Why is convex conjugate defined on functions taking values on extended real line?
$begingroup$
Recall a definition of convex conjugate (taken from Wiki):
Let $X$ be a real topological vector space, and let $X^*$ be the dual space to $X.$
Denote the dual pairing by
$$langle cdot,cdot rangle :X^*times Xtomathbb{R}.$$
For a function $f:Xto mathbb{R}cup {-infty,infty}$
taking values on the extended real number line, the convex conjugate
$$f^*:X^*tomathbb{R}cup{-infty,infty}$$
is defined in terms of the supremum by
$$f^*(x^*) =sup{ langle x^*,x rangle -f(x)| xin X },$$
or, equivalently, in terms of the infimum by
$$f^*(x^*) =-inf{ f(x) -langle x^*,x rangle | xin X },$$
Do we really need codomain of $f$ to be $mathbb{R}cup{-infty,infty}?$
Would anything go wrong if codomain of $f$ is just $mathbb{R}?$
real-analysis convex-analysis definition
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
$endgroup$
add a comment |
$begingroup$
Recall a definition of convex conjugate (taken from Wiki):
Let $X$ be a real topological vector space, and let $X^*$ be the dual space to $X.$
Denote the dual pairing by
$$langle cdot,cdot rangle :X^*times Xtomathbb{R}.$$
For a function $f:Xto mathbb{R}cup {-infty,infty}$
taking values on the extended real number line, the convex conjugate
$$f^*:X^*tomathbb{R}cup{-infty,infty}$$
is defined in terms of the supremum by
$$f^*(x^*) =sup{ langle x^*,x rangle -f(x)| xin X },$$
or, equivalently, in terms of the infimum by
$$f^*(x^*) =-inf{ f(x) -langle x^*,x rangle | xin X },$$
Do we really need codomain of $f$ to be $mathbb{R}cup{-infty,infty}?$
Would anything go wrong if codomain of $f$ is just $mathbb{R}?$
real-analysis convex-analysis definition
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
$endgroup$
$begingroup$
Since $mathbb R$ is a subset of $mathbb Rcupleft{-infty,inftyright}$, you can of course take any $fcolon Xtomathbb R$ as well. However, $f^*$ might still take the value $+infty$ for some $x^*in X^*$.
$endgroup$
– Christoph
Dec 5 '18 at 9:56
2
$begingroup$
In convex analysis, it's convenient to allow functions to take on extended real values. For example, even if the convex functions $f_i$ are finite everywhere, the function $f(x) = sup_i f_i(x)$ might take on the value $+infty$ for some values of $x$. Also, in convex optimization it's often useful to enforce hard constraints using convex indicator functions. A constraint $x in C$ in an optimization problem can be enforced by including the term $I_C(x)$ in the objective function (where $I_C(x) = 0$ if $x in C$ and $I_C(x) = infty$ otherwise).
$endgroup$
– littleO
Dec 5 '18 at 9:57
$begingroup$
@littleO It seems that you are answering why the convex conjugate can take values on extended real line. But I am curious whether the original function $f$ must take value on extended real line..
$endgroup$
– Idonknow
Dec 6 '18 at 0:37
$begingroup$
Here's one observation: One of the most important facts about the conjugate is that if $f$ is a closed convex function then $(f^*)^* = f$. In order to state this fact, we must be able to take the conjugate of $f^*$, which might not be finite everywhere.
$endgroup$
– littleO
Dec 6 '18 at 0:58
$begingroup$
@littleO I see. I have a clearer picture now. Thanks.
