7.6 Uniform Convexity Of Lp
To really understand the Lp spaces, it is necessary to at some point go through the Clarkson inequalities.
These are used to show that Lp is uniformly convex. This term says that the unit ball is to some
extent like a circle. It described the “roundness” of the unit ball. Here is the definition. In the
following definition, consider the unit disk in ℝ2 and observe how the definition holds for this
Definition 7.6.1 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 inequality,
. The idea in these inequalities is to consider the case that
This inequality is easy to remember if you just take Holder’s inequality and turn it around in the
case that 0 < p <
Lemma 7.6.2 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. Then
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 1∕
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 7.6.3 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 7.6.4 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
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 7.6.5 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 7.19 equals
The derivative is
This equals 0 when θ
or when θ
The point at π
has the derivative positive for θ
and θ < π
and negative for θ > π.
Thus a local maximum occurs at π
. At 0 the derivative is positive for
0 and negative for θ >
0 so a local maximum also occurs at 0 and 2π
. The other two points don’t yield
a local maximum. Thus, from the above lemma,
With this corollary, here is the easy Clarkson inequality.
Theorem 7.6.6 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
Lemma 7.6.7 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 get
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. Then
Thus these alternate between positive and negative with
0 for all k
. What about
= 0 it is positive. When k
= 1 it is also positive. When k
= 2 it equals
is positive when
is odd and is negative when k
Now return to 7.21. The left side equals
The k = 0 terms add to 0. Then this reduces to
From the above observation about the binomial coefficients, the above is larger than
It remains to show the kth term in the above sum is nonnegative. Now q
for all k ≥
2. Then since 0 < s <
0. This uses the identity
As before, this leads to the following corollary.
Corollary 7.6.8 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 7.6.5, for t = reiθ can be used here
From this the hard Clarkson inequality follows. The two Clarkson inequalities are summarized in the
Theorem 7.6.9 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 7.6.8,
Now with these Clarkson inequalities, it is not hard to show that all the Lp spaces are uniformly
Theorem 7.6.10 The Lp spaces are uniformly convex.
Proof: First suppose p ≥ 2. Suppose
Then from the first
Next suppose 1 < p < 2 and
1. Then from the second Clarkson inequality
which shows that
+ class=”left” align=”middle”(K, ℝn)7.7. EXERCISES