Joint Measurability from Weak Measurability and Continuity
$begingroup$
Suppose I have a function $f:Xtimes H_1to H_2$, where $X$ is a measurable space with some associated $sigma$-algebra, $mathcal{X}$, and $H_1$ and $H_2$ are both separable Hilbert spaces (not necessarily the same). Assume I know that there is a constant $C$ such that $f$ is Lipschitz in the $h$ argument,
$$
|f(x,u) - f(x,v)|_{H_2}leq C |u-v|_{H_1}
$$
and that for all $u$ and $h$, the real valued mapping $xmapsto (f(x,u),h)_{H_2}$ is measurable.
I would like to conclude something about the joint measurability of $f$, in particular that $f$ is $(Xtimes H_1, mathcal{X}times mathcal{B}(H_1))to (H_2, mathcal{B}(H_2))$, but don't quite see how to do it.
I think this can reduced to the problem to proving measurability of
$$
(x,u)mapsto(f(x,u),h_n)_{H_2}
$$
where ${h_n}$ form an orthonormal basis of $H_2$.
general-topology functional-analysis measure-theory
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
$endgroup$
add a comment |
$begingroup$
Suppose I have a function $f:Xtimes H_1to H_2$, where $X$ is a measurable space with some associated $sigma$-algebra, $mathcal{X}$, and $H_1$ and $H_2$ are both separable Hilbert spaces (not necessarily the same). Assume I know that there is a constant $C$ such that $f$ is Lipschitz in the $h$ argument,
$$
|f(x,u) - f(x,v)|_{H_2}leq C |u-v|_{H_1}
$$
and that for all $u$ and $h$, the real valued mapping $xmapsto (f(x,u),h)_{H_2}$ is measurable.
I would like to conclude something about the joint measurability of $f$, in particular that $f$ is $(Xtimes H_1, mathcal{X}times mathcal{B}(H_1))to (H_2, mathcal{B}(H_2))$, but don't quite see how to do it.
I think this can reduced to the problem to proving measurability of
$$
(x,u)mapsto(f(x,u),h_n)_{H_2}
$$
where ${h_n}$ form an orthonormal basis of $H_2$.
general-topology functional-analysis measure-theory
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
$endgroup$
add a comment |
$begingroup$
Suppose I have a function $f:Xtimes H_1to H_2$, where $X$ is a measurable space with some associated $sigma$-algebra, $mathcal{X}$, and $H_1$ and $H_2$ are both separable Hilbert spaces (not necessarily the same). Assume I know that there is a constant $C$ such that $f$ is Lipschitz in the $h$ argument,
$$
|f(x,u) - f(x,v)|_{H_2}leq C |u-v|_{H_1}
$$
and that for all $u$ and $h$, the real valued mapping $xmapsto (f(x,u),h)_{H_2}$ is measurable.
I would like to conclude something about the joint measurability of $f$, in particular that $f$ is $(Xtimes H_1, mathcal{X}times mathcal{B}(H_1))to (H_2, mathcal{B}(H_2))$, but don't quite see how to do it.
I think this can reduced to the problem to proving measurability of
$$
(x,u)mapsto(f(x,u),h_n)_{H_2}
$$
where ${h_n}$ form an orthonormal basis of $H_2$.
general-topology functional-analysis measure-theory
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
$endgroup$
Suppose I have a function $f:Xtimes H_1to H_2$, where $X$ is a measurable space with some associated $sigma$-algebra, $mathcal{X}$, and $H_1$ and $H_2$ are both separable Hilbert spaces (not necessarily the same). Assume I know that there is a constant $C$ such that $f$ is Lipschitz in the $h$ argument,
$$
|f(x,u) - f(x,v)|_{H_2}leq C |u-v|_{H_1}
$$
and that for all $u$ and $h$, the real valued mapping $xmapsto (f(x,u),h)_{H_2}$ is measurable.
I would like to conclude something about the joint measurability of $f$, in particular that $f$ is $(Xtimes H_1, mathcal{X}times mathcal{B}(H_1))to (H_2, mathcal{B}(H_2))$, but don't quite see how to do it.
I think this can reduced to the problem to proving measurability of
$$
(x,u)mapsto(f(x,u),h_n)_{H_2}
$$
where ${h_n}$ form an orthonormal basis of $H_2$.
general-topology functional-analysis measure-theory
general-topology functional-analysis measure-theory
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
asked Dec 2 '18 at 21:35
user2379888user2379888
24019
24019
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I realize now this is really just an instance of a Caratheodory functions. Once we define
$$
f_n(x,u) = (f(x,u),h_n)_{H_2}
$$
we have a function which is continuous in one argument and measurable in the other on the separable and measurable spaces, so it is jointly measurable. Then using separability again,
$$
|f(x,u)-y|_{H_2} < rLeftrightarrow sup_n |(f(x,u)-y,h_n)_{H_2}|<r
$$
and the result follows by the countability of the $h_n$.
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
I realize now this is really just an instance of a Caratheodory functions. Once we define
$$
f_n(x,u) = (f(x,u),h_n)_{H_2}
$$
we have a function which is continuous in one argument and measurable in the other on the separable and measurable spaces, so it is jointly measurable. Then using separability again,
$$
|f(x,u)-y|_{H_2} < rLeftrightarrow sup_n |(f(x,u)-y,h_n)_{H_2}|<r
$$
and the result follows by the countability of the $h_n$.
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
$endgroup$
add a comment |
$begingroup$
I realize now this is really just an instance of a Caratheodory functions. Once we define
$$
f_n(x,u) = (f(x,u),h_n)_{H_2}
$$
we have a function which is continuous in one argument and measurable in the other on the separable and measurable spaces, so it is jointly measurable. Then using separability again,
$$
|f(x,u)-y|_{H_2} < rLeftrightarrow sup_n |(f(x,u)-y,h_n)_{H_2}|<r
$$
and the result follows by the countability of the $h_n$.
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
$endgroup$
add a comment |
$begingroup$
I realize now this is really just an instance of a Caratheodory functions. Once we define
$$
f_n(x,u) = (f(x,u),h_n)_{H_2}
$$
we have a function which is continuous in one argument and measurable in the other on the separable and measurable spaces, so it is jointly measurable. Then using separability again,
$$
|f(x,u)-y|_{H_2} < rLeftrightarrow sup_n |(f(x,u)-y,h_n)_{H_2}|<r
$$
and the result follows by the countability of the $h_n$.
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
$endgroup$
I realize now this is really just an instance of a Caratheodory functions. Once we define
$$
f_n(x,u) = (f(x,u),h_n)_{H_2}
$$
we have a function which is continuous in one argument and measurable in the other on the separable and measurable spaces, so it is jointly measurable. Then using separability again,
$$
|f(x,u)-y|_{H_2} < rLeftrightarrow sup_n |(f(x,u)-y,h_n)_{H_2}|<r
$$
and the result follows by the countability of the $h_n$.
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
answered Dec 3 '18 at 1:48
user2379888user2379888
24019
24019
add a comment |
add a comment |
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