$endgroup$
– Idonknow
Dec 6 '18 at 1:24
add a comment |
$begingroup$
Recall a definition of convex conjugate (taken from Wiki):
Let $X$ be a real topological vector space, and let $X^*$ be the dual space to $X.$
Denote the dual pairing by
$$langle cdot,cdot rangle :X^*times Xtomathbb{R}.$$
For a function $f:Xto mathbb{R}cup {-infty,infty}$
taking values on the extended real number line, the convex conjugate
$$f^*:X^*tomathbb{R}cup{-infty,infty}$$
is defined in terms of the supremum by
$$f^*(x^*) =sup{ langle x^*,x rangle -f(x)| xin X },$$
or, equivalently, in terms of the infimum by
$$f^*(x^*) =-inf{ f(x) -langle x^*,x rangle | xin X },$$
Do we really need codomain of $f$ to be $mathbb{R}cup{-infty,infty}?$
Would anything go wrong if codomain of $f$ is just $mathbb{R}?$
real-analysis convex-analysis definition
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
$endgroup$
Recall a definition of convex conjugate (taken from Wiki):
Let $X$ be a real topological vector space, and let $X^*$ be the dual space to $X.$
Denote the dual pairing by
$$langle cdot,cdot rangle :X^*times Xtomathbb{R}.$$
For a function $f:Xto mathbb{R}cup {-infty,infty}$
taking values on the extended real number line, the convex conjugate
$$f^*:X^*tomathbb{R}cup{-infty,infty}$$
is defined in terms of the supremum by
$$f^*(x^*) =sup{ langle x^*,x rangle -f(x)| xin X },$$
or, equivalently, in terms of the infimum by
$$f^*(x^*) =-inf{ f(x) -langle x^*,x rangle | xin X },$$
Do we really need codomain of $f$ to be $mathbb{R}cup{-infty,infty}?$
Would anything go wrong if codomain of $f$ is just $mathbb{R}?$
real-analysis convex-analysis definition
real-analysis convex-analysis definition
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
asked Dec 5 '18 at 9:49
IdonknowIdonknow
2,400750114
2,400750114
$begingroup$
Since $mathbb R$ is a subset of $mathbb Rcupleft{-infty,inftyright}$, you can of course take any $fcolon Xtomathbb R$ as well. However, $f^*$ might still take the value $+infty$ for some $x^*in X^*$.
$endgroup$
– Christoph
Dec 5 '18 at 9:56
2
$begingroup$
In convex analysis, it's convenient to allow functions to take on extended real values. For example, even if the convex functions $f_i$ are finite everywhere, the function $f(x) = sup_i f_i(x)$ might take on the value $+infty$ for some values of $x$. Also, in convex optimization it's often useful to enforce hard constraints using convex indicator functions. A constraint $x in C$ in an optimization problem can be enforced by including the term $I_C(x)$ in the objective function (where $I_C(x) = 0$ if $x in C$ and $I_C(x) = infty$ otherwise).
$endgroup$
– littleO
Dec 5 '18 at 9:57
$begingroup$
@littleO It seems that you are answering why the convex conjugate can take values on extended real line. But I am curious whether the original function $f$ must take value on extended real line..
$endgroup$
– Idonknow
Dec 6 '18 at 0:37
$begingroup$
Here's one observation: One of the most important facts about the conjugate is that if $f$ is a closed convex function then $(f^*)^* = f$. In order to state this fact, we must be able to take the conjugate of $f^*$, which might not be finite everywhere.
$endgroup$
– littleO
Dec 6 '18 at 0:58
$begingroup$
@littleO I see. I have a clearer picture now. Thanks.
$endgroup$
– Idonknow
Dec 6 '18 at 1:24
add a comment |
$begingroup$
Since $mathbb R$ is a subset of $mathbb Rcupleft{-infty,inftyright}$, you can of course take any $fcolon Xtomathbb R$ as well. However, $f^*$ might still take the value $+infty$ for some $x^*in X^*$.
$endgroup$
– Christoph
Dec 5 '18 at 9:56
2
$begingroup$
In convex analysis, it's convenient to allow functions to take on extended real values. For example, even if the convex functions $f_i$ are finite everywhere, the function $f(x) = sup_i f_i(x)$ might take on the value $+infty$ for some values of $x$. Also, in convex optimization it's often useful to enforce hard constraints using convex indicator functions. A constraint $x in C$ in an optimization problem can be enforced by including the term $I_C(x)$ in the objective function (where $I_C(x) = 0$ if $x in C$ and $I_C(x) = infty$ otherwise).
$endgroup$
– littleO
Dec 5 '18 at 9:57
$begingroup$
@littleO It seems that you are answering why the convex conjugate can take values on extended real line. But I am curious whether the original function $f$ must take value on extended real line..
$endgroup$
– Idonknow
Dec 6 '18 at 0:37
$begingroup$
Here's one observation: One of the most important facts about the conjugate is that if $f$ is a closed convex function then $(f^*)^* = f$. In order to state this fact, we must be able to take the conjugate of $f^*$, which might not be finite everywhere.
$endgroup$
– littleO
Dec 6 '18 at 0:58
$begingroup$
@littleO I see. I have a clearer picture now. Thanks.
$endgroup$
– Idonknow
Dec 6 '18 at 1:24
$begingroup$
Since $mathbb R$ is a subset of $mathbb Rcupleft{-infty,inftyright}$, you can of course take any $fcolon Xtomathbb R$ as well. However, $f^*$ might still take the value $+infty$ for some $x^*in X^*$.
