Mixing
How to prove that a delta function belongs to the Besov space $B^{-1}$?
$begingroup$
$defR{mathbb{R}}$
$DeclareMathOperator{supp}{supp}$
I am trying to understand the definition of the Besov space and to prove that the delta function in $R^1$. However I got stuck.
First, let us recall that a sequence of smooth functions ${phi_i}_{ige-1}$ is called a partition of unity if $sum_{i=-1}^infty phi_i=1$, $supp(phi_{-1})subset{xcolon |x|le 2}$; $supp(phi_{i})subset{xcolon 2^{i-1}le |x|le 2^{i+1}}$.
The Besov space $B^s_{p,q}$, where $sinR$, $p,qge1$ is defined as the collection of all functions $f$ such that
$$
|f|_{p,q}^s:=|(2^{sj}|F^{-1}phi_j Ff|_{L_p})_{jge-1}|_{l_q}<infty.
$$
Here $F$ is the Fourier transform operator.
I would like to prove that for any fixed $xinR$ the delta function $delta_x$ is in $B^{-1}_{infty,infty}$. Clearly,
$$
Fdelta_x(lambda)=e^{ilambda x}.
$$
But now I am confused. How should we estimate
$$
|F^{-1}phi_j e^{ilambda x}|_{L_infty}?
$$
functional-analysis dirac-delta besov-space
asked Oct 11 '18 at 14:30
OlegOleg
301212
$endgroup$
add a comment |
$begingroup$
$defR{mathbb{R}}$
$DeclareMathOperator{supp}{supp}$
I am trying to understand the definition of the Besov space and to prove that the delta function in $R^1$. However I got stuck.
First, let us recall that a sequence of smooth functions ${phi_i}_{ige-1}$ is called a partition of unity if $sum_{i=-1}^infty phi_i=1$, $supp(phi_{-1})subset{xcolon |x|le 2}$; $supp(phi_{i})subset{xcolon 2^{i-1}le |x|le 2^{i+1}}$.
The Besov space $B^s_{p,q}$, where $sinR$, $p,qge1$ is defined as the collection of all functions $f$ such that
$$
|f|_{p,q}^s:=|(2^{sj}|F^{-1}phi_j Ff|_{L_p})_{jge-1}|_{l_q}<infty.
$$
Here $F$ is the Fourier transform operator.
I would like to prove that for any fixed $xinR$ the delta function $delta_x$ is in $B^{-1}_{infty,infty}$. Clearly,
$$
Fdelta_x(lambda)=e^{ilambda x}.
$$
But now I am confused. How should we estimate
$$
|F^{-1}phi_j e^{ilambda x}|_{L_infty}?
$$
functional-analysis dirac-delta besov-space
asked Oct 11 '18 at 14:30
OlegOleg
301212
$endgroup$
$begingroup$
I would appreciate even partial solutions or just some useful hints.
$endgroup$
– Oleg
Oct 15 '18 at 14:11
$begingroup$
You've been around but you have let the bounty expire and you haven't accepted my answer. Can I ask you to tell me how to improve it to your satisfaction?
$endgroup$
– Calvin Khor
Oct 24 '18 at 9:34
add a comment |
$begingroup$
$defR{mathbb{R}}$
$DeclareMathOperator{supp}{supp}$
I am trying to understand the definition of the Besov space and to prove that the delta function in $R^1$. However I got stuck.
First, let us recall that a sequence of smooth functions ${phi_i}_{ige-1}$ is called a partition of unity if $sum_{i=-1}^infty phi_i=1$, $supp(phi_{-1})subset{xcolon |x|le 2}$; $supp(phi_{i})subset{xcolon 2^{i-1}le |x|le 2^{i+1}}$.
The Besov space $B^s_{p,q}$, where $sinR$, $p,qge1$ is defined as the collection of all functions $f$ such that
$$
|f|_{p,q}^s:=|(2^{sj}|F^{-1}phi_j Ff|_{L_p})_{jge-1}|_{l_q}<infty.
$$
Here $F$ is the Fourier transform operator.
