TMA4230 Functional analysis
2004
Sequence spaces and Lp spaces
Harald Hanche-Olsen
[email protected]
Sequence spaces
A sequence space is a subspace of the set of all sequences x = (x1 , x2 , . . .) =
(xk )∞
k=1 . For the sake of brevity, we shall simply write x = (xk )k .
We shall be interested in normed sequence spaces. We shall usually consider
sequences of complex numbers, though almost everything we shall say works
equally well if we restrict our attention to real sequences.
All the sequence spaces we shall be concerned with in this note consist of
bounded sequences, i.e., those for which
kxk∞ = sup |xk | < ∞.
k
We write `∞ for the space of bounded sequences, equipped with the norm
k · k∞ .
1 Proposition. `∞ is complete.
∞
Proof: Consider a Cauchy sequence (xn )∞
n=1 in ` . Note carefully that each
xn is itself a sequence. Write xn = (xnk )k = (xn1 , xn2 , . . .). If we fix some k,
then the sequence (xnk )∞
n=1 is a Cauchy sequence of complex numbers, because
|xmk −xnk | ≤ kxm −xn k∞ . Since C is complete this sequence has a limit, which
we shall call yk . We shall show that the limit sequence y = (yk )k is bounded,
and ky − xn k∞ → 0 when n → ∞.
In fact, given ε > 0, let N be so that kxm − xn k∞ < ε whenever m, n ≥ N .
Then, in particular, |xmk −xnk ≤ kxm −xn k∞ < ε for any k. Let m → ∞ to
get |yk − xnk | ≤ ε. As this holds for every k and n ≥ N , we get ky − xn k∞ ≤ ε
for every n ≥ N . Thus y is in fact bounded, and we have proved the desired
convergence.
Two interesting subspaces are c0 ⊂ c ⊂ `∞ , where c is the set of all convergent
sequences and c0 is the set of all sequences in c with limit zero.
∞
2 Proposition. c and c0 are closed subspaces of ` .
Sequence spaces and Lp spaces
2
Proof: We first show that c is closed. So let xn ∈ c for n = 1, 2, . . ., and
assume xn → y with y ∈ `∞ . We need to show that y ∈ c. Let us write xn∞ =
limk→∞ xnk and note that |xm∞ − xn∞ | ≤ kxm − xn k∞ . Since the sequence
(xn )n converges in `∞ , it is also Cauchy, and from the above inequality we
conclude that the sequence (xn∞ )n is Cauchy in C, and hence convergent.
Write w = limn→∞ xn∞ .
We claim that yk → w when k → ∞. Fix ε > 0, and pick N so that
ky − xn k∞ < ε whenever n ≥ N . By making N larger, if necessary, we can also
ensure that |w − xn∞ | < ε for n ≥ N . Finally, pick K so that |xN k − xN ∞ | < ε
whenever k ≥ K. For such k, then,
|yk − w| ≤ |yk − xN k | + |xN k − xN ∞ | + |xN ∞ − w|
≤ ky − xN k∞ + |xN k − xN ∞ | + |xN ∞ − w|
< 3ε
which completes the proof.
Next, we show that c0 is closed. To this end, define the linear functional f∞
on c by
f∞ (x) = lim xk
(x ∈ c).
k→∞
We note that f∞ is in fact bounded, with norm 1. Hence it is continuous, so
its null space c0 is closed.
Of course c0 and c, being closed subspaces of a Banach space `∞ , are themselves
Banach spaces. We shall want to identify their dual spaces next.
`1 is the space of absolutely summable sequcences, i.e., the space of sequences
x for which
∞
X
kxk1 =
|xk | < ∞.
k=1
3 Proposition. Whenever x ∈ `1 and y ∈ `∞ then
∞
X
|xk yk | ≤ kxk1 kyk∞ .
k=1
Thus the sum
P∞
k=1
xk yk is absolutely convergent, and
∞
X
x
y
k k ≤ kxk1 kyk∞ .
k=1
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Sequence spaces and Lp spaces
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In particular, any x ∈ `1 defines a bounded linear functional x̃ on `∞ , and any
y ∈ `∞ defines a bounded linear functional ỹ on `1 by
x̃(y) = ỹ(x) =
∞
X
Sequence spaces and Lp spaces
4
Proof: Let f be a bounded linear functional on `1 . Define ek ∈ `1 as above,
and let yk = f (ek ). Then |yk | ≤ kf k kek k1 = kf k, so y ∈ `∞ .
