To really understand the L^{p} spaces, it is necessary to at some point go through the Clarkson inequalities.
These are used to show that L^{p} 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 ℝ^{2} and observe how the definition holds for this
example.
Definition 7.6.1A 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,
-p-
p−1
= q = p^{′}. The idea in these inequalities is to consider the case that
p ∈
(0,1)
. 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 7.6.2Let 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.
dμ = ∞, there is nothing to prove. Hence assume this is finite. Then
∫ ∫
|f|pdμ = |g|−p|fg|p 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∕
(p− 1)
< 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∕
(1− p)
also
larger than 1 with p +
(1 − p)
= 1,
∫ ( ∫ ) ( ∫ ( ) )1− p
p p -1- 1∕(1−p)
|f| dμ ≤ |fg|dμ |g|p dμ
( ∫ )p( ∫ )1−p
= |fg|dμ |g|p∕p−1 dμ
Now divide by
(∫ )
|g|p∕p− 1dμ
^{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.
. The point at π has the derivative positive for θ close to
π and θ < π and negative for θ > π. Thus a local maximum occurs at π. At 0 the derivative is positive for
θ < 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,