consists of functions f : Ω → F such that f
is measurable and
∫
|f (ω)|dμ < ∞
Ω
The L^{p} spaces are defined as follows.
Definition 7.3.1Let
(Ω,F,μ)
be a measure space. Then L^{p}
(Ω )
consists of those measurable functions fsuch that∫_{Ω}
|f|
^{p}dμ < ∞. Here it is assumed that p > 1. Also define the conjugate exponent q assatisfying
1 + 1= 1
p q
In case p = 1, we let q = ∞ and give a special meaning to L^{∞}
(Ω )
discussed later.
Here we assume p > 1. There is an essential inequality which makes possible the study of
L^{p}
(Ω )
.
Proposition 7.3.2Let 0 ≤ a,b. Then
ap bq
p + q ≥ ab (7.6)
(7.6)
Proof:Let b ≥ 0 be fixed and let f
(a)
≡
ap
p
+
bq
q
− ab. Then f
(0)
=
bq-
q
≥ 0 and f^{′}
(a)
= a^{p−1}− b. If
b = 0 the desired inequality is obvious. If b > 0, then f^{′}
(a)
< 0 for a close to 0 and f^{′}
(a )
> 0 if a^{p−1}> b.
Thus f has a minimum at the point where a^{p−1} = b. But p − 1 = p∕q and so, at this point a^{p} = b^{q}.
Therefore, at this point, f
(a)
=
ap
p-
+
ap
q-
− aa^{p−1} = a^{p}− a^{p} = 0. Therefore, f
(a)
≥ 0 for all a ≥ 0 and it
equals 0 exactly when a^{p} = b^{q}. ■
This implies the following major result, Holder’s inequality.
Theorem 7.3.3Let f,g be measurable and nonnegative functions. Then
∫ (∫ p )1∕p(∫ q )1∕q
Ω fgdμ ≤ Ω f dμ Ω g dμ
Proof: If either
(∫ )
Ω fpdμ
^{1∕p} or
(∫ )
Ωgqdμ
^{1∕q} is 0, then there is nothing to show because
if
(∫Ω fpdμ)
^{1∕p} = 0, then ∫_{Ω}f^{p}dμ = 0 and you could let A_{n}≡
{ω : fp(ω) ≥ 1∕n}
. Then
0 = ∫_{Ω}f^{p}dμ ≥∫_{An}f^{p}dμ ≥
(1∕n)
μ
(An )
and so μ
(An )
= 0. Therefore,
{ω : f (ω ) ⁄= 0}
= ∪_{n=1}^{∞}A_{n} and
each of these sets in the union has measure zero. It follows that
{ω : f (ω ) ⁄= 0}
has measure zero.
Therefore, ∫_{Ω}fgdμ = 0 and so indeed, there is nothing left to show. The situation is the same if
(∫Ω gqdμ)
^{1∕q} = 0. Thus assume both of the factors on the right in the inequality are nonzero. Then let
A ≡
_{p} = 0, it does not follow that f = 0.
What can be concluded if
∥f ∥
_{p} = 0? From the first part of the argument in Theorem 7.3.3, it follows that
if
∥f∥
_{p} = 0, then f
(ω)
= 0 for a.e.ω.
Definition 7.3.6L^{p}
(Ω)
is a normedvector space (normed linear space) if we agree to identifyany two functions in L^{p}
(Ω )
which are equal off a set of measure zero and let
∥f∥
_{p}≡
(∫ p)
Ω |f|
^{1∕p}.More precisely, L^{p}
(Ω)
consists of a vector space of equivalence classes of functions, the equivalencerelation being that the functions are equal a.e.
The big result about L^{p}
(Ω)
is that it is a complete space. Recall that this means that every Cauchy
sequence converges. Recall Theorem 2.2.21 which said that if a subsequence of a Cauchy sequence in
ℝ^{p} converges then the original Cauchy sequence converges. Have a look a that theorem and
notice that the specific context is completely irrelevant. The same argument shows that in
an arbitrary normed linear space, if a subsequence of a Cauchy sequence converges, then the
original Cauchy sequence converges. Also note that Theorem 2.2.20 which said that Cauchy
sequences are bounded also does not depend on the context. It holds for an arbitrary normed linear
space.
To show L^{p}
(Ω)
is complete, I will show that a Cauchy sequence has a subsequence which converges for
a.e. ω. Then an appeal to limit theorems will show L^{p}
(Ω )
is complete.