I would like to prove that for any fixed $xinR$ the delta function $delta_x$ is in $B^{-1}_{infty,infty}$. Clearly,
$$
Fdelta_x(lambda)=e^{ilambda x}.
$$
But now I am confused. How should we estimate
$$
|F^{-1}phi_j e^{ilambda x}|_{L_infty}?
$$
functional-analysis dirac-delta besov-space
asked Oct 11 '18 at 14:30
OlegOleg
301212
$endgroup$
$defR{mathbb{R}}$
$DeclareMathOperator{supp}{supp}$
I am trying to understand the definition of the Besov space and to prove that the delta function in $R^1$. However I got stuck.
First, let us recall that a sequence of smooth functions ${phi_i}_{ige-1}$ is called a partition of unity if $sum_{i=-1}^infty phi_i=1$, $supp(phi_{-1})subset{xcolon |x|le 2}$; $supp(phi_{i})subset{xcolon 2^{i-1}le |x|le 2^{i+1}}$.
The Besov space $B^s_{p,q}$, where $sinR$, $p,qge1$ is defined as the collection of all functions $f$ such that
$$
|f|_{p,q}^s:=|(2^{sj}|F^{-1}phi_j Ff|_{L_p})_{jge-1}|_{l_q}<infty.
$$
Here $F$ is the Fourier transform operator.
I would like to prove that for any fixed $xinR$ the delta function $delta_x$ is in $B^{-1}_{infty,infty}$. Clearly,
$$
Fdelta_x(lambda)=e^{ilambda x}.
$$
But now I am confused. How should we estimate
$$
|F^{-1}phi_j e^{ilambda x}|_{L_infty}?
$$
functional-analysis dirac-delta besov-space
functional-analysis dirac-delta besov-space
asked Oct 11 '18 at 14:30
OlegOleg
301212
asked Oct 11 '18 at 14:30
OlegOleg
301212
edited Oct 15 '18 at 14:10
Oleg
asked Oct 11 '18 at 14:30
OlegOleg
301212
asked Oct 11 '18 at 14:30
OlegOleg
301212
asked Oct 11 '18 at 14:30
OlegOleg
301212
301212
$begingroup$
I would appreciate even partial solutions or just some useful hints.
$endgroup$
– Oleg
Oct 15 '18 at 14:11
$begingroup$
You've been around but you have let the bounty expire and you haven't accepted my answer. Can I ask you to tell me how to improve it to your satisfaction?
$endgroup$
– Calvin Khor
Oct 24 '18 at 9:34
add a comment |
$begingroup$
I would appreciate even partial solutions or just some useful hints.
$endgroup$
– Oleg
Oct 15 '18 at 14:11
$begingroup$
You've been around but you have let the bounty expire and you haven't accepted my answer. Can I ask you to tell me how to improve it to your satisfaction?
$endgroup$
– Calvin Khor
Oct 24 '18 at 9:34
$begingroup$
I would appreciate even partial solutions or just some useful hints.
$endgroup$
– Oleg
Oct 15 '18 at 14:11
$begingroup$
I would appreciate even partial solutions or just some useful hints.
$endgroup$
– Oleg
Oct 15 '18 at 14:11
$begingroup$
You've been around but you have let the bounty expire and you haven't accepted my answer. Can I ask you to tell me how to improve it to your satisfaction?
$endgroup$
– Calvin Khor
Oct 24 '18 at 9:34
$begingroup$
You've been around but you have let the bounty expire and you haven't accepted my answer. Can I ask you to tell me how to improve it to your satisfaction?
$endgroup$
– Calvin Khor
Oct 24 '18 at 9:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Firstly, the norm doesn't depend on $x$, due to the shift property $ mathscr F_y [u(y-x)](lambda)=mathscr F[u](lambda)e^{ilambda x},$ i.e. $$mathscr F^{-1}_lambda[phi_j(lambda) e^{ilambda x}](y)=mathscr F^{-1}_lambda [phi_j(lambda)](y-x).$$ So
$$|delta_x|_{B^s_{infty,infty}} = sup_j 2^{sj}| mathscr F^{-1}phi_j(y-x) |_{L^infty_y(mathbb R)} = sup_{j} 2^{sj}|F^{-1}phi_j|_{L^infty(mathbb R)}. $$
Secondly, recall that the $phi_j$ for a Littlewood-Paley decomposition are defined (for $j> -1$) via rescalings of a smooth function $phi$ supported on some annulus,
$$ phi_j(lambda) = phi(2^{-j}lambda)$$
Then the scaling property of the Fourier transform gives $F^{-1}[phi_j](y) = 2^{j} F^{-1}[phi](2^{j} y)$, and so
$$ |delta_x|_{B^s_{infty,infty}} = max left( |F^{-1}[phi_{-1}]|_{L^infty}2^{-s} , |F^{-1}[phi]|_{L^infty}sup_{j> -1} 2^{(s+1)j}right)$$
this is finite as long as $sle- 1$; therefore, $delta_x in B^s_{infty,infty}$ for $sle -1$.
