How do I show a linear operator is bounded?
up vote
3
down vote
favorite
Let $ A:L_2[0,pi] to L_2[0,pi] $ is a linear operator such that
$$(Ax)(t)=sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)$$
How do I work out that $A$ is a bounded linear operator for this specific example?
My Attempt:
I must show $ |Ax|leq M.|x| $ there exists some $ M geq 0 $ .
$$|Ax|^{2} =int_{0}^{pi}|sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)|^{2}dt\ =int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}|int_{0}^{pi}|x(s)sin(ns)ds|^{2}dt\ < int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}[underbrace{ int_{0}^{pi}|x(s)|^{2}ds}_{=|x|^{2}}int_{0}^{pi}|sin(ns)|^{2}ds]dt$$
functional-analysis operator-theory
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
asked Nov 28 at 21:57
os.
183
add a comment |
up vote
3
down vote
favorite
Let $ A:L_2[0,pi] to L_2[0,pi] $ is a linear operator such that
$$(Ax)(t)=sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)$$
How do I work out that $A$ is a bounded linear operator for this specific example?
My Attempt:
I must show $ |Ax|leq M.|x| $ there exists some $ M geq 0 $ .
$$|Ax|^{2} =int_{0}^{pi}|sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)|^{2}dt\ =int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}|int_{0}^{pi}|x(s)sin(ns)ds|^{2}dt\ < int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}[underbrace{ int_{0}^{pi}|x(s)|^{2}ds}_{=|x|^{2}}int_{0}^{pi}|sin(ns)|^{2}ds]dt$$
functional-analysis operator-theory
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
asked Nov 28 at 21:57
os.
183
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $ A:L_2[0,pi] to L_2[0,pi] $ is a linear operator such that
$$(Ax)(t)=sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)$$
How do I work out that $A$ is a bounded linear operator for this specific example?
My Attempt:
I must show $ |Ax|leq M.|x| $ there exists some $ M geq 0 $ .
$$|Ax|^{2} =int_{0}^{pi}|sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)|^{2}dt\ =int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}|int_{0}^{pi}|x(s)sin(ns)ds|^{2}dt\ < int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}[underbrace{ int_{0}^{pi}|x(s)|^{2}ds}_{=|x|^{2}}int_{0}^{pi}|sin(ns)|^{2}ds]dt$$
functional-analysis operator-theory
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
asked Nov 28 at 21:57
os.
183
Let $ A:L_2[0,pi] to L_2[0,pi] $ is a linear operator such that
$$(Ax)(t)=sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)$$
How do I work out that $A$ is a bounded linear operator for this specific example?
My Attempt:
I must show $ |Ax|leq M.|x| $ there exists some $ M geq 0 $ .
$$|Ax|^{2} =int_{0}^{pi}|sum_{n=1}^{infty}[int_{0}^{pi}x(s)sin(ns)ds]cos(nt)|^{2}dt\ =int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}|int_{0}^{pi}|x(s)sin(ns)ds|^{2}dt\ < int_{0}^{pi}sum_{n=1}^{infty}|cos(nt)|^{2}[underbrace{ int_{0}^{pi}|x(s)|^{2}ds}_{=|x|^{2}}int_{0}^{pi}|sin(ns)|^{2}ds]dt$$
functional-analysis operator-theory
functional-analysis operator-theory
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
asked Nov 28 at 21:57
os.
183
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
asked Nov 28 at 21:57
os.
183
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
edited Nov 28 at 22:03
Davide Giraudo
124k16150259
124k16150259
asked Nov 28 at 21:57
os.
183
asked Nov 28 at 21:57
os.
183
asked Nov 28 at 21:57
os.
