23.3 Uniform Convexity Of Lp
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
||xn||,||yn||≤ 1 and ||xn + yn||→ 2, it follows that ||xn − yn||→ 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 Lp 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
. 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 <
Lemma 23.3.2 Let 0 < p < 1 and let f,g be measurable functions. Also
Then the following backwards Holder inequality holds.
Proof: If ∫
there is nothing to prove. Hence assume this is finite.
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
is the set where g
= 0 and
then note that nothing changes in the resulting inequality if you replace Ω ∖ Z
Then by the usual Holder inequality, one of the exponents being 1∕p > 1, the other being
also larger than 1 with
Now divide by
and then take the pth
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 ∫
pdμ < ∞ for h
= f,g. Then
Proof: If ∫
= 0 then there is nothing to prove since this implies
= 0 a.e. so assume this is not zero.
Since p < 1,
Hence the backward Holder inequality applies and it follows that
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 ∈
Proof: It is clear that, since p ≥ 2, the inequality holds for t = 0 and t = 1.Thus it
suffices to consider only t ∈
Then, dividing by 1∕tp,
the inequality holds if
and only if
for all x ≥ 1. Let
= 0 and
Since p − 1 ≥ 1, f
is convex. Hence its graph is like a smile. Thus
0 for all x ≥
Corollary 23.3.5 If z,w ∈ ℂ and p ≥ 2, then
Proof: One of
is larger. Say
Then dividing both sides of
the proposed inequality by
it suffices to verify that for all complex t
Say t = reiθ where r ≤ 1.Then consider the expression
It suffices to show that this is no larger than
The function on the left in 23.8 equals
The derivative is
This equals 0 when θ
or when θ
At the last two values, the value of the
function in 23.9
This follows from convexity of y = xp∕2 for p ≥ 2. Thus
At 0 or π, the value of the function in 23.9 is
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
Proof: This follows right away from the above corollary.
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 ∈
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 →
. Then you
Multiplying both sides by
this is equivalent to showing that for all s ∈
This is the same as establishing
where p − 1 = p∕q due to the definition of q above.
1. What is the sign of
? Recall that 1
< p <
2 so the sign is positive if
What about l
so this is negative.
is positive. Thus these alternate between positive and negative with
0 for all k
. What about
= 0 it is positive. When
= 1 it is also positive. When k
= 2 it equals
Then when k
is positive when
is odd and is negative when k
Now return to 23.10. The left side equals
The first term equals 0. Then this reduces to
From the above observation about the binomial coefficients, the above is larger
It remains to show the kth term in the above sum is nonnegative. Now q
1 because q >
2. Then since 0 < s <
However, this is nonnegative because it equals
As before, this leads to the following corollary.
Corollary 23.3.8 Let z,w ∈ ℂ. Then for p ∈
Proof: One of
is larger. Say
Then dividing by
showing the above inequality is equivalent to showing that for all t ∈ ℂ
Now q > 2 and so by the same argument given in proving Corollary 23.3.5, for t = reiθ, the
left side of the above inequality is maximized when θ = 0. Hence, from Lemma
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
Let 1 < p < 2. then for 1∕p + 1∕q = 1,
Proof: The first was established above.
Now p∕q < 1 and so the backwards Minkowski inequality applies. Thus
From Corollary 23.3.8,
Now with these Clarkson inequalities, it is not hard to show that all the Lp spaces are
Theorem 23.3.10 The Lp spaces are uniformly convex.
Proof: First suppose p ≥ 2. Suppose
from the first Clarkson inequality,
Next suppose 1 < p < 2 and
1. Then from the second Clarkson
which shows that