and so, letting x^{∗} be a Hahn Banach extension, it follows x^{∗}∈ F
(x)
. Also, F
(x )
is closed
and convex. It is clearly closed because if x_{n}^{∗}→ x^{∗}, the condition on the norm clearly
holds and also the other one does too. It is convex because
^{p−1}
which shows ξ does everyting it needs to do to equal F
(x)
and so it is F
(x)
. Since this
conclusion follows for any convergent sequence, it follows that F
(x+ ty)
converges to
F
(x)
weakly as t → 0. This is what it means to be hemicontinuous. This proves the
following theorem. One can show also that F is demicontinuous which means
strongly convergent sequences go to weakly convergent sequences. Here is a
proof for the case where p = 2. You can clearly do the same thing for arbitrary
p.
Lemma 23.2.2Let F be a duality map for p = 2 where X,X^{′}are reflexive andhave strictly convex norms. (If X is reflexive, there is always an equivalent strictlyconvex norm [?].) Then F is demicontinuous.
Proof: Say x_{n}→ x. Then does it follow that Fx_{n}⇀ Fx? Suppose not. Then there is
a subsequence, still denoted as x_{n} such that x_{n}→ x but Fx_{n}⇀ y≠Fx where here ⇀
denotes weak convergence. This follows from the Eberlein Smulian theorem.
Then
〈y,x〉 = lim 〈F x ,x 〉 = lim ∥x ∥2 = ∥x∥2
n→ ∞ n n n→ ∞ n
. It follows from the above construction of Fx, that
y = Fx after all, a contradiction. ■
Theorem 23.2.3Let X be a reflexive Banach space with X^{′}having strictly convexnorm^{1}.Then for p > 1, there exists a mapping F : X → X^{′}which is bounded, monotone,hemicontinuous, coercive in the sense that lim_{}
|x|
→∞
〈Fx,x〉
∕
|x|
= ∞, which alsosatisfies the inequalities
1∕p′ 1∕p
|〈Fx,y〉| ≤ |〈F x,x〉| |〈Fy,y〉|
Note that these conclusions about duality maps show that they map onto the dual
space.
The duality map was onto and it was monotone. This was shown above. Consider the
form of a duality map for the L^{p} spaces. Let F : L^{p}→
Now here is an interesting inequality which I will only consider in the case where the
quantities are real valued.
Lemma 23.2.4Let p ≥ 2. Then for a,b real numbers,
( p−2 p−2 ) p
|a| a− |b| b (a − b) ≥ C |a− b|
for some constant C independent of a,b.
Proof:There is nothing to show if a = b. Without loss of generality, assume a > b.
Also assume p > 2. There is nothing to show if p = 2. I want to show that there exists a
constant C such that for a > b,
|a|p−2a − |b|p−2 b
---|a−-b|p−1----≥ C (23.2.9)
(23.2.9)
First assume also that b ≥ 0. Now it is clear that as a →∞, the quotient above converges
to 1. Take the derivative of this quotient. This yields
Since b ≥ 0, this is negative and so 1 would be a lower bound. Now suppose b < 0. Then
the above derivative is negative for b < a ≤−b and then it is positive for a > −b. It
equals 0 when a = −b. Therefore the quotient in 23.2.9 achieves its minimum value when
a = −b. This value is
Therefore, the conclusion holds whenever p ≥ 2. That is
( )
|a|p−2a − |b|p−2 b (a− b) ≥-p1−2 |a − b|p.
2
This proves the lemma.
However, in the context of strictly convex norms on the reflexive Banach space X, the
following important result holds. I will give it first for the case where p = 2 since this is
the case of most interest.
Theorem 23.2.5Let X be areflexive Banach space and X,X^{′}have strictly convexnorms as discussed above. Let F be the duality map with p = 2. Then F is strictlymonotone. This means
〈F u− F v,u − v〉 ≥ 0
and it equals 0 if and only if u − v.
Proof:First why is it monotone? By definition of F,
As mentioned, this will hold for any p > 1. Here is a proof in the case that the
Banach space is real which is the usual case of interest. First here is a simple
observation.
Observation 23.2.6Let p > 1. Then x →
|x|
^{p−2}x is strictly monotone. Herex ∈ ℝ.
To verify this observation,
( )
-d- (x2)p−22x = 1-(p− 1)(x2)12p > 0
dx x2
Theorem 23.2.7Let X be a realreflexive Banach space and X,X^{′}have strictly convexnorms as discussed above. Let F be the duality map for p > 1. Then F is strictlymonotone. This means
〈F u− F v,u − v〉 ≥ 0
and it equals 0 if and only if u − v.
Proof:First why is it monotone? By definition of F,
Thus the desired result holds in the case that one vector is a multiple of the other.
The other case is that neither vector is a multiple of the other. Thus, in particular,
x∕