In fact, since the scaling property of the Fourier transform in $d$ dimensions is
$$mathscr F [u(Ky)](lambda) = frac1{K^d}mathscr F [u](lambda/K) $$
we can also verify with virtually no additional effort that
$$|delta_x|_{B^s_{p,infty}}=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , sup_{j> -1}|F^{-1}[phi](2^jy)|_{L^p_y} 2^{(s+d)j}right)\=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , |F^{-1}[phi]|_{L^p}sup_{j> -1} 2^{(s+d-d/p)j}right)$$
using the scaling property of $L^p$ norms
$$ int_{mathbb R^d} |u(ky)|^p dy = k^{-d}int_{mathbb R^d} |u|^p $$That is, $ delta_x in B^{-d+d/p}_{p,infty}(mathbb R^d)$ for every $pin[1,infty]$, as can be found in e.g. the introduction of this paper by Prof. Hairer.
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
(i missed a minus sign that gave a silly result...this should be fixed now)
$endgroup$
– Calvin Khor
Oct 15 '18 at 17:06
add a comment |
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$begingroup$
Firstly, the norm doesn't depend on $x$, due to the shift property $ mathscr F_y [u(y-x)](lambda)=mathscr F[u](lambda)e^{ilambda x},$ i.e. $$mathscr F^{-1}_lambda[phi_j(lambda) e^{ilambda x}](y)=mathscr F^{-1}_lambda [phi_j(lambda)](y-x).$$ So
$$|delta_x|_{B^s_{infty,infty}} = sup_j 2^{sj}| mathscr F^{-1}phi_j(y-x) |_{L^infty_y(mathbb R)} = sup_{j} 2^{sj}|F^{-1}phi_j|_{L^infty(mathbb R)}. $$
Secondly, recall that the $phi_j$ for a Littlewood-Paley decomposition are defined (for $j> -1$) via rescalings of a smooth function $phi$ supported on some annulus,
$$ phi_j(lambda) = phi(2^{-j}lambda)$$
Then the scaling property of the Fourier transform gives $F^{-1}[phi_j](y) = 2^{j} F^{-1}[phi](2^{j} y)$, and so
$$ |delta_x|_{B^s_{infty,infty}} = max left( |F^{-1}[phi_{-1}]|_{L^infty}2^{-s} , |F^{-1}[phi]|_{L^infty}sup_{j> -1} 2^{(s+1)j}right)$$
this is finite as long as $sle- 1$; therefore, $delta_x in B^s_{infty,infty}$ for $sle -1$.
In fact, since the scaling property of the Fourier transform in $d$ dimensions is
$$mathscr F [u(Ky)](lambda) = frac1{K^d}mathscr F [u](lambda/K) $$
we can also verify with virtually no additional effort that
$$|delta_x|_{B^s_{p,infty}}=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , sup_{j> -1}|F^{-1}[phi](2^jy)|_{L^p_y} 2^{(s+d)j}right)\=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , |F^{-1}[phi]|_{L^p}sup_{j> -1} 2^{(s+d-d/p)j}right)$$
using the scaling property of $L^p$ norms
$$ int_{mathbb R^d} |u(ky)|^p dy = k^{-d}int_{mathbb R^d} |u|^p $$That is, $ delta_x in B^{-d+d/p}_{p,infty}(mathbb R^d)$ for every $pin[1,infty]$, as can be found in e.g. the introduction of this paper by Prof. Hairer.
