23.3 Uniform Convexity Of L ^{p}
These terms refer roughly to how round the unit ball is. Here is the definition.
Definition 23.3.1 A Banach space is uniformly convex if whenever
|| x _{n} || , || y _{n} ||≤ 1 and || x _{n} + y _{n} ||→ 2, it follows that || x _{n} − y _{n} ||→ 0.
You can show that uniform convexity implies strict convexity. There are various other
things which can also be shown. See the exercises for some of these. In this section, it will be
shown that the L ^{p} spaces are examples of uniformly convex spaces. This involves some
inequalities known as Clarkson’s inequalities. Before presenting these, here are the backwards
Holder inequality and the backwards Minkowski inequality. Recall that in the Holder
inequality,
=
q =
p ^{′} . The idea in these inequalities is to consider the case that
p ∈ .
This inequality is easy to remember if you just take Holder’s inequality and turn it around in
the case that 0
< p < 1
.
Lemma 23.3.2 Let 0 < p < 1 and let f,g be measurable functions. Also
∫ ∫
|g|p∕(p−1)dμ < ∞, |f|pdμ < ∞
Ω Ω
Then the following backwards Holder inequality holds.
∫ (∫ p )1∕p( ∫ p∕(p− 1) )(p−1)∕p
|fg|dμ ≥ |f| dμ |g| dμ
Ω Ω Ω
Proof: If ∫
dμ =
∞ , there is nothing to prove. Hence assume this is finite.
Then
∫ ∫
p −p p
|f| dμ = |g| |fg| dμ
This makes sense because, due to the hypothesis on g it must be the case that g
equals 0 only on a set of measure zero, since p∕
< 0. Alternatively, you could
carry out the following maipulations on Ω
∖ Z where
Z is the set where
g = 0 and
then note that nothing changes in the resulting inequality if you replace Ω
∖ Z with
Ω.
Then by the usual Holder inequality, one of the exponents being 1∕p > 1, the other being
1∕
also larger than 1 with
p +
= 1
,
∫ ( ∫ )p ( ∫ ( 1 )1∕(1−p) )1−p
|f|pdμ ≤ |f g|dμ --p dμ
|g|
( ∫ )p (∫ p∕p−1 )1−p
= |f g|dμ |g| dμ
Now divide by
^{1−p} and then take the
p ^{th} root.
■
Here is the backwards Minkowski inequality. It looks just like the ordinary Minkowski
inequality except the inequality is turned around.
Corollary 23.3.3 Let 0 < p < 1 and suppose ∫
^{p} dμ < ∞ for h =
f,g . Then
(∫ p )1∕p ( ∫ p )1∕p (∫ p )1∕p
(|f|+ |g|) dμ ≥ |f|dμ + |g| dμ
Proof: If ∫
^{p} dμ = 0 then there is nothing to prove since this implies
=
= 0 a.e. so assume this is not zero.
∫ ∫
(|f |+ |g|)pdμ = (|f|+ |g|)p−1(|f|+ |g|)dμ
Since p < 1,
^{p} ≤ ^{p} +
^{p} and so
∫ ( )
(|f|+ |g|)p−1 p∕p−1dμ < ∞.
Hence the backward Holder inequality applies and it follows that
∫ ∫ ∫
(|f|+ |g|)pdμ = (|f|+ |g|)p− 1|f|dμ + (|f|+ |g|)p−1|g|dμ
[ ]
(∫ ( p− 1)p∕p−1 ) (p−1)∕p (∫ p )1∕p ( ∫ p )1∕p
≥ (|f|+ |g|) dμ |f| dμ + |g| dμ
(∫ ) [( ∫ ) ( ∫ ) ]
p (p−1)∕p p 1∕p p 1∕p
= (|f|+ |g|) dμ |f| dμ + |g|dμ
and so, dividing gives the desired inequality.
■
Consider the easy Clarkson inequalities.
