Density of test functions in the space of distributions — a clarification The Next CEO of Stack OverflowTopologies on the space $mathcal D'(U)$ of distributionsProblem on multiplication of distributions with test functionsDo tempered distributions form a topological subspace of the space of distributions?Definition of the convolution with tempered distributions and Schwartz functionMetrizability of space test $mathcalD(Omega)$, and of distributions spaces $mathcalD'(Omega)$convergence is space of test functions and distributionsDensity of tensor product of test functionsPlancherel for tempered distributions, and when does equality as tempered distributions imply equality as functions.Showing a subset of tempered distributions equals the space of complex Radon measuresHahn-Banach in dual space
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Density of test functions in the space of distributions — a clarification
The Next CEO of Stack OverflowTopologies on the space $mathcal D'(U)$ of distributionsProblem on multiplication of distributions with test functionsDo tempered distributions form a topological subspace of the space of distributions?Definition of the convolution with tempered distributions and Schwartz functionMetrizability of space test $mathcalD(Omega)$, and of distributions spaces $mathcalD'(Omega)$convergence is space of test functions and distributionsDensity of tensor product of test functionsPlancherel for tempered distributions, and when does equality as tempered distributions imply equality as functions.Showing a subset of tempered distributions equals the space of complex Radon measuresHahn-Banach in dual space
$begingroup$
Let $U subseteq mathbbR^n$ be open and denote by $mathcalD(U)$ the space of all compactly supported smooth functions $U to mathbbR$. Let $mathcalD^prime(U)$ be the space of all distributions $mathcalD(U) to mathbbR$ with the standard topology.
Given a distribution $T$, I would like to prove that there exists a sequence $(psi_n)$ in $mathcalD(U)$ such that
beginequationlabeleq:1tag$ast$
lim_n to infty leftlangle psi_n, varphi rightrangle = leftlangle T , varphirightrangle
endequation
for all $varphi in mathcalD(U)$. I became interested in this question while the following paragraph from this Wikipedia article:
The test functions are themselves locally integrable, and so define distributions. As such they are dense in $mathcalD^prime(U)$ with respect to the topology on $mathcalD^prime(U)$ in the sense that for any distribution $T in mathcalD^prime(U)$, there is a sequence $psi_n in mathcalD(U)$ such that
$$
leftlangle psi_n, varphi rightrangle to leftlangle T, varphi rightrangle
$$
for all $varphi in mathcalD(U)$. This fact follows from the Hahn-Banach theorem, since the dual of $mathcalD^prime(U)$ with its weak*-topology is the space $mathcalD(U)$.
My question is as follows: how does this follow from the Hahn-Banach theorem? I understand why $(mathcalD^prime(U))^ast cong mathcalD(U)$ when the former is given the weak*-topology, but I fail to see how eqrefeq:1 follows from the Hahn-Banach theorem.
real-analysis functional-analysis distribution-theory topological-vector-spaces
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
$endgroup$
add a comment |
$begingroup$
Let $U subseteq mathbbR^n$ be open and denote by $mathcalD(U)$ the space of all compactly supported smooth functions $U to mathbbR$. Let $mathcalD^prime(U)$ be the space of all distributions $mathcalD(U) to mathbbR$ with the standard topology.
Given a distribution $T$, I would like to prove that there exists a sequence $(psi_n)$ in $mathcalD(U)$ such that
beginequationlabeleq:1tag$ast$
lim_n to infty leftlangle psi_n, varphi rightrangle = leftlangle T , varphirightrangle
endequation
for all $varphi in mathcalD(U)$. I became interested in this question while the following paragraph from this Wikipedia article:
The test functions are themselves locally integrable, and so define distributions. As such they are dense in $mathcalD^prime(U)$ with respect to the topology on $mathcalD^prime(U)$ in the sense that for any distribution $T in mathcalD^prime(U)$, there is a sequence $psi_n in mathcalD(U)$ such that
$$
leftlangle psi_n, varphi rightrangle to leftlangle T, varphi rightrangle
$$
for all $varphi in mathcalD(U)$. This fact follows from the Hahn-Banach theorem, since the dual of $mathcalD^prime(U)$ with its weak*-topology is the space $mathcalD(U)$.