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
(i missed a minus sign that gave a silly result...this should be fixed now)
$endgroup$
– Calvin Khor
Oct 15 '18 at 17:06
add a comment |
$begingroup$
Firstly, the norm doesn't depend on $x$, due to the shift property $ mathscr F_y [u(y-x)](lambda)=mathscr F[u](lambda)e^{ilambda x},$ i.e. $$mathscr F^{-1}_lambda[phi_j(lambda) e^{ilambda x}](y)=mathscr F^{-1}_lambda [phi_j(lambda)](y-x).$$ So
$$|delta_x|_{B^s_{infty,infty}} = sup_j 2^{sj}| mathscr F^{-1}phi_j(y-x) |_{L^infty_y(mathbb R)} = sup_{j} 2^{sj}|F^{-1}phi_j|_{L^infty(mathbb R)}. $$
Secondly, recall that the $phi_j$ for a Littlewood-Paley decomposition are defined (for $j> -1$) via rescalings of a smooth function $phi$ supported on some annulus,
$$ phi_j(lambda) = phi(2^{-j}lambda)$$
Then the scaling property of the Fourier transform gives $F^{-1}[phi_j](y) = 2^{j} F^{-1}[phi](2^{j} y)$, and so
$$ |delta_x|_{B^s_{infty,infty}} = max left( |F^{-1}[phi_{-1}]|_{L^infty}2^{-s} , |F^{-1}[phi]|_{L^infty}sup_{j> -1} 2^{(s+1)j}right)$$
this is finite as long as $sle- 1$; therefore, $delta_x in B^s_{infty,infty}$ for $sle -1$.
In fact, since the scaling property of the Fourier transform in $d$ dimensions is
$$mathscr F [u(Ky)](lambda) = frac1{K^d}mathscr F [u](lambda/K) $$
we can also verify with virtually no additional effort that
$$|delta_x|_{B^s_{p,infty}}=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , sup_{j> -1}|F^{-1}[phi](2^jy)|_{L^p_y} 2^{(s+d)j}right)\=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , |F^{-1}[phi]|_{L^p}sup_{j> -1} 2^{(s+d-d/p)j}right)$$
using the scaling property of $L^p$ norms
$$ int_{mathbb R^d} |u(ky)|^p dy = k^{-d}int_{mathbb R^d} |u|^p $$That is, $ delta_x in B^{-d+d/p}_{p,infty}(mathbb R^d)$ for every $pin[1,infty]$, as can be found in e.g. the introduction of this paper by Prof. Hairer.
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
(i missed a minus sign that gave a silly result...this should be fixed now)
$endgroup$
– Calvin Khor
Oct 15 '18 at 17:06
add a comment |
$begingroup$
Firstly, the norm doesn't depend on $x$, due to the shift property $ mathscr F_y [u(y-x)](lambda)=mathscr F[u](lambda)e^{ilambda x},$ i.e. $$mathscr F^{-1}_lambda[phi_j(lambda) e^{ilambda x}](y)=mathscr F^{-1}_lambda [phi_j(lambda)](y-x).$$ So
$$|delta_x|_{B^s_{infty,infty}} = sup_j 2^{sj}| mathscr F^{-1}phi_j(y-x) |_{L^infty_y(mathbb R)} = sup_{j} 2^{sj}|F^{-1}phi_j|_{L^infty(mathbb R)}. $$
Secondly, recall that the $phi_j$ for a Littlewood-Paley decomposition are defined (for $j> -1$) via rescalings of a smooth function $phi$ supported on some annulus,
$$ phi_j(lambda) = phi(2^{-j}lambda)$$
Then the scaling property of the Fourier transform gives $F^{-1}[phi_j](y) = 2^{j} F^{-1}[phi](2^{j} y)$, and so
$$ |delta_x|_{B^s_{infty,infty}} = max left( |F^{-1}[phi_{-1}]|_{L^infty}2^{-s} , |F^{-1}[phi]|_{L^infty}sup_{j> -1} 2^{(s+1)j}right)$$
this is finite as long as $sle- 1$; therefore, $delta_x in B^s_{infty,infty}$ for $sle -1$.
