Mixing
Continuous distribution-valued function induces distribution
$begingroup$
Suppose that the map $mathbb{R}^n to mathcal{D}'(mathbb{R}^n), hspace{3mm}etamapsto E_eta$ is continuous. Furthermore let $mu$ be a Radon-measure with compact support.
I'm having trouble showing that the functional $E$ on $mathcal{D}(mathbb{R}^n)$, whose action on a test function $psi$ is defined by
begin{equation}
langle E,psirangle = int_{mathbb{R}^n} langle E_eta, psi rangle dmu(eta),
end{equation}
is in fact a distribution.
functional-analysis lebesgue-integral distribution-theory
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
$endgroup$
add a comment |
$begingroup$
Suppose that the map $mathbb{R}^n to mathcal{D}'(mathbb{R}^n), hspace{3mm}etamapsto E_eta$ is continuous. Furthermore let $mu$ be a Radon-measure with compact support.
I'm having trouble showing that the functional $E$ on $mathcal{D}(mathbb{R}^n)$, whose action on a test function $psi$ is defined by
begin{equation}
langle E,psirangle = int_{mathbb{R}^n} langle E_eta, psi rangle dmu(eta),
end{equation}
is in fact a distribution.
functional-analysis lebesgue-integral distribution-theory
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
$endgroup$
add a comment |
$begingroup$
Suppose that the map $mathbb{R}^n to mathcal{D}'(mathbb{R}^n), hspace{3mm}etamapsto E_eta$ is continuous. Furthermore let $mu$ be a Radon-measure with compact support.
I'm having trouble showing that the functional $E$ on $mathcal{D}(mathbb{R}^n)$, whose action on a test function $psi$ is defined by
begin{equation}
langle E,psirangle = int_{mathbb{R}^n} langle E_eta, psi rangle dmu(eta),
end{equation}
is in fact a distribution.
functional-analysis lebesgue-integral distribution-theory
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
$endgroup$
Suppose that the map $mathbb{R}^n to mathcal{D}'(mathbb{R}^n), hspace{3mm}etamapsto E_eta$ is continuous. Furthermore let $mu$ be a Radon-measure with compact support.
I'm having trouble showing that the functional $E$ on $mathcal{D}(mathbb{R}^n)$, whose action on a test function $psi$ is defined by
begin{equation}
langle E,psirangle = int_{mathbb{R}^n} langle E_eta, psi rangle dmu(eta),
end{equation}
is in fact a distribution.
functional-analysis lebesgue-integral distribution-theory
functional-analysis lebesgue-integral distribution-theory
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
edited Jan 22 at 16:44
Abdelmalek Abdesselam
647311
647311
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
asked Jan 18 at 12:20
Joseph ExpoJoseph Expo
465
465
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The question is a bit ambiguous because the topology on $mathcal{D}'(mathbb{R}^n)$ is not specified when mentioning that $etarightarrow E_{eta}$ is continuous.
I don't see immediately if $E$ is indeed continuous when the chosen topology is the weak-$ast$ topology. However, if the topology is the strong one (a better choice anyway) then the answer is yes as is shown below.
Firstly, by writing $mu$ as a difference of positive measures one can reduce the problem to the case where $mu$ is positive. Moreover, by assumption there is a compact set $K$ on which it is supported so
$$
langle E,psirangle=int_K langle E_eta,psirangle dmu(eta) .
$$
Let $A$ be a bounded set in $mathcal{D}(mathbb{R}^n)$ and for a distribution $T$ let us use the notation
$$
||T||_A=sup_{psiin A} |langle T,psirangle| .
$$
These seminorms define the strong topology on $mathcal{D}'(mathbb{R}^n)$ and are thus continuous. By composition, the map $Krightarrow [0,infty)$, $etamapsto ||E_eta||_A$ is continuous and therefore, by compactness, is bounded by some constant $M$.
Now let $(psi_n)$ be a sequence that converges to $psi$ in $mathcal{D}(mathbb{R}^n)$. Let $A={psi_n : ninmathbb{N}}$ which is bounded.
We have $|langle E_{eta},psi_nrangle|le M$ for all $ninmathbb{N}$ and $etain K$. Moreover, for fixed $eta$, $langle E_eta,psi_nranglerightarrow
langle E_eta,psirangle$. Therefore, by dominated convergence
$$
langle E,psi_nranglerightarrow langle E,psirangle
$$
and $E$ is a distribution.
Addendum recalling some prerequisites:
Here the space of smooth compactly supported functions $mathcal{D}(mathbb{R}^n)$ is a topological vector space. Its topology is defined by an uncountable collection of seminorms which are described explicitly here. As for the notion of bounded set in $mathcal{D}(mathbb{R}^n)$, the following are equivalent, for $Asubsetmathcal{D}(mathbb{R}^n)$:
$A$ is bounded- For every open set $Vsubsetmathcal{D}(mathbb{R}^n)$ which contains the origin, there exists $lambda>0$ such that $lambda Asubset V$.