Now f and ỹ take the same value on every vector ek . But if x ∈ `1 then
xk yk .
x=
k=1
∞
X
xk ek ,
(1)
k=1
We have, in fact,
kx̃k = kxk1
and
kỹk = kyk∞ .
the sum being convergent in `1 , and so because f is bounded,
Proof: We find, using |yk | ≤ kyk∞ ,
∞
X
|xk yk | ≤
∞
X
f (x) =
|xk | kyk∞ = kxk1 kyk∞ ,
which proves the first inequality. The second follows immediately from the
triangle inequality for infinite sums, and the bounds
kx̃k ≤ kxk1
and
∞
X
k=1
xk yk =
∞
X
xk sgn xk =
k=1
∞
X
|xk | = kxk1 ,
j=k+1
because
P∞
j=1
|xj | < ∞.
In brief, we state the above result by saying that the dual space of `1 is `∞ .
5 Proposition. Every bounded linear functional on c0 is of the form
ỹ(x) =
Then ek ∈ `1 , kek k1 = 1, and ỹ(ek ) = yk . Thus kỹk ≥ |yk | for every k, and
taking the supremum over all k we get kỹk ≥ kyk∞ .
4 Proposition. Every bounded linear functional on `1 is of the form
ỹ(x) =
∞
X
xk yk
k=1
for some y ∈ `∞ .
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xk yk = ỹ(x).
k=1
k=1
so that kx̃k ≥ kxk1 .
Similarly, if y ∈ `∞ , for any k let ek be the sequence defined by
(
1 if j = k,
ekj =
0 if j =
6 k.
∞
X
Pk
It remains to prove (1). For any partial sum sk = j=1 xj ej , the vector x − sk
has j-component 0 for j ≤ k, while the other components are those of x itself.
So
∞
X
kx − sk k1 =
|xj | → 0
(k → ∞)
kỹk ≤ kyk∞
are also immediate.
If x ∈ `1 , let yk = sgn xk . Then y ∈ `∞ , in fact kyk∞ = 1, and
x̃(y) =
xk f (ek ) =
k=1
k=1
k=1
∞
X
∞
X
xk yk
k=1
for some y ∈ `1 .
Proof: Just like in the preceding proof, define y by yk = f (ek ). Note that (1)
holds in c0 as well, with convergence in c0 (i.e., in the norm k · k∞ ) – although
for a very different reason, namely that xk → 0 when k → ∞ and x ∈ c0 . (You
should work out the details for yourself.)
P∞
Then the same argument shows that f (x) = k=1 xk yk for every x ∈ c0 .
It only remains to show that y ∈ `1 and kyk1 = kf k. For each k, let
(
sgn yj , if j ≤ k,
xkj =
0
otherwise.
Pk
Pk
Then xk ∈ c0 and kxk k∞ = 1, and f (x) = j=1 |yj |. Thus j=1 |yj | ≤ kf k.
Letting k → ∞, we get y ∈ `1 and kyk1 ≤ kf k.
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Sequence spaces and Lp spaces
5
On the other hand,
∞
∞
∞
X
X
X
|f (x)| = xk yk ≤
|xk yk | ≤
kxk∞ |yk | = kxk∞ kyk1
k=1
k=1
Sequence spaces and Lp spaces
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so that we get
kf k ≥ |f (xk )| ≥
k=1
k
X
|yj | −
j=1
proves the opposite inequality kf k ≤ kyk1 .
∞
X
|yj | + |α|.
j=k+1
Now let k → ∞ to get kf k ≥ kyk1 + |α|.
6 Proposition. Every bounded linear functional on c is of the form
ỹ(x) + αf∞ (x) =
∞
X
Lp spaces
xk yk + α lim xk
k→∞
k=1
In this section µ is a positive, σ-finite measure on a measure space Ω.1 2
Whenever we talk about functions on Ω, we shall only consider measurable
complex functions. For any function u and real number p > 0, define
Z
1/p
kukp =
|u|p dµ
.
for some y ∈ `1 and scalar α. Moreover,
kf k = kyk1 + |α|.
This means that, if we write z1 = α and put zk+1 = yk for k = 1, 2, . . ., then
z ∈ `1 and kf k = kzk1 . Thus, the dual of c is also `1 , so this is an example of
distinct spaces having the same dual.
