The following notation and notions are used, where
R
:
X
⇉
Y
{\displaystyle {\mathcal {R}}:X\rightrightarrows Y}
is a set-valued function and
S
{\displaystyle S}
is a non-empty subset of a topological vector space
X
{\displaystyle X}
:
the affine span of
S
{\displaystyle S}
is denoted by
aff
S
{\displaystyle \operatorname {aff} S}
and the linear span is denoted by
span
S
.
{\displaystyle \operatorname {span} S.}
S
i
:=
aint
X
S
{\displaystyle S^{i}:=\operatorname {aint} _{X}S}
denotes the algebraic interior of
S
{\displaystyle S}
in
X
.
{\displaystyle X.}
i
S
:=
aint
aff
(
S
−
S
)
S
{\displaystyle {}^{i}S:=\operatorname {aint} _{\operatorname {aff} (S-S)}S}
denotes the relative algebraic interior of
S
{\displaystyle S}
(i.e. the algebraic interior of
S
{\displaystyle S}
in
aff
(
S
−
S
)
{\displaystyle \operatorname {aff} (S-S)}
).
i
b
S
:=
i
S
{\displaystyle {}^{ib}S:={}^{i}S}
if
span
(
S
−
s
0
)
{\displaystyle \operatorname {span} \left(S-s_{0}\right)}
is barreled for some/every
s
0
∈
S
{\displaystyle s_{0}\in S}
while
i
b
S
:=
∅
{\displaystyle {}^{ib}S:=\varnothing }
otherwise.
If
S
{\displaystyle S}
is convex then it can be shown that for any
x
∈
X
,
{\displaystyle x\in X,}
x
∈
i
b
S
{\displaystyle x\in {}^{ib}S}
if and only if the cone generated by
S
−
x
{\displaystyle S-x}
is a barreled linear subspace of
X
{\displaystyle X}
or equivalently, if and only if
∪
n
∈
N
n
(
S
−
x
)
{\displaystyle \cup _{n\in \mathbb {N} }n(S-x)}
is a barreled linear subspace of
X
{\displaystyle X}
The domain of
R
{\displaystyle {\mathcal {R}}}
is
Dom
R
:=
{
x
∈
X
:
R
(
x
)
≠
∅
}
.
{\displaystyle \operatorname {Dom} {\mathcal {R}}:=\{x\in X:{\mathcal {R}}(x)\neq \varnothing \}.}
The image of
R
{\displaystyle {\mathcal {R}}}
is
Im
R
:=
∪
x
∈
X
R
(
x
)
.
{\displaystyle \operatorname {Im} {\mathcal {R}}:=\cup _{x\in X}{\mathcal {R}}(x).}
For any subset
A
⊆
X
,
{\displaystyle A\subseteq X,}
R
(
A
)
:=
∪
x
∈
A
R
(
x
)
.
{\displaystyle {\mathcal {R}}(A):=\cup _{x\in A}{\mathcal {R}}(x).}
The graph of
R
{\displaystyle {\mathcal {R}}}
is
gr
R
:=
{
(
x
,
y
)
∈
X
×
Y
:
y
∈
R
(
x
)
}
.
{\displaystyle \operatorname {gr} {\mathcal {R}}:=\{(x,y)\in X\times Y:y\in {\mathcal {R}}(x)\}.}
R
{\displaystyle {\mathcal {R}}}
is closed (respectively, convex ) if the graph of
R
{\displaystyle {\mathcal {R}}}
is closed (resp. convex) in
X
×
Y
.
{\displaystyle X\times Y.}
Note that
R
{\displaystyle {\mathcal {R}}}
is convex if and only if for all
x
0
,
x
1
∈
X
{\displaystyle x_{0},x_{1}\in X}
and all
r
∈
[
0
,
1
]
,
{\displaystyle r\in [0,1],}
r
R
(
x
0
)
+
(
1
−
r
)
R
(
x
1
)
⊆
R
(
r
x
0
+
(
1
−
r
)
x
1
)
.
{\displaystyle r{\mathcal {R}}\left(x_{0}\right)+(1-r){\mathcal {R}}\left(x_{1}\right)\subseteq {\mathcal {R}}\left(rx_{0}+(1-r)x_{1}\right).}
The inverse of
R
{\displaystyle {\mathcal {R}}}
is the set-valued function
R
−
1
:
Y
⇉
X
{\displaystyle {\mathcal {R}}^{-1}:Y\rightrightarrows X}
defined by
R
−
1
(
y
)
:=
{
x
∈
X
:
y
∈
R
(
x
)
}
.
{\displaystyle {\mathcal {R}}^{-1}(y):=\{x\in X:y\in {\mathcal {R}}(x)\}.}
For any subset
B
⊆
Y
,
{\displaystyle B\subseteq Y,}
R
−
1
(
B
)
:=
∪
y
∈
B
R
−
1
(
y
)
.