- For every continuous seminorm $||cdot||$ on $mathcal{D}(mathbb{R}^n)$, we have $sup_{psiin A}||psi||<infty$.
- For every seminorm $||cdot||$ in a defining set of seminorms (e.g., the ones in the MO link I gave above), we have $sup_{psiin A}||psi||<infty$.
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
$endgroup$
$begingroup$
Thanks for the reply. Could you specify what you mean when a set is bounded in $mathcal{D}(mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms?
$endgroup$
– Joseph Expo
Jan 24 at 9:23
$begingroup$
I edited my answer to recall what a bounded set is. If by supremum norm you mean $sup_{xinmathbb{R}^n}|psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $mathbb{R}^n$.
$endgroup$
– Abdelmalek Abdesselam
Jan 24 at 15:37
add a comment |
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$begingroup$
The question is a bit ambiguous because the topology on $mathcal{D}'(mathbb{R}^n)$ is not specified when mentioning that $etarightarrow E_{eta}$ is continuous.
I don't see immediately if $E$ is indeed continuous when the chosen topology is the weak-$ast$ topology. However, if the topology is the strong one (a better choice anyway) then the answer is yes as is shown below.
Firstly, by writing $mu$ as a difference of positive measures one can reduce the problem to the case where $mu$ is positive. Moreover, by assumption there is a compact set $K$ on which it is supported so
$$
langle E,psirangle=int_K langle E_eta,psirangle dmu(eta) .
$$
Let $A$ be a bounded set in $mathcal{D}(mathbb{R}^n)$ and for a distribution $T$ let us use the notation
$$
||T||_A=sup_{psiin A} |langle T,psirangle| .
$$
These seminorms define the strong topology on $mathcal{D}'(mathbb{R}^n)$ and are thus continuous. By composition, the map $Krightarrow [0,infty)$, $etamapsto ||E_eta||_A$ is continuous and therefore, by compactness, is bounded by some constant $M$.
Now let $(psi_n)$ be a sequence that converges to $psi$ in $mathcal{D}(mathbb{R}^n)$. Let $A={psi_n : ninmathbb{N}}$ which is bounded.
We have $|langle E_{eta},psi_nrangle|le M$ for all $ninmathbb{N}$ and $etain K$. Moreover, for fixed $eta$, $langle E_eta,psi_nranglerightarrow
langle E_eta,psirangle$. Therefore, by dominated convergence
$$
langle E,psi_nranglerightarrow langle E,psirangle
$$
and $E$ is a distribution.
Addendum recalling some prerequisites:
Here the space of smooth compactly supported functions $mathcal{D}(mathbb{R}^n)$ is a topological vector space. Its topology is defined by an uncountable collection of seminorms which are described explicitly here. As for the notion of bounded set in $mathcal{D}(mathbb{R}^n)$, the following are equivalent, for $Asubsetmathcal{D}(mathbb{R}^n)$:
$A$ is bounded- For every open set $Vsubsetmathcal{D}(mathbb{R}^n)$ which contains the origin, there exists $lambda>0$ such that $lambda Asubset V$.
- For every continuous seminorm $||cdot||$ on $mathcal{D}(mathbb{R}^n)$, we have $sup_{psiin A}||psi||<infty$.
- For every seminorm $||cdot||$ in a defining set of seminorms (e.g., the ones in the MO link I gave above), we have $sup_{psiin A}||psi||<infty$.
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
$endgroup$
$begingroup$
Thanks for the reply. Could you specify what you mean when a set is bounded in $mathcal{D}(mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms?
$endgroup$
– Joseph Expo
Jan 24 at 9:23
$begingroup$
I edited my answer to recall what a bounded set is. If by supremum norm you mean $sup_{xinmathbb{R}^n}|psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $mathbb{R}^n$.
$endgroup$
– Abdelmalek Abdesselam
Jan 24 at 15:37
add a comment |
$begingroup$
The question is a bit ambiguous because the topology on $mathcal{D}'(mathbb{R}^n)$ is not specified when mentioning that $etarightarrow E_{eta}$ is continuous.
I don't see immediately if $E$ is indeed continuous when the chosen topology is the weak-$ast$ topology. However, if the topology is the strong one (a better choice anyway) then the answer is yes as is shown below.
