Showing a subset of tempered distributions equals the space of complex Radon measures
$begingroup$
The following question is part a) of exercise 29 in the Distributions chapter of Folland's Real Analysis.
Let $S'$ denote the space of tempered distributions on $mathbb{R}^n$. That is, $S'$ is the space of distributions that act on the space of Schwartz functions $S$.
For $1 leq p < infty$, let $C_p$ be the set of all $F in S'$ for which there exists $C geq 0$ such that $||F ast phi||_p leq C||phi||_p$ for all $phi in S$, so that the map $phi mapsto F ast phi $ extends to a bounded operator on $L^p$
a) Show that $C_1 = M(mathbb{R}^n)$ where $M(mathbb{R}^n)$ is the space of complex Radon measures on $mathbb{R}^n$.
Folland provides a hint which reads: "If $F in C_1$, consider $F ast phi_t$ where ${phi_t}$ is an approximate identity, and apply Alaoglu's theorem."
So, I am actually pretty stuck on where to even begin. It seems as though the Riesz representation theorem should be used, but I am not completely sure.
Appealing to the hint, I have noticed that if we let $F_t = F ast phi_t$, then we have that $F_t longrightarrow F$ in $S'$:
$$ langle F_t, psi rangle = langle F ast phi_t, psi rangle = langle F, widetilde{phi_t} ast psi rangle longrightarrow langle F, psi rangle text{ as } t longrightarrow 0 $$
Where the convergence follows by properties of approximations to the identity.
If anyone could provide a step in the right direction, that'd be very helpful. I don't see how Alaoglu's theorem applies.
real-analysis distribution-theory
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
$endgroup$
add a comment |
$begingroup$
The following question is part a) of exercise 29 in the Distributions chapter of Folland's Real Analysis.
Let $S'$ denote the space of tempered distributions on $mathbb{R}^n$. That is, $S'$ is the space of distributions that act on the space of Schwartz functions $S$.
For $1 leq p < infty$, let $C_p$ be the set of all $F in S'$ for which there exists $C geq 0$ such that $||F ast phi||_p leq C||phi||_p$ for all $phi in S$, so that the map $phi mapsto F ast phi $ extends to a bounded operator on $L^p$
a) Show that $C_1 = M(mathbb{R}^n)$ where $M(mathbb{R}^n)$ is the space of complex Radon measures on $mathbb{R}^n$.
Folland provides a hint which reads: "If $F in C_1$, consider $F ast phi_t$ where ${phi_t}$ is an approximate identity, and apply Alaoglu's theorem."
So, I am actually pretty stuck on where to even begin. It seems as though the Riesz representation theorem should be used, but I am not completely sure.
Appealing to the hint, I have noticed that if we let $F_t = F ast phi_t$, then we have that $F_t longrightarrow F$ in $S'$:
$$ langle F_t, psi rangle = langle F ast phi_t, psi rangle = langle F, widetilde{phi_t} ast psi rangle longrightarrow langle F, psi rangle text{ as } t longrightarrow 0 $$
Where the convergence follows by properties of approximations to the identity.
If anyone could provide a step in the right direction, that'd be very helpful. I don't see how Alaoglu's theorem applies.
real-analysis distribution-theory
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
$endgroup$
add a comment |
$begingroup$
The following question is part a) of exercise 29 in the Distributions chapter of Folland's Real Analysis.
Let $S'$ denote the space of tempered distributions on $mathbb{R}^n$. That is, $S'$ is the space of distributions that act on the space of Schwartz functions $S$.
For $1 leq p < infty$, let $C_p$ be the set of all $F in S'$ for which there exists $C geq 0$ such that $||F ast phi||_p leq C||phi||_p$ for all $phi in S$, so that the map $phi mapsto F ast phi $ extends to a bounded operator on $L^p$
a) Show that $C_1 = M(mathbb{R}^n)$ where $M(mathbb{R}^n)$ is the space of complex Radon measures on $mathbb{R}^n$.
Folland provides a hint which reads: "If $F in C_1$, consider $F ast phi_t$ where ${phi_t}$ is an approximate identity, and apply Alaoglu's theorem."
So, I am actually pretty stuck on where to even begin. It seems as though the Riesz representation theorem should be used, but I am not completely sure.
