Mixing
Why is ( U, $ Vert . Vert $ ) a Banach Space? The space of uniformly convergent Fourier series
$begingroup$
Consider $ U = { f in mathcal{C}_{2pi}, : S_n(f) : $converges uniformly on $ mathbb{R} } $ with $ Vert f Vert = $sup$_n Vert S_n (f) Vert_{infty } $
Show that $ (U, Vert .Vert ) $ is a Banach Space.
I am not sure how to go about the first step : how does f has an existing limit, and if f is equal to its Fourier series or not. Thanks for any help on this.
$ forall epsilon > 0 : exists N > 0, : forall p,q > N : $sup$_n Vert S_n (f_p - f_q) Vert_{infty } < epsilon $
How to show $ exists f, : f_n rightarrow f $ ?
Then after we need to prove $ f in U$
functional-analysis fourier-analysis banach-spaces fourier-series
asked Jan 20 at 16:42
PsylexPsylex
1068
$endgroup$
add a comment |
$begingroup$
Consider $ U = { f in mathcal{C}_{2pi}, : S_n(f) : $converges uniformly on $ mathbb{R} } $ with $ Vert f Vert = $sup$_n Vert S_n (f) Vert_{infty } $
Show that $ (U, Vert .Vert ) $ is a Banach Space.
I am not sure how to go about the first step : how does f has an existing limit, and if f is equal to its Fourier series or not. Thanks for any help on this.
$ forall epsilon > 0 : exists N > 0, : forall p,q > N : $sup$_n Vert S_n (f_p - f_q) Vert_{infty } < epsilon $
How to show $ exists f, : f_n rightarrow f $ ?
Then after we need to prove $ f in U$
functional-analysis fourier-analysis banach-spaces fourier-series
asked Jan 20 at 16:42
PsylexPsylex
1068
$endgroup$
$begingroup$
(I made a mistake in my first comment, here is the ediited version): The "first step" is an ambiguous term. Please clarify what you mean. I understand that you want to prove that uniform boundedness of the partial sums is equivalent to uniform convergence of the Fourier series. You can use convergence in a "good" dense subspace, plus the uniform condition.
$endgroup$
– Adrián González-Pérez
Jan 21 at 12:48
$begingroup$
Thanks for your answer. I tried the method of "finding" a suitable limit for a sequence $ f_j $ but as you did it's better to prove that absolute convergence of a series in U implies simple convergence and this proves that U is indeed a Banach.
$endgroup$
– Psylex
Jan 22 at 7:31
add a comment |
$begingroup$
Consider $ U = { f in mathcal{C}_{2pi}, : S_n(f) : $converges uniformly on $ mathbb{R} } $ with $ Vert f Vert = $sup$_n Vert S_n (f) Vert_{infty } $
Show that $ (U, Vert .Vert ) $ is a Banach Space.
I am not sure how to go about the first step : how does f has an existing limit, and if f is equal to its Fourier series or not. Thanks for any help on this.
$ forall epsilon > 0 : exists N > 0, : forall p,q > N : $sup$_n Vert S_n (f_p - f_q) Vert_{infty } < epsilon $
How to show $ exists f, : f_n rightarrow f $ ?
Then after we need to prove $ f in U$
functional-analysis fourier-analysis banach-spaces fourier-series
asked Jan 20 at 16:42
PsylexPsylex
1068
$endgroup$
Consider $ U = { f in mathcal{C}_{2pi}, : S_n(f) : $converges uniformly on $ mathbb{R} } $ with $ Vert f Vert = $sup$_n Vert S_n (f) Vert_{infty } $
Show that $ (U, Vert .Vert ) $ is a Banach Space.
I am not sure how to go about the first step : how does f has an existing limit, and if f is equal to its Fourier series or not. Thanks for any help on this.
$ forall epsilon > 0 : exists N > 0, : forall p,q > N : $sup$_n Vert S_n (f_p - f_q) Vert_{infty } < epsilon $
How to show $ exists f, : f_n rightarrow f $ ?
Then after we need to prove $ f in U$
functional-analysis fourier-analysis banach-spaces fourier-series
functional-analysis fourier-analysis banach-spaces fourier-series
asked Jan 20 at 16:42
PsylexPsylex
1068
asked Jan 20 at 16:42
PsylexPsylex
1068
asked Jan 20 at 16:42
PsylexPsylex
1068
asked Jan 20 at 16:42
PsylexPsylex
1068
asked Jan 20 at 16:42
PsylexPsylex
1068
1068
$begingroup$
(I made a mistake in my first comment, here is the ediited version): The "first step" is an ambiguous term. Please clarify what you mean. I understand that you want to prove that uniform boundedness of the partial sums is equivalent to uniform convergence of the Fourier series. You can use convergence in a "good" dense subspace, plus the uniform condition.
