Mixing
Positivity of $int_0^1 sin^2(kx)f(x) dx$
Clash Royale CLAN TAG#URR8PPP
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I'm interested in proving restrictions on $f$ necessary for the integral
beginequation*
int_0^1 sin^2 (kx) f(x) dx
endequation*
to be positive for all $k > 0$. Obviously a nonnegative $f$ would suffice (provided it was positive on a set of nonzero measure). I'm wondering if there are less restrictive conditions that may be placed on $f$. Thanks in advance.
integration analysis definite-integrals
edited Aug 6 at 22:54
Batominovski
23.4k22779
asked Aug 6 at 20:00
John
1588
add a comment |Â
up vote
5
down vote
favorite
I'm interested in proving restrictions on $f$ necessary for the integral
beginequation*
int_0^1 sin^2 (kx) f(x) dx
endequation*
to be positive for all $k > 0$. Obviously a nonnegative $f$ would suffice (provided it was positive on a set of nonzero measure). I'm wondering if there are less restrictive conditions that may be placed on $f$. Thanks in advance.
integration analysis definite-integrals
edited Aug 6 at 22:54
Batominovski
23.4k22779
asked Aug 6 at 20:00
John
1588
What kind of conditions are you looking for?
– Pjonin
Aug 6 at 20:09
The limits of this integral are from zero to one? Not zero to, I don't know, $pi$ or something?
– Hamed
Aug 6 at 20:23
2
If for any $zin(0,1)$ you are able to approximate $delta(z)$ with a linear combination (with non-negative coefficients) of functions of the form $sin^2(kz)=frac12left(1-cos(2kz)right)$, then $fgeq 0$ is necessary.
– Jack D'Aurizio♦
Aug 6 at 20:25
What regularity assumptions do we have on $f$? $fin C^0(0,1)$ or $fin L^2(0,1)$, by chance?
– Jack D'Aurizio♦
Aug 6 at 20:26
Actually my initial comment was completely misleading, but I am going to leave it for documentation purposes. I suggest to add the harmonic-analysis tag to the main question, since it seems pretty clear that Fourier series are the way to go, here.
– Jack D'Aurizio♦
Aug 6 at 20:57
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm interested in proving restrictions on $f$ necessary for the integral
beginequation*
int_0^1 sin^2 (kx) f(x) dx
endequation*
to be positive for all $k > 0$. Obviously a nonnegative $f$ would suffice (provided it was positive on a set of nonzero measure). I'm wondering if there are less restrictive conditions that may be placed on $f$. Thanks in advance.
integration analysis definite-integrals
edited Aug 6 at 22:54
Batominovski
23.4k22779
asked Aug 6 at 20:00
John
1588
I'm interested in proving restrictions on $f$ necessary for the integral
beginequation*
int_0^1 sin^2 (kx) f(x) dx
endequation*
to be positive for all $k > 0$. Obviously a nonnegative $f$ would suffice (provided it was positive on a set of nonzero measure). I'm wondering if there are less restrictive conditions that may be placed on $f$. Thanks in advance.
integration analysis definite-integrals
edited Aug 6 at 22:54
Batominovski
23.4k22779
asked Aug 6 at 20:00
John
1588
edited Aug 6 at 22:54
Batominovski
23.4k22779
edited Aug 6 at 22:54
Batominovski
23.4k22779
edited Aug 6 at 22:54
Batominovski
23.4k22779
23.4k22779
asked Aug 6 at 20:00
John
1588
asked Aug 6 at 20:00
John
1588
asked Aug 6 at 20:00
John
1588
1588
What kind of conditions are you looking for?
– Pjonin
Aug 6 at 20:09
The limits of this integral are from zero to one? Not zero to, I don't know, $pi$ or something?
– Hamed
Aug 6 at 20:23
2
If for any $zin(0,1)$ you are able to approximate $delta(z)$ with a linear combination (with non-negative coefficients) of functions of the form $sin^2(kz)=frac12left(1-cos(2kz)right)$, then $fgeq 0$ is necessary.