My question is as follows: how does this follow from the Hahn-Banach theorem? I understand why $(mathcalD^prime(U))^ast cong mathcalD(U)$ when the former is given the weak*-topology, but I fail to see how eqrefeq:1 follows from the Hahn-Banach theorem.
real-analysis functional-analysis distribution-theory topological-vector-spaces
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
$endgroup$
3
$begingroup$
The "best" $C^infty_c$ approximations to $T$ are $T_k = (k^nphi(k.) ast T) phi(./k)$ where $phi in C^infty_c,phi ge 0, phi(0)=1,int phi = 1$. The approximation is uniformly continuous when restricted to distributions of order $le M$ supported on $|x| le r$.
$endgroup$
– reuns
Mar 19 at 10:25
2
$begingroup$
This simply shows you can't believe everything on wikipedia. The appeal to the Hahn-Banach theorem for this density result is math garbage. I agree with reun that the explicit and constructive approach with convolution is "the best" way to prove this density. Moreover, you get sequential density which is stronger than just density as pointed out by Jochen.
$endgroup$
– Abdelmalek Abdesselam
Mar 19 at 14:58
add a comment |
$begingroup$
Let $U subseteq mathbbR^n$ be open and denote by $mathcalD(U)$ the space of all compactly supported smooth functions $U to mathbbR$. Let $mathcalD^prime(U)$ be the space of all distributions $mathcalD(U) to mathbbR$ with the standard topology.
Given a distribution $T$, I would like to prove that there exists a sequence $(psi_n)$ in $mathcalD(U)$ such that
beginequationlabeleq:1tag$ast$
lim_n to infty leftlangle psi_n, varphi rightrangle = leftlangle T , varphirightrangle
endequation
for all $varphi in mathcalD(U)$. I became interested in this question while the following paragraph from this Wikipedia article:
The test functions are themselves locally integrable, and so define distributions. As such they are dense in $mathcalD^prime(U)$ with respect to the topology on $mathcalD^prime(U)$ in the sense that for any distribution $T in mathcalD^prime(U)$, there is a sequence $psi_n in mathcalD(U)$ such that
$$
leftlangle psi_n, varphi rightrangle to leftlangle T, varphi rightrangle
$$
for all $varphi in mathcalD(U)$. This fact follows from the Hahn-Banach theorem, since the dual of $mathcalD^prime(U)$ with its weak*-topology is the space $mathcalD(U)$.
My question is as follows: how does this follow from the Hahn-Banach theorem? I understand why $(mathcalD^prime(U))^ast cong mathcalD(U)$ when the former is given the weak*-topology, but I fail to see how eqrefeq:1 follows from the Hahn-Banach theorem.
real-analysis functional-analysis distribution-theory topological-vector-spaces
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
$endgroup$
Let $U subseteq mathbbR^n$ be open and denote by $mathcalD(U)$ the space of all compactly supported smooth functions $U to mathbbR$. Let $mathcalD^prime(U)$ be the space of all distributions $mathcalD(U) to mathbbR$ with the standard topology.
Given a distribution $T$, I would like to prove that there exists a sequence $(psi_n)$ in $mathcalD(U)$ such that
beginequationlabeleq:1tag$ast$
lim_n to infty leftlangle psi_n, varphi rightrangle = leftlangle T , varphirightrangle
endequation
for all $varphi in mathcalD(U)$. I became interested in this question while the following paragraph from this Wikipedia article:
The test functions are themselves locally integrable, and so define distributions. As such they are dense in $mathcalD^prime(U)$ with respect to the topology on $mathcalD^prime(U)$ in the sense that for any distribution $T in mathcalD^prime(U)$, there is a sequence $psi_n in mathcalD(U)$ such that
$$
leftlangle psi_n, varphi rightrangle to leftlangle T, varphi rightrangle
$$
for all $varphi in mathcalD(U)$. This fact follows from the Hahn-Banach theorem, since the dual of $mathcalD^prime(U)$ with its weak*-topology is the space $mathcalD(U)$.
