Mixing
Is there a way to read a summability property of a family of functions from the behavior of coefficients of...
$begingroup$
If $(K_N)_{Ninmathbb{N}}$ (or indexed by any other directed set) is a family of functions in $L^1(mathbb{T})$ defined on the 1-torus $mathbb{T}$ such that:
$$forall pin[1,+infty), forall fin L^p(mathbb{T}), |K_N*f-f|_pto0, Nto+infty$$
is there a way to read this property from the family of its Fourier transforms?
Some examples:
$F_N(t):=frac {1}{N+1}left (frac{sinleft(frac{N+1}{2}tright)}{sinleft(frac{t}{2}right)}right)^2$ satisfies this property for $Nto+infty$ and $mathcal{F}(F_N)(n)=(1-frac{|n|}{N})chi_{[-N,N]}(n);$
$P_r(t):= frac{1-r^2}{|1-r e^{it}|^2}$ satisfies this property for $rto 1^-$ and $mathcal{F}(P_r)(n)=r^{|n|};$
$K_s(t):=sum_{ninmathbb{Z}}e^{-sn^2}e^{int}$ satisfies this property for $sto0^+$ and obviously $mathcal{F}(K_s)(n)=e^{-sn^2}$;
$D_N(t):=frac{sinleft(left(N+frac{1}{2}right)tright)}{sinleft(frac{t}{2}right)}$ satisfies this property for $Nto +infty$ and $1<p<+infty$ but not for $p=1$ and $mathcal{F}(D_N)(n)=chi_{[-N,N]}(n)$.
Now, in all these examples, the family of Fourier transforms converges boundedly to $1$ as the index goes to the proper limit, but I can't see any obvious (sufficient) condition where we can read why for the first three examples the condition is satisfied while it is not for the last one. Any ideas?
fourier-series summability-theory
asked Jan 30 at 23:44
BobBob
1,7001725
$endgroup$
add a comment |
$begingroup$
If $(K_N)_{Ninmathbb{N}}$ (or indexed by any other directed set) is a family of functions in $L^1(mathbb{T})$ defined on the 1-torus $mathbb{T}$ such that:
$$forall pin[1,+infty), forall fin L^p(mathbb{T}), |K_N*f-f|_pto0, Nto+infty$$
is there a way to read this property from the family of its Fourier transforms?
Some examples:
$F_N(t):=frac {1}{N+1}left (frac{sinleft(frac{N+1}{2}tright)}{sinleft(frac{t}{2}right)}right)^2$ satisfies this property for $Nto+infty$ and $mathcal{F}(F_N)(n)=(1-frac{|n|}{N})chi_{[-N,N]}(n);$
$P_r(t):= frac{1-r^2}{|1-r e^{it}|^2}$ satisfies this property for $rto 1^-$ and $mathcal{F}(P_r)(n)=r^{|n|};$
$K_s(t):=sum_{ninmathbb{Z}}e^{-sn^2}e^{int}$ satisfies this property for $sto0^+$ and obviously $mathcal{F}(K_s)(n)=e^{-sn^2}$;
$D_N(t):=frac{sinleft(left(N+frac{1}{2}right)tright)}{sinleft(frac{t}{2}right)}$ satisfies this property for $Nto +infty$ and $1<p<+infty$ but not for $p=1$ and $mathcal{F}(D_N)(n)=chi_{[-N,N]}(n)$.
Now, in all these examples, the family of Fourier transforms converges boundedly to $1$ as the index goes to the proper limit, but I can't see any obvious (sufficient) condition where we can read why for the first three examples the condition is satisfied while it is not for the last one. Any ideas?
fourier-series summability-theory
asked Jan 30 at 23:44
BobBob
1,7001725
$endgroup$
add a comment |
$begingroup$
If $(K_N)_{Ninmathbb{N}}$ (or indexed by any other directed set) is a family of functions in $L^1(mathbb{T})$ defined on the 1-torus $mathbb{T}$ such that:
$$forall pin[1,+infty), forall fin L^p(mathbb{T}), |K_N*f-f|_pto0, Nto+infty$$
is there a way to read this property from the family of its Fourier transforms?
