Convex functions: twice-differentiability property after partitioning domain into uncountable set
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Suppose $f colon C rightarrow [0,1]$ is a convex function, where $C subseteq mathbb{R}^n$ is a full-dimensional compact convex set. A classical theorem of Alexandrov states that $f$ has second derivatives almost everywhere. That is, the set of points in int$(C)$ where $f$ is not twice differentiable forms a set of measure zero.
For all $epsilon in [0,1]$, let $C_{epsilon} := {x in C colon f(x) = epsilon}$. Certainly, $C = bigcup_{epsilon in [0,1]} C_{epsilon}$. Is the following statement true?
For almost every $epsilon in [0,1]$, $f$ has second derivatives almost everywhere in $C_{epsilon}$.
This question is probably best tackled using results from measure theory. However, each $C_{epsilon}$ could be of dimension less than $n$ (we partition $C$ into an uncountably infinite number of sets), so many standard results from measure theory do not apply.
To avoid some possibly trivial cases, it's fine if $C_{epsilon} neq emptyset$ for all $epsilon in [0,1]$.
measure-theory convex-analysis
asked Jan 3 at 20:58
ectoecto
11
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add a comment |
$begingroup$
Suppose $f colon C rightarrow [0,1]$ is a convex function, where $C subseteq mathbb{R}^n$ is a full-dimensional compact convex set. A classical theorem of Alexandrov states that $f$ has second derivatives almost everywhere. That is, the set of points in int$(C)$ where $f$ is not twice differentiable forms a set of measure zero.
For all $epsilon in [0,1]$, let $C_{epsilon} := {x in C colon f(x) = epsilon}$. Certainly, $C = bigcup_{epsilon in [0,1]} C_{epsilon}$. Is the following statement true?
For almost every $epsilon in [0,1]$, $f$ has second derivatives almost everywhere in $C_{epsilon}$.
This question is probably best tackled using results from measure theory. However, each $C_{epsilon}$ could be of dimension less than $n$ (we partition $C$ into an uncountably infinite number of sets), so many standard results from measure theory do not apply.
To avoid some possibly trivial cases, it's fine if $C_{epsilon} neq emptyset$ for all $epsilon in [0,1]$.
measure-theory convex-analysis
asked Jan 3 at 20:58
ectoecto
11
$endgroup$
add a comment |
$begingroup$
Suppose $f colon C rightarrow [0,1]$ is a convex function, where $C subseteq mathbb{R}^n$ is a full-dimensional compact convex set. A classical theorem of Alexandrov states that $f$ has second derivatives almost everywhere. That is, the set of points in int$(C)$ where $f$ is not twice differentiable forms a set of measure zero.
For all $epsilon in [0,1]$, let $C_{epsilon} := {x in C colon f(x) = epsilon}$. Certainly, $C = bigcup_{epsilon in [0,1]} C_{epsilon}$. Is the following statement true?
For almost every $epsilon in [0,1]$, $f$ has second derivatives almost everywhere in $C_{epsilon}$.
This question is probably best tackled using results from measure theory. However, each $C_{epsilon}$ could be of dimension less than $n$ (we partition $C$ into an uncountably infinite number of sets), so many standard results from measure theory do not apply.
To avoid some possibly trivial cases, it's fine if $C_{epsilon} neq emptyset$ for all $epsilon in [0,1]$.
measure-theory convex-analysis
asked Jan 3 at 20:58
ectoecto
11
$endgroup$
Suppose $f colon C rightarrow [0,1]$ is a convex function, where $C subseteq mathbb{R}^n$ is a full-dimensional compact convex set. A classical theorem of Alexandrov states that $f$ has second derivatives almost everywhere. That is, the set of points in int$(C)$ where $f$ is not twice differentiable forms a set of measure zero.
For all $epsilon in [0,1]$, let $C_{epsilon} := {x in C colon f(x) = epsilon}$. Certainly, $C = bigcup_{epsilon in [0,1]} C_{epsilon}$. Is the following statement true?
For almost every $epsilon in [0,1]$, $f$ has second derivatives almost everywhere in $C_{epsilon}$.
This question is probably best tackled using results from measure theory. However, each $C_{epsilon}$ could be of dimension less than $n$ (we partition $C$ into an uncountably infinite number of sets), so many standard results from measure theory do not apply.
To avoid some possibly trivial cases, it's fine if $C_{epsilon} neq emptyset$ for all $epsilon in [0,1]$.
measure-theory convex-analysis
measure-theory convex-analysis
asked Jan 3 at 20:58
ectoecto
11
asked Jan 3 at 20:58
ectoecto
11
asked Jan 3 at 20:58
ectoecto
11
asked Jan 3 at 20:58
ectoecto
11
asked Jan 3 at 20:58
ectoecto
11
11
add a comment |
add a comment |
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