Mixing
Existence minimizer for total variation over measures
$begingroup$
I want to prove that there exists a minimizer to the following problem
$$
min || mu ||_{text{TV}} text{ such that } mathcal{F} mu = y
$$
where $mu in mathcal{M}([0,1])$, the space of Radon measures in the interval $[0,1]$, and $mathcal{F}: mathcal{M}([0,1]) to mathbb{C}^n$ is some linear operator, to which we know that $mathcal{F}^{-1} y neq { emptyset }$.
Typically for convex problems like this, I would approach it by using convexity and lower semi-continuity of the total variation norm. But since the underlying space is (I think) not reflexive, there is no weak compacity for closed, convex, bounded sets.
measure-theory optimization convex-optimization total-variation
asked Jan 22 at 10:32
panchopancho
531210
$endgroup$
add a comment |
$begingroup$
I want to prove that there exists a minimizer to the following problem
$$
min || mu ||_{text{TV}} text{ such that } mathcal{F} mu = y
$$
where $mu in mathcal{M}([0,1])$, the space of Radon measures in the interval $[0,1]$, and $mathcal{F}: mathcal{M}([0,1]) to mathbb{C}^n$ is some linear operator, to which we know that $mathcal{F}^{-1} y neq { emptyset }$.
Typically for convex problems like this, I would approach it by using convexity and lower semi-continuity of the total variation norm. But since the underlying space is (I think) not reflexive, there is no weak compacity for closed, convex, bounded sets.
measure-theory optimization convex-optimization total-variation
asked Jan 22 at 10:32
panchopancho
531210
$endgroup$
add a comment |
$begingroup$
I want to prove that there exists a minimizer to the following problem
$$
min || mu ||_{text{TV}} text{ such that } mathcal{F} mu = y
$$
where $mu in mathcal{M}([0,1])$, the space of Radon measures in the interval $[0,1]$, and $mathcal{F}: mathcal{M}([0,1]) to mathbb{C}^n$ is some linear operator, to which we know that $mathcal{F}^{-1} y neq { emptyset }$.
Typically for convex problems like this, I would approach it by using convexity and lower semi-continuity of the total variation norm. But since the underlying space is (I think) not reflexive, there is no weak compacity for closed, convex, bounded sets.
measure-theory optimization convex-optimization total-variation
asked Jan 22 at 10:32
panchopancho
531210
$endgroup$
I want to prove that there exists a minimizer to the following problem
$$
min || mu ||_{text{TV}} text{ such that } mathcal{F} mu = y
$$
where $mu in mathcal{M}([0,1])$, the space of Radon measures in the interval $[0,1]$, and $mathcal{F}: mathcal{M}([0,1]) to mathbb{C}^n$ is some linear operator, to which we know that $mathcal{F}^{-1} y neq { emptyset }$.
Typically for convex problems like this, I would approach it by using convexity and lower semi-continuity of the total variation norm. But since the underlying space is (I think) not reflexive, there is no weak compacity for closed, convex, bounded sets.
measure-theory optimization convex-optimization total-variation
measure-theory optimization convex-optimization total-variation
asked Jan 22 at 10:32
panchopancho
531210
asked Jan 22 at 10:32
panchopancho
531210
edited Jan 22 at 12:40
pancho
asked Jan 22 at 10:32
panchopancho
531210
asked Jan 22 at 10:32
panchopancho
531210
asked Jan 22 at 10:32
panchopancho
531210
531210
add a comment |
add a comment |
1 Answer
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Typically, you can use that $mathcal M(([0,1])$ is the dual space of the separable space $C([0,1])$. Hence, for every bounded sequence you can extract weak-$*$ convergent subsequences. If $mathcal F$ is also continuous w.r.t. the weak-$*$ convergent sequences, then you can do the usual arguments. If you do not know this (additional) continuity of $mathcal F$, the existence of minimizers may fail.
answered Jan 24 at 7:59
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Indeed, $mathcal{F}$ is in particular defined as applying the measures to a family of predefined $C([0,1])$, therefore weak-* continuity is also present. Thank you !
$endgroup$
– pancho
Jan 24 at 9:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Typically, you can use that $mathcal M(([0,1])$ is the dual space of the separable space $C([0,1])$. Hence, for every bounded sequence you can extract weak-$*$ convergent subsequences. If $mathcal F$ is also continuous w.r.t. the weak-$*$ convergent sequences, then you can do the usual arguments. If you do not know this (additional) continuity of $mathcal F$, the existence of minimizers may fail.
answered Jan 24 at 7:59
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Indeed, $mathcal{F}$ is in particular defined as applying the measures to a family of predefined $C([0,1])$, therefore weak-* continuity is also present. Thank you !
$endgroup$
– pancho
Jan 24 at 9:40
add a comment |
$begingroup$
Typically, you can use that $mathcal M(([0,1])$ is the dual space of the separable space $C([0,1])$. Hence, for every bounded sequence you can extract weak-$*$ convergent subsequences. If $mathcal F$ is also continuous w.r.t. the weak-$*$ convergent sequences, then you can do the usual arguments. If you do not know this (additional) continuity of $mathcal F$, the existence of minimizers may fail.
answered Jan 24 at 7:59
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Indeed, $mathcal{F}$ is in particular defined as applying the measures to a family of predefined $C([0,1])$, therefore weak-* continuity is also present. Thank you !
$endgroup$
– pancho
Jan 24 at 9:40
add a comment |
$begingroup$
Typically, you can use that $mathcal M(([0,1])$ is the dual space of the separable space $C([0,1])$. Hence, for every bounded sequence you can extract weak-$*$ convergent subsequences. If $mathcal F$ is also continuous w.r.t. the weak-$*$ convergent sequences, then you can do the usual arguments. If you do not know this (additional) continuity of $mathcal F$, the existence of minimizers may fail.
answered Jan 24 at 7:59
gerwgerw
19.6k11334
$endgroup$
Typically, you can use that $mathcal M(([0,1])$ is the dual space of the separable space $C([0,1])$. Hence, for every bounded sequence you can extract weak-$*$ convergent subsequences. If $mathcal F$ is also continuous w.r.t. the weak-$*$ convergent sequences, then you can do the usual arguments. If you do not know this (additional) continuity of $mathcal F$, the existence of minimizers may fail.
answered Jan 24 at 7:59
gerwgerw
19.6k11334
answered Jan 24 at 7:59
gerwgerw
19.6k11334
answered Jan 24 at 7:59
gerwgerw
19.6k11334
answered Jan 24 at 7:59
gerwgerw
19.6k11334
19.6k11334
$begingroup$
Indeed, $mathcal{F}$ is in particular defined as applying the measures to a family of predefined $C([0,1])$, therefore weak-* continuity is also present. Thank you !
$endgroup$
– pancho
Jan 24 at 9:40
add a comment |
$begingroup$
Indeed, $mathcal{F}$ is in particular defined as applying the measures to a family of predefined $C([0,1])$, therefore weak-* continuity is also present. Thank you !
$endgroup$
– pancho
Jan 24 at 9:40
$begingroup$
Indeed, $mathcal{F}$ is in particular defined as applying the measures to a family of predefined $C([0,1])$, therefore weak-* continuity is also present. Thank you !
$endgroup$
– pancho
Jan 24 at 9:40
$begingroup$
Indeed, $mathcal{F}$ is in particular defined as applying the measures to a family of predefined $C([0,1])$, therefore weak-* continuity is also present. Thank you !
$endgroup$
– pancho
Jan 24 at 9:40
add a comment |
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