In fact, since the scaling property of the Fourier transform in $d$ dimensions is
$$mathscr F [u(Ky)](lambda) = frac1{K^d}mathscr F [u](lambda/K) $$
we can also verify with virtually no additional effort that
$$|delta_x|_{B^s_{p,infty}}=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , sup_{j> -1}|F^{-1}[phi](2^jy)|_{L^p_y} 2^{(s+d)j}right)\=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , |F^{-1}[phi]|_{L^p}sup_{j> -1} 2^{(s+d-d/p)j}right)$$
using the scaling property of $L^p$ norms
$$ int_{mathbb R^d} |u(ky)|^p dy = k^{-d}int_{mathbb R^d} |u|^p $$That is, $ delta_x in B^{-d+d/p}_{p,infty}(mathbb R^d)$ for every $pin[1,infty]$, as can be found in e.g. the introduction of this paper by Prof. Hairer.
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
Firstly, the norm doesn't depend on $x$, due to the shift property $ mathscr F_y [u(y-x)](lambda)=mathscr F[u](lambda)e^{ilambda x},$ i.e. $$mathscr F^{-1}_lambda[phi_j(lambda) e^{ilambda x}](y)=mathscr F^{-1}_lambda [phi_j(lambda)](y-x).$$ So
$$|delta_x|_{B^s_{infty,infty}} = sup_j 2^{sj}| mathscr F^{-1}phi_j(y-x) |_{L^infty_y(mathbb R)} = sup_{j} 2^{sj}|F^{-1}phi_j|_{L^infty(mathbb R)}. $$
Secondly, recall that the $phi_j$ for a Littlewood-Paley decomposition are defined (for $j> -1$) via rescalings of a smooth function $phi$ supported on some annulus,
$$ phi_j(lambda) = phi(2^{-j}lambda)$$
Then the scaling property of the Fourier transform gives $F^{-1}[phi_j](y) = 2^{j} F^{-1}[phi](2^{j} y)$, and so
$$ |delta_x|_{B^s_{infty,infty}} = max left( |F^{-1}[phi_{-1}]|_{L^infty}2^{-s} , |F^{-1}[phi]|_{L^infty}sup_{j> -1} 2^{(s+1)j}right)$$
this is finite as long as $sle- 1$; therefore, $delta_x in B^s_{infty,infty}$ for $sle -1$.
In fact, since the scaling property of the Fourier transform in $d$ dimensions is
$$mathscr F [u(Ky)](lambda) = frac1{K^d}mathscr F [u](lambda/K) $$
we can also verify with virtually no additional effort that
$$|delta_x|_{B^s_{p,infty}}=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , sup_{j> -1}|F^{-1}[phi](2^jy)|_{L^p_y} 2^{(s+d)j}right)\=max left( |F^{-1}[phi_{-1}]|_{L^p}2^{-s} , |F^{-1}[phi]|_{L^p}sup_{j> -1} 2^{(s+d-d/p)j}right)$$
using the scaling property of $L^p$ norms
$$ int_{mathbb R^d} |u(ky)|^p dy = k^{-d}int_{mathbb R^d} |u|^p $$That is, $ delta_x in B^{-d+d/p}_{p,infty}(mathbb R^d)$ for every $pin[1,infty]$, as can be found in e.g. the introduction of this paper by Prof. Hairer.
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
edited Oct 17 '18 at 16:27
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
answered Oct 15 '18 at 14:18
Calvin KhorCalvin Khor
12.4k21439
12.4k21439
$begingroup$
(i missed a minus sign that gave a silly result...this should be fixed now)
$endgroup$
– Calvin Khor
Oct 15 '18 at 17:06
add a comment |
$begingroup$
(i missed a minus sign that gave a silly result...this should be fixed now)
$endgroup$
– Calvin Khor
Oct 15 '18 at 17:06
$begingroup$
(i missed a minus sign that gave a silly result...this should be fixed now)
$endgroup$
– Calvin Khor
Oct 15 '18 at 17:06
$begingroup$
(i missed a minus sign that gave a silly result...this should be fixed now)
$endgroup$
– Calvin Khor
Oct 15 '18 at 17:06
add a comment |
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$begingroup$
I would appreciate even partial solutions or just some useful hints.
$endgroup$
– Oleg
Oct 15 '18 at 14:11
$begingroup$
You've been around but you have let the bounty expire and you haven't accepted my answer. Can I ask you to tell me how to improve it to your satisfaction?
$endgroup$
– Calvin Khor
Oct 24 '18 at 9:34