Proof: Let f be a bounded linear functional on c. The restriction of f to c0
equals ỹ for some y ∈ `1 according to our previous result. Let e = (1, 1, 1, . . .) ∈
c. We find that x − f∞ (x)e ∈ c0 whenever x ∈ c, so that
∞
X
f x − f∞ (x)e = ỹ x − f∞ (x)e =
xk − f∞ (x) yk
k=1
and hence
f (x) =
∞
X
xk yk + αf∞ (x),
α = f (e) −
k=1
∞
X
yk .
Ω
We also define
kuk∞ = ess. sup |u(t)| = min{M : |u(t)| ≤ M for a.e. t ∈ Ω}.
t∈Ω
(The latter equality is the definition of the essential supremum. In this definition, one should first replace the minimum by an infimum, then use a bit
of measure theorey to show that the infimum is in fact attained, so that the
minimum is defined.)
We put kuk∞ = ∞ if there is no real number M so that |u| ≤ M almost
everywhere. To sum up:
|u| ≤ kuk∞ a.e.,
µ{t ∈ Ω : |u(t)| > M } > 0 if M < kuk∞ .
k=1
The estimate kf k ≤ kuk1 + |α| is immediate. To prove the opposite inequality,
for each k define xk ∈ c by setting
(
sgn yj , if j ≤ k,
xkj =
sgn α, if j > k.
Then kxkj k = 1, f∞ (xk ) = α, and
f (xk ) =
k
X
j=1
|yj | +
∞
X
sgn α yj + |α|,
j=k+1
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Exercise: Prove that
lim kukp = kuk∞
p→∞
if µ(Ω) < ∞.
When 0 < p < 1, k·kp is not a norm (the triangle inequality is not satisfied).
This case is just too strange in many ways, though it is sometimes encountered.
1 We do not bother to name the σ-algebra, but simply talk about measurable sets when
we do need them.
2 We may be able to get away with less than σ-finiteness: The most important property
is that there are no atomic sets of infinite measure. An atomic set is a measurable subset
A ⊆ Ω so that, whenever B ⊆ A is measurable, then either µ(B) = 0 or µ(B) = µ(A).
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Sequence spaces and Lp spaces
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In all cases, the homogeneity kαukp = |α| kukp is obvious when α ∈ C, but the
triangle inequality is harder to prove. The triangle inequality for k · kp is called
Minkowski’s inequality, and will be proved later. However, as an easy exercise
you are encouraged to give direct proofs for the cases p = 1 and p = ∞.
We shall say that real numbers p and q are conjugate exponents if any (hence
all) of the following equivalent conditions hold:
1 1
+ = 1,
p q
p + q = pq,
(p − 1)(q − 1) = 1.
7 Proposition. (Young’s inequality) For a ≥ 0, b ≥ 0, and conjugate exponents p, q with 1 < p < ∞,
ab ≤
ap
bq
+ .
p
q
Equality holds if and only if ap = bq .
Proof:
The sum of the two shaded areas in the figure
is an overestimate of the area ab of the indicated rectangle. (The convexity properties of
the curve varies with p, but all we shall need
is that the y is a strictly increasing function of
x, and (0, 0) is on the curve.) We compute the
area under the curve by an x integral and the
area to the left of the curve by a y integral:
Z
0
a
xp−1 dx +
Z
0
8 Proposition. (Hölder’s inequality) Let p, q be conjugate exponents with
1 ≤ p ≤ ∞. Then
Z
|uv| dµ ≤ kukp kvkq .
Ω
for any two measurable functions u and v. In particular, when the righthand
side is finite then uv is integrable, and
Z
uv dµ ≤ kukp kvkq .
Ω
In addition to these, we allow as special cases p = 1 and q = ∞, or p = ∞ and
q = 1.
ab ≤
Sequence spaces and Lp spaces
8
b
y q−1 dy =
If 0 < kukp kvkq < ∞ and 1 < p < ∞, equality holds in the first inequality if
and only if there is a scalar α so that |u|p = α|v q | almost everywhere.