{\displaystyle {\mathcal {R}}^{-1}(B):=\cup _{y\in B}{\mathcal {R}}^{-1}(y).}
If
f
:
X
→
Y
{\displaystyle f:X\to Y}
is a function, then its inverse is the set-valued function
f
−
1
:
Y
⇉
X
{\displaystyle f^{-1}:Y\rightrightarrows X}
obtained from canonically identifying
f
{\displaystyle f}
with the set-valued function
f
:
X
⇉
Y
{\displaystyle f:X\rightrightarrows Y}
defined by
x
↦
{
f
(
x
)
}
.
{\displaystyle x\mapsto \{f(x)\}.}
int
T
S
{\displaystyle \operatorname {int} _{T}S}
is the topological interior of
S
{\displaystyle S}
with respect to
T
,
{\displaystyle T,}
where
S
⊆
T
.
{\displaystyle S\subseteq T.}
rint
S
:=
int
aff
S
S
{\displaystyle \operatorname {rint} S:=\operatorname {int} _{\operatorname {aff} S}S}
is the interior of
S
{\displaystyle S}
with respect to
aff
S
.
{\displaystyle \operatorname {aff} S.}
Theorem (Ursescu ) — Let
X
{\displaystyle X}
be a complete semi-metrizable locally convex topological vector space and
R
:
X
⇉
Y
{\displaystyle {\mathcal {R}}:X\rightrightarrows Y}
be a closed convex multifunction with non-empty domain.
Assume that
span
(
Im
R
−
y
)
{\displaystyle \operatorname {span} (\operatorname {Im} {\mathcal {R}}-y)}
is a barrelled space for some/every
y
∈
Im
R
.
{\displaystyle y\in \operatorname {Im} {\mathcal {R}}.}
Assume that
y
0
∈
i
(
Im
R
)
{\displaystyle y_{0}\in {}^{i}(\operatorname {Im} {\mathcal {R}})}
and let
x
0
∈
R
−
1
(
y
0
)
{\displaystyle x_{0}\in {\mathcal {R}}^{-1}\left(y_{0}\right)}
(so that
y
0
∈
R
(
x
0
)
{\displaystyle y_{0}\in {\mathcal {R}}\left(x_{0}\right)}
).
Then for every neighborhood
U
{\displaystyle U}
of
x
0
{\displaystyle x_{0}}
in
X
,
{\displaystyle X,}
y
0
{\displaystyle y_{0}}
belongs to the relative interior of
R
(
U
)
{\displaystyle {\mathcal {R}}(U)}
in
aff
(
Im
R
)
{\displaystyle \operatorname {aff} (\operatorname {Im} {\mathcal {R}})}
(that is,
y
0
∈
int
aff
(
Im
R
)
R
(
U
)
{\displaystyle y_{0}\in \operatorname {int} _{\operatorname {aff} (\operatorname {Im} {\mathcal {R}})}{\mathcal {R}}(U)}
).
In particular, if
i
b
(
Im
R
)
≠
∅
{\displaystyle {}^{ib}(\operatorname {Im} {\mathcal {R}})\neq \varnothing }
then
i
b
(
Im
R
)
=
i
(
Im
R
)
=
rint
(
Im
R
)
.
{\displaystyle {}^{ib}(\operatorname {Im} {\mathcal {R}})={}^{i}(\operatorname {Im} {\mathcal {R}})=\operatorname {rint} (\operatorname {Im} {\mathcal {R}}).}
Closed graph theorem
edit
Closed graph theorem — Let
X
{\displaystyle X}
and
Y
{\displaystyle Y}
be Fréchet spaces and
T
:
X
→
Y
{\displaystyle T:X\to Y}
be a linear map. Then
T
{\displaystyle T}
is continuous if and only if the graph of
T
{\displaystyle T}
is closed in
X
×
Y
.
{\displaystyle X\times Y.}
Proof
For the non-trivial direction, assume that the graph of
T
{\displaystyle T}
is closed and let
R
:=
T
−
1
:
Y
⇉
X
.
{\displaystyle {\mathcal {R}}:=T^{-1}:Y\rightrightarrows X.}
It is easy to see that
gr
R
{\displaystyle \operatorname {gr} {\mathcal {R}}}
is closed and convex and that its image is
X
.
{\displaystyle X.}
Given
x
∈
X
,
{\displaystyle x\in X,}
(
T
x
,
x
)
{\displaystyle (Tx,x)}
belongs to
Y
×
X
{\displaystyle Y\times X}
so that for every open neighborhood
V
{\displaystyle V}
of
T
x
{\displaystyle Tx}
in
Y
,
{\displaystyle Y,}
R
(
V
)
=
T
−
1
(
V
)
{\displaystyle {\mathcal {R}}(V)=T^{-1}(V)}
is a neighborhood of
x
{\displaystyle x}
in
X
.