Firstly, by writing $mu$ as a difference of positive measures one can reduce the problem to the case where $mu$ is positive. Moreover, by assumption there is a compact set $K$ on which it is supported so
$$
langle E,psirangle=int_K langle E_eta,psirangle dmu(eta) .
$$
Let $A$ be a bounded set in $mathcal{D}(mathbb{R}^n)$ and for a distribution $T$ let us use the notation
$$
||T||_A=sup_{psiin A} |langle T,psirangle| .
$$
These seminorms define the strong topology on $mathcal{D}'(mathbb{R}^n)$ and are thus continuous. By composition, the map $Krightarrow [0,infty)$, $etamapsto ||E_eta||_A$ is continuous and therefore, by compactness, is bounded by some constant $M$.
Now let $(psi_n)$ be a sequence that converges to $psi$ in $mathcal{D}(mathbb{R}^n)$. Let $A={psi_n : ninmathbb{N}}$ which is bounded.
We have $|langle E_{eta},psi_nrangle|le M$ for all $ninmathbb{N}$ and $etain K$. Moreover, for fixed $eta$, $langle E_eta,psi_nranglerightarrow
langle E_eta,psirangle$. Therefore, by dominated convergence
$$
langle E,psi_nranglerightarrow langle E,psirangle
$$
and $E$ is a distribution.
Addendum recalling some prerequisites:
Here the space of smooth compactly supported functions $mathcal{D}(mathbb{R}^n)$ is a topological vector space. Its topology is defined by an uncountable collection of seminorms which are described explicitly here. As for the notion of bounded set in $mathcal{D}(mathbb{R}^n)$, the following are equivalent, for $Asubsetmathcal{D}(mathbb{R}^n)$:
$A$ is bounded- For every open set $Vsubsetmathcal{D}(mathbb{R}^n)$ which contains the origin, there exists $lambda>0$ such that $lambda Asubset V$.
- For every continuous seminorm $||cdot||$ on $mathcal{D}(mathbb{R}^n)$, we have $sup_{psiin A}||psi||<infty$.
- For every seminorm $||cdot||$ in a defining set of seminorms (e.g., the ones in the MO link I gave above), we have $sup_{psiin A}||psi||<infty$.
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
$endgroup$
$begingroup$
Thanks for the reply. Could you specify what you mean when a set is bounded in $mathcal{D}(mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms?
$endgroup$
– Joseph Expo
Jan 24 at 9:23
$begingroup$
I edited my answer to recall what a bounded set is. If by supremum norm you mean $sup_{xinmathbb{R}^n}|psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $mathbb{R}^n$.
$endgroup$
– Abdelmalek Abdesselam
Jan 24 at 15:37
add a comment |
$begingroup$
The question is a bit ambiguous because the topology on $mathcal{D}'(mathbb{R}^n)$ is not specified when mentioning that $etarightarrow E_{eta}$ is continuous.
I don't see immediately if $E$ is indeed continuous when the chosen topology is the weak-$ast$ topology. However, if the topology is the strong one (a better choice anyway) then the answer is yes as is shown below.
Firstly, by writing $mu$ as a difference of positive measures one can reduce the problem to the case where $mu$ is positive. Moreover, by assumption there is a compact set $K$ on which it is supported so
$$
langle E,psirangle=int_K langle E_eta,psirangle dmu(eta) .
$$
Let $A$ be a bounded set in $mathcal{D}(mathbb{R}^n)$ and for a distribution $T$ let us use the notation
$$
||T||_A=sup_{psiin A} |langle T,psirangle| .
$$
These seminorms define the strong topology on $mathcal{D}'(mathbb{R}^n)$ and are thus continuous. By composition, the map $Krightarrow [0,infty)$, $etamapsto ||E_eta||_A$ is continuous and therefore, by compactness, is bounded by some constant $M$.
Now let $(psi_n)$ be a sequence that converges to $psi$ in $mathcal{D}(mathbb{R}^n)$. Let $A={psi_n : ninmathbb{N}}$ which is bounded.
We have $|langle E_{eta},psi_nrangle|le M$ for all $ninmathbb{N}$ and $etain K$. Moreover, for fixed $eta$, $langle E_eta,psi_nranglerightarrow
langle E_eta,psirangle$. Therefore, by dominated convergence
$$
langle E,psi_nranglerightarrow langle E,psirangle
$$
and $E$ is a distribution.
Addendum recalling some prerequisites:
Here the space of smooth compactly supported functions $mathcal{D}(mathbb{R}^n)$ is a topological vector space. Its topology is defined by an uncountable collection of seminorms which are described explicitly here. As for the notion of bounded set in $mathcal{D}(mathbb{R}^n)$, the following are equivalent, for $Asubsetmathcal{D}(mathbb{R}^n)$:
$A$ is bounded- For every open set $Vsubsetmathcal{D}(mathbb{R}^n)$ which contains the origin, there exists $lambda>0$ such that $lambda Asubset V$.