$endgroup$
– Adrián González-Pérez
Jan 21 at 12:48
$begingroup$
Thanks for your answer. I tried the method of "finding" a suitable limit for a sequence $ f_j $ but as you did it's better to prove that absolute convergence of a series in U implies simple convergence and this proves that U is indeed a Banach.
$endgroup$
– Psylex
Jan 22 at 7:31
add a comment |
$begingroup$
(I made a mistake in my first comment, here is the ediited version): The "first step" is an ambiguous term. Please clarify what you mean. I understand that you want to prove that uniform boundedness of the partial sums is equivalent to uniform convergence of the Fourier series. You can use convergence in a "good" dense subspace, plus the uniform condition.
$endgroup$
– Adrián González-Pérez
Jan 21 at 12:48
$begingroup$
Thanks for your answer. I tried the method of "finding" a suitable limit for a sequence $ f_j $ but as you did it's better to prove that absolute convergence of a series in U implies simple convergence and this proves that U is indeed a Banach.
$endgroup$
– Psylex
Jan 22 at 7:31
$begingroup$
(I made a mistake in my first comment, here is the ediited version): The "first step" is an ambiguous term. Please clarify what you mean. I understand that you want to prove that uniform boundedness of the partial sums is equivalent to uniform convergence of the Fourier series. You can use convergence in a "good" dense subspace, plus the uniform condition.
$endgroup$
– Adrián González-Pérez
Jan 21 at 12:48
$begingroup$
(I made a mistake in my first comment, here is the ediited version): The "first step" is an ambiguous term. Please clarify what you mean. I understand that you want to prove that uniform boundedness of the partial sums is equivalent to uniform convergence of the Fourier series. You can use convergence in a "good" dense subspace, plus the uniform condition.
$endgroup$
– Adrián González-Pérez
Jan 21 at 12:48
$begingroup$
Thanks for your answer. I tried the method of "finding" a suitable limit for a sequence $ f_j $ but as you did it's better to prove that absolute convergence of a series in U implies simple convergence and this proves that U is indeed a Banach.
$endgroup$
– Psylex
Jan 22 at 7:31
$begingroup$
Thanks for your answer. I tried the method of "finding" a suitable limit for a sequence $ f_j $ but as you did it's better to prove that absolute convergence of a series in U implies simple convergence and this proves that U is indeed a Banach.
$endgroup$
– Psylex
Jan 22 at 7:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, we are going to see that the condition
$$
| f |_U := sup_N | S_N(f) |_infty < infty tag{U}
$$
is equivalent to the fact that $S_N(f)$ converge to $f$ uniformly. First, if $S_N(f) to f$ uniformly, then, by continuity of the suppremum norm $|S_N(f)|_infty to |f|_infty$ and therefore the sequence $(| S_N(f) |_infty)_N$ is bounded. For the other direction we need to use that, for every $epsilon > 0$ and function $f$ satisfying (U), there is a $g in C[0,2pi]$ such that
- $quad displaystyle{sup_N | S_N(g -f)|_infty < epsilon}$
$quad S_N(g) to g$ in the $| |_infty$-norm.
Think of $g$ as a Schwartz class function approximating $f$. Notice also that, for every continuous function $h$, we have that
$$
| h |_infty leq sup_N |S_N(h) |_infty.
$$
Indeed, that holds for every trig. polynomial and you can extend by the Stone-Wierstrass theorem. The identity above implies that $|f-g|_infty < epsilon$.
Now, for every $f$ satisfying (U), you can pick $g_epsilon$ satisfying the two properties above, therefore:
begin{eqnarray}
& & |S_N(f) - f |_infty \
& leq & | S_N(f - g_epsilon) |_infty + | S_N(g_epsilon) - g_epsilon |_infty + | g_epsilon - f |_infty\
& leq & epsilon + | S_N(g_epsilon) - g_epsilon |_infty + epsilon to 2 , epsilon.
end{eqnarray}
Therefore the limit of the quantity above is smaller that $ 2 epsilon$, but since $epsilon > 0$ is arbitrary we have that the limit is $0$.
Obtaining that $U$ is a Banach space is a corollary, since for every sequence $f_j$ satisfying that
$$
sum_{j=1} | f_j |_U < infty
$$
it also holds that the sum above converge uniformly. Therefore there is a limit $f = sum_{j} f_j$. And checking that $f in U$ is a trivial calculation.