Proof: The integrability of uv and the second inequality follow immediately
from the first and the definition of integrability together with the triangle
inequality for the integral, so we only need prove the first part. If p = 1 or
p = ∞ the first part is easily proved (exercise: convince yourself of this), so
we need only consider the case 1 < p < ∞. We may also assume without loss
of generality that u ≥ 0 and v ≥ 0. Finally, the inequality is obvious if the
righthand side is either 0 or ∞. So we may assume it is neither. But then, since
both sides are homogeneous in u and v, we may assume that
kukp = kvkq = 1.
y
Now we apply Young’s inequality and integrate:
Z
Z p
u
vq kukp
kvkq
1 1
uv dµ ≤
+
dµ =
+
= + = 1,
p
q
p
q
p
q
Ω
Ω
y = xp−1
x = y q−1
b
ap
bq
+ .
p
q
x
a
Clearly, we have equality if and only if the point
(a, b) lies on the curve, i.e., if b = ap−1 – or, equivalently, if bq = a(p−1)q =
apq−q = ap .
In the picture, ap > bq . If ap < bq we get a similar picture with the excess
area on the other side of the curve.
The above proof works for a > 0 and b > 0. Otherwise, if a = 0 or b = 0
the conclusion is obvious.
which proves the inequality.
The final statement on equality follows from the condition on equality in
Young’s inequality. We need to have equality almost everywhere in the above
integral, which in the general case (remember that we have rescaled u and v)
becomes (u/kukp )p = (v/kvkq )q .
9 Corollary. Let p and q be conjugate exponents, 1 ≤ p ≤ ∞. For any measurable function u,
Z
kukp = sup
|uv| dµ.
kvkq =1
Ω
If 1 ≤ p < ∞, and kukp < ∞, there is some v with kvkq = 1 and
Z
uv dµ = kukp .
Ω
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Proof: You may wonder why I call this a corollary when the proof is so long.
The reason is that the proof, though long, contains no deep or difficult ideas.
We prove the final part first; this will take care of most cases for the first
statement.
For the case p = 1, if kuk1 < ∞ let v = sgn u. Then kvk∞ = 1 and
Z
Z
uv dµ =
|u| dµ = kuk1 .
Ω
Ω
Next, if 1 < p < ∞ and kukp < ∞, note that if kukp = 0 there is nothing to
prove; otherwise, let
p/q
.
v = sgn u |u|/kukp
Then kvkq = 1, uv > 0 and the conditions for equality in Hölder’s inequality
hold, so that
Z
Z
uv dµ =
|uv| dµ = kukp kvkq = kukp .
Ω
Sequence spaces and Lp spaces
10
can ensure that µ(E) < ∞ as well. Let v = χE /µ(E). Then v ∈ L1 , kvk1 , and
Z
Z
|u|
|uv| dµ =
dµ ≥ M.
µ(E)
Ω
E
In other words,
Z
|uv| dµ ≥ M
sup
kvk1 =1
whenever M < kuk∞ .
Ω
Letting M → kuk∞ from below, we conclude that the supremum on the left
is at least kuk∞ . But by Hölder’s inequality it can be no bigger, so we have
equality.
10 Proposition. (Minkowski’s inequality) Whenever 1 ≤ p ≤ ∞,
Ω
ku + vkp ≤ kukp + kvkp .
The proof of the second part is now done.
To prove the first half, note that the second half (together with Hölder’s inequality) proves the first half whenever 1 ≤ p < ∞ and kukp < ∞, and in fact
the supremum is attained in these cases. We must show the remaining cases.
Recall that Ω is assumed to be σ-finite. Hence we can find measurable sets
E1 ⊂ E2 ⊂ · · · ⊂ Ω, each with finite measure, so that E1 ∪ E2 ∪ · · · = Ω.
Assume 1 ≤ p < ∞ and kukp = ∞. Write Dk = {t ∈ Ek : |u(t)| < k}.
Then D1 ∪ D2 ∪ · · · = Ω as well. Write and uk = uχDk .3 Then uk ∈ Lp .
(In fact, kuk kpp ≤ k pRµ(Dk ) since |uk | ≤ k.) Now there is some function vk
with kvk kq = 1 and Ω uk vk dµ = kuRk kp . This function
must in fact be zero
R
almost everywhere outside Dk . Thus Ω uvk dµ = Ω uk vk dµ = kuk kp . But the
monotone convergence theorem implies
Z
Z
|uk |p dµ →
|u|p dµ = ∞
(k → ∞),
Ω
Ω
since |uk | increases
pointwise to u. Thus kuk kp → ∞, so we can find v with
R
kvkq = 1 and Ω |uv| dµ as large as we may wish.