- For every continuous seminorm $||cdot||$ on $mathcal{D}(mathbb{R}^n)$, we have $sup_{psiin A}||psi||<infty$.
- For every seminorm $||cdot||$ in a defining set of seminorms (e.g., the ones in the MO link I gave above), we have $sup_{psiin A}||psi||<infty$.
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
$endgroup$
The question is a bit ambiguous because the topology on $mathcal{D}'(mathbb{R}^n)$ is not specified when mentioning that $etarightarrow E_{eta}$ is continuous.
I don't see immediately if $E$ is indeed continuous when the chosen topology is the weak-$ast$ topology. However, if the topology is the strong one (a better choice anyway) then the answer is yes as is shown below.
Firstly, by writing $mu$ as a difference of positive measures one can reduce the problem to the case where $mu$ is positive. Moreover, by assumption there is a compact set $K$ on which it is supported so
$$
langle E,psirangle=int_K langle E_eta,psirangle dmu(eta) .
$$
Let $A$ be a bounded set in $mathcal{D}(mathbb{R}^n)$ and for a distribution $T$ let us use the notation
$$
||T||_A=sup_{psiin A} |langle T,psirangle| .
$$
These seminorms define the strong topology on $mathcal{D}'(mathbb{R}^n)$ and are thus continuous. By composition, the map $Krightarrow [0,infty)$, $etamapsto ||E_eta||_A$ is continuous and therefore, by compactness, is bounded by some constant $M$.
Now let $(psi_n)$ be a sequence that converges to $psi$ in $mathcal{D}(mathbb{R}^n)$. Let $A={psi_n : ninmathbb{N}}$ which is bounded.
We have $|langle E_{eta},psi_nrangle|le M$ for all $ninmathbb{N}$ and $etain K$. Moreover, for fixed $eta$, $langle E_eta,psi_nranglerightarrow
langle E_eta,psirangle$. Therefore, by dominated convergence
$$
langle E,psi_nranglerightarrow langle E,psirangle
$$
and $E$ is a distribution.
Addendum recalling some prerequisites:
Here the space of smooth compactly supported functions $mathcal{D}(mathbb{R}^n)$ is a topological vector space. Its topology is defined by an uncountable collection of seminorms which are described explicitly here. As for the notion of bounded set in $mathcal{D}(mathbb{R}^n)$, the following are equivalent, for $Asubsetmathcal{D}(mathbb{R}^n)$:
$A$ is bounded- For every open set $Vsubsetmathcal{D}(mathbb{R}^n)$ which contains the origin, there exists $lambda>0$ such that $lambda Asubset V$.
- For every continuous seminorm $||cdot||$ on $mathcal{D}(mathbb{R}^n)$, we have $sup_{psiin A}||psi||<infty$.
- For every seminorm $||cdot||$ in a defining set of seminorms (e.g., the ones in the MO link I gave above), we have $sup_{psiin A}||psi||<infty$.
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
edited Jan 24 at 15:34
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
answered Jan 22 at 15:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
647311
647311
$begingroup$
Thanks for the reply. Could you specify what you mean when a set is bounded in $mathcal{D}(mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms?
$endgroup$
– Joseph Expo
Jan 24 at 9:23
$begingroup$
I edited my answer to recall what a bounded set is. If by supremum norm you mean $sup_{xinmathbb{R}^n}|psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $mathbb{R}^n$.
$endgroup$
– Abdelmalek Abdesselam
Jan 24 at 15:37
add a comment |
$begingroup$
Thanks for the reply. Could you specify what you mean when a set is bounded in $mathcal{D}(mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms?
$endgroup$
– Joseph Expo
Jan 24 at 9:23
$begingroup$
I edited my answer to recall what a bounded set is. If by supremum norm you mean $sup_{xinmathbb{R}^n}|psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $mathbb{R}^n$.
$endgroup$
– Abdelmalek Abdesselam
Jan 24 at 15:37
$begingroup$
Thanks for the reply. Could you specify what you mean when a set is bounded in $mathcal{D}(mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms?
$endgroup$
– Joseph Expo
Jan 24 at 9:23
$begingroup$
Thanks for the reply. Could you specify what you mean when a set is bounded in $mathcal{D}(mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms?
$endgroup$
– Joseph Expo
Jan 24 at 9:23
$begingroup$
I edited my answer to recall what a bounded set is. If by supremum norm you mean $sup_{xinmathbb{R}^n}|psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $mathbb{R}^n$.
$endgroup$
– Abdelmalek Abdesselam
Jan 24 at 15:37
$begingroup$
I edited my answer to recall what a bounded set is. If by supremum norm you mean $sup_{xinmathbb{R}^n}|psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $mathbb{R}^n$.
$endgroup$
– Abdelmalek Abdesselam
Jan 24 at 15:37
add a comment |
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