Some examples:
$F_N(t):=frac {1}{N+1}left (frac{sinleft(frac{N+1}{2}tright)}{sinleft(frac{t}{2}right)}right)^2$ satisfies this property for $Nto+infty$ and $mathcal{F}(F_N)(n)=(1-frac{|n|}{N})chi_{[-N,N]}(n);$
$P_r(t):= frac{1-r^2}{|1-r e^{it}|^2}$ satisfies this property for $rto 1^-$ and $mathcal{F}(P_r)(n)=r^{|n|};$
$K_s(t):=sum_{ninmathbb{Z}}e^{-sn^2}e^{int}$ satisfies this property for $sto0^+$ and obviously $mathcal{F}(K_s)(n)=e^{-sn^2}$;
$D_N(t):=frac{sinleft(left(N+frac{1}{2}right)tright)}{sinleft(frac{t}{2}right)}$ satisfies this property for $Nto +infty$ and $1<p<+infty$ but not for $p=1$ and $mathcal{F}(D_N)(n)=chi_{[-N,N]}(n)$.
Now, in all these examples, the family of Fourier transforms converges boundedly to $1$ as the index goes to the proper limit, but I can't see any obvious (sufficient) condition where we can read why for the first three examples the condition is satisfied while it is not for the last one. Any ideas?
fourier-series summability-theory
asked Jan 30 at 23:44
BobBob
1,7001725
$endgroup$
If $(K_N)_{Ninmathbb{N}}$ (or indexed by any other directed set) is a family of functions in $L^1(mathbb{T})$ defined on the 1-torus $mathbb{T}$ such that:
$$forall pin[1,+infty), forall fin L^p(mathbb{T}), |K_N*f-f|_pto0, Nto+infty$$
is there a way to read this property from the family of its Fourier transforms?
Some examples:
$F_N(t):=frac {1}{N+1}left (frac{sinleft(frac{N+1}{2}tright)}{sinleft(frac{t}{2}right)}right)^2$ satisfies this property for $Nto+infty$ and $mathcal{F}(F_N)(n)=(1-frac{|n|}{N})chi_{[-N,N]}(n);$
$P_r(t):= frac{1-r^2}{|1-r e^{it}|^2}$ satisfies this property for $rto 1^-$ and $mathcal{F}(P_r)(n)=r^{|n|};$
$K_s(t):=sum_{ninmathbb{Z}}e^{-sn^2}e^{int}$ satisfies this property for $sto0^+$ and obviously $mathcal{F}(K_s)(n)=e^{-sn^2}$;
$D_N(t):=frac{sinleft(left(N+frac{1}{2}right)tright)}{sinleft(frac{t}{2}right)}$ satisfies this property for $Nto +infty$ and $1<p<+infty$ but not for $p=1$ and $mathcal{F}(D_N)(n)=chi_{[-N,N]}(n)$.
Now, in all these examples, the family of Fourier transforms converges boundedly to $1$ as the index goes to the proper limit, but I can't see any obvious (sufficient) condition where we can read why for the first three examples the condition is satisfied while it is not for the last one. Any ideas?
fourier-series summability-theory
fourier-series summability-theory
asked Jan 30 at 23:44
BobBob
1,7001725
asked Jan 30 at 23:44
BobBob
1,7001725
edited Jan 30 at 23:58
Bob
asked Jan 30 at 23:44
BobBob
1,7001725
asked Jan 30 at 23:44
BobBob
1,7001725
asked Jan 30 at 23:44
BobBob
1,7001725
1,7001725
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In fact
(i) $||f-K_N*f||_2to0$ for every $fin L^2$ if and only if (ii) $sup_{n,N}|widehat{K_N}(n)|<infty$ and $lim_{Ntoinfty}widehat{K_N}(n)=1$ for all $n$.