– Jack D'Aurizio♦
Aug 6 at 20:25
What regularity assumptions do we have on $f$? $fin C^0(0,1)$ or $fin L^2(0,1)$, by chance?
– Jack D'Aurizio♦
Aug 6 at 20:26
Actually my initial comment was completely misleading, but I am going to leave it for documentation purposes. I suggest to add the harmonic-analysis tag to the main question, since it seems pretty clear that Fourier series are the way to go, here.
– Jack D'Aurizio♦
Aug 6 at 20:57
add a comment |Â
What kind of conditions are you looking for?
– Pjonin
Aug 6 at 20:09
The limits of this integral are from zero to one? Not zero to, I don't know, $pi$ or something?
– Hamed
Aug 6 at 20:23
2
If for any $zin(0,1)$ you are able to approximate $delta(z)$ with a linear combination (with non-negative coefficients) of functions of the form $sin^2(kz)=frac12left(1-cos(2kz)right)$, then $fgeq 0$ is necessary.
– Jack D'Aurizio♦
Aug 6 at 20:25
What regularity assumptions do we have on $f$? $fin C^0(0,1)$ or $fin L^2(0,1)$, by chance?
– Jack D'Aurizio♦
Aug 6 at 20:26
Actually my initial comment was completely misleading, but I am going to leave it for documentation purposes. I suggest to add the harmonic-analysis tag to the main question, since it seems pretty clear that Fourier series are the way to go, here.
– Jack D'Aurizio♦
Aug 6 at 20:57
What kind of conditions are you looking for?
– Pjonin
Aug 6 at 20:09
What kind of conditions are you looking for?
– Pjonin
Aug 6 at 20:09
The limits of this integral are from zero to one? Not zero to, I don't know, $pi$ or something?
– Hamed
Aug 6 at 20:23
The limits of this integral are from zero to one? Not zero to, I don't know, $pi$ or something?
– Hamed
Aug 6 at 20:23
2
2
If for any $zin(0,1)$ you are able to approximate $delta(z)$ with a linear combination (with non-negative coefficients) of functions of the form $sin^2(kz)=frac12left(1-cos(2kz)right)$, then $fgeq 0$ is necessary.
– Jack D'Aurizio♦
Aug 6 at 20:25
If for any $zin(0,1)$ you are able to approximate $delta(z)$ with a linear combination (with non-negative coefficients) of functions of the form $sin^2(kz)=frac12left(1-cos(2kz)right)$, then $fgeq 0$ is necessary.
– Jack D'Aurizio♦
Aug 6 at 20:25
What regularity assumptions do we have on $f$? $fin C^0(0,1)$ or $fin L^2(0,1)$, by chance?
– Jack D'Aurizio♦
Aug 6 at 20:26
What regularity assumptions do we have on $f$? $fin C^0(0,1)$ or $fin L^2(0,1)$, by chance?
– Jack D'Aurizio♦
Aug 6 at 20:26
Actually my initial comment was completely misleading, but I am going to leave it for documentation purposes. I suggest to add the harmonic-analysis tag to the main question, since it seems pretty clear that Fourier series are the way to go, here.
– Jack D'Aurizio♦
Aug 6 at 20:57
Actually my initial comment was completely misleading, but I am going to leave it for documentation purposes. I suggest to add the harmonic-analysis tag to the main question, since it seems pretty clear that Fourier series are the way to go, here.
– Jack D'Aurizio♦
Aug 6 at 20:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
A partial answer: the question is less trivial than I initially suspected.
It is practical to consider that $sin^2(kx)=frac12left(1-cos(2kx)right)$.
Assuming $fin L^2(0,1)$ we are allowed to write
$$ f(x) stackrelL^2(0,1)= c_0 + sum_ngeq 1 c_n cos(2pi n x)+sum_ngeq 1 s_n sin(2pi n x) $$
where $c_0$ has to be non-negative as a consequence of the Riemann-Lebesgue lemma.
