Is the Legendre transform connected to identity in any sense












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 16:54







tonydo

















asked Dec 13 '18 at 16:42









tonydotonydo

1497




1497












  • $begingroup$
    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










  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















2












$begingroup$

Yes, this is possible using the proximal average. The link to the official paper is https://epubs.siam.org/doi/10.1137/060664513






share|cite|improve this answer









$endgroup$













  • $begingroup$
    exactly what I was looking for
    $endgroup$
    – tonydo
    Jan 14 at 18:08



















1












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Yes, this is possible using the proximal average. The link to the official paper is https://epubs.siam.org/doi/10.1137/060664513






share|cite|improve this answer









$endgroup$













  • $begingroup$
    exactly what I was looking for
    $endgroup$
    – tonydo
    Jan 14 at 18:08
















2












$begingroup$

Yes, this is possible using the proximal average. The link to the official paper is https://epubs.siam.org/doi/10.1137/060664513






share|cite|improve this answer









$endgroup$













  • $begingroup$
    exactly what I was looking for
    $endgroup$
    – tonydo
    Jan 14 at 18:08














2












2








2





$begingroup$

Yes, this is possible using the proximal average. The link to the official paper is https://epubs.siam.org/doi/10.1137/060664513






share|cite|improve this answer









$endgroup$



Yes, this is possible using the proximal average. The link to the official paper is https://epubs.siam.org/doi/10.1137/060664513







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 7:37









max_zornmax_zorn

3,43061429




3,43061429












  • $begingroup$
    exactly what I was looking for
    $endgroup$
    – tonydo
    Jan 14 at 18:08


















  • $begingroup$
    exactly what I was looking for
    $endgroup$
    – tonydo
    Jan 14 at 18:08
















$begingroup$
exactly what I was looking for
$endgroup$
– tonydo
Jan 14 at 18:08




$begingroup$
exactly what I was looking for
$endgroup$
– tonydo
Jan 14 at 18:08











1












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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).






share|cite|improve this answer









$endgroup$



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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 17:04









BigbearZzzBigbearZzz

8,93521652




8,93521652












  • $begingroup$
    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












  • $begingroup$
    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


















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