(i) implies (ii): Define $T_N:L^2to L^2$ by $T_Nf=K_n*f$. Then $||T_N||=sup_n|widehat{K_N}(n)|$ and uniform boundedness implies that $sup_N||T_N||<infty$. If $e_n(t)=e^{int}$ then $T_Ne_n=widehat{K_N}(n)e_n$, hence $lim_Nwidehat{K_N}(n)=1$.
(ii) implies (i): (ii) shows that $||K_N*P-P||_2to0$ for every trigonometric polynomial $P$, and (i) follows since $||T_N||$ is bounded.
I don't believe there is any such simple characterization for $pne2$, just because if $T_N:L^pto L^p$ is defined by $T_Nf=K_N*f$ nobody knows exactly what $||T_N||$ is in general (this would be a characterization of "Fourier multipliers" for $L^p$, which does not exist). Of course if $1le p<infty$ the same argument gives this:
$||K_N*f-f||_pto0$ for every $fin L^p$ if and only if $||T_N||$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for every $n$,
and you can often deduce a yes or no for a specific $K_N$ from that, noting that $$sup_n|widehat{K_N}(n)|le||T_N||le||K_N||_1.$$
Of course if $p=infty$ it's clear that $||T_N||=||K_N||_1$, but there cannot be any such result for $p=infty$ because $||K_N*f-f||_inftyto0$ is impossible, since $K_N*f$ is continuous (the proof fails because the trigonometric polynomials are not dense in $L^infty$). Note however
$K_N*fto f$ uniformly for every $fin C(Bbb T)$ if and only if $||K_N||_1$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for all $n$;
I believe that's the only other nice clean result in this direction that exists.
This is enough information to settle many of the special cases you mention. For example, yes for the Fejer kernel and the Poisson kernel because the $L^1$ norm is bounded; no for the Dirichlet kernel operating on $C(Bbb T)$ because the $L^1$ norm is not bounded. But for example nothing above settles the case of the Dirichlet kernel on $L^p$, $1<p<infty$; that one happens to be a yes, but that's a theorem, not something we can just see by looking at the coefficients.
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094292%2fis-there-a-way-to-read-a-summability-property-of-a-family-of-functions-from-the%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In fact
(i) $||f-K_N*f||_2to0$ for every $fin L^2$ if and only if (ii) $sup_{n,N}|widehat{K_N}(n)|<infty$ and $lim_{Ntoinfty}widehat{K_N}(n)=1$ for all $n$.
(i) implies (ii): Define $T_N:L^2to L^2$ by $T_Nf=K_n*f$. Then $||T_N||=sup_n|widehat{K_N}(n)|$ and uniform boundedness implies that $sup_N||T_N||<infty$. If $e_n(t)=e^{int}$ then $T_Ne_n=widehat{K_N}(n)e_n$, hence $lim_Nwidehat{K_N}(n)=1$.
(ii) implies (i): (ii) shows that $||K_N*P-P||_2to0$ for every trigonometric polynomial $P$, and (i) follows since $||T_N||$ is bounded.
I don't believe there is any such simple characterization for $pne2$, just because if $T_N:L^pto L^p$ is defined by $T_Nf=K_N*f$ nobody knows exactly what $||T_N||$ is in general (this would be a characterization of "Fourier multipliers" for $L^p$, which does not exist). Of course if $1le p<infty$ the same argument gives this:
$||K_N*f-f||_pto0$ for every $fin L^p$ if and only if $||T_N||$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for every $n$,
and you can often deduce a yes or no for a specific $K_N$ from that, noting that $$sup_n|widehat{K_N}(n)|le||T_N||le||K_N||_1.$$
Of course if $p=infty$ it's clear that $||T_N||=||K_N||_1$, but there cannot be any such result for $p=infty$ because $||K_N*f-f||_inftyto0$ is impossible, since $K_N*f$ is continuous (the proof fails because the trigonometric polynomials are not dense in $L^infty$). Note however
$K_N*fto f$ uniformly for every $fin C(Bbb T)$ if and only if $||K_N||_1$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for all $n$;
I believe that's the only other nice clean result in this direction that exists.