By considering $k=pi n$ with $ninmathbbN^+$ we get that each $c_n$ has to be $leq 2c_0$.
This leaves enough room for plenty of somewhere-negative functions fulfilling the constraints for each $kin pimathbbN$, like
$$ pi^2 x(1-x)+1-fracpi^26 = 1-sum_ngeq 1fraccos(2pi n x)n^2 $$
which can be shown to fulfill the constraints for any $k>0$, ouch!
So my actual conjecture is that the claim holds iff the coefficients of the Fourier cosine series of $f(x)+f(-x)$ over $(0,1)$ fulfill some set of inequalities to be discovered. $fgeq 0$ is definitely not necessary.
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
Something simple like Cauchy-Schwarz can give a sufficient condition on $(sum_n=1^infty c_n^2)^1/2$ (seems to be something like $leq 1.98 c_0$).
– Winther
Aug 7 at 0:13
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
A partial answer: the question is less trivial than I initially suspected.
It is practical to consider that $sin^2(kx)=frac12left(1-cos(2kx)right)$.
Assuming $fin L^2(0,1)$ we are allowed to write
$$ f(x) stackrelL^2(0,1)= c_0 + sum_ngeq 1 c_n cos(2pi n x)+sum_ngeq 1 s_n sin(2pi n x) $$
where $c_0$ has to be non-negative as a consequence of the Riemann-Lebesgue lemma.
By considering $k=pi n$ with $ninmathbbN^+$ we get that each $c_n$ has to be $leq 2c_0$.
This leaves enough room for plenty of somewhere-negative functions fulfilling the constraints for each $kin pimathbbN$, like
$$ pi^2 x(1-x)+1-fracpi^26 = 1-sum_ngeq 1fraccos(2pi n x)n^2 $$
which can be shown to fulfill the constraints for any $k>0$, ouch!
So my actual conjecture is that the claim holds iff the coefficients of the Fourier cosine series of $f(x)+f(-x)$ over $(0,1)$ fulfill some set of inequalities to be discovered. $fgeq 0$ is definitely not necessary.
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
Something simple like Cauchy-Schwarz can give a sufficient condition on $(sum_n=1^infty c_n^2)^1/2$ (seems to be something like $leq 1.98 c_0$).
– Winther
Aug 7 at 0:13
add a comment |Â
up vote
3
down vote
A partial answer: the question is less trivial than I initially suspected.
It is practical to consider that $sin^2(kx)=frac12left(1-cos(2kx)right)$.
Assuming $fin L^2(0,1)$ we are allowed to write
$$ f(x) stackrelL^2(0,1)= c_0 + sum_ngeq 1 c_n cos(2pi n x)+sum_ngeq 1 s_n sin(2pi n x) $$
where $c_0$ has to be non-negative as a consequence of the Riemann-Lebesgue lemma.
By considering $k=pi n$ with $ninmathbbN^+$ we get that each $c_n$ has to be $leq 2c_0$.
This leaves enough room for plenty of somewhere-negative functions fulfilling the constraints for each $kin pimathbbN$, like
$$ pi^2 x(1-x)+1-fracpi^26 = 1-sum_ngeq 1fraccos(2pi n x)n^2 $$
which can be shown to fulfill the constraints for any $k>0$, ouch!
So my actual conjecture is that the claim holds iff the coefficients of the Fourier cosine series of $f(x)+f(-x)$ over $(0,1)$ fulfill some set of inequalities to be discovered. $fgeq 0$ is definitely not necessary.
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
Something simple like Cauchy-Schwarz can give a sufficient condition on $(sum_n=1^infty c_n^2)^1/2$ (seems to be something like $leq 1.98 c_0$).
– Winther
Aug 7 at 0:13
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A partial answer: the question is less trivial than I initially suspected.
It is practical to consider that $sin^2(kx)=frac12left(1-cos(2kx)right)$.
Assuming $fin L^2(0,1)$ we are allowed to write
$$ f(x) stackrelL^2(0,1)= c_0 + sum_ngeq 1 c_n cos(2pi n x)+sum_ngeq 1 s_n sin(2pi n x) $$
where $c_0$ has to be non-negative as a consequence of the Riemann-Lebesgue lemma.
