Is the Legendre transform connected to identity in any sense
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Is the Legendre transform connected to identity in any sense? Is there a "good" way to interpolate continuously between a convex function and its Legendre transform in the sense that
$gamma: [0,1] rightarrow { text{mappings of convex functions into convex functions} }$
such that
$gamma(0) = id quad text{and} quad gamma(1) = text{Legendre transform}$
convex-analysis
asked Dec 13 '18 at 16:42
tonydotonydo
1477
$endgroup$
add a comment |
$begingroup$
Is the Legendre transform connected to identity in any sense? Is there a "good" way to interpolate continuously between a convex function and its Legendre transform in the sense that
$gamma: [0,1] rightarrow { text{mappings of convex functions into convex functions} }$
such that
$gamma(0) = id quad text{and} quad gamma(1) = text{Legendre transform}$
convex-analysis
asked Dec 13 '18 at 16:42
tonydotonydo
1477
$endgroup$
$begingroup$
What do you mean by "connected to identity"?
$endgroup$
– BigbearZzz
Dec 13 '18 at 16:47
$begingroup$
clarified a little
$endgroup$
– tonydo
Dec 13 '18 at 16:55
add a comment |
$begingroup$
Is the Legendre transform connected to identity in any sense? Is there a "good" way to interpolate continuously between a convex function and its Legendre transform in the sense that
$gamma: [0,1] rightarrow { text{mappings of convex functions into convex functions} }$
such that
$gamma(0) = id quad text{and} quad gamma(1) = text{Legendre transform}$
convex-analysis
asked Dec 13 '18 at 16:42
tonydotonydo
1477
$endgroup$
Is the Legendre transform connected to identity in any sense? Is there a "good" way to interpolate continuously between a convex function and its Legendre transform in the sense that
$gamma: [0,1] rightarrow { text{mappings of convex functions into convex functions} }$
such that
$gamma(0) = id quad text{and} quad gamma(1) = text{Legendre transform}$
convex-analysis
convex-analysis
asked Dec 13 '18 at 16:42
tonydotonydo
1477
asked Dec 13 '18 at 16:42
tonydotonydo
1477
edited Dec 13 '18 at 16:54
tonydo
asked Dec 13 '18 at 16:42
tonydotonydo
1477
asked Dec 13 '18 at 16:42
tonydotonydo
1477
asked Dec 13 '18 at 16:42
tonydotonydo
1477
1477
$begingroup$
What do you mean by "connected to identity"?
$endgroup$
– BigbearZzz
Dec 13 '18 at 16:47
$begingroup$
clarified a little
$endgroup$
– tonydo
Dec 13 '18 at 16:55
add a comment |
$begingroup$
What do you mean by "connected to identity"?
$endgroup$
– BigbearZzz
Dec 13 '18 at 16:47
$begingroup$
clarified a little
$endgroup$
– tonydo
Dec 13 '18 at 16:55
$begingroup$
What do you mean by "connected to identity"?
$endgroup$
– BigbearZzz
Dec 13 '18 at 16:47
$begingroup$
What do you mean by "connected to identity"?
$endgroup$
– BigbearZzz
Dec 13 '18 at 16:47
$begingroup$
clarified a little
$endgroup$
– tonydo
Dec 13 '18 at 16:55
$begingroup$
clarified a little
$endgroup$
– tonydo
Dec 13 '18 at 16:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I doubt there's anything like what you're seeking. Let $X$ be a Banach space, for a proper function $f:Xto Bbb Rcup {infty}$, its Legendre-Fenchel transform $f^*:X^*toBbb Rcup{infty}$ is defined on a totally different space.
Given that you want $gamma$ such that $gamma(0)=f$ and $gamma(1)=f^*$, the domain of the function $gamma(t)$ for $tin (0,1)$ is already a problem. Do you want it to be $X,X^*$ or something else entirely? You'd need to think of a continuous way to transform $X$ into $X^*$ (and I'm not even sure how to make that precise).
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
$endgroup$
$begingroup$
doesn't need to be general; what about special cases for $X=[text{min}(f),infty]$?
$endgroup$
– tonydo
Dec 13 '18 at 17:14
$begingroup$
It just happens to be the case that $(Bbb R^n)^*simeqBbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^*$ so you can't expect anything "in between".
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:18
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
I doubt there's anything like what you're seeking. Let $X$ be a Banach space, for a proper function $f:Xto Bbb Rcup {infty}$, its Legendre-Fenchel transform $f^*:X^*toBbb Rcup{infty}$ is defined on a totally different space.