This is enough information to settle many of the special cases you mention. For example, yes for the Fejer kernel and the Poisson kernel because the $L^1$ norm is bounded; no for the Dirichlet kernel operating on $C(Bbb T)$ because the $L^1$ norm is not bounded. But for example nothing above settles the case of the Dirichlet kernel on $L^p$, $1<p<infty$; that one happens to be a yes, but that's a theorem, not something we can just see by looking at the coefficients.
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
$begingroup$
In fact
(i) $||f-K_N*f||_2to0$ for every $fin L^2$ if and only if (ii) $sup_{n,N}|widehat{K_N}(n)|<infty$ and $lim_{Ntoinfty}widehat{K_N}(n)=1$ for all $n$.
(i) implies (ii): Define $T_N:L^2to L^2$ by $T_Nf=K_n*f$. Then $||T_N||=sup_n|widehat{K_N}(n)|$ and uniform boundedness implies that $sup_N||T_N||<infty$. If $e_n(t)=e^{int}$ then $T_Ne_n=widehat{K_N}(n)e_n$, hence $lim_Nwidehat{K_N}(n)=1$.
(ii) implies (i): (ii) shows that $||K_N*P-P||_2to0$ for every trigonometric polynomial $P$, and (i) follows since $||T_N||$ is bounded.
I don't believe there is any such simple characterization for $pne2$, just because if $T_N:L^pto L^p$ is defined by $T_Nf=K_N*f$ nobody knows exactly what $||T_N||$ is in general (this would be a characterization of "Fourier multipliers" for $L^p$, which does not exist). Of course if $1le p<infty$ the same argument gives this:
$||K_N*f-f||_pto0$ for every $fin L^p$ if and only if $||T_N||$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for every $n$,
and you can often deduce a yes or no for a specific $K_N$ from that, noting that $$sup_n|widehat{K_N}(n)|le||T_N||le||K_N||_1.$$
Of course if $p=infty$ it's clear that $||T_N||=||K_N||_1$, but there cannot be any such result for $p=infty$ because $||K_N*f-f||_inftyto0$ is impossible, since $K_N*f$ is continuous (the proof fails because the trigonometric polynomials are not dense in $L^infty$). Note however
$K_N*fto f$ uniformly for every $fin C(Bbb T)$ if and only if $||K_N||_1$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for all $n$;
I believe that's the only other nice clean result in this direction that exists.
This is enough information to settle many of the special cases you mention. For example, yes for the Fejer kernel and the Poisson kernel because the $L^1$ norm is bounded; no for the Dirichlet kernel operating on $C(Bbb T)$ because the $L^1$ norm is not bounded. But for example nothing above settles the case of the Dirichlet kernel on $L^p$, $1<p<infty$; that one happens to be a yes, but that's a theorem, not something we can just see by looking at the coefficients.
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
$begingroup$
In fact
(i) $||f-K_N*f||_2to0$ for every $fin L^2$ if and only if (ii) $sup_{n,N}|widehat{K_N}(n)|<infty$ and $lim_{Ntoinfty}widehat{K_N}(n)=1$ for all $n$.
(i) implies (ii): Define $T_N:L^2to L^2$ by $T_Nf=K_n*f$. Then $||T_N||=sup_n|widehat{K_N}(n)|$ and uniform boundedness implies that $sup_N||T_N||<infty$. If $e_n(t)=e^{int}$ then $T_Ne_n=widehat{K_N}(n)e_n$, hence $lim_Nwidehat{K_N}(n)=1$.
(ii) implies (i): (ii) shows that $||K_N*P-P||_2to0$ for every trigonometric polynomial $P$, and (i) follows since $||T_N||$ is bounded.