Only the case p = ∞ remains. Whenever M < kuk∞ there is some measurable set E with µ(E) > 0 and |u| ≥ M on E. Using the σ-finiteness of µ, we
3χ
Dk
is the characteristic function of Dk : It takes the value 1 on Dk and 0 outside Dk .
(Statisticians often use the term indicator function because “characteristic function” has a
different meaning in statistics.)
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Proof: Let q be the conjugate exponent.
Z
ku + vkp = sup
|(u + v)w| dµ
kwkq =1
Ω
Z
≤ sup
kwkq =1
Z
|uw| dµ +
Ω
|vw| dµ
Ω
≤ kukp + kvkp
where we used the ordinary triangle inequality in the second line and Hölder’s
inequality in the final line.
Now that we know that Lp is indeed a normed space when p ≥ 1, it is time to
tackle completeness. But first, a lemma, and then a word of caution.
11 Lemma. A normed space is complete if, and only if, every absolutely convergent series in the space is convergent.
In other words, the crierion for completeness is
if
∞
X
k=1
kxk k < ∞ then
∞
X
xk converges.
k=1
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Sequence spaces and Lp spaces
11
P∞
Proof: First, if the
k=1 kxk k < ∞, consider the
Pn space is complete and
partial sums sn = k=1 xk . By the triangle inequality, for m < n we get
ksn − sm k ≤
n
X
kxk k ≤
∞
X
Z X
∞
P∞
P∞
follows. Thus k=1 |uk | < ∞ almost everywhere, and so k=1 uk (t) converges
for almost every t ∈ Ω. Let the sum be s(t).
Now let ε > 0 Repeating the above computation with the sum starting at
k = m + 1, we find instead
Z X
∞
if m is big enough. But
1
k=1
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k=m
∞
∞
m
X
X
X
|uk (t)|,
uk (t) ≤
uk (t) = s(t) −
so
k=m+1
k=m+1
k=1
∞
m
X
X
|uk | < ε
uk ≤ s −
k=m+1
k=1
for large enough m. Thus the sum converges in Lp to the limit s.
The case p = ∞ is similar, but simpler: Now
∞
X
tn −
1
n
tn
tn +
1
n
∞
X
|uk (t)| ≤
k=1
kuk k∞ ≤
k=1
P∞
for almost every t ∈ Ω, so the sum k=1 uk (t) is absolutely convergent, hence
convergent, almost everywhere. Again let s(t) be the sum. But now
m
∞
∞
∞
X
X
X
X
uk (t) = uk (t) ≤
|uk (t)| ≤
kuk k∞ < ε
s(t) −
12 Proposition. Lp is complete (hence a Banach space) for 1 ≤ p ≤ ∞.
P∞
Proof: We first prove this for 1 ≤ p < ∞. Let uk ∈ Lp , k=1 kuk kp = M <
∞. By the Minkowski inequality,
Z X
n
n
n
∞
X
p X
p
p X
p
|uk | dµ = |uk | ≤
kuk k ≤
kuk k = M p .
k=1
∞
p
X
p
|uk | dµ ≤
kuk k < ε
Ω k=m+1
k=1
Ω k=1
p
|uk | dµ ≤ M
Ω k=1
if m is big enough, which shows that the sequence (sn )n is Cauchy and hence
convergent.
Conversely, assume that every absolutely convergent series is convergent,
and consider a Cauchy sequence (un )n . Pick successively k1 < k2 < + · · · so
that, kum −un k < 2−j whenever
n ≥ kj . Put x1 = un1 and xj = ukj+1 −ukj .
Pm,
∞
Then kxj k < 2−j for j ≥ 2, so j=1 xj is absolutely convergent, and therefore
convergent. Since ukj = x1 +x2 +· · ·+xj , the sequence (ukj )j is convergent. We
have shown that any Cauchy sequence has a convergent subsequence, which is
enough to prove completeness.
Here is an example to demonstrate
why showing completeness by considering general Cauchy sequences in Lp is
difficult. Such sequences may converge in
Lp , and yet diverge pointwise on a set of
positive measure: Let un be the function
shown on the right. Clearly, kun kp → 0
as n → ∞ for any p < ∞. And yet, we
can chose the center points tn so that
un (t) does not converge to zero for any
t ∈ (0, 1)!
let tn be the fractional
PNamely,
n
part of k=1 1/n. (I.e., so that the sum
is an integer plus tn , with 0 ≤ tn < 1.)