Given that you want $gamma$ such that $gamma(0)=f$ and $gamma(1)=f^*$, the domain of the function $gamma(t)$ for $tin (0,1)$ is already a problem. Do you want it to be $X,X^*$ or something else entirely? You'd need to think of a continuous way to transform $X$ into $X^*$ (and I'm not even sure how to make that precise).
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
$endgroup$
$begingroup$
doesn't need to be general; what about special cases for $X=[text{min}(f),infty]$?
$endgroup$
– tonydo
Dec 13 '18 at 17:14
$begingroup$
It just happens to be the case that $(Bbb R^n)^*simeqBbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^*$ so you can't expect anything "in between".
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:18
add a comment |
$begingroup$
I doubt there's anything like what you're seeking. Let $X$ be a Banach space, for a proper function $f:Xto Bbb Rcup {infty}$, its Legendre-Fenchel transform $f^*:X^*toBbb Rcup{infty}$ is defined on a totally different space.
Given that you want $gamma$ such that $gamma(0)=f$ and $gamma(1)=f^*$, the domain of the function $gamma(t)$ for $tin (0,1)$ is already a problem. Do you want it to be $X,X^*$ or something else entirely? You'd need to think of a continuous way to transform $X$ into $X^*$ (and I'm not even sure how to make that precise).
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
$endgroup$
$begingroup$
doesn't need to be general; what about special cases for $X=[text{min}(f),infty]$?
$endgroup$
– tonydo
Dec 13 '18 at 17:14
$begingroup$
It just happens to be the case that $(Bbb R^n)^*simeqBbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^*$ so you can't expect anything "in between".
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:18
add a comment |
$begingroup$
I doubt there's anything like what you're seeking. Let $X$ be a Banach space, for a proper function $f:Xto Bbb Rcup {infty}$, its Legendre-Fenchel transform $f^*:X^*toBbb Rcup{infty}$ is defined on a totally different space.
Given that you want $gamma$ such that $gamma(0)=f$ and $gamma(1)=f^*$, the domain of the function $gamma(t)$ for $tin (0,1)$ is already a problem. Do you want it to be $X,X^*$ or something else entirely? You'd need to think of a continuous way to transform $X$ into $X^*$ (and I'm not even sure how to make that precise).
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
$endgroup$
I doubt there's anything like what you're seeking. Let $X$ be a Banach space, for a proper function $f:Xto Bbb Rcup {infty}$, its Legendre-Fenchel transform $f^*:X^*toBbb Rcup{infty}$ is defined on a totally different space.
Given that you want $gamma$ such that $gamma(0)=f$ and $gamma(1)=f^*$, the domain of the function $gamma(t)$ for $tin (0,1)$ is already a problem. Do you want it to be $X,X^*$ or something else entirely? You'd need to think of a continuous way to transform $X$ into $X^*$ (and I'm not even sure how to make that precise).
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
answered Dec 13 '18 at 17:04
BigbearZzzBigbearZzz
7,56821650
7,56821650
$begingroup$
doesn't need to be general; what about special cases for $X=[text{min}(f),infty]$?
$endgroup$
– tonydo
Dec 13 '18 at 17:14
$begingroup$
It just happens to be the case that $(Bbb R^n)^*simeqBbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^*$ so you can't expect anything "in between".
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:18
add a comment |
$begingroup$
doesn't need to be general; what about special cases for $X=[text{min}(f),infty]$?
$endgroup$
– tonydo
Dec 13 '18 at 17:14
$begingroup$
It just happens to be the case that $(Bbb R^n)^*simeqBbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^*$ so you can't expect anything "in between".
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:18
$begingroup$
doesn't need to be general; what about special cases for $X=[text{min}(f),infty]$?
$endgroup$
– tonydo
Dec 13 '18 at 17:14
$begingroup$
doesn't need to be general; what about special cases for $X=[text{min}(f),infty]$?
$endgroup$
– tonydo
Dec 13 '18 at 17:14
$begingroup$
It just happens to be the case that $(Bbb R^n)^*simeqBbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^*$ so you can't expect anything "in between".
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:18
$begingroup$
It just happens to be the case that $(Bbb R^n)^*simeqBbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^*$ so you can't expect anything "in between".
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:18
add a comment |
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$begingroup$
What do you mean by "connected to identity"?
$endgroup$
– BigbearZzz
Dec 13 '18 at 16:47
$begingroup$
clarified a little
$endgroup$
– tonydo
Dec 13 '18 at 16:55