I don't believe there is any such simple characterization for $pne2$, just because if $T_N:L^pto L^p$ is defined by $T_Nf=K_N*f$ nobody knows exactly what $||T_N||$ is in general (this would be a characterization of "Fourier multipliers" for $L^p$, which does not exist). Of course if $1le p<infty$ the same argument gives this:
$||K_N*f-f||_pto0$ for every $fin L^p$ if and only if $||T_N||$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for every $n$,
and you can often deduce a yes or no for a specific $K_N$ from that, noting that $$sup_n|widehat{K_N}(n)|le||T_N||le||K_N||_1.$$
Of course if $p=infty$ it's clear that $||T_N||=||K_N||_1$, but there cannot be any such result for $p=infty$ because $||K_N*f-f||_inftyto0$ is impossible, since $K_N*f$ is continuous (the proof fails because the trigonometric polynomials are not dense in $L^infty$). Note however
$K_N*fto f$ uniformly for every $fin C(Bbb T)$ if and only if $||K_N||_1$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for all $n$;
I believe that's the only other nice clean result in this direction that exists.
This is enough information to settle many of the special cases you mention. For example, yes for the Fejer kernel and the Poisson kernel because the $L^1$ norm is bounded; no for the Dirichlet kernel operating on $C(Bbb T)$ because the $L^1$ norm is not bounded. But for example nothing above settles the case of the Dirichlet kernel on $L^p$, $1<p<infty$; that one happens to be a yes, but that's a theorem, not something we can just see by looking at the coefficients.
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
In fact
(i) $||f-K_N*f||_2to0$ for every $fin L^2$ if and only if (ii) $sup_{n,N}|widehat{K_N}(n)|<infty$ and $lim_{Ntoinfty}widehat{K_N}(n)=1$ for all $n$.
(i) implies (ii): Define $T_N:L^2to L^2$ by $T_Nf=K_n*f$. Then $||T_N||=sup_n|widehat{K_N}(n)|$ and uniform boundedness implies that $sup_N||T_N||<infty$. If $e_n(t)=e^{int}$ then $T_Ne_n=widehat{K_N}(n)e_n$, hence $lim_Nwidehat{K_N}(n)=1$.
(ii) implies (i): (ii) shows that $||K_N*P-P||_2to0$ for every trigonometric polynomial $P$, and (i) follows since $||T_N||$ is bounded.
I don't believe there is any such simple characterization for $pne2$, just because if $T_N:L^pto L^p$ is defined by $T_Nf=K_N*f$ nobody knows exactly what $||T_N||$ is in general (this would be a characterization of "Fourier multipliers" for $L^p$, which does not exist). Of course if $1le p<infty$ the same argument gives this:
$||K_N*f-f||_pto0$ for every $fin L^p$ if and only if $||T_N||$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for every $n$,
and you can often deduce a yes or no for a specific $K_N$ from that, noting that $$sup_n|widehat{K_N}(n)|le||T_N||le||K_N||_1.$$
Of course if $p=infty$ it's clear that $||T_N||=||K_N||_1$, but there cannot be any such result for $p=infty$ because $||K_N*f-f||_inftyto0$ is impossible, since $K_N*f$ is continuous (the proof fails because the trigonometric polynomials are not dense in $L^infty$). Note however
$K_N*fto f$ uniformly for every $fin C(Bbb T)$ if and only if $||K_N||_1$ is bounded and $lim_Nwidehat{K_N}(n)=1$ for all $n$;
I believe that's the only other nice clean result in this direction that exists.
This is enough information to settle many of the special cases you mention. For example, yes for the Fejer kernel and the Poisson kernel because the $L^1$ norm is bounded; no for the Dirichlet kernel operating on $C(Bbb T)$ because the $L^1$ norm is not bounded. But for example nothing above settles the case of the Dirichlet kernel on $L^p$, $1<p<infty$; that one happens to be a yes, but that's a theorem, not something we can just see by looking at the coefficients.
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
edited Jan 31 at 14:13
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 31 at 14:04
David C. UllrichDavid C. Ullrich
61.6k43995
61.6k43995
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094292%2fis-there-a-way-to-read-a-summability-property-of-a-family-of-functions-from-the%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
