As before, X will be a Banach space in what follows. Sometimes it will be a
reflexive Banach space and in this case, it will be assumed that the norm is strictly
convex.
Definition 23.7.44Let ϕ : X → (−∞,∞]. Then ϕ is convex if whenevert ∈
[0,1]
,x,y ∈ X,
ϕ(tx + (1− t)y) ≤ tϕ(x)+ (1− t)ϕ(y)
The epigraph of ϕ is defined by
epi(ϕ) ≡ {(x,y) : y ≥ ϕ(x)}
Whenepi
(ϕ)
is closed in X × (−∞,∞], we say that ϕ is lower semicontinuous, l.s.c.The function is called proper if ϕ
(x)
< ∞ for some x. The collection of all such x iscalled D
(ϕ)
, the domain of ϕ.
This definition of lower semicontinuity is equivalent to the usual definition.
Lemma 23.7.45The above definition of lower semicontinuity is equivalent to theassertion that whenever x_{n}→ x, it follows that ϕ
(x)
≤ liminf _{n→∞}ϕ
(xn)
. In casethat ϕ is convex, lower semicontinuity is equivalent to weak lower semicontinuity.That isepi
(ϕ)
is closed if and only ifepi
(ϕ )
is weakly closed. In this case, the limitcondition: If x_{x}→ x weakly, then ϕ
(x)
≤ liminf _{n→∞}ϕ
(xn)
is valid.
Proof:Suppose the limit condition holds. Why is epi
a contradiction. It follows that there exists r > 0 such that B
(x,r)
×
(α− δ,α + δ)
2 2
∩epi
(ϕ)
= ∅.
Since epi
(ϕ)
^{C} is open, it follows that epi
(ϕ )
is closed.
Next suppose epi
(ϕ)
is closed. Why does the limit condition hold? Suppose x_{n}→ x.
Then
(xn,ϕ (xn ))
∈epi
(ϕ)
. There is a subsequence such that
α ≡ limni→nf∞ ϕ(xn) = kli→m∞ϕ (xnk)
and so
(xnk,ϕ (xnk))
→
(x,α)
. Since epi
(ϕ)
is closed, this means
(x,α)
∈epi
(ϕ)
.
Hence
α ≡ lim inf ϕ (x ) ≥ ϕ(x).
n→∞ n
Consider the last claim. In this case, epi
(ϕ)
is convex. If it is closed, then it is weakly
closed thanks to separation theorems: If
(x,α)
∈epi
(ϕ)
^{C}, then α < ∞ and so there
exists
(x∗,β )
∈
(X × ℝ)
^{′} and l such that for all
(t,γ)
∈epi
(ϕ)
,
∗ ∗
x (t)+ βγ > l > x (x)+ αβ
Then B_{}
(x∗,β)
((x,α),δ)
is a weakly open set containing
(x,α)
. For δ small enough, it does
not intersect epi
(ϕ)
since if not so, there would exist
(t ,γ )
n n
∈epi
(ϕ )
∩B_{ ∗
(x ,β)
}
((x,α), 1 )
n
and so
x ∗(tn)+ βγn → x∗ (x )+ αβ
contrary to the above inequality. Thus epi
(ϕ)
is weakly closed. Also, if epi
(ϕ)
is weakly
closed, then it is obviously strongly closed.
What of the limit condition using weak convergence instead of strong
convergence? Say x_{n}→ x weakly. Does it follow that if epi
(ϕ)
is weakly closed that
ϕ
(x )
≤ liminf _{n→∞}ϕ
(xn)
? It is just as above. There is a subsequence such
that
α ≡ limni→nf∞ ϕ(xn) = kli→m∞ϕ (xnk)
and so
(x ,ϕ (x ))
nk nk
→
(x,α)
weakly. Since epi
(ϕ)
is weakly closed, this means
(x,α)
∈epi
(ϕ )
. Hence
α ≡ lim inf ϕ (xn) ≥ ϕ(x). ■
n→ ∞
There is also another convenient characterization of what it means for a function to
be lower semicontinuous.
Lemma 23.7.46Let ϕ : X → (−∞,∞]. Then ϕ is lower semicontinuous if andonly if ϕ^{−1}
((a,∞ ])
is open for any a ∈ ℝ.
