also give a version of the Vitali convergence theorem.
Definition 19.5.6Let A⊆ L^{1}
(Ω;X )
. Then A is said to be uniformly integrable if forevery ε > 0 there exists δ > 0 such that whenever μ
(E )
< δ, it follows
∫
||f||X dμ < ε
E
for all f ∈A. It is bounded if
∫
sfu∈pA Ω||f||X dμ < ∞.
Theorem 19.5.7Let
(Ω,ℱ,μ )
be a finite measure space and let X be a separableBanach space. Let
{fn}
⊆ L^{1}
(Ω;X )
be uniformly integrable and bounded such thatf_{n}
(ω )
→ f
(ω)
for each ω ∈ Ω. Then f ∈ L^{1}
(Ω; X)
and
∫
lim ||fn − f||X dμ = 0.
n→ ∞ Ω
Proof: Let ε > 0 be given. Then by uniform integrability there exists δ > 0 such that
if μ
(E )
< δ then
∫
||fn||dμ < ε∕3.
E
By Fatou’s lemma the same inequality holds for f. Also Fatou’s lemma shows
f ∈ L^{1}