Mixing
$F(nc)=0$ then $F=0$
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Let $fin L^1(0,infty)$ and $F(x):=int_0^infty xe^-xyint_0^yf(t)dtdy$ on$(0,infty)$
Suppose there exists $cin mathbbR$, for any $nin mathbbN$, $F(nc)=0$. Then $F=0$
I checked $F$ is well-defined on $(0,infty)$
I tried to use uniqueness of Fourier's transformation, but I couldn't. Someone know how to prove $F=0$?
real-analysis analysis lebesgue-integral
edited 2 days ago
user254433
2,072612
asked 2 days ago
Yan Wen Lie
696
add a comment |Â
up vote
1
down vote
favorite
Let $fin L^1(0,infty)$ and $F(x):=int_0^infty xe^-xyint_0^yf(t)dtdy$ on$(0,infty)$
Suppose there exists $cin mathbbR$, for any $nin mathbbN$, $F(nc)=0$. Then $F=0$
I checked $F$ is well-defined on $(0,infty)$
I tried to use uniqueness of Fourier's transformation, but I couldn't. Someone know how to prove $F=0$?
real-analysis analysis lebesgue-integral
edited 2 days ago
user254433
2,072612
asked 2 days ago
Yan Wen Lie
696
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $fin L^1(0,infty)$ and $F(x):=int_0^infty xe^-xyint_0^yf(t)dtdy$ on$(0,infty)$
Suppose there exists $cin mathbbR$, for any $nin mathbbN$, $F(nc)=0$. Then $F=0$
I checked $F$ is well-defined on $(0,infty)$
I tried to use uniqueness of Fourier's transformation, but I couldn't. Someone know how to prove $F=0$?
real-analysis analysis lebesgue-integral
edited 2 days ago
user254433
2,072612
asked 2 days ago
Yan Wen Lie
696
Let $fin L^1(0,infty)$ and $F(x):=int_0^infty xe^-xyint_0^yf(t)dtdy$ on$(0,infty)$
Suppose there exists $cin mathbbR$, for any $nin mathbbN$, $F(nc)=0$. Then $F=0$
I checked $F$ is well-defined on $(0,infty)$
I tried to use uniqueness of Fourier's transformation, but I couldn't. Someone know how to prove $F=0$?
real-analysis analysis lebesgue-integral
edited 2 days ago
user254433
2,072612
asked 2 days ago
Yan Wen Lie
696
edited 2 days ago
user254433
2,072612
edited 2 days ago
user254433
2,072612
edited 2 days ago
user254433
2,072612
2,072612
asked 2 days ago
Yan Wen Lie
696
asked 2 days ago
Yan Wen Lie
696
asked 2 days ago
Yan Wen Lie
696
696
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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Observe that $xe^-xy=-fracpartialpartial ye^-xy$, so that we can apply integration by parts:
$$
F(x)=lim_Ytoinfty-e^-xYint_0^Yf(t)dt+int_0^infty e^-xyf(y)dy.
$$
The first term is zero since $|int_0^Yf(t)dt|le |f|_L^1$, so in fact
$$
F(x)=int_0^infty e^-xyf(y)dy
$$
is just the Laplace transform of $f$. Let us try to avoid using complex analysis to invert this transform.
The condition that $F(nc)=0$ says that if $E_n(x)=e^-ncx$, then $int_0^infty E_n(x)f(x)dx=0$ for all $n$, which means this problem is analogous to proving $phi(theta)=0$ if the Fourier coefficients $int_0^2pie^inthetaphi(theta)dtheta=0$ for all $n$. One way to do this is the Stone-Weierstrass theorem: since the algebra generated by $e^itheta$, which we write as $<e^itheta>$, is an algebra of continuous functions on $C(mathbb S^1)$ which separates points of $mathbb S^1$, it is dense in $C(mathbb S^1)$. This means $int_0^2pig(theta)phi(theta)dtheta=0$ for all $gin C$, which obviously implies $phi(theta)equiv 0$.
