Prove $int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$
Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$
Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.
calculus
asked Nov 24 at 4:37
Jaqen Chou
406110
add a comment |
Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$
Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.
calculus
asked Nov 24 at 4:37
Jaqen Chou
406110
3
This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39
@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45
@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03
add a comment |
Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$
Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.
calculus
asked Nov 24 at 4:37
Jaqen Chou
406110
Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$
Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.
calculus
calculus
asked Nov 24 at 4:37
Jaqen Chou
406110
asked Nov 24 at 4:37
Jaqen Chou
406110
asked Nov 24 at 4:37
Jaqen Chou
406110
asked Nov 24 at 4:37
Jaqen Chou
406110
asked Nov 24 at 4:37
Jaqen Chou
406110
406110
3
This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39
@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45
@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03
add a comment |
3
This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39
@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45
@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03
3
3
This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39
This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39
@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45
@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45
@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03
@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03
add a comment |
1 Answer
1
active
oldest
votes
Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.
answered Nov 24 at 5:08
Seewoo Lee
6,165826
[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24
why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10
@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29
Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30
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%2f3011187%2fprove-int-01fx2dx-le-frac14-int-01fx2dx%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
Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.
answered Nov 24 at 5:08
Seewoo Lee
6,165826
[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24
why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10
@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29
Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30
add a comment |
Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.
answered Nov 24 at 5:08
Seewoo Lee
6,165826
[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24
why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10
@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29
Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30
add a comment |
Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.
answered Nov 24 at 5:08
Seewoo Lee
6,165826
Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.
answered Nov 24 at 5:08
Seewoo Lee
6,165826
edited Nov 24 at 20:30
answered Nov 24 at 5:08
Seewoo Lee
6,165826
answered Nov 24 at 5:08
Seewoo Lee
6,165826
answered Nov 24 at 5:08
Seewoo Lee
6,165826
6,165826
[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24
why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10
@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29
Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30
add a comment |
[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24
why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10
@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29
Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30
[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24
[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24
why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10
why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10
@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29
@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29
Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30
Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30
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%2f3011187%2fprove-int-01fx2dx-le-frac14-int-01fx2dx%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
3
This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39
@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45
@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03