15.3 Uniform Convexity Of Lp
These terms refer roughly to how round the unit ball is. Here is the definition.
Definition 15.3.1 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.
Lemma 15.3.2 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∕
Now divide and then take the pth
Here is the backwards Minkowski inequality.
Corollary 15.3.3 Let 0 < p < 1 and suppose ∫
pdμ < ∞ for h
Proof: If ∫
= 0 then there is nothing to prove so assume this is not
Hence the backward Holder inequality applies and it follows that
and so, dividing gives the desired inequality. ■
Consider the easy Clarkson inequalities.
Lemma 15.3.4 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,
holds if and only if
for all x ≥ 1. Let
= 0 and
Since p − 1 ≥ 1, by convexity of f
0 for all x ≥
Corollary 15.3.5 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 is 2−p times
a continuous periodic function for θ ∈ ℝ
which achieves its maximum value when
This follows from the first derivative test from calculus. Therefore, for
by the above lemma. ■
With this corollary, here is the easy Clarkson inequality.
Theorem 15.3.6 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 15.3.7 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
. Then you
Multiplying both sides by
this is equivalent to showing that for all
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
What about l
so this is negative. Then
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 15.3.8. 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
all k ≥
1 because q >
2. Then since 0 < s <
However, this is nonnegative because it equals
As before, this leads to the following corollary.
Corollary 15.3.8 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 15.3.5, 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 15.3.9 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 15.3.8,
Now with these Clarkson inequalities, it is not hard to show that all the Lp spaces are
Theorem 15.3.10 The Lp spaces are uniformly convex.
Proof: First suppose p ≥ 2. Suppose
Then from the first Clarkson inequality,
Next suppose 1 < p < 2 and
1. Then from the second Clarkson
which shows that