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Is a convex function always continuous?
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It is well known that a convex function defined on $mathbbR$ is continuous (it is even left and right differentiable. Now you can define a convex function for any normed vector space $E$ : $f : Emapsto mathbbR$ is convex iff $$fbig(lambda x + (1-lambda)ybig) le lambda f(x)+(1-lambda)f(y)$$
I know that such a function is not necessarily continuous if $E$ has infinite dimension: $f$ can be a discontinuous linear form. For instance, if $E = ell^2(mathbbN)$ the space of square summable sequences (endowed with its natural norm), and $f(u) = sum limits_i ge 1 fracu_ii$, then $f$ is linear, this convex, yet it ie well-known that $f$ is not continuous.
Now my question is: what about finite dimensions? Does there exist a convex function $f : mathbbR^2 to mathbbR which is not continuous?
I know that there are discontinuous functions from $mathbbR^2$ to $mathbbR$ that have derivatives in every direction (that's a good start since this is a necessary condition !) but I don't know any that is convex.
functional-analysis convex-analysis
asked 48 mins ago
Charles Madeline
3,2081836
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up vote
2
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It is well known that a convex function defined on $mathbbR$ is continuous (it is even left and right differentiable. Now you can define a convex function for any normed vector space $E$ : $f : Emapsto mathbbR$ is convex iff $$fbig(lambda x + (1-lambda)ybig) le lambda f(x)+(1-lambda)f(y)$$
I know that such a function is not necessarily continuous if $E$ has infinite dimension: $f$ can be a discontinuous linear form. For instance, if $E = ell^2(mathbbN)$ the space of square summable sequences (endowed with its natural norm), and $f(u) = sum limits_i ge 1 fracu_ii$, then $f$ is linear, this convex, yet it ie well-known that $f$ is not continuous.
Now my question is: what about finite dimensions? Does there exist a convex function $f : mathbbR^2 to mathbbR which is not continuous?
I know that there are discontinuous functions from $mathbbR^2$ to $mathbbR$ that have derivatives in every direction (that's a good start since this is a necessary condition !) but I don't know any that is convex.
functional-analysis convex-analysis
asked 48 mins ago
Charles Madeline
3,2081836
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
It is well known that a convex function defined on $mathbbR$ is continuous (it is even left and right differentiable. Now you can define a convex function for any normed vector space $E$ : $f : Emapsto mathbbR$ is convex iff $$fbig(lambda x + (1-lambda)ybig) le lambda f(x)+(1-lambda)f(y)$$
I know that such a function is not necessarily continuous if $E$ has infinite dimension: $f$ can be a discontinuous linear form. For instance, if $E = ell^2(mathbbN)$ the space of square summable sequences (endowed with its natural norm), and $f(u) = sum limits_i ge 1 fracu_ii$, then $f$ is linear, this convex, yet it ie well-known that $f$ is not continuous.
Now my question is: what about finite dimensions? Does there exist a convex function $f : mathbbR^2 to mathbbR which is not continuous?
I know that there are discontinuous functions from $mathbbR^2$ to $mathbbR$ that have derivatives in every direction (that's a good start since this is a necessary condition !) but I don't know any that is convex.
functional-analysis convex-analysis
asked 48 mins ago
Charles Madeline
3,2081836
It is well known that a convex function defined on $mathbbR$ is continuous (it is even left and right differentiable. Now you can define a convex function for any normed vector space $E$ : $f : Emapsto mathbbR$ is convex iff $$fbig(lambda x + (1-lambda)ybig) le lambda f(x)+(1-lambda)f(y)$$
I know that such a function is not necessarily continuous if $E$ has infinite dimension: $f$ can be a discontinuous linear form. For instance, if $E = ell^2(mathbbN)$ the space of square summable sequences (endowed with its natural norm), and $f(u) = sum limits_i ge 1 fracu_ii$, then $f$ is linear, this convex, yet it ie well-known that $f$ is not continuous.
Now my question is: what about finite dimensions? Does there exist a convex function $f : mathbbR^2 to mathbbR which is not continuous?
I know that there are discontinuous functions from $mathbbR^2$ to $mathbbR$ that have derivatives in every direction (that's a good start since this is a necessary condition !) but I don't know any that is convex.
functional-analysis convex-analysis
functional-analysis convex-analysis
asked 48 mins ago
Charles Madeline
3,2081836
asked 48 mins ago
Charles Madeline
3,2081836
asked 48 mins ago
Charles Madeline
3,2081836
asked 48 mins ago
Charles Madeline
3,2081836
asked 48 mins ago
Charles Madeline
3,2081836
3,2081836
add a comment |Â
add a comment |Â
1 Answer
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Corollary 10.1.1 of Convex Analysis by Rockafellar says all convex functions from $mathbb R^n$ to $mathbb R$ are continuous.
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Corollary 10.1.1 of Convex Analysis by Rockafellar says all convex functions from $mathbb R^n$ to $mathbb R$ are continuous.
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
add a comment |Â
up vote
4
down vote
Corollary 10.1.1 of Convex Analysis by Rockafellar says all convex functions from $mathbb R^n$ to $mathbb R$ are continuous.
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Corollary 10.1.1 of Convex Analysis by Rockafellar says all convex functions from $mathbb R^n$ to $mathbb R$ are continuous.
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
Corollary 10.1.1 of Convex Analysis by Rockafellar says all convex functions from $mathbb R^n$ to $mathbb R$ are continuous.
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
answered 45 mins ago
Kavi Rama Murthy
31.9k31543
31.9k31543
add a comment |Â
add a comment |Â
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