9.3. THE Lp SPACES 227
and so one of the terms on the right side in 9.6 is ∞. Therefore again 9.6 is obvious.Otherwise divide (
∫Ω| f +g|p dµ)1/q on both sides to obtain 9.6. ■
By induction, you have(∫Ω
∣∣∣∣∣ n
∑k=1
fi
∣∣∣∣∣p
dµ
)1/p
≤n
∑k=1
(∫Ω
| fi|p dµ
)1/p
Observation 9.3.5 If f ,g are in Lp and α,β are scalars, then α f +βg ∈ Lp also.
To see this, note that α f +βg is measurable thanks to Proposition 6.1.8. Is |α f +βg|pin L1?(∫
Ω
|α f +βg|p dµ
)1/p
≤(∫
Ω
|α f |p dµ
)1/p
+
(∫Ω
|βg|p dµ
)1/p
= |α|(∫
Ω
| f |p dµ
)1/p
+ |β |(∫
Ω
|g|p dµ
)1/p
< ∞
and so it follows that Lp (Ω) is a vector space of functions. If ∥ f∥p ≡ (∫
Ω| f |p dµ)1/p , then
the above computation shows that ∥·∥p is a norm except for one problem. If ∥ f∥p = 0, itdoes not follow that f = 0. What can be concluded if ∥ f∥p = 0? From the first part of theargument in Theorem 9.3.3, it follows that if ∥ f∥p = 0, then f (ω) = 0 for a.e.ω.
Definition 9.3.6 Lp (Ω) is a normed vector space (normed linear space) if we agreeto identify any two functions in Lp (Ω) which are equal off a set of measure zero and let∥ f∥p ≡ (
∫Ω| f |p)1/p. More precisely, Lp (Ω) consists of a vector space of equivalence
classes of functions, the equivalence relation being that the functions are equal a.e.
The big result about Lp (Ω) is that it is a complete space. Recall that this means thatevery Cauchy sequence converges. Recall Theorem 2.3.3 which said that if a subsequenceof a Cauchy sequence in Rp converges then the original Cauchy sequence converges. Havea look a that theorem and notice that the specific context is completely irrelevant. The sameargument shows that in an arbitrary normed linear space, if a subsequence of a Cauchysequence converges, then the original Cauchy sequence converges. Also note that Theorem2.3.2 which said that Cauchy sequences are bounded also does not depend on the context.It holds for an arbitrary normed linear space.
To show Lp (Ω) is complete, I will show that a Cauchy sequence has a subsequencewhich converges for a.e. ω . Then an appeal to limit theorems will show Lp (Ω) is complete.
Theorem 9.3.7 Let { fn}∞
n=1 be a Cauchy sequence in Lp (Ω) . Then there exists g ∈Lp (Ω) and
{fnk
}∞
k=1 such that fnk (ω)→ g(ω) a.e. ω and
limn→∞∥ fn−g∥p = 0.
Proof: First note that there exists M such that∥∥ fnk
∥∥pp < M by Theorem 2.3.2 (Cauchy
sequences are bounded.) applied to this normed linear space. (Same argument) Select asubsequence
{fnk
}such that if m≥ nk,
∥∥ fnk − fm∥∥p
p < 4−k. Let
Bk ≡{
ω :∣∣ fnk+1 (ω)− fnk (ω)
∣∣p > 2−k}.