Proving that $f(x)=0$ in all points of continuity if $f$ is orthogonal to all polynomials
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
ModeGen
333
add a comment |
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
ModeGen
333
I'm not sure if it would work, but have you tried using repeated integration by parts?
– Seth
Nov 25 '18 at 22:54
For integration by parts $f$ have to be differentiable but it is not.
– ModeGen
Nov 25 '18 at 23:15
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
– Seth
Nov 25 '18 at 23:16
It will not lead to anything.
– ModeGen
Nov 25 '18 at 23:19
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.
– ModeGen
Nov 25 '18 at 23:22
add a comment |
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
ModeGen
333
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
ModeGen
333
asked Nov 25 '18 at 22:51
ModeGen
333
asked Nov 25 '18 at 22:51
ModeGen
333
asked Nov 25 '18 at 22:51
ModeGen
333
asked Nov 25 '18 at 22:51
ModeGen
333
333
I'm not sure if it would work, but have you tried using repeated integration by parts?
– Seth
Nov 25 '18 at 22:54
For integration by parts $f$ have to be differentiable but it is not.
– ModeGen
Nov 25 '18 at 23:15
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
– Seth
Nov 25 '18 at 23:16
It will not lead to anything.
– ModeGen
Nov 25 '18 at 23:19
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.
– ModeGen
Nov 25 '18 at 23:22
add a comment |
I'm not sure if it would work, but have you tried using repeated integration by parts?
– Seth
Nov 25 '18 at 22:54
For integration by parts $f$ have to be differentiable but it is not.
– ModeGen
Nov 25 '18 at 23:15
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
– Seth
Nov 25 '18 at 23:16
It will not lead to anything.
– ModeGen
Nov 25 '18 at 23:19
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.
– ModeGen
Nov 25 '18 at 23:22
I'm not sure if it would work, but have you tried using repeated integration by parts?
– Seth
Nov 25 '18 at 22:54
I'm not sure if it would work, but have you tried using repeated integration by parts?
– Seth
Nov 25 '18 at 22:54
For integration by parts $f$ have to be differentiable but it is not.
– ModeGen
Nov 25 '18 at 23:15
For integration by parts $f$ have to be differentiable but it is not.
– ModeGen
Nov 25 '18 at 23:15
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
– Seth
Nov 25 '18 at 23:16
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
– Seth
Nov 25 '18 at 23:16
It will not lead to anything.
– ModeGen
Nov 25 '18 at 23:19
It will not lead to anything.
– ModeGen
Nov 25 '18 at 23:19
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.
– ModeGen
Nov 25 '18 at 23:22
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.
– ModeGen
Nov 25 '18 at 23:22
add a comment |
3 Answers
3
active
oldest
votes
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.
71.6k43075
add a comment |
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 Argerami
124k1176174
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
– ModeGen
Nov 26 '18 at 7:47
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?
– Martin Argerami
Nov 26 '18 at 14:23
I know the basics of the measure theory.
– ModeGen
Nov 26 '18 at 14:39
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
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
DisintegratingByParts
58.6k42579
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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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.
71.6k43075
add a comment |
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.
71.6k43075
add a comment |
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.
71.6k43075
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.
71.6k43075
answered Nov 26 '18 at 18:48
zhw.
71.6k43075
answered Nov 26 '18 at 18:48
zhw.
71.6k43075
answered Nov 26 '18 at 18:48
zhw.
71.6k43075
71.6k43075
add a comment |
add a comment |
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 Argerami
124k1176174
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
– ModeGen
Nov 26 '18 at 7:47
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?
– Martin Argerami
Nov 26 '18 at 14:23
I know the basics of the measure theory.
– ModeGen
Nov 26 '18 at 14:39
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
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 Argerami
124k1176174
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
– ModeGen
Nov 26 '18 at 7:47
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?
– Martin Argerami
Nov 26 '18 at 14:23
I know the basics of the measure theory.
– ModeGen
Nov 26 '18 at 14:39
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
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 Argerami
124k1176174
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 Argerami
124k1176174
edited Nov 26 '18 at 22:29
answered Nov 26 '18 at 2:42
Martin Argerami
124k1176174
answered Nov 26 '18 at 2:42
Martin Argerami
124k1176174
answered Nov 26 '18 at 2:42
Martin Argerami
124k1176174
124k1176174
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
– ModeGen
Nov 26 '18 at 7:47
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?
– Martin Argerami
Nov 26 '18 at 14:23
I know the basics of the measure theory.
– ModeGen
Nov 26 '18 at 14:39
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
– ModeGen
Nov 26 '18 at 7:47
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?
– Martin Argerami
Nov 26 '18 at 14:23
I know the basics of the measure theory.
– ModeGen
Nov 26 '18 at 14:39
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
– Martin Argerami
Nov 26 '18 at 14:59
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
– ModeGen
Nov 26 '18 at 7:47
We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
– ModeGen
Nov 26 '18 at 7:47
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?
– Martin Argerami
Nov 26 '18 at 14:23
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?
– Martin Argerami
Nov 26 '18 at 14:23
I know the basics of the measure theory.
– ModeGen
Nov 26 '18 at 14:39
I know the basics of the measure theory.
– ModeGen
Nov 26 '18 at 14:39
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
– Martin Argerami
Nov 26 '18 at 14:59
In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
– Martin Argerami
Nov 26 '18 at 14:59
add a comment |
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
DisintegratingByParts
58.6k42579
add a comment |
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
DisintegratingByParts
58.6k42579
add a comment |
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
DisintegratingByParts
58.6k42579
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
DisintegratingByParts
58.6k42579
answered Nov 26 '18 at 15:16
DisintegratingByParts
58.6k42579
answered Nov 26 '18 at 15:16
DisintegratingByParts
58.6k42579
answered Nov 26 '18 at 15:16
DisintegratingByParts
58.6k42579
58.6k42579
add a comment |
add a comment |
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I'm not sure if it would work, but have you tried using repeated integration by parts?
– Seth
Nov 25 '18 at 22:54
For integration by parts $f$ have to be differentiable but it is not.
– ModeGen
Nov 25 '18 at 23:15
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
– Seth
Nov 25 '18 at 23:16
It will not lead to anything.
– ModeGen
Nov 25 '18 at 23:19
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.
– ModeGen
Nov 25 '18 at 23:22