Since the harmonic series diverges, we
can see that un (t) > 21 for infinitely many
n (namely for n = m or n = m + 1 where
tm ≤ t ≤ tm+1 ).
By the monotone convergence theorem,
kxk k < ε
k=m+1
k=m+1
Sequence spaces and Lp spaces
12
k=m+1
k=m+1
k=m+1
almost everywhere when m is big enough, which implies
m
X
uk s(t) −
k=1
∞
≤ ε.
This finishes the proof.
k=1
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13 Lemma. Assume that µ is a finite measure. Then, if 1 ≤ p < p0 ≤ ∞,
0
µ(Ω)−1/p kukp ≤ µ(Ω)−1/p kukp0 .
When p0 = ∞, we write 1/p0 = 0. Apart from the adjustment by powers of
µ(Ω), this roughly states that k · kp is an increasing function of p. In particular,
0
when p < p0 then Lp ⊆ Lp : Lp decreases when p increases.
Proof: Assume 1 ≤ p < p0 < ∞. Write p0 = rp, so that r > 1. Let s be the
exponent conjugate to r. For simplicity, assume u ≥ 0. Then
Z
Z
kukpp =
up dµ =
up · 1 dµ
Ω
p
Ω
Z
≤ ku kr k1ks =
u
Ω
rp
1/r
dµ
µ(Ω)1/s = kukpp0 µ(Ω)1/s
14
Sequence spaces and Lp spaces
Proof: We prove this first for n = 2. We take the natural logarithm of
(ap + bp )1/p and differentiate:
d 1
1
ap ln a + bp ln b
d
ln (ap + bp )1/p =
ln(ap + bp ) = − 2 ln(ap + bp ) +
.
dp
dp p
p
pap + pbp
To show that this is negative is the same as showing that
pap ln a + pbp ln b < ap ln(ap + bp ) + bp ln(ap + bp ).
But ap ln(ap + bp ) > ap ln ap = pap ln a, so the first term on the righthand side
is bigger than the first term on the lefthand side. The same thing happens to
the second terms, so we are done with the case n = 2.
The general case can be proved in the same way, or one can proceed by
induction. Consider the case n = 3: We can write
1/p
(ap + bp + cp )1/p = [(ap + bp )1/p ]p + cp
Raise this to the 1/pth power and note that
1
1 1
1
1
=
1−
= − 0,
ps
p
r
p p
and a simple rearrangement finishes the proof.
When the measure space Ω is the set of natural numbers {1, 2, 3, . . .} and µ is
the counting measure, the Lp spaces become sequence spaces `p , with norms
kxkp =
∞
X
|xk |p
1/p
.
Increasing p and pretending that the expression in square brackets is unchanged, we get a smaller value from the case n = 2. But then the square
bracket also decreases, again from the case n = 2, and so the total expression decreases even more. The general induction step from n to n + 1 terms is
similar, using the 2-term case and the n-term case.
15 Proposition. For a sequence x = (xk )k , kxkp is a decreasing function of
p, for 0 < p < ∞. It is strictly decreasing provided at least two entries are
nonzero, except where the norm is infinite.
k=1
The Hölder and Minkowski inequalities, and completeness, for these spaces are
just special cases of the same properties for Lp spaces.
The inequalities between norms for different p get reversed however, compared to the case for finite measures.
14 Lemma. Let n ≥ 2 and a1 , . . . , an be positive real numbers. Then
(ap1
+
ap2
+ ··· +
apn )1/p
Uniform convexity. A normed space is called uniformly convex if for every
ε > 0 there is a δ > 0 so that whenever x and y are vectors with kxk = kyk = 1,
kx + yk > 2 − δ implies kx − yk < ε.
Our interest in uniformly convexity stems from the following result. We shall
prove it later, as it requires some results we have not covered yet.
16 Theorem. A uniformly convex Banach space is reflexive.
17 Lemma. Let 2 ≤ p < ∞. For u, v ∈ Lp we have
is a strictly decreasing function of p, for 0 < p < ∞.
ku + vkpp + ku − vkpp ≤ 2p/2 kuk2p + kvk2p
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p/2
.
