Proving that $f(x)=0$ in all points of continuity if $f$ is orthogonal to all polynomials
$begingroup$
Suppose that the function $f$ is:
1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;
2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).
Prove that $f(x) = 0$ in all points of continuity $f$.
real-analysis functional-analysis riemann-integration
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
$endgroup$
add a comment |
$begingroup$
Suppose that the function $f$ is:
1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;
2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).
Prove that $f(x) = 0$ in all points of continuity $f$.
real-analysis functional-analysis riemann-integration
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
$endgroup$
$begingroup$
I'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54
$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15
$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16
$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19
$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22
add a comment |
$begingroup$
Suppose that the function $f$ is:
1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;
2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).
Prove that $f(x) = 0$ in all points of continuity $f$.
real-analysis functional-analysis riemann-integration
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
$endgroup$
Suppose that the function $f$ is:
1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;
2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).
Prove that $f(x) = 0$ in all points of continuity $f$.
real-analysis functional-analysis riemann-integration
real-analysis functional-analysis riemann-integration
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
asked Nov 25 '18 at 22:51
ModeGenModeGen
333
333
$begingroup$
I'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54
$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15
$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16
$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19
$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22
add a comment |
$begingroup$
I'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54
$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15
$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16
$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19
$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22
$begingroup$
I'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54
$begingroup$
I'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54
$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15
$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15
$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16
$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16
$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19
$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19
$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22
$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$
Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$
Define $p_n(x) = sqrt n(1-x^2)^n.$ Then
$$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$
The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$
On the other hand, for large $n$ we have
$$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$
This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
$endgroup$
add a comment |
$begingroup$
Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.
Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
$$
|f(x_0)|-|f(x)|<|f(x_0)|/2,
$$
so $|f(x)|>|f(x_0)|/2$. Then
$$
int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
$$
a contradiction. So $f(x_0)=0$.
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
$begingroup$
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
$endgroup$
– ModeGen
Nov 26 '18 at 7:47
$begingroup$
Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:23
$begingroup$
I know the basics of the measure theory.
$endgroup$
– ModeGen
Nov 26 '18 at 14:39
$begingroup$
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
$begingroup$
If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
$$
int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
$$
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
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%2f3013529%2fproving-that-fx-0-in-all-points-of-continuity-if-f-is-orthogonal-to-all-po%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$
Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$
Define $p_n(x) = sqrt n(1-x^2)^n.$ Then
$$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$
The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$
On the other hand, for large $n$ we have
$$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$
This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
$endgroup$
add a comment |
$begingroup$
To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$
Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$
Define $p_n(x) = sqrt n(1-x^2)^n.$ Then
$$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$
The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$
On the other hand, for large $n$ we have
$$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$
This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
$endgroup$
add a comment |
$begingroup$
To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$
Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$
Define $p_n(x) = sqrt n(1-x^2)^n.$ Then
$$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$
The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$
On the other hand, for large $n$ we have
$$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$
This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
$endgroup$
To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$
Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$
Define $p_n(x) = sqrt n(1-x^2)^n.$ Then
$$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$
The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$
On the other hand, for large $n$ we have
$$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$
This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
answered Nov 26 '18 at 18:48
zhw.zhw.
71.9k43075
71.9k43075
add a comment |
add a comment |
$begingroup$
Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.
Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
$$
|f(x_0)|-|f(x)|<|f(x_0)|/2,
$$
so $|f(x)|>|f(x_0)|/2$. Then
$$
int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
$$
a contradiction. So $f(x_0)=0$.
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
$begingroup$
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
$endgroup$
– ModeGen
Nov 26 '18 at 7:47
$begingroup$
Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:23
$begingroup$
I know the basics of the measure theory.
$endgroup$
– ModeGen
Nov 26 '18 at 14:39
$begingroup$
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
$begingroup$
Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.
Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
$$
|f(x_0)|-|f(x)|<|f(x_0)|/2,
$$
so $|f(x)|>|f(x_0)|/2$. Then
$$
int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
$$
a contradiction. So $f(x_0)=0$.
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
$begingroup$
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
$endgroup$
– ModeGen
Nov 26 '18 at 7:47
$begingroup$
Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:23
$begingroup$
I know the basics of the measure theory.
$endgroup$
– ModeGen
Nov 26 '18 at 14:39
$begingroup$
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
$begingroup$
Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.
Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
$$
|f(x_0)|-|f(x)|<|f(x_0)|/2,
$$
so $|f(x)|>|f(x_0)|/2$. Then
$$
int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
$$
a contradiction. So $f(x_0)=0$.
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.
Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
$$
|f(x_0)|-|f(x)|<|f(x_0)|/2,
$$
so $|f(x)|>|f(x_0)|/2$. Then
$$
int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
$$
a contradiction. So $f(x_0)=0$.
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
edited Nov 26 '18 at 22:29
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
answered Nov 26 '18 at 2:42
Martin ArgeramiMartin Argerami
125k1178178
125k1178178
$begingroup$
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
$endgroup$
– ModeGen
Nov 26 '18 at 7:47
$begingroup$
Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:23
$begingroup$
I know the basics of the measure theory.
$endgroup$
– ModeGen
Nov 26 '18 at 14:39
$begingroup$
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
$begingroup$
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
$endgroup$
– ModeGen
Nov 26 '18 at 7:47
$begingroup$
Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:23
$begingroup$
I know the basics of the measure theory.
$endgroup$
– ModeGen
Nov 26 '18 at 14:39
$begingroup$
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:59
$begingroup$
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
$endgroup$
– ModeGen
Nov 26 '18 at 7:47
$begingroup$
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
$endgroup$
– ModeGen
Nov 26 '18 at 7:47
$begingroup$
Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:23
$begingroup$
Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:23
$begingroup$
I know the basics of the measure theory.
$endgroup$
– ModeGen
Nov 26 '18 at 14:39
$begingroup$
I know the basics of the measure theory.
$endgroup$
– ModeGen
Nov 26 '18 at 14:39
$begingroup$
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:59
$begingroup$
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
$endgroup$
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
$begingroup$
If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
$$
int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
$$
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
add a comment |
$begingroup$
If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
$$
int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
$$
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
add a comment |
$begingroup$
If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
$$
int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
$$
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
$$
int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
$$
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Nov 26 '18 at 15:16
DisintegratingByPartsDisintegratingByParts
58.9k42580
58.9k42580
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%2f3013529%2fproving-that-fx-0-in-all-points-of-continuity-if-f-is-orthogonal-to-all-po%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'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54
$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15
$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16
$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19
$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22