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
$endgroup$
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%2f3080798%2fwhy-is-u-vert-vert-a-banach-space-the-space-of-uniformly-convergen%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
$begingroup$
First, we are going to see that the condition
$$
| f |_U := sup_N | S_N(f) |_infty < infty tag{U}
$$
is equivalent to the fact that $S_N(f)$ converge to $f$ uniformly. First, if $S_N(f) to f$ uniformly, then, by continuity of the suppremum norm $|S_N(f)|_infty to |f|_infty$ and therefore the sequence $(| S_N(f) |_infty)_N$ is bounded. For the other direction we need to use that, for every $epsilon > 0$ and function $f$ satisfying (U), there is a $g in C[0,2pi]$ such that
- $quad displaystyle{sup_N | S_N(g -f)|_infty < epsilon}$
$quad S_N(g) to g$ in the $| |_infty$-norm.
Think of $g$ as a Schwartz class function approximating $f$. Notice also that, for every continuous function $h$, we have that
$$
| h |_infty leq sup_N |S_N(h) |_infty.
$$
Indeed, that holds for every trig. polynomial and you can extend by the Stone-Wierstrass theorem. The identity above implies that $|f-g|_infty < epsilon$.
Now, for every $f$ satisfying (U), you can pick $g_epsilon$ satisfying the two properties above, therefore:
begin{eqnarray}
& & |S_N(f) - f |_infty \
& leq & | S_N(f - g_epsilon) |_infty + | S_N(g_epsilon) - g_epsilon |_infty + | g_epsilon - f |_infty\
& leq & epsilon + | S_N(g_epsilon) - g_epsilon |_infty + epsilon to 2 , epsilon.
end{eqnarray}
Therefore the limit of the quantity above is smaller that $ 2 epsilon$, but since $epsilon > 0$ is arbitrary we have that the limit is $0$.
Obtaining that $U$ is a Banach space is a corollary, since for every sequence $f_j$ satisfying that
$$
sum_{j=1} | f_j |_U < infty
$$
it also holds that the sum above converge uniformly. Therefore there is a limit $f = sum_{j} f_j$. And checking that $f in U$ is a trivial calculation.
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
$endgroup$
add a comment |
$begingroup$
First, we are going to see that the condition
$$
| f |_U := sup_N | S_N(f) |_infty < infty tag{U}
$$
is equivalent to the fact that $S_N(f)$ converge to $f$ uniformly. First, if $S_N(f) to f$ uniformly, then, by continuity of the suppremum norm $|S_N(f)|_infty to |f|_infty$ and therefore the sequence $(| S_N(f) |_infty)_N$ is bounded. For the other direction we need to use that, for every $epsilon > 0$ and function $f$ satisfying (U), there is a $g in C[0,2pi]$ such that
- $quad displaystyle{sup_N | S_N(g -f)|_infty < epsilon}$
$quad S_N(g) to g$ in the $| |_infty$-norm.
Think of $g$ as a Schwartz class function approximating $f$. Notice also that, for every continuous function $h$, we have that
$$
| h |_infty leq sup_N |S_N(h) |_infty.
$$
Indeed, that holds for every trig. polynomial and you can extend by the Stone-Wierstrass theorem. The identity above implies that $|f-g|_infty < epsilon$.
Now, for every $f$ satisfying (U), you can pick $g_epsilon$ satisfying the two properties above, therefore:
begin{eqnarray}
& & |S_N(f) - f |_infty \
& leq & | S_N(f - g_epsilon) |_infty + | S_N(g_epsilon) - g_epsilon |_infty + | g_epsilon - f |_infty\
& leq & epsilon + | S_N(g_epsilon) - g_epsilon |_infty + epsilon to 2 , epsilon.
end{eqnarray}
Therefore the limit of the quantity above is smaller that $ 2 epsilon$, but since $epsilon > 0$ is arbitrary we have that the limit is $0$.
Obtaining that $U$ is a Banach space is a corollary, since for every sequence $f_j$ satisfying that
$$
sum_{j=1} | f_j |_U < infty
$$
it also holds that the sum above converge uniformly. Therefore there is a limit $f = sum_{j} f_j$. And checking that $f in U$ is a trivial calculation.