Sequence spaces and Lp spaces
15
This inequality is a weaker form of one of four inequalities called Clarkson’s
inequalities. This one has the advantage that it has a very short proof, which
Clarkson’s inequalities do not.
Note that for p = 2, the inequality becomes an equality: It is then the
parallellogram law.
Proof: From Lemma 14 with 2 and p replacing p and p0 we get (ap + bp )1/p ≤
(a2 + b2 )1/2 , which we write as ap + bp ≤ (a2 + b2 )p/2 . This expains the first
inequality used in the following proof. The second is Minkowski’s inequality
for k · kp/2 . (Note that p/2 ≥ 1, so this is OK.) All the equalities are simple
calculations.
Z
ku + vkpp + ku − vkpp =
|u + v|p + |u − v|p dµ
ZΩ
p/2
≤
|u + v|2 + |u − v|2
dµ
Ω
Z
p/2
p/2
= 2p/2
|u|2 + |v|2
dµ = 2p/2 |u|2 + |v|2 ≤2
ku2 kp/2 + kv 2 kp/2
p/2
= 2p/2 kuk2p + kvk2p
p/2
,
and the proof is complete.
18 Proposition. Lp is uniformly convex for 2 ≤ p < ∞.
In fact, Lp is uniformly convex whenever 1 < p < ∞. We do not prove this
more general case, which relies on Clarkson’s inequalities in the same way that
the proof below builds on the inequality of the previous lemma.
Proof: Let kukp = kvkp = 1 in the previous lemma. We find
ku + vkpp + ku − vkpp ≤ 2p .
If ku + vkp > 2 − δ then
ku − vkp < 2p − (2 − δ)p
1/p
= 2 1p − (1 − 21 δ)p
1/p
.
Given ε > 0 we have
2 1p − (1 − 12 δ)p
1/p
19 Corollary. Lp is reflexive for 2 ≤ p < ∞.
In fact, Lp is reflexive for 1 < p < ∞. But we prefer not to state a stronger
result than we have proved.
It will often be useful to know that all Lp spaces share a common dense subspace. First, we prove a result showing that a Lp function cannot have very
large on sets of very large measure.
20 Lemma. If u ∈ Lp where 1 ≤ p < ∞ then
µ{x ∈ Ω : |u(x)| ≥ ε} ≤ ε−p kukpp
(ε > 0).
Proof: The proof is almost trivial, if we start at the righthand side:
Z
Z
p
p
kukp =
|u| dµ ≥
|u|p dµ ≥ εp µ(E),
Ω
E
where we used the fact that |u|p ≥ εp on E = {x ∈ Ω : |u(x)| ≥ ε}.
p/2
Ω
p/2
Sequence spaces and Lp spaces
16
1/p < ε ⇔ δ ≤ 2 1 − 1 − ( 12 ε)p
.
21 Proposition. The space of functions v ∈ L∞ which vanish outside a set of
finite measure, i.e., for which µ{t ∈ Ω : v(t) 6= 0} < ∞, is dense in Lp whenever
1 ≤ p < ∞.
We may further restrict the space to simple functions. This is an exercise in
integration theory which we leave to the reader.
Proof: Let u ∈ Lp . For a given n, define vn by
(
u(t), 1/n ≤ |u(t)| ≤ n,
vn (t) =
0
otherwise.
Then vn belongs to the space in question, thanks to the previous lemma.
Furthermore, |u−vn |p | ≤ |u|p and |u−vn |p → 0 pointwise as n → ∞. It follows
from the definition of the norm and the dominated convergence theorem that
ku − vn kpp → 0.
Whenever Y ⊆ X ∗ is a subset of the dual space of a normed space X, we
define its pre-annihilator as
Y⊥ = {x ∈ X : f (x) = 0 for all f ∈ Y }.
Thus, given ε > 0, we chose δ by the above formula, and the required implication is clear.
22 Lemma. Let X be a reflexive Banach space, and let Y ⊆ X ∗ be a closed
subspace with Y⊥ = {0}. Then Y = X ∗ .
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Sequence spaces and Lp spaces
17
Proof: Assume Y 6= X ∗ . It is a simple consequence of the Hahn–Banach
theorem that there is a nonzero bounded functional ξ on X ∗ which vanishes
on Y ; see Kreyszig Lemma 4.6-7 (page 243). (It is important for this that Y
is closed.)