Lemma 23.3.4 For any p ≥ 2 the following inequality holds for any t ∈
,
| | | |
||1+-t||p ||1−-t||p 1 p
| 2 | + | 2 | ≤ 2 (|t| + 1)
Proof: It is clear that, since p ≥ 2, the inequality holds for t = 0 and t = 1. Thus it
suffices to consider only t ∈
. Let
x = 1
∕t. Then, dividing by 1
∕t ^{p} , the inequality holds if
and only if
(x + 1)p ( x − 1 )p 1
--2-- + --2-- ≤ 2 (1+ xp)
for all x ≥ 1. Let
( ( )p ( )p)
f (x) = 1(1+ xp)− x-+-1 + x-−-1
2 2 2
Then f
= 0 and
( )
′ p p−1 p (x-+-1)p−1 p( x−-1)p− 1
f(x) = 2x − 2 2 + 2 2
Since p − 1 ≥ 1, f
=
x ^{p−1} is convex. Hence its graph is like a smile. Thus
≥ f and so
( )
′ p p−1 -x+12-+-x−21 p−1 p p− 1 (x-)p−1
f (x) ≥ 2x − p 2 = 2 x − p 2 ≥ 0
Hence f
≥ 0 for all
x ≥ 1
. ■
Corollary 23.3.5 If z,w ∈ ℂ and p ≥ 2, then
||z + w ||p ||z − w ||p 1 p p
||--2--|| + ||--2--|| ≤ 2 (|z|+ |w| ) (23.7)
(23.7)
Proof: One of
, is larger. Say
≥ . Then dividing both sides of
the proposed inequality by
^{p} it suffices to verify that for all complex
t having
≤ 1
,
| | | |
||1+-t||p ||1−-t||p 1 p
| 2 | + | 2 | ≤ 2 (|t| + 1) (23.8)
(23.8)
Say t = re ^{iθ} where r ≤ 1. Then consider the expression
| iθ|p | iθ|p
||1-+-re-|| + ||1−-re-||
| 2 | | 2 |
It suffices to show that this is no larger than
The function on the left in 23.8 equals
1 ( 2 2 2 )p∕2 ( 2 2 2 )p∕2
2p (1+ rcosθ) + r sin (θ) + (1− rcosθ) + r sin (θ)
1 ( )p∕2 ( )p∕2
= 2p 1+ r2 + 2rcosθ + 1+ r2 − 2rcosθ , (23.9)
The derivative is
times
( p−2 p−2)
p (1+ r2 + 2rcosθ)-2-(− 2r sinθ)+ (2rsin θ)(1+ r2 − 2rcosθ)-2
2 ( )
p ( 2 )p−22- ( 2 )p−22
= 2 (2rsinθ) 1 +r − 2rcosθ − 1+ r + 2rcosθ
This equals 0 when
θ = 0
,π, 2
π or when
θ =
, . At the last two values, the value of the
function in
23.9 is
--1- (1+ r2)p∕2 ≤ 1(rp + 1).
2p−1 2
This follows from convexity of y = x ^{p∕2} for p ≥ 2. Thus
1 ( 2)p∕2 2p∕2( 1+ r2)p∕2 2p∕21 p
2p−-1 1+ r = 2p−1 --2--- ≤ 2p−12 (1 + r)
At 0 or π, the value of the function in 23.9 is
( ) ( )p ( )p
(1+ r2 − 2r)p∕2 + (1+ r2 + 2r)p∕2 1 = 1-+-r + 1−-r-
2p 2 2
and from the above lemma, this is no larger than
.
■
With this corollary, here is the easy Clarkson inequality.
Theorem 23.3.6 Let p ≥ 2. Then
∥∥f + g∥∥p ∥∥f − g∥∥p 1 p p
∥∥--2--∥∥ p + ∥∥-2--∥∥ p ≤ 2 (∥f∥Lp + ∥g∥Lp)
L L
Proof: This follows right away from the above corollary.
∫ ||f +-g||p ∫ ||f −-g-||p 1 ∫ p p
Ω || 2 || dμ + Ω|| 2 ||dμ ≤ 2 Ω(|f| + |g|)dμ ■
Now it remains to consider the hard Clarkson inequalities. These pertain to p < 2. First is
the following elementary inequality.
Lemma 23.3.7 For 1 < p < 2, the following inequality holds for all t ∈
.
| |q | |q ( )q∕p
||1+-t|| + ||1−-t|| ≤ 1+ 1 |t|p
| 2 | | 2 | 2 2
where here 1∕p + 1∕q = 1 so q > 2.
Proof: First note that if t = 0 or 1, the inequality holds. Next observe that the
map s →
maps
onto
. Replace
t with
∕ . Then you
get
|| 1 ||q || s ||q ( 1 1||1− s||p)q∕p
||s+-1|| + ||s+-1|| ≤ 2 + 2||s+-1||
Multiplying both sides by
^{q} , this is equivalent to showing that for all
s ∈ ,
q p q∕p (1 1||1 − s||p)q∕p
1+ s ≤ ((1 + s) ) 2 + 2||s-+-1||
( )q∕p
= 1 ((1 + s)p + (1− s)p)q∕p
2
This is the same as establishing
1 ((1 + s)p + (1− s)p)− (1+ sq)p−1 ≥ 0 (23.10)
2
(23.10)
where p − 1 = p∕q due to the definition of q above.
( )
p ≡ p(p−-1)⋅⋅⋅(p-−-k+-1), l ≥ 1
l l!
and
≡ 1. What is the sign of
? Recall that 1
< p < 2 so the sign is positive if
l = 0
,l = 1
,l = 2
. What about
l = 3?
=
so this is negative.
Then
is positive. Thus these alternate between positive and negative with
> 0 for all
k . What about
? When
k = 0 it is positive. When
k = 1 it is also positive. When
k = 2 it equals
< 0
. Then when
k = 3
,
> 0
. Thus
is positive when
k is odd and is negative when
k is
even.
Now return to 23.10 . The left side equals
( ∞ ( ) ∞ ( ) ) ∞ ( )
1 ∑ p sk + ∑ p (− s)k −∑ p− 1 sqk.