My question is as follows: how does this follow from the Hahn-Banach theorem? I understand why $(mathcalD^prime(U))^ast cong mathcalD(U)$ when the former is given the weak*-topology, but I fail to see how eqrefeq:1 follows from the Hahn-Banach theorem.
real-analysis functional-analysis distribution-theory topological-vector-spaces
real-analysis functional-analysis distribution-theory topological-vector-spaces
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
asked Mar 18 at 20:53
ImNotThereRightNow_ImNotThereRightNow_
938
938
3
$begingroup$
The "best" $C^infty_c$ approximations to $T$ are $T_k = (k^nphi(k.) ast T) phi(./k)$ where $phi in C^infty_c,phi ge 0, phi(0)=1,int phi = 1$. The approximation is uniformly continuous when restricted to distributions of order $le M$ supported on $|x| le r$.
$endgroup$
– reuns
Mar 19 at 10:25
2
$begingroup$
This simply shows you can't believe everything on wikipedia. The appeal to the Hahn-Banach theorem for this density result is math garbage. I agree with reun that the explicit and constructive approach with convolution is "the best" way to prove this density. Moreover, you get sequential density which is stronger than just density as pointed out by Jochen.
$endgroup$
– Abdelmalek Abdesselam
Mar 19 at 14:58
add a comment |
3
$begingroup$
The "best" $C^infty_c$ approximations to $T$ are $T_k = (k^nphi(k.) ast T) phi(./k)$ where $phi in C^infty_c,phi ge 0, phi(0)=1,int phi = 1$. The approximation is uniformly continuous when restricted to distributions of order $le M$ supported on $|x| le r$.
$endgroup$
– reuns
Mar 19 at 10:25
2
$begingroup$
This simply shows you can't believe everything on wikipedia. The appeal to the Hahn-Banach theorem for this density result is math garbage. I agree with reun that the explicit and constructive approach with convolution is "the best" way to prove this density. Moreover, you get sequential density which is stronger than just density as pointed out by Jochen.
$endgroup$
– Abdelmalek Abdesselam
Mar 19 at 14:58
3
3
$begingroup$
The "best" $C^infty_c$ approximations to $T$ are $T_k = (k^nphi(k.) ast T) phi(./k)$ where $phi in C^infty_c,phi ge 0, phi(0)=1,int phi = 1$. The approximation is uniformly continuous when restricted to distributions of order $le M$ supported on $|x| le r$.
$endgroup$
– reuns
Mar 19 at 10:25
$begingroup$
The "best" $C^infty_c$ approximations to $T$ are $T_k = (k^nphi(k.) ast T) phi(./k)$ where $phi in C^infty_c,phi ge 0, phi(0)=1,int phi = 1$. The approximation is uniformly continuous when restricted to distributions of order $le M$ supported on $|x| le r$.
$endgroup$
– reuns
Mar 19 at 10:25
2
2
$begingroup$
This simply shows you can't believe everything on wikipedia. The appeal to the Hahn-Banach theorem for this density result is math garbage. I agree with reun that the explicit and constructive approach with convolution is "the best" way to prove this density. Moreover, you get sequential density which is stronger than just density as pointed out by Jochen.
$endgroup$
– Abdelmalek Abdesselam
Mar 19 at 14:58
$begingroup$
This simply shows you can't believe everything on wikipedia. The appeal to the Hahn-Banach theorem for this density result is math garbage. I agree with reun that the explicit and constructive approach with convolution is "the best" way to prove this density. Moreover, you get sequential density which is stronger than just density as pointed out by Jochen.