Proof:Suppose first that epi
(ϕ)
is closed. Consider x ∈ ϕ^{−1}
((a,∞ ])
. Thus
ϕ
(x )
> a. Thus
(x,a)
∈epi
(ϕ)
^{C} because a < ϕ
(x)
. Since epi
(ϕ)
is closed, there exists
r,ε > 0 such that
B (x,r)× (a− ε,a +ε) ⊆ epi(ϕ)C
Hence if y ∈ B
(x,r)
, it follows that ϕ
(y)
≥ a + ε since otherwise there would be a point
of epi
(ϕ)
^{C} in this open set B
(x,r)
×
(a − ε,a+ ε)
. Hence B
(x,r)
⊆ ϕ^{−1}
((a,∞ ])
.
Conversely, suppose ϕ^{−1}
((a,∞ ])
is open for any a and let
(x,b)
∈epi
(ϕ)
^{C}. Then
ϕ
(x )
> b. Thus there exists B
(x,r)
such that for y ∈ B
(x,r)
, it follows that ϕ
(y)
> b.
That is, y ∈ ϕ^{−1}
((b,∞ ])
. So consider B
(x,r)
× (−∞,b). If
(y,α)
∈ B
(x,r)
× (−∞,b),
then since ϕ
(y)
> b,α < ϕ
(y)
and so there is no point of intersection between epi
(ϕ )
and
this open set B
(x,r)
× (−∞,b).■
Of course one can define upper semicontinuous the same way that ϕ^{−1}
(− ∞, a)
is
open. Thus a function is continuous if and only if it is both upper and lower
semicontinuous.
In case X is reflexive, the limit condition implies that epi
(ϕ )
is weakly closed.
Suppose
(x,α)
is a weak limit point of epi
(ϕ)
. Then by the Eberlein Smulian theorem,
there is a subsequence of points of X,
(x ,α )
n n
which converges weakly to
(x,α)
. Thus if
the limit condition holds,
ϕ(x) ≤ lim inf ϕ(x ) ≤ lim inf α = α
n→ ∞ n n→ ∞ n
and so
(x,α)
∈epi
(ϕ )
. If X is not reflexive, this isn’t all that clear because it is not clear
that a limit point is the limit of a sequence. However, one could consider a limit condition
involving nets and get a similar result.
Definition 23.7.47Let ϕ : X → (−∞,∞] be convex lower semicontinuous, and proper.Then
∗ ∗
∂ϕ (x) ≡ {x : ϕ (y)− ϕ(x) ≥ 〈x ,y− x〉 for all y}
The domain of ∂ϕ, denoted as D
(∂ϕ)
is just the set of all x for which ∂ϕ
(x )
≠∅. Notethat D
(∂ϕ)
⊆ D
(ϕ)
since if x
∕∈
D
(ϕ)
, the defining inequality could not hold for all ybecause the left side would be −∞ for some y.
Theorem 23.7.48For X a real Banach space, let ϕ
(x)
≡
1
2
||x||
^{2}. ThenF
(x)
= ∂ϕ
(x )
. Here F was the set valued map satisfying x^{∗}∈ Fx means
The next result gives conditions under which the subgradient is onto. This means that
if y^{∗}∈ X^{′}, then there exists x ∈ X such that y^{∗}∈ ∂ϕ
(x)
.
Theorem 23.7.49Suppose X is a reflexive Banach space and supposeϕ : X → (−∞,∞] is convex, proper, l.s.c., and for all y^{∗}∈ X^{′}, x → ϕ
(x)
−
〈y∗,x〉
iscoercive,
lim ϕ (x) − 〈y∗,x〉 = ∞
||x||→∞
Then ∂ϕ is onto.
Proof:The function x → ϕ
(x)
−y^{∗}
(x)
≡ ψ
(x )
is convex, proper, l.s.c., and coercive.
Let
∗
λ ≡ inf{ϕ(x)− 〈y ,x〉 : x ∈ X}
and let
{xn}
be a minimizing sequence satisfying
∗
λ = nli→m∞ ϕ (xn) − 〈y ,xn〉
By coercivity,
lim ϕ (x) − 〈y∗,x〉 = ∞
||x||→∞
and so this minimizing sequence is bounded. By the Eberlein Smulian theorem,
Theorem 15.5.12, there is a weakly convergent subsequence x_{nk}→ x. By Lemma
23.7.45,
Now let ϕ be a convex proper lower semicontinuous function defined on X where X is
a reflexive Banach space with strictly convex norm. Consider ∂ϕ. Is it maximal
monotone? Is it the case that F + ∂ϕ is onto? First of all, is ∂ϕ monotone? Let
x^{∗}∈ ∂ϕ