Theorem 7.3.7Let
{fn}
_{n=1}^{∞}be a Cauchy sequence in L^{p}
(Ω )
. Then there exists g ∈ L^{p}
(Ω)
anda subsequence
{fnk}
_{k=1}^{∞}such that f_{nk}
(ω)
→ g
(ω)
a.e. ω and lim_{n→∞}
∥fn − g∥
_{p} = 0.
Proof:First note that there exists M such that
∥fn ∥
k
_{p}^{p}< M by Theorem 2.2.20 applied to this normed
linear space. (Same argument) Select a subsequence
{fn }
k
such that if m ≥ n_{k},
∥fn − fm ∥
k
_{p}^{p}< 4^{−k}. Let
B_{k}≡
{ω : ||fn (ω)− fn (ω)||p > 2−k}
k+1 k
. Then
∫
2−kμ(B ) ≤ ||f (ω )− f (ω)||pdμ < 4−k
k Bk nk+1 nk
and so μ
(Bk )
< 2^{−k}. Now if f_{nk}
(ω )
fails to be a Cauchy sequence, then ω ∈ B_{k} for infinitely many k. In
other words,
ω ∈ ∩∞n=1 ∪k≥n Bk ≡ B
This measurable set B has measure zero because
∑∞ -1--
μ(B ) ≤ μ(∪k≥nBk ) ≤ μ (Bk) < 2n− 1 for every n ∈ ℕ
k=n
Therefore, for ω
∕∈
B,
{fnk}
_{k=1}^{∞} is a Cauchy sequence. Let g
(ω)
≡ 0 on B and let g
(ω )
≡ lim_{k→∞}f_{nk}
(ω)
if ω
∕∈
B. Why is g ∈ L^{p} and why does f_{n} converge to g in L^{p}? g is the limit of the measurable functions
f_{nk}X_{BC} and so it is measurable. By Fatou’s lemma,
and each of the sets in the union has measure zero. ■
Note that this implies that if
∥f∥
_{∞} = 0, then f = 0 a.e. so f is regarded as 0. Thus, to say that
∥f − g∥
_{∞} = 0 is to say that the two functions inside the norm are equal except for a set of measure
zero, and the convention is that when this happens, we regard them as the same function.
If α = 0 then
∥αf ∥
_{∞} = 0 = 0
∥f∥
_{∞}. If α≠0, then M ≥
|f (ω)|
implies
|α|
M ≥
|αf (ω)|
.
In particular,
|α|
( )
∥f∥∞ + 1n
≥
|αf (ω )|
for all ω not in the union of the sets of measure
zero corresponding to each
∥f∥
_{∞} +
1n
. Thus there is a set of measure zero N such that for
ω
∕∈
N,
( )
|α | ∥f∥ + 1- ≥ |αf (ω)| for all n
∞ n
Therefore, for ω
∕∈
N,
|α|
∥f∥
_{∞}≥
∥αf∥
_{∞}. This implies that
∥f ∥
_{∞} =
∥ ∥
∥ 1ααf∥
_{∞}≤
|α1|
∥αf ∥
_{∞} and so
∥αf∥
_{∞}≥
|α|
∥f ∥
_{∞} also. This shows that this acts like a norm relative to multiplication by scalars. What
of the triangle inequality? Let M_{n}↓
∥f∥
_{∞} and N_{n}↓
∥g∥
_{∞}. Thus for each n, there is an exceptional set of
measure zero such that off this set M_{n}≥
|f (ω)|
and a similar condition holding for g and N_{n}. Let N be
the union of all the exceptional sets for f and g for each n. Then for ω
∕∈
N, the following holds for all
ω
∕∈
N
Mn + Nn ≥ |f (ω )|+ |g(ω)| ≥ |f (ω)+ g(ω)|
So take a limit of both sides and find that
∥f∥∞ + ∥g∥∞ ≥ |f (ω)+ g(ω)|
for all ω off a set of measure zero. Therefore,
∥f∥∞ + ∥g∥∞ ≥ ∥f + g∥∞
Theorem 7.3.10L^{∞}
(Ω )
is complete.
Proof: Let
{fn}
be a Cauchy sequence. Let N be the union of all sets where it is not the case that
|fn(ω) − fm (ω)|
≤
∥fn − fm ∥
_{∞}. By Proposition 7.3.9, there is such an exceptional set M_{mn} for each
choice of m,n. Thus N is the countable union of these sets of measure zero. Therefore, for
ω