{\displaystyle X.}
Thus
T
{\displaystyle T}
is continuous at
x
.
{\displaystyle x.}
Q.E.D.
Uniform boundedness principle — Let
X
{\displaystyle X}
and
Y
{\displaystyle Y}
be Fréchet spaces and
T
:
X
→
Y
{\displaystyle T:X\to Y}
be a bijective linear map. Then
T
{\displaystyle T}
is continuous if and only if
T
−
1
:
Y
→
X
{\displaystyle T^{-1}:Y\to X}
is continuous. Furthermore, if
T
{\displaystyle T}
is continuous then
T
{\displaystyle T}
is an isomorphism of Fréchet spaces .
Proof
Apply the closed graph theorem to
T
{\displaystyle T}
and
T
−
1
.
{\displaystyle T^{-1}.}
Q.E.D.
Open mapping theorem
edit
Proof
Clearly,
T
{\displaystyle T}
is a closed and convex relation whose image is
Y
.
{\displaystyle Y.}
Let
U
{\displaystyle U}
be a non-empty open subset of
X
,
{\displaystyle X,}
let
y
{\displaystyle y}
be in
T
(
U
)
,
{\displaystyle T(U),}
and let
x
{\displaystyle x}
in
U
{\displaystyle U}
be such that
y
=
T
x
.
{\displaystyle y=Tx.}
From the Ursescu theorem it follows that
T
(
U
)
{\displaystyle T(U)}
is a neighborhood of
y
.
{\displaystyle y.}
Q.E.D.
Additional corollaries
edit
The following notation and notions are used for these corollaries, where
R
:
X
⇉
Y
{\displaystyle {\mathcal {R}}:X\rightrightarrows Y}
is a set-valued function,
S
{\displaystyle S}
is a non-empty subset of a topological vector space
X
{\displaystyle X}
:
a convex series with elements of
S
{\displaystyle S}
is a series of the form
∑
i
=
1
∞
r
i
s
i
{\textstyle \sum _{i=1}^{\infty }r_{i}s_{i}}
where all
s
i
∈
S
{\displaystyle s_{i}\in S}
and
∑
i
=
1
∞
r
i
=
1
{\textstyle \sum _{i=1}^{\infty }r_{i}=1}
is a series of non-negative numbers. If
∑
i
=
1
∞
r
i
s
i
{\textstyle \sum _{i=1}^{\infty }r_{i}s_{i}}
converges then the series is called convergent while if
(
s
i
)
i
=
1
∞
{\displaystyle \left(s_{i}\right)_{i=1}^{\infty }}
is bounded then the series is called bounded and b-convex .
S
{\displaystyle S}
is ideally convex if any convergent b-convex series of elements of
S
{\displaystyle S}
has its sum in
S
.
{\displaystyle S.}
S
{\displaystyle S}
is lower ideally convex if there exists a Fréchet space
Y
{\displaystyle Y}
such that
S
{\displaystyle S}
is equal to the projection onto
X
{\displaystyle X}
of some ideally convex subset B of
X
×
Y
.
{\displaystyle X\times Y.}
Every ideally convex set is lower ideally convex.
Corollary — Let
X
{\displaystyle X}
be a barreled first countable space and let
C
{\displaystyle C}
be a subset of
X
.
{\displaystyle X.}
Then:
If
C
{\displaystyle C}
is lower ideally convex then
C
i
=
int
C
.
{\displaystyle C^{i}=\operatorname {int} C.}
If
C
{\displaystyle C}
is ideally convex then
C
i
=
int
C
=
int
(
cl
C
)
=
(
cl
C
)
i
.
{\displaystyle C^{i}=\operatorname {int} C=\operatorname {int} \left(\operatorname {cl} C\right)=\left(\operatorname {cl} C\right)^{i}.}
Simons' theorem — Let
X
{\displaystyle X}
and
Y
{\displaystyle Y}
be first countable with
X
{\displaystyle X}
locally convex. Suppose that
R
:
X
⇉
Y
{\displaystyle {\mathcal {R}}:X\rightrightarrows Y}
is a multimap with non-empty domain that satisfies condition (Hwx ) or else assume that
X
{\displaystyle X}
is a Fréchet space and that
R
{\displaystyle {\mathcal {R}}}
is lower ideally convex .
Assume that
span
(
Im
R
−
y
)
{\displaystyle \operatorname {span} (\operatorname {Im} {\mathcal {R}}-y)}
is barreled for some/every
y
∈
Im
R
.
{\displaystyle y\in \operatorname {Im} {\mathcal {R}}.}
Assume that
y
0
∈
i
(
Im
R
)
{\displaystyle y_{0}\in {}^{i}(\operatorname {Im} {\mathcal {R}})}
and let
x
0
∈
R
−
1
(
y
0
)
.