There is a version of the Stone-Weierstrass theorem for locally compact Hausdorff spaces, such as $[0,infty)$. It is clear that the algebra $<e^-cx>$ separates points of $[0,infty)$ and vanishes nowhere, so it is dense in $C_0([0,infty))$. As before, this means $int_0^infty g(x)f(x)dx=0$ for all $gin C_0$, which gives $fequiv 0$.
If I'm not mistaken, one might prove the locally compact version of Stone-Weierstrass by successively "multiplying the algebra" by a bump function $psi(x/k)$, applying compact SW, and letting $ktoinfty$.
answered 2 days ago
user254433
2,072612
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Observe that $xe^-xy=-fracpartialpartial ye^-xy$, so that we can apply integration by parts:
$$
F(x)=lim_Ytoinfty-e^-xYint_0^Yf(t)dt+int_0^infty e^-xyf(y)dy.
$$
The first term is zero since $|int_0^Yf(t)dt|le |f|_L^1$, so in fact
$$
F(x)=int_0^infty e^-xyf(y)dy
$$
is just the Laplace transform of $f$. Let us try to avoid using complex analysis to invert this transform.
The condition that $F(nc)=0$ says that if $E_n(x)=e^-ncx$, then $int_0^infty E_n(x)f(x)dx=0$ for all $n$, which means this problem is analogous to proving $phi(theta)=0$ if the Fourier coefficients $int_0^2pie^inthetaphi(theta)dtheta=0$ for all $n$. One way to do this is the Stone-Weierstrass theorem: since the algebra generated by $e^itheta$, which we write as $<e^itheta>$, is an algebra of continuous functions on $C(mathbb S^1)$ which separates points of $mathbb S^1$, it is dense in $C(mathbb S^1)$. This means $int_0^2pig(theta)phi(theta)dtheta=0$ for all $gin C$, which obviously implies $phi(theta)equiv 0$.
There is a version of the Stone-Weierstrass theorem for locally compact Hausdorff spaces, such as $[0,infty)$. It is clear that the algebra $<e^-cx>$ separates points of $[0,infty)$ and vanishes nowhere, so it is dense in $C_0([0,infty))$. As before, this means $int_0^infty g(x)f(x)dx=0$ for all $gin C_0$, which gives $fequiv 0$.
If I'm not mistaken, one might prove the locally compact version of Stone-Weierstrass by successively "multiplying the algebra" by a bump function $psi(x/k)$, applying compact SW, and letting $ktoinfty$.
answered 2 days ago
user254433
2,072612
add a comment |Â
up vote
1
down vote
accepted
Observe that $xe^-xy=-fracpartialpartial ye^-xy$, so that we can apply integration by parts:
$$
F(x)=lim_Ytoinfty-e^-xYint_0^Yf(t)dt+int_0^infty e^-xyf(y)dy.
$$
The first term is zero since $|int_0^Yf(t)dt|le |f|_L^1$, so in fact
$$
F(x)=int_0^infty e^-xyf(y)dy
$$
is just the Laplace transform of $f$. Let us try to avoid using complex analysis to invert this transform.
The condition that $F(nc)=0$ says that if $E_n(x)=e^-ncx$, then $int_0^infty E_n(x)f(x)dx=0$ for all $n$, which means this problem is analogous to proving $phi(theta)=0$ if the Fourier coefficients $int_0^2pie^inthetaphi(theta)dtheta=0$ for all $n$. One way to do this is the Stone-Weierstrass theorem: since the algebra generated by $e^itheta$, which we write as $<e^itheta>$, is an algebra of continuous functions on $C(mathbb S^1)$ which separates points of $mathbb S^1$, it is dense in $C(mathbb S^1)$. This means $int_0^2pig(theta)phi(theta)dtheta=0$ for all $gin C$, which obviously implies $phi(theta)equiv 0$.
There is a version of the Stone-Weierstrass theorem for locally compact Hausdorff spaces, such as $[0,infty)$. It is clear that the algebra $<e^-cx>$ separates points of $[0,infty)$ and vanishes nowhere, so it is dense in $C_0([0,infty))$. As before, this means $int_0^infty g(x)f(x)dx=0$ for all $gin C_0$, which gives $fequiv 0$.