183
183
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
It's a bit easier to do the problem in abstract. Suppose that ${e_n}$ and ${f_n}$ are orthonormal sets in a Hilbert space $H$. Define
$$tag1
Ax=sum_n langle x,e_nrangle,f_n.
$$
This is well-defined because for any finite sum we have, since ${f_n}$ is orthonormal,
$$tag2
left|sum_{n=1}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=1}^N|langle x,e_nrangle|^2leq|x|^2,
$$
where the inequality is Bessel's inequality. Similarly,
$$
left|sum_{n=M}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=M}^N|langle x,e_nrangle|^2,
$$
showing that the sequence of partial sums is Cauchy. So the series in $(1)$ exists in $H$, and by $(2)$ we have $|Ax|leq|x|$.
answered Nov 28 at 23:43
Martin Argerami
123k1176174
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%2f3017778%2fhow-do-i-show-a-linear-operator-is-bounded%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
up vote
1
down vote
accepted
It's a bit easier to do the problem in abstract. Suppose that ${e_n}$ and ${f_n}$ are orthonormal sets in a Hilbert space $H$. Define
$$tag1
Ax=sum_n langle x,e_nrangle,f_n.
$$
This is well-defined because for any finite sum we have, since ${f_n}$ is orthonormal,
$$tag2
left|sum_{n=1}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=1}^N|langle x,e_nrangle|^2leq|x|^2,
$$
where the inequality is Bessel's inequality. Similarly,
$$
left|sum_{n=M}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=M}^N|langle x,e_nrangle|^2,
$$
showing that the sequence of partial sums is Cauchy. So the series in $(1)$ exists in $H$, and by $(2)$ we have $|Ax|leq|x|$.
answered Nov 28 at 23:43
Martin Argerami
123k1176174
add a comment |
up vote
1
down vote
accepted
It's a bit easier to do the problem in abstract. Suppose that ${e_n}$ and ${f_n}$ are orthonormal sets in a Hilbert space $H$. Define
$$tag1
Ax=sum_n langle x,e_nrangle,f_n.
$$
This is well-defined because for any finite sum we have, since ${f_n}$ is orthonormal,
$$tag2
left|sum_{n=1}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=1}^N|langle x,e_nrangle|^2leq|x|^2,
$$
where the inequality is Bessel's inequality. Similarly,
$$
left|sum_{n=M}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=M}^N|langle x,e_nrangle|^2,
$$
showing that the sequence of partial sums is Cauchy. So the series in $(1)$ exists in $H$, and by $(2)$ we have $|Ax|leq|x|$.
answered Nov 28 at 23:43
Martin Argerami
123k1176174
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It's a bit easier to do the problem in abstract. Suppose that ${e_n}$ and ${f_n}$ are orthonormal sets in a Hilbert space $H$. Define
$$tag1
Ax=sum_n langle x,e_nrangle,f_n.
$$
This is well-defined because for any finite sum we have, since ${f_n}$ is orthonormal,
$$tag2
left|sum_{n=1}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=1}^N|langle x,e_nrangle|^2leq|x|^2,
$$
where the inequality is Bessel's inequality. Similarly,
$$
left|sum_{n=M}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=M}^N|langle x,e_nrangle|^2,
$$
showing that the sequence of partial sums is Cauchy. So the series in $(1)$ exists in $H$, and by $(2)$ we have $|Ax|leq|x|$.
answered Nov 28 at 23:43
Martin Argerami
123k1176174
It's a bit easier to do the problem in abstract. Suppose that ${e_n}$ and ${f_n}$ are orthonormal sets in a Hilbert space $H$. Define
$$tag1
Ax=sum_n langle x,e_nrangle,f_n.
$$
This is well-defined because for any finite sum we have, since ${f_n}$ is orthonormal,
$$tag2
left|sum_{n=1}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=1}^N|langle x,e_nrangle|^2leq|x|^2,
$$
where the inequality is Bessel's inequality. Similarly,
$$
left|sum_{n=M}^Nlangle x,e_nrangle,f_nright|^2=sum_{n=M}^N|langle x,e_nrangle|^2,
$$
showing that the sequence of partial sums is Cauchy. So the series in $(1)$ exists in $H$, and by $(2)$ we have $|Ax|leq|x|$.
answered Nov 28 at 23:43
Martin Argerami
123k1176174
answered Nov 28 at 23:43
Martin Argerami
123k1176174
answered Nov 28 at 23:43
Martin Argerami
123k1176174
answered Nov 28 at 23:43
Martin Argerami
123k1176174
123k1176174
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3017778%2fhow-do-i-show-a-linear-operator-is-bounded%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