$endgroup$
– Christoph
Dec 5 '18 at 9:56
$begingroup$
Since $mathbb R$ is a subset of $mathbb Rcupleft{-infty,inftyright}$, you can of course take any $fcolon Xtomathbb R$ as well. However, $f^*$ might still take the value $+infty$ for some $x^*in X^*$.
$endgroup$
– Christoph
Dec 5 '18 at 9:56
2
2
$begingroup$
In convex analysis, it's convenient to allow functions to take on extended real values. For example, even if the convex functions $f_i$ are finite everywhere, the function $f(x) = sup_i f_i(x)$ might take on the value $+infty$ for some values of $x$. Also, in convex optimization it's often useful to enforce hard constraints using convex indicator functions. A constraint $x in C$ in an optimization problem can be enforced by including the term $I_C(x)$ in the objective function (where $I_C(x) = 0$ if $x in C$ and $I_C(x) = infty$ otherwise).
$endgroup$
– littleO
Dec 5 '18 at 9:57
$begingroup$
In convex analysis, it's convenient to allow functions to take on extended real values. For example, even if the convex functions $f_i$ are finite everywhere, the function $f(x) = sup_i f_i(x)$ might take on the value $+infty$ for some values of $x$. Also, in convex optimization it's often useful to enforce hard constraints using convex indicator functions. A constraint $x in C$ in an optimization problem can be enforced by including the term $I_C(x)$ in the objective function (where $I_C(x) = 0$ if $x in C$ and $I_C(x) = infty$ otherwise).
$endgroup$
– littleO
Dec 5 '18 at 9:57
$begingroup$
@littleO It seems that you are answering why the convex conjugate can take values on extended real line. But I am curious whether the original function $f$ must take value on extended real line..
$endgroup$
– Idonknow
Dec 6 '18 at 0:37
$begingroup$
@littleO It seems that you are answering why the convex conjugate can take values on extended real line. But I am curious whether the original function $f$ must take value on extended real line..
$endgroup$
– Idonknow
Dec 6 '18 at 0:37
$begingroup$
Here's one observation: One of the most important facts about the conjugate is that if $f$ is a closed convex function then $(f^*)^* = f$. In order to state this fact, we must be able to take the conjugate of $f^*$, which might not be finite everywhere.
$endgroup$
– littleO
Dec 6 '18 at 0:58
$begingroup$
Here's one observation: One of the most important facts about the conjugate is that if $f$ is a closed convex function then $(f^*)^* = f$. In order to state this fact, we must be able to take the conjugate of $f^*$, which might not be finite everywhere.
$endgroup$
– littleO
Dec 6 '18 at 0:58
$begingroup$
@littleO I see. I have a clearer picture now. Thanks.
$endgroup$
– Idonknow
Dec 6 '18 at 1:24
$begingroup$
@littleO I see. I have a clearer picture now. Thanks.
$endgroup$
– Idonknow
Dec 6 '18 at 1:24
add a comment |
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$begingroup$
Since $mathbb R$ is a subset of $mathbb Rcupleft{-infty,inftyright}$, you can of course take any $fcolon Xtomathbb R$ as well. However, $f^*$ might still take the value $+infty$ for some $x^*in X^*$.
$endgroup$
– Christoph
Dec 5 '18 at 9:56
2
$begingroup$
In convex analysis, it's convenient to allow functions to take on extended real values. For example, even if the convex functions $f_i$ are finite everywhere, the function $f(x) = sup_i f_i(x)$ might take on the value $+infty$ for some values of $x$. Also, in convex optimization it's often useful to enforce hard constraints using convex indicator functions. A constraint $x in C$ in an optimization problem can be enforced by including the term $I_C(x)$ in the objective function (where $I_C(x) = 0$ if $x in C$ and $I_C(x) = infty$ otherwise).
$endgroup$
– littleO
Dec 5 '18 at 9:57
$begingroup$
@littleO It seems that you are answering why the convex conjugate can take values on extended real line. But I am curious whether the original function $f$ must take value on extended real line..
$endgroup$
– Idonknow
Dec 6 '18 at 0:37
$begingroup$
Here's one observation: One of the most important facts about the conjugate is that if $f$ is a closed convex function then $(f^*)^* = f$. In order to state this fact, we must be able to take the conjugate of $f^*$, which might not be finite everywhere.
$endgroup$
– littleO
Dec 6 '18 at 0:58
$begingroup$
@littleO I see. I have a clearer picture now. Thanks.
$endgroup$
– Idonknow
Dec 6 '18 at 1:24