If I'm not mistaken, one might prove the locally compact version of Stone-Weierstrass by successively "multiplying the algebra" by a bump function $psi(x/k)$, applying compact SW, and letting $ktoinfty$.
answered 2 days ago
user254433
2,072612
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Observe that $xe^-xy=-fracpartialpartial ye^-xy$, so that we can apply integration by parts:
$$
F(x)=lim_Ytoinfty-e^-xYint_0^Yf(t)dt+int_0^infty e^-xyf(y)dy.
$$
The first term is zero since $|int_0^Yf(t)dt|le |f|_L^1$, so in fact
$$
F(x)=int_0^infty e^-xyf(y)dy
$$
is just the Laplace transform of $f$. Let us try to avoid using complex analysis to invert this transform.
The condition that $F(nc)=0$ says that if $E_n(x)=e^-ncx$, then $int_0^infty E_n(x)f(x)dx=0$ for all $n$, which means this problem is analogous to proving $phi(theta)=0$ if the Fourier coefficients $int_0^2pie^inthetaphi(theta)dtheta=0$ for all $n$. One way to do this is the Stone-Weierstrass theorem: since the algebra generated by $e^itheta$, which we write as $<e^itheta>$, is an algebra of continuous functions on $C(mathbb S^1)$ which separates points of $mathbb S^1$, it is dense in $C(mathbb S^1)$. This means $int_0^2pig(theta)phi(theta)dtheta=0$ for all $gin C$, which obviously implies $phi(theta)equiv 0$.
There is a version of the Stone-Weierstrass theorem for locally compact Hausdorff spaces, such as $[0,infty)$. It is clear that the algebra $<e^-cx>$ separates points of $[0,infty)$ and vanishes nowhere, so it is dense in $C_0([0,infty))$. As before, this means $int_0^infty g(x)f(x)dx=0$ for all $gin C_0$, which gives $fequiv 0$.
If I'm not mistaken, one might prove the locally compact version of Stone-Weierstrass by successively "multiplying the algebra" by a bump function $psi(x/k)$, applying compact SW, and letting $ktoinfty$.
answered 2 days ago
user254433
2,072612
Observe that $xe^-xy=-fracpartialpartial ye^-xy$, so that we can apply integration by parts:
$$
F(x)=lim_Ytoinfty-e^-xYint_0^Yf(t)dt+int_0^infty e^-xyf(y)dy.
$$
The first term is zero since $|int_0^Yf(t)dt|le |f|_L^1$, so in fact
$$
F(x)=int_0^infty e^-xyf(y)dy
$$
is just the Laplace transform of $f$. Let us try to avoid using complex analysis to invert this transform.
The condition that $F(nc)=0$ says that if $E_n(x)=e^-ncx$, then $int_0^infty E_n(x)f(x)dx=0$ for all $n$, which means this problem is analogous to proving $phi(theta)=0$ if the Fourier coefficients $int_0^2pie^inthetaphi(theta)dtheta=0$ for all $n$. One way to do this is the Stone-Weierstrass theorem: since the algebra generated by $e^itheta$, which we write as $<e^itheta>$, is an algebra of continuous functions on $C(mathbb S^1)$ which separates points of $mathbb S^1$, it is dense in $C(mathbb S^1)$. This means $int_0^2pig(theta)phi(theta)dtheta=0$ for all $gin C$, which obviously implies $phi(theta)equiv 0$.
There is a version of the Stone-Weierstrass theorem for locally compact Hausdorff spaces, such as $[0,infty)$. It is clear that the algebra $<e^-cx>$ separates points of $[0,infty)$ and vanishes nowhere, so it is dense in $C_0([0,infty))$. As before, this means $int_0^infty g(x)f(x)dx=0$ for all $gin C_0$, which gives $fequiv 0$.
If I'm not mistaken, one might prove the locally compact version of Stone-Weierstrass by successively "multiplying the algebra" by a bump function $psi(x/k)$, applying compact SW, and letting $ktoinfty$.
answered 2 days ago
user254433
2,072612
edited 2 days ago
answered 2 days ago
user254433
2,072612
answered 2 days ago
user254433
2,072612
answered 2 days ago
user254433
2,072612
2,072612
add a comment |Â
add a comment |Â
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