A sufficient condition for subdifferentiability

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Corollary 9 in here (page 31) states that a proper convex function $g:Yrightarrow mathbb{R}cup{infty}$ (not necessarily continuous) on a locally convex space $Y$ is subdifferentiable on a point $y$ in the quasi-relative interior of its domain whenever $(y,g(y))$ does not belong to the quasi-relative interior of the epigraph of $g$.
I was wondering when it is possible to omit the last condition about $(y,g(y))$ and still have the subdifferentiablility at $y$.
In particular I may assume one or more of the following:
$Y$ is a dual space of a Banach space.
$g$ is lower semicontinuous.
$g$ is sublinear.
$g$ is continuous on its domain.
functional-analysis convex-analysis locally-convex-spaces
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up vote
2
down vote
favorite
Corollary 9 in here (page 31) states that a proper convex function $g:Yrightarrow mathbb{R}cup{infty}$ (not necessarily continuous) on a locally convex space $Y$ is subdifferentiable on a point $y$ in the quasi-relative interior of its domain whenever $(y,g(y))$ does not belong to the quasi-relative interior of the epigraph of $g$.
I was wondering when it is possible to omit the last condition about $(y,g(y))$ and still have the subdifferentiablility at $y$.
In particular I may assume one or more of the following:
$Y$ is a dual space of a Banach space.
$g$ is lower semicontinuous.
$g$ is sublinear.
$g$ is continuous on its domain.
functional-analysis convex-analysis locally-convex-spaces
the link to the paper is not working
– daw
2 days ago
Fixed. thanks. .
– MOMO
2 days ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Corollary 9 in here (page 31) states that a proper convex function $g:Yrightarrow mathbb{R}cup{infty}$ (not necessarily continuous) on a locally convex space $Y$ is subdifferentiable on a point $y$ in the quasi-relative interior of its domain whenever $(y,g(y))$ does not belong to the quasi-relative interior of the epigraph of $g$.
I was wondering when it is possible to omit the last condition about $(y,g(y))$ and still have the subdifferentiablility at $y$.
In particular I may assume one or more of the following:
$Y$ is a dual space of a Banach space.
$g$ is lower semicontinuous.
$g$ is sublinear.
$g$ is continuous on its domain.
functional-analysis convex-analysis locally-convex-spaces
Corollary 9 in here (page 31) states that a proper convex function $g:Yrightarrow mathbb{R}cup{infty}$ (not necessarily continuous) on a locally convex space $Y$ is subdifferentiable on a point $y$ in the quasi-relative interior of its domain whenever $(y,g(y))$ does not belong to the quasi-relative interior of the epigraph of $g$.
I was wondering when it is possible to omit the last condition about $(y,g(y))$ and still have the subdifferentiablility at $y$.
In particular I may assume one or more of the following:
$Y$ is a dual space of a Banach space.
$g$ is lower semicontinuous.
$g$ is sublinear.
$g$ is continuous on its domain.
functional-analysis convex-analysis locally-convex-spaces
functional-analysis convex-analysis locally-convex-spaces
edited 10 hours ago
asked 2 days ago
MOMO
32519
32519
the link to the paper is not working
– daw
2 days ago
Fixed. thanks. .
– MOMO
2 days ago
add a comment |
the link to the paper is not working
– daw
2 days ago
Fixed. thanks. .
– MOMO
2 days ago
the link to the paper is not working
– daw
2 days ago
the link to the paper is not working
– daw
2 days ago
Fixed. thanks. .
– MOMO
2 days ago
Fixed. thanks. .
– MOMO
2 days ago
add a comment |
1 Answer
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You need additional assumptions: Let $Y$ be an infinite dimensional Banach space and let $g : V to mathbb{R}$ be a linear, unbounded functional. Then it is clear that $g$ has an empty subdifferential everywhere.
If you add continuity (in $y$), then $partial g(y)$ is not empty (see, e.g., the book by Bauschke&Combettes, Prop. 16.14). However, I feel that this already gives you the condition on $(y,g(y))$.
I am interested in the more general case where $g$ is not continuous. I edited the post so it would be more clear.
– MOMO
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You need additional assumptions: Let $Y$ be an infinite dimensional Banach space and let $g : V to mathbb{R}$ be a linear, unbounded functional. Then it is clear that $g$ has an empty subdifferential everywhere.
If you add continuity (in $y$), then $partial g(y)$ is not empty (see, e.g., the book by Bauschke&Combettes, Prop. 16.14). However, I feel that this already gives you the condition on $(y,g(y))$.
I am interested in the more general case where $g$ is not continuous. I edited the post so it would be more clear.
– MOMO
yesterday
add a comment |
up vote
0
down vote
You need additional assumptions: Let $Y$ be an infinite dimensional Banach space and let $g : V to mathbb{R}$ be a linear, unbounded functional. Then it is clear that $g$ has an empty subdifferential everywhere.
If you add continuity (in $y$), then $partial g(y)$ is not empty (see, e.g., the book by Bauschke&Combettes, Prop. 16.14). However, I feel that this already gives you the condition on $(y,g(y))$.
I am interested in the more general case where $g$ is not continuous. I edited the post so it would be more clear.
– MOMO
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
You need additional assumptions: Let $Y$ be an infinite dimensional Banach space and let $g : V to mathbb{R}$ be a linear, unbounded functional. Then it is clear that $g$ has an empty subdifferential everywhere.
If you add continuity (in $y$), then $partial g(y)$ is not empty (see, e.g., the book by Bauschke&Combettes, Prop. 16.14). However, I feel that this already gives you the condition on $(y,g(y))$.
You need additional assumptions: Let $Y$ be an infinite dimensional Banach space and let $g : V to mathbb{R}$ be a linear, unbounded functional. Then it is clear that $g$ has an empty subdifferential everywhere.
If you add continuity (in $y$), then $partial g(y)$ is not empty (see, e.g., the book by Bauschke&Combettes, Prop. 16.14). However, I feel that this already gives you the condition on $(y,g(y))$.
answered yesterday
gerw
18.6k11133
18.6k11133
I am interested in the more general case where $g$ is not continuous. I edited the post so it would be more clear.
– MOMO
yesterday
add a comment |
I am interested in the more general case where $g$ is not continuous. I edited the post so it would be more clear.
– MOMO
yesterday
I am interested in the more general case where $g$ is not continuous. I edited the post so it would be more clear.
– MOMO
yesterday
I am interested in the more general case where $g$ is not continuous. I edited the post so it would be more clear.
– MOMO
yesterday
add a comment |
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X lHAnRWikpGVa7,Xu
the link to the paper is not working
– daw
2 days ago
Fixed. thanks. .
– MOMO
2 days ago