$endgroup$
– Abdelmalek Abdesselam
Mar 19 at 14:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your scepticism ist justified. A consequence of the Hahn-Banach theorem is that a continuous linear map $f:Xto Y$ between two locally convex Hausdorff spaces has dense range if and only if the transposed $f^t: Y'to X'$ is injective. Applying this to the inclusion $f:mathscr D hookrightarrow mathscr D'$ one gets that $mathscr D$ is weak$^*$-dense in $mathscr D'$ -- but in general this does not imply sequential density.
answered Mar 19 at 8:28
JochenJochen
7,0431023
$endgroup$
add a comment |
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$begingroup$
Your scepticism ist justified. A consequence of the Hahn-Banach theorem is that a continuous linear map $f:Xto Y$ between two locally convex Hausdorff spaces has dense range if and only if the transposed $f^t: Y'to X'$ is injective. Applying this to the inclusion $f:mathscr D hookrightarrow mathscr D'$ one gets that $mathscr D$ is weak$^*$-dense in $mathscr D'$ -- but in general this does not imply sequential density.
answered Mar 19 at 8:28
JochenJochen
7,0431023
$endgroup$
add a comment |
$begingroup$
Your scepticism ist justified. A consequence of the Hahn-Banach theorem is that a continuous linear map $f:Xto Y$ between two locally convex Hausdorff spaces has dense range if and only if the transposed $f^t: Y'to X'$ is injective. Applying this to the inclusion $f:mathscr D hookrightarrow mathscr D'$ one gets that $mathscr D$ is weak$^*$-dense in $mathscr D'$ -- but in general this does not imply sequential density.
answered Mar 19 at 8:28
JochenJochen
7,0431023
$endgroup$
add a comment |
$begingroup$
Your scepticism ist justified. A consequence of the Hahn-Banach theorem is that a continuous linear map $f:Xto Y$ between two locally convex Hausdorff spaces has dense range if and only if the transposed $f^t: Y'to X'$ is injective. Applying this to the inclusion $f:mathscr D hookrightarrow mathscr D'$ one gets that $mathscr D$ is weak$^*$-dense in $mathscr D'$ -- but in general this does not imply sequential density.
answered Mar 19 at 8:28
JochenJochen
7,0431023
$endgroup$
Your scepticism ist justified. A consequence of the Hahn-Banach theorem is that a continuous linear map $f:Xto Y$ between two locally convex Hausdorff spaces has dense range if and only if the transposed $f^t: Y'to X'$ is injective. Applying this to the inclusion $f:mathscr D hookrightarrow mathscr D'$ one gets that $mathscr D$ is weak$^*$-dense in $mathscr D'$ -- but in general this does not imply sequential density.
answered Mar 19 at 8:28
JochenJochen
7,0431023
answered Mar 19 at 8:28
JochenJochen
7,0431023
answered Mar 19 at 8:28
JochenJochen
7,0431023
answered Mar 19 at 8:28
JochenJochen
7,0431023
7,0431023
add a comment |
add a comment |
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$begingroup$
The "best" $C^infty_c$ approximations to $T$ are $T_k = (k^nphi(k.) ast T) phi(./k)$ where $phi in C^infty_c,phi ge 0, phi(0)=1,int phi = 1$. The approximation is uniformly continuous when restricted to distributions of order $le M$ supported on $|x| le r$.
$endgroup$
– reuns
Mar 19 at 10:25
2
$begingroup$
This simply shows you can't believe everything on wikipedia. The appeal to the Hahn-Banach theorem for this density result is math garbage. I agree with reun that the explicit and constructive approach with convolution is "the best" way to prove this density. Moreover, you get sequential density which is stronger than just density as pointed out by Jochen.
$endgroup$
– Abdelmalek Abdesselam
Mar 19 at 14:58