But X is reflexive, so there is some x ∈ X so that ξ(g) = g(x) for all g ∈ X ∗ .
But then g(x) = ξ(g) = 0 for all g ∈ Y , so x ∈ Y⊥ = {0}. But this contradicts
ξ 6= 0.
Sequence spaces and Lp spaces
18
It remains to prove the result for p = 1. We assume first that µ(Ω) < ∞. Let
f ∈ (L1 )∗ . Since then kuk1 ≤ µ(Ω)1/2 kuk2 , we find for u ∈ L2 that u ∈ L1 as
well, and |f (u)| ≤ kf k kuk1 ≤ µ(Ω)1/2 kf k kuk2 . So f defines a bounded linear
functional on L2 as well, and there is some v ∈ L2 so that
Z
f (u) =
uv dµ
(u ∈ L2 ).
Ω
∞
∗
23 Corollary. If Z is a Banach space and Z is reflexive, then Z is reflexive.
Proof: Apply the lemma with X = Z ∗ and Y the image of Z under the
canonical map Z → Z ∗∗ = X ∗ : This maps z ∈ Z to ẑ ∈ Z ∗∗ defined by
ẑ(f ) = f (z) where f ∈ Z ∗ . This mapping is isometric, so the image is closed.
If f ∈ Y⊥ then f (z) = ẑ(f ) = 0 whenever z ∈ Z, so f = 0. Thus the conditions
of the lemma are fulfilled.
24 Proposition. Let p and q be conjugate exponents with 1 ≤ p < ∞. Then
every bounded linear functional on Lp has the form
Z
ṽ(u) =
uv dµ
Ω
q
where v ∈ L .
This result is already known for p = 2, since L2 is a Hilbert space, and this is
just the Riesz representation theorem.
Proof: We prove this first for p ≥ 2. Then Lp is reflexive. The space of all
functionals ṽ on Lp , where v ∈ Lq , satifisies the conditions of the previous
lemma. This completes the proof when p ≥ 2.
It would complete the proof for p > 1, if we had proven the uniform convexity,
and hence reflexivity, of Lp for p > 1.
Furthermore, since v 7→ ṽ is an isometry of Lq onto (Lp )∗ , any bounded
linear functional on Lq has the form v 7→ η(ṽ) where η ∈ (LP )∗∗ . But since
Lp is reflexive for p ≥ 2, we can write η(f ) = f (y) for some y ∈ Lp . Thus our
bounded linear functional on Lq has the form
Z
v 7→ ṽ(y) =
vy dµ = ỹ(v).
We shall prove that v ∈ L . Since L2 is dense in L1 , the above equality will
then extend to all u ∈ L1 by continuity, and we are done.
Assume now that kvk∞ > M > kf k. Then |v| ≥ M on a measurable set E
with µ(E) > 0. Let u = χE sgn v/µ(E). Then kuk1 = 1 and u ∈ L2 as well,
since it is bounded. Thus
Z
Z
1
|v| dµ ≥ M,
kf k ≥ f (u) =
uv dµ =
µ(E) E
Ω
which is a contradiction. This finishes the proof for µ(Ω) < ∞.
Otherwise, if µ(Ω) = ∞ but µ is σ-finite, write Ω = E1 ∪ E2 ∪ · · · where
each Ej has finite measure and all the sets Ej are pairwise
disjoint. Use what
R
we just proved to find vj ∈ L∞ (Ej ) with f (u) = Ej uv dµ when u ∈ L1 Ej ,
and kvj k∞ ≤ kf k. Define v on Ω by setting v(t) = vj (t) when t ∈ Ej . Then,
for u ∈ L1 ,
∞
X
u=
uχEj
j=1
(convergent in L1 ), so
f (u) =
∞
X
j=1
f (uχEj ) =
∞ Z
X
j=1
Z
uv dµ =
Ej
This finishes the proof. We may add that the above proof technique works for
any p ≤ 2; so for 1 < p < 2, we now have two independent proofs. (Three if
we had proved uniform convexity of Lp for 1 < p < ∞.)
25 Corollary. Lp is reflexive for 1 < p < ∞.
Ω
Since q varies through all values between 1 and 2 when p varies between 2 and
∞, the result is thus proved also for 1 < p < 2.
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uv dµ.
Ω
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