2 k=0 k k=0 k k=0 k
The first term equals 0. Then this reduces to
∞∑ ( p ) 2k ( p− 1 ) q2k ( p − 1 ) q(2k−1)
2k s − 2k s − 2k− 1 s
k=1
From the above observation about the binomial coefficients, the above is larger
than
∑∞ ( ) ( )
p s2k − p− 1 sq(2k−1)
k=1 2k 2k − 1
It remains to show the k ^{th} term in the above sum is nonnegative. Now q
> 2
k for all
k ≥ 1 because
q > 2. Then since 0
< s < 1
( ) ( ) (( ) ( ))
p s2k − p− 1 sq(2k−1) ≥ s2k p − p − 1
2k 2k − 1 2k 2k− 1
However, this is nonnegative because it equals
( >0 )
◜----------◞◟-----------◝
s2k|| p(p-− 1)⋅⋅⋅(p−-2k+-1) − (p−-1)(p−-2)⋅⋅⋅(p-−-2k-+-1)||
( (2k)! (2k − 1)! )
2k( p(p−-1)⋅⋅⋅(p−-2k+-1) (p-−-1)(p-−-2)⋅⋅⋅(p−-2k-+-1))
≥ s (2k)! − (2k)!
(p− 1)(p− 2)⋅⋅⋅(p− 2k+ 1)
= s2k-------------------------(p − 1) > 0.■
(2k)!
As before, this leads to the following corollary.
Corollary 23.3.8 Let z,w ∈ ℂ . Then for p ∈
,
| | | | ( )
||z-+-w||q ||z −-w-||q 1 p 1 p q∕p
| 2 | + | 2 | ≤ 2 |z|+ 2 |w|
Proof: One of
, is larger. Say
≥ . Then dividing by
^{q} , for
t =
z∕w, showing the above inequality is equivalent to showing that for all
t ∈ ℂ ,
≤ 1
,
| |q | |q ( )q∕p
||t+-1|| + ||1−-t|| ≤ 1|t|p + 1
| 2 | | 2 | 2 2
Now q > 2 and so by the same argument given in proving Corollary 23.3.5 , for t = re ^{iθ} , the
left side of the above inequality is maximized when θ = 0. Hence, from Lemma
23.3.7 ,
| | | | | | | |
||t+-1||q ||1−-t||q |||t|+-1||q ||1−-|t|||q
| 2 | + | 2 | ≤ | 2 | + | 2 |
(1 p 1)q∕p
≤ 2 |t| + 2 . ■
From this the hard Clarkson inequality follows. The two Clarkson inequalities are
summarized in the following theorem.
Theorem 23.3.9 Let 2 ≤ p. Then
|||| ||||p |||| ||||p
||||f +-g|||| + ||||f-−-g|||| ≤ 1(||f||pLp + ||g||pLp)
2 Lp 2 Lp 2
Let 1 < p < 2. then for 1∕p + 1∕q = 1,
|| || || || ( )
||||f +-g-||||q ||||f −-g||||q 1 p 1 p q∕p
|| 2 ||Lp + || 2 ||Lp ≤ 2 ||f||Lp + 2 ||g||Lp
Proof: The first was established above.
|| ||q || ||q
||||f +-g|||| + ||||f-−-g|||| ≤
|| 2 ||Lp || 2 ||Lp
( ∫ ||f + g||p )q∕p (∫ ||f − g||p )q ∕p
||--2--|| dμ + ||-2--|| dμ
Ω Ω
( ∫ (|| ||q)p∕q )q∕p ( ∫ (|| ||q)p ∕q )q∕p
= ||f-+-g|| dμ + ||f-−-g|| dμ
Ω 2 Ω 2
Now p∕q < 1 and so the backwards Minkowski inequality applies. Thus
(∫ (||f + g||q ||f − g||q)p∕q )q∕p
≤ ||-----|| + ||----|| dμ
Ω 2 2
From Corollary 23.3.8 ,
( )
∫ ( (1 1 )q∕p)p∕q q∕p
≤ ( - |f |p + -|g|p dμ)
Ω 2 2
(∫ ( ) )q∕p ( )q∕p
= 1|f|p + 1|g|p dμ = 1 ||f||pLp + 1 ||g||pLp ■
Ω 2 2 2 2
Now with these Clarkson inequalities, it is not hard to show that all the L ^{p} spaces are
uniformly convex.
Theorem 23.3.10 The L ^{p} spaces are uniformly convex.
Proof: First suppose p ≥ 2. Suppose
_{Lp} , _{Lp} ≤ 1 and
_{Lp} → 1
. Then
from the first Clarkson inequality,
|| || || ||
||||fn-+-gn-||||p ||||fn −-gn||||p 1 p p
|| 2 ||Lp + || 2 ||Lp ≤ 2 (||fn||Lp + ||gn||Lp) ≤ 1
and so
_{Lp} → 0
.
Next suppose 1 < p < 2 and
_{Lp} → 1. Then from the second Clarkson
inequality
|| ||q || ||q ( )q∕p
||||fn +-gn|||| + ||||fn-− gn-|||| ≤ 1||f ||p + 1 ||g ||p ≤ 1
|| 2 ||Lp || 2 ||Lp 2 n Lp 2 n Lp
which shows that
_{Lp} → 0.
■