{\displaystyle x_{0}\in {\mathcal {R}}^{-1}\left(y_{0}\right).}
Then for every neighborhood
U
{\displaystyle U}
of
x
0
{\displaystyle x_{0}}
in
X
,
{\displaystyle X,}
y
0
{\displaystyle y_{0}}
belongs to the relative interior of
R
(
U
)
{\displaystyle {\mathcal {R}}(U)}
in
aff
(
Im
R
)
{\displaystyle \operatorname {aff} (\operatorname {Im} {\mathcal {R}})}
(i.e.
y
0
∈
int
aff
(
Im
R
)
R
(
U
)
{\displaystyle y_{0}\in \operatorname {int} _{\operatorname {aff} (\operatorname {Im} {\mathcal {R}})}{\mathcal {R}}(U)}
).
In particular, if
i
b
(
Im
R
)
≠
∅
{\displaystyle {}^{ib}(\operatorname {Im} {\mathcal {R}})\neq \varnothing }
then
i
b
(
Im
R
)
=
i
(
Im
R
)
=
rint
(
Im
R
)
.
{\displaystyle {}^{ib}(\operatorname {Im} {\mathcal {R}})={}^{i}(\operatorname {Im} {\mathcal {R}})=\operatorname {rint} (\operatorname {Im} {\mathcal {R}}).}
Robinson–Ursescu theorem
edit
The implication (1)
⟹
{\displaystyle \implies }
(2) in the following theorem is known as the Robinson–Ursescu theorem.
Robinson–Ursescu theorem — Let
(
X
,
‖
⋅
‖
)
{\displaystyle (X,\|\,\cdot \,\|)}
and
(
Y
,
‖
⋅
‖
)
{\displaystyle (Y,\|\,\cdot \,\|)}
be normed spaces and
R
:
X
⇉
Y
{\displaystyle {\mathcal {R}}:X\rightrightarrows Y}
be a multimap with non-empty domain.
Suppose that
Y
{\displaystyle Y}
is a barreled space , the graph of
R
{\displaystyle {\mathcal {R}}}
verifies condition condition (Hwx ) , and that
(
x
0
,
y
0
)
∈
gr
R
.
{\displaystyle (x_{0},y_{0})\in \operatorname {gr} {\mathcal {R}}.}
Let
C
X
{\displaystyle C_{X}}
(resp.
C
Y
{\displaystyle C_{Y}}
) denote the closed unit ball in
X
{\displaystyle X}
(resp.
Y
{\displaystyle Y}
) (so
C
X
=
{
x
∈
X
:
‖
x
‖
≤
1
}
{\displaystyle C_{X}=\{x\in X:\|x\|\leq 1\}}
).
Then the following are equivalent:
y
0
{\displaystyle y_{0}}
belongs to the algebraic interior of
Im
R
.
{\displaystyle \operatorname {Im} {\mathcal {R}}.}
y
0
∈
int
R
(
x
0
+
C
X
)
.
{\displaystyle y_{0}\in \operatorname {int} {\mathcal {R}}\left(x_{0}+C_{X}\right).}
There exists
B
>
0
{\displaystyle B>0}
such that for all
0
≤
r
≤
1
,
{\displaystyle 0\leq r\leq 1,}
y
0
+
B
r
C
Y
⊆
R
(
x
0
+
r
C
X
)
.
{\displaystyle y_{0}+BrC_{Y}\subseteq {\mathcal {R}}\left(x_{0}+rC_{X}\right).}
There exist
A
>
0
{\displaystyle A>0}
and
B
>
0
{\displaystyle B>0}
such that for all
x
∈
x
0
+
A
C
X
{\displaystyle x\in x_{0}+AC_{X}}
and all
y
∈
y
0
+
A
C
Y
,
{\displaystyle y\in y_{0}+AC_{Y},}
d
(
x
,
R
−
1
(
y
)
)
≤
B
⋅
d
(
y
,
R
(
x
)
)
.
{\displaystyle d\left(x,{\mathcal {R}}^{-1}(y)\right)\leq B\cdot d(y,{\mathcal {R}}(x)).}
There exists
B
>
0
{\displaystyle B>0}
such that for all
x
∈
X
{\displaystyle x\in X}
and all
y
∈
y
0
+
B
C
Y
,
{\displaystyle y\in y_{0}+BC_{Y},}
d
(
x
,
R
−
1
(
y
)
)
≤
1
+
‖
x
−
x
0
‖
B
−
‖
y
−
y
0
‖
⋅
d
(
y
,
R
(
x
)
)
.
{\displaystyle d\left(x,{\mathcal {R}}^{-1}(y)\right)\leq {\frac {1+\left\|x-x_{0}\right\|}{B-\left\|y-y_{0}\right\|}}\cdot d(y,{\mathcal {R}}(x)).}