Mixing
For $u in C[0,1]$ with $u(0)=u(1)=0$ show $int_0^1 u^2(x) dx leq Cint_0^1 (u'(x))^2 dx$
$begingroup$
Let $u$ be a continously differentiable function of the interval $[0,1]$ s.t $u(0)=u(1)=0$ show there exists a constant $C$ so the following hold $int_0^1 u^2(x) dx leq Cint_0^1 (u'(x))^2 dx$.
I'm not sure where to go with this problem I tried starting by using the fundamental theorem of calculus and could get anywhere.
integration functional-analysis
asked Jan 20 at 17:01
RogerRoger
847
$endgroup$
add a comment |
$begingroup$
Let $u$ be a continously differentiable function of the interval $[0,1]$ s.t $u(0)=u(1)=0$ show there exists a constant $C$ so the following hold $int_0^1 u^2(x) dx leq Cint_0^1 (u'(x))^2 dx$.
I'm not sure where to go with this problem I tried starting by using the fundamental theorem of calculus and could get anywhere.
integration functional-analysis
asked Jan 20 at 17:01
RogerRoger
847
$endgroup$
2
$begingroup$
It is called Wirtinger's inequality. See "second version" here: en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions
$endgroup$
– Zeekless
Jan 20 at 17:05
1
$begingroup$
Hint: Use fourier expansion and apply parseval's theorem
$endgroup$
– TheOscillator
Jan 20 at 17:19
add a comment |
$begingroup$
Let $u$ be a continously differentiable function of the interval $[0,1]$ s.t $u(0)=u(1)=0$ show there exists a constant $C$ so the following hold $int_0^1 u^2(x) dx leq Cint_0^1 (u'(x))^2 dx$.
I'm not sure where to go with this problem I tried starting by using the fundamental theorem of calculus and could get anywhere.
integration functional-analysis
asked Jan 20 at 17:01
RogerRoger
847
$endgroup$
Let $u$ be a continously differentiable function of the interval $[0,1]$ s.t $u(0)=u(1)=0$ show there exists a constant $C$ so the following hold $int_0^1 u^2(x) dx leq Cint_0^1 (u'(x))^2 dx$.
I'm not sure where to go with this problem I tried starting by using the fundamental theorem of calculus and could get anywhere.
integration functional-analysis
integration functional-analysis
asked Jan 20 at 17:01
RogerRoger
847
asked Jan 20 at 17:01
RogerRoger
847
asked Jan 20 at 17:01
RogerRoger
847
asked Jan 20 at 17:01
RogerRoger
847
asked Jan 20 at 17:01
RogerRoger
847
847
2
$begingroup$
It is called Wirtinger's inequality. See "second version" here: en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions
$endgroup$
– Zeekless
Jan 20 at 17:05
1
$begingroup$
Hint: Use fourier expansion and apply parseval's theorem
$endgroup$
– TheOscillator
Jan 20 at 17:19
add a comment |
2
$begingroup$
It is called Wirtinger's inequality. See "second version" here: en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions
$endgroup$
– Zeekless
Jan 20 at 17:05
1
$begingroup$
Hint: Use fourier expansion and apply parseval's theorem
$endgroup$
– TheOscillator
Jan 20 at 17:19
2
2
$begingroup$
It is called Wirtinger's inequality. See "second version" here: en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions
$endgroup$
– Zeekless
Jan 20 at 17:05
$begingroup$
It is called Wirtinger's inequality. See "second version" here: en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions
$endgroup$
– Zeekless
Jan 20 at 17:05
1
1
$begingroup$
Hint: Use fourier expansion and apply parseval's theorem
$endgroup$
– TheOscillator
Jan 20 at 17:19
$begingroup$
Hint: Use fourier expansion and apply parseval's theorem
$endgroup$
– TheOscillator
Jan 20 at 17:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As Zeekless said, you can get $C=frac{1}{pi^2}$ using Wirtinger's inequality.
On the other hand, it is possible to bound it (by a worse constant) using FTC. Writing
$$ u(x)^2=left( int_0^x u'(t) mathrm{d} t right)^2 leq left(int_0^x u'(t)^2 mathrm{d} tright) left( int_0^x 1^2 mathrm{d} tright) leq int_0^1 u'(t)^2 mathrm{d}t,$$
you get $C=1$, after integrating from $x=0$ to $1$.
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
$endgroup$
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%2f3080822%2ffor-u-in-c0-1-with-u0-u1-0-show-int-01-u2x-dx-leq-c-int-01-u%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
$begingroup$
As Zeekless said, you can get $C=frac{1}{pi^2}$ using Wirtinger's inequality.
On the other hand, it is possible to bound it (by a worse constant) using FTC. Writing
$$ u(x)^2=left( int_0^x u'(t) mathrm{d} t right)^2 leq left(int_0^x u'(t)^2 mathrm{d} tright) left( int_0^x 1^2 mathrm{d} tright) leq int_0^1 u'(t)^2 mathrm{d}t,$$
you get $C=1$, after integrating from $x=0$ to $1$.
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
$endgroup$
add a comment |
$begingroup$
As Zeekless said, you can get $C=frac{1}{pi^2}$ using Wirtinger's inequality.
On the other hand, it is possible to bound it (by a worse constant) using FTC. Writing
$$ u(x)^2=left( int_0^x u'(t) mathrm{d} t right)^2 leq left(int_0^x u'(t)^2 mathrm{d} tright) left( int_0^x 1^2 mathrm{d} tright) leq int_0^1 u'(t)^2 mathrm{d}t,$$
you get $C=1$, after integrating from $x=0$ to $1$.
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
$endgroup$
add a comment |
$begingroup$
As Zeekless said, you can get $C=frac{1}{pi^2}$ using Wirtinger's inequality.
On the other hand, it is possible to bound it (by a worse constant) using FTC. Writing
$$ u(x)^2=left( int_0^x u'(t) mathrm{d} t right)^2 leq left(int_0^x u'(t)^2 mathrm{d} tright) left( int_0^x 1^2 mathrm{d} tright) leq int_0^1 u'(t)^2 mathrm{d}t,$$
you get $C=1$, after integrating from $x=0$ to $1$.
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
$endgroup$
As Zeekless said, you can get $C=frac{1}{pi^2}$ using Wirtinger's inequality.
On the other hand, it is possible to bound it (by a worse constant) using FTC. Writing
$$ u(x)^2=left( int_0^x u'(t) mathrm{d} t right)^2 leq left(int_0^x u'(t)^2 mathrm{d} tright) left( int_0^x 1^2 mathrm{d} tright) leq int_0^1 u'(t)^2 mathrm{d}t,$$
you get $C=1$, after integrating from $x=0$ to $1$.
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
answered Jan 20 at 17:20
Nicolás VilchesNicolás Vilches
55638
55638
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%2f3080822%2ffor-u-in-c0-1-with-u0-u1-0-show-int-01-u2x-dx-leq-c-int-01-u%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
2
$begingroup$
It is called Wirtinger's inequality. See "second version" here: en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions
$endgroup$
– Zeekless
Jan 20 at 17:05
1
$begingroup$
Hint: Use fourier expansion and apply parseval's theorem
$endgroup$
– TheOscillator
Jan 20 at 17:19