By considering $k=pi n$ with $ninmathbbN^+$ we get that each $c_n$ has to be $leq 2c_0$.
This leaves enough room for plenty of somewhere-negative functions fulfilling the constraints for each $kin pimathbbN$, like
$$ pi^2 x(1-x)+1-fracpi^26 = 1-sum_ngeq 1fraccos(2pi n x)n^2 $$
which can be shown to fulfill the constraints for any $k>0$, ouch!
So my actual conjecture is that the claim holds iff the coefficients of the Fourier cosine series of $f(x)+f(-x)$ over $(0,1)$ fulfill some set of inequalities to be discovered. $fgeq 0$ is definitely not necessary.
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
A partial answer: the question is less trivial than I initially suspected.
It is practical to consider that $sin^2(kx)=frac12left(1-cos(2kx)right)$.
Assuming $fin L^2(0,1)$ we are allowed to write
$$ f(x) stackrelL^2(0,1)= c_0 + sum_ngeq 1 c_n cos(2pi n x)+sum_ngeq 1 s_n sin(2pi n x) $$
where $c_0$ has to be non-negative as a consequence of the Riemann-Lebesgue lemma.
By considering $k=pi n$ with $ninmathbbN^+$ we get that each $c_n$ has to be $leq 2c_0$.
This leaves enough room for plenty of somewhere-negative functions fulfilling the constraints for each $kin pimathbbN$, like
$$ pi^2 x(1-x)+1-fracpi^26 = 1-sum_ngeq 1fraccos(2pi n x)n^2 $$
which can be shown to fulfill the constraints for any $k>0$, ouch!
So my actual conjecture is that the claim holds iff the coefficients of the Fourier cosine series of $f(x)+f(-x)$ over $(0,1)$ fulfill some set of inequalities to be discovered. $fgeq 0$ is definitely not necessary.
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
edited Aug 6 at 21:06
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
answered Aug 6 at 20:57
Jack D'Aurizio♦
270k31266630
270k31266630
Something simple like Cauchy-Schwarz can give a sufficient condition on $(sum_n=1^infty c_n^2)^1/2$ (seems to be something like $leq 1.98 c_0$).
– Winther
Aug 7 at 0:13
add a comment |Â
Something simple like Cauchy-Schwarz can give a sufficient condition on $(sum_n=1^infty c_n^2)^1/2$ (seems to be something like $leq 1.98 c_0$).
– Winther
Aug 7 at 0:13
Something simple like Cauchy-Schwarz can give a sufficient condition on $(sum_n=1^infty c_n^2)^1/2$ (seems to be something like $leq 1.98 c_0$).
– Winther
Aug 7 at 0:13
Something simple like Cauchy-Schwarz can give a sufficient condition on $(sum_n=1^infty c_n^2)^1/2$ (seems to be something like $leq 1.98 c_0$).
– Winther
Aug 7 at 0:13
add a comment |Â
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What kind of conditions are you looking for?
– Pjonin
Aug 6 at 20:09
The limits of this integral are from zero to one? Not zero to, I don't know, $pi$ or something?
– Hamed
Aug 6 at 20:23
2
If for any $zin(0,1)$ you are able to approximate $delta(z)$ with a linear combination (with non-negative coefficients) of functions of the form $sin^2(kz)=frac12left(1-cos(2kz)right)$, then $fgeq 0$ is necessary.
– Jack D'Aurizio♦
Aug 6 at 20:25
What regularity assumptions do we have on $f$? $fin C^0(0,1)$ or $fin L^2(0,1)$, by chance?
– Jack D'Aurizio♦
Aug 6 at 20:26
Actually my initial comment was completely misleading, but I am going to leave it for documentation purposes. I suggest to add the harmonic-analysis tag to the main question, since it seems pretty clear that Fourier series are the way to go, here.
– Jack D'Aurizio♦
Aug 6 at 20:57