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.










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  • the link to the paper is not working
    – daw
    2 days ago










  • Fixed. thanks. .
    – MOMO
    2 days ago















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.










share|cite|improve this question
























  • the link to the paper is not working
    – daw
    2 days ago










  • Fixed. thanks. .
    – MOMO
    2 days ago













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.










share|cite|improve this question















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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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


















  • 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










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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 $(y,g(y))$.






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  • 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













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active

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1 Answer
1






active

oldest

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active

oldest

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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 $(y,g(y))$.






share|cite|improve this answer





















  • 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

















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 $(y,g(y))$.






share|cite|improve this answer





















  • 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















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 $(y,g(y))$.






share|cite|improve this answer












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 $(y,g(y))$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • 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




















 

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