Appealing to the hint, I have noticed that if we let $F_t = F ast phi_t$, then we have that $F_t longrightarrow F$ in $S'$:
$$ langle F_t, psi rangle = langle F ast phi_t, psi rangle = langle F, widetilde{phi_t} ast psi rangle longrightarrow langle F, psi rangle text{ as } t longrightarrow 0 $$
Where the convergence follows by properties of approximations to the identity.
If anyone could provide a step in the right direction, that'd be very helpful. I don't see how Alaoglu's theorem applies.
real-analysis distribution-theory
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
$endgroup$
The following question is part a) of exercise 29 in the Distributions chapter of Folland's Real Analysis.
Let $S'$ denote the space of tempered distributions on $mathbb{R}^n$. That is, $S'$ is the space of distributions that act on the space of Schwartz functions $S$.
For $1 leq p < infty$, let $C_p$ be the set of all $F in S'$ for which there exists $C geq 0$ such that $||F ast phi||_p leq C||phi||_p$ for all $phi in S$, so that the map $phi mapsto F ast phi $ extends to a bounded operator on $L^p$
a) Show that $C_1 = M(mathbb{R}^n)$ where $M(mathbb{R}^n)$ is the space of complex Radon measures on $mathbb{R}^n$.
Folland provides a hint which reads: "If $F in C_1$, consider $F ast phi_t$ where ${phi_t}$ is an approximate identity, and apply Alaoglu's theorem."
So, I am actually pretty stuck on where to even begin. It seems as though the Riesz representation theorem should be used, but I am not completely sure.
Appealing to the hint, I have noticed that if we let $F_t = F ast phi_t$, then we have that $F_t longrightarrow F$ in $S'$:
$$ langle F_t, psi rangle = langle F ast phi_t, psi rangle = langle F, widetilde{phi_t} ast psi rangle longrightarrow langle F, psi rangle text{ as } t longrightarrow 0 $$
Where the convergence follows by properties of approximations to the identity.
If anyone could provide a step in the right direction, that'd be very helpful. I don't see how Alaoglu's theorem applies.
real-analysis distribution-theory
real-analysis distribution-theory
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
asked Jan 13 at 4:18
Nicholas RobertsNicholas Roberts
115112
115112
add a comment |
add a comment |
1 Answer
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In fact $$||F*phi_t||_{C_0(Bbb R^n)^*}
=||F*phi_t||_1le C||phi_t||_1=C;$$apply Alaoglu in $C_0^*$.
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
$begingroup$
So do we use Alaoglu to obtain a convergent subsequence?
$endgroup$
– Nicholas Roberts
Jan 13 at 17:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In fact $$||F*phi_t||_{C_0(Bbb R^n)^*}
=||F*phi_t||_1le C||phi_t||_1=C;$$apply Alaoglu in $C_0^*$.
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
$begingroup$
So do we use Alaoglu to obtain a convergent subsequence?
$endgroup$
– Nicholas Roberts
Jan 13 at 17:10
add a comment |
$begingroup$
In fact $$||F*phi_t||_{C_0(Bbb R^n)^*}
=||F*phi_t||_1le C||phi_t||_1=C;$$apply Alaoglu in $C_0^*$.
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
$begingroup$
So do we use Alaoglu to obtain a convergent subsequence?
$endgroup$
– Nicholas Roberts
Jan 13 at 17:10
add a comment |
$begingroup$
In fact $$||F*phi_t||_{C_0(Bbb R^n)^*}
=||F*phi_t||_1le C||phi_t||_1=C;$$apply Alaoglu in $C_0^*$.
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
$endgroup$
In fact $$||F*phi_t||_{C_0(Bbb R^n)^*}
=||F*phi_t||_1le C||phi_t||_1=C;$$apply Alaoglu in $C_0^*$.
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
answered Jan 13 at 13:46
David C. UllrichDavid C. Ullrich
60.1k43994
60.1k43994
$begingroup$
So do we use Alaoglu to obtain a convergent subsequence?
$endgroup$
– Nicholas Roberts
Jan 13 at 17:10
add a comment |
$begingroup$
So do we use Alaoglu to obtain a convergent subsequence?
$endgroup$
– Nicholas Roberts
Jan 13 at 17:10
$begingroup$
So do we use Alaoglu to obtain a convergent subsequence?
$endgroup$
– Nicholas Roberts
Jan 13 at 17:10
$begingroup$
So do we use Alaoglu to obtain a convergent subsequence?
$endgroup$
– Nicholas Roberts
Jan 13 at 17:10
add a comment |
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