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
$endgroup$
add a comment |
$begingroup$
First, we are going to see that the condition
$$
| f |_U := sup_N | S_N(f) |_infty < infty tag{U}
$$
is equivalent to the fact that $S_N(f)$ converge to $f$ uniformly. First, if $S_N(f) to f$ uniformly, then, by continuity of the suppremum norm $|S_N(f)|_infty to |f|_infty$ and therefore the sequence $(| S_N(f) |_infty)_N$ is bounded. For the other direction we need to use that, for every $epsilon > 0$ and function $f$ satisfying (U), there is a $g in C[0,2pi]$ such that
- $quad displaystyle{sup_N | S_N(g -f)|_infty < epsilon}$
$quad S_N(g) to g$ in the $| |_infty$-norm.
Think of $g$ as a Schwartz class function approximating $f$. Notice also that, for every continuous function $h$, we have that
$$
| h |_infty leq sup_N |S_N(h) |_infty.
$$
Indeed, that holds for every trig. polynomial and you can extend by the Stone-Wierstrass theorem. The identity above implies that $|f-g|_infty < epsilon$.
Now, for every $f$ satisfying (U), you can pick $g_epsilon$ satisfying the two properties above, therefore:
begin{eqnarray}
& & |S_N(f) - f |_infty \
& leq & | S_N(f - g_epsilon) |_infty + | S_N(g_epsilon) - g_epsilon |_infty + | g_epsilon - f |_infty\
& leq & epsilon + | S_N(g_epsilon) - g_epsilon |_infty + epsilon to 2 , epsilon.
end{eqnarray}
Therefore the limit of the quantity above is smaller that $ 2 epsilon$, but since $epsilon > 0$ is arbitrary we have that the limit is $0$.
Obtaining that $U$ is a Banach space is a corollary, since for every sequence $f_j$ satisfying that
$$
sum_{j=1} | f_j |_U < infty
$$
it also holds that the sum above converge uniformly. Therefore there is a limit $f = sum_{j} f_j$. And checking that $f in U$ is a trivial calculation.
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
$endgroup$
First, we are going to see that the condition
$$
| f |_U := sup_N | S_N(f) |_infty < infty tag{U}
$$
is equivalent to the fact that $S_N(f)$ converge to $f$ uniformly. First, if $S_N(f) to f$ uniformly, then, by continuity of the suppremum norm $|S_N(f)|_infty to |f|_infty$ and therefore the sequence $(| S_N(f) |_infty)_N$ is bounded. For the other direction we need to use that, for every $epsilon > 0$ and function $f$ satisfying (U), there is a $g in C[0,2pi]$ such that
- $quad displaystyle{sup_N | S_N(g -f)|_infty < epsilon}$
$quad S_N(g) to g$ in the $| |_infty$-norm.
Think of $g$ as a Schwartz class function approximating $f$. Notice also that, for every continuous function $h$, we have that
$$
| h |_infty leq sup_N |S_N(h) |_infty.
$$
Indeed, that holds for every trig. polynomial and you can extend by the Stone-Wierstrass theorem. The identity above implies that $|f-g|_infty < epsilon$.
Now, for every $f$ satisfying (U), you can pick $g_epsilon$ satisfying the two properties above, therefore:
begin{eqnarray}
& & |S_N(f) - f |_infty \
& leq & | S_N(f - g_epsilon) |_infty + | S_N(g_epsilon) - g_epsilon |_infty + | g_epsilon - f |_infty\
& leq & epsilon + | S_N(g_epsilon) - g_epsilon |_infty + epsilon to 2 , epsilon.
end{eqnarray}
Therefore the limit of the quantity above is smaller that $ 2 epsilon$, but since $epsilon > 0$ is arbitrary we have that the limit is $0$.
Obtaining that $U$ is a Banach space is a corollary, since for every sequence $f_j$ satisfying that
$$
sum_{j=1} | f_j |_U < infty
$$
it also holds that the sum above converge uniformly. Therefore there is a limit $f = sum_{j} f_j$. And checking that $f in U$ is a trivial calculation.
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
answered Jan 21 at 13:44
Adrián González-PérezAdrián González-Pérez
1,196139
1,196139
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%2f3080798%2fwhy-is-u-vert-vert-a-banach-space-the-space-of-uniformly-convergen%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
$begingroup$
(I made a mistake in my first comment, here is the ediited version): The "first step" is an ambiguous term. Please clarify what you mean. I understand that you want to prove that uniform boundedness of the partial sums is equivalent to uniform convergence of the Fourier series. You can use convergence in a "good" dense subspace, plus the uniform condition.
$endgroup$
– Adrián González-Pérez
Jan 21 at 12:48
$begingroup$
Thanks for your answer. I tried the method of "finding" a suitable limit for a sequence $ f_j $ but as you did it's better to prove that absolute convergence of a series in U implies simple convergence and this proves that U is indeed a Banach.
$endgroup$
– Psylex
Jan 22 at 7:31