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Examples of complete families of functions forming an absolutely convergent series
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I am searching for some examples of complete families of functions $left phi_m(t) right_m = 1^infty$ on $t in [0, T]$ that form an absolutely convergent series:
$$
sum_m = 1^infty |phi_m(t)| < infty, ~~~~ t in [0, T].
$$
According to the Weierstrass M-test, $phi_m$ must satisfy
$$
|phi_m(t)| leq c_m, ~~ t in [0, T],
$$
such that
$$
sum_m = 1^infty c_m < infty.
$$
Any idea or references would be highly appreciated.
sequences-and-series convergence orthogonality absolute-convergence
asked Aug 6 at 12:57
Asatur Khurshudyan
395111
add a comment |Â
up vote
1
down vote
favorite
I am searching for some examples of complete families of functions $left phi_m(t) right_m = 1^infty$ on $t in [0, T]$ that form an absolutely convergent series:
$$
sum_m = 1^infty |phi_m(t)| < infty, ~~~~ t in [0, T].
$$
According to the Weierstrass M-test, $phi_m$ must satisfy
$$
|phi_m(t)| leq c_m, ~~ t in [0, T],
$$
such that
$$
sum_m = 1^infty c_m < infty.
$$
Any idea or references would be highly appreciated.
sequences-and-series convergence orthogonality absolute-convergence
asked Aug 6 at 12:57
Asatur Khurshudyan
395111
Do you work in the Hilbert space $L^2([0,T])$? What is the precise definition of a complete family?
– Paul Frost
Aug 6 at 13:44
Yes, let the space be $L^2$. By complete family I mean encyclopediaofmath.org/index.php/Complete_system_of_functions
– Asatur Khurshudyan
Aug 6 at 14:07
1
So you work in $L^2([0,T])$, but the $phi_m$ must be $L^infty([0,T]) cap L^2([0,T])$.
– Paul Frost
Aug 6 at 14:28
1
Weirstrass $M$-test is an if theorem. You can have a series $sum f_n$ that converges absolutely uniformly but $sumsup|f_n|=infty$.
– Julián Aguirre
Aug 6 at 14:50
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am searching for some examples of complete families of functions $left phi_m(t) right_m = 1^infty$ on $t in [0, T]$ that form an absolutely convergent series:
$$
sum_m = 1^infty |phi_m(t)| < infty, ~~~~ t in [0, T].
$$
According to the Weierstrass M-test, $phi_m$ must satisfy
$$
|phi_m(t)| leq c_m, ~~ t in [0, T],
$$
such that
$$
sum_m = 1^infty c_m < infty.
$$
Any idea or references would be highly appreciated.
sequences-and-series convergence orthogonality absolute-convergence
asked Aug 6 at 12:57
Asatur Khurshudyan
395111
I am searching for some examples of complete families of functions $left phi_m(t) right_m = 1^infty$ on $t in [0, T]$ that form an absolutely convergent series:
$$
sum_m = 1^infty |phi_m(t)| < infty, ~~~~ t in [0, T].
$$
According to the Weierstrass M-test, $phi_m$ must satisfy
$$
|phi_m(t)| leq c_m, ~~ t in [0, T],
$$
such that
$$
sum_m = 1^infty c_m < infty.
$$
Any idea or references would be highly appreciated.
sequences-and-series convergence orthogonality absolute-convergence
asked Aug 6 at 12:57
Asatur Khurshudyan
395111
edited Aug 8 at 13:11
asked Aug 6 at 12:57
Asatur Khurshudyan
395111
asked Aug 6 at 12:57
Asatur Khurshudyan
395111
asked Aug 6 at 12:57
Asatur Khurshudyan
395111
395111
Do you work in the Hilbert space $L^2([0,T])$? What is the precise definition of a complete family?
– Paul Frost
Aug 6 at 13:44
Yes, let the space be $L^2$. By complete family I mean encyclopediaofmath.org/index.php/Complete_system_of_functions
– Asatur Khurshudyan
Aug 6 at 14:07
1
So you work in $L^2([0,T])$, but the $phi_m$ must be $L^infty([0,T]) cap L^2([0,T])$.
– Paul Frost
Aug 6 at 14:28
1
Weirstrass $M$-test is an if theorem. You can have a series $sum f_n$ that converges absolutely uniformly but $sumsup|f_n|=infty$.
– Julián Aguirre
Aug 6 at 14:50
add a comment |Â
Do you work in the Hilbert space $L^2([0,T])$? What is the precise definition of a complete family?
– Paul Frost
Aug 6 at 13:44
Yes, let the space be $L^2$. By complete family I mean encyclopediaofmath.org/index.php/Complete_system_of_functions
– Asatur Khurshudyan
Aug 6 at 14:07
1
So you work in $L^2([0,T])$, but the $phi_m$ must be $L^infty([0,T]) cap L^2([0,T])$.
– Paul Frost
Aug 6 at 14:28
1
Weirstrass $M$-test is an if theorem. You can have a series $sum f_n$ that converges absolutely uniformly but $sumsup|f_n|=infty$.
– Julián Aguirre
Aug 6 at 14:50
Do you work in the Hilbert space $L^2([0,T])$? What is the precise definition of a complete family?
– Paul Frost
Aug 6 at 13:44
Do you work in the Hilbert space $L^2([0,T])$? What is the precise definition of a complete family?
– Paul Frost
Aug 6 at 13:44
Yes, let the space be $L^2$. By complete family I mean encyclopediaofmath.org/index.php/Complete_system_of_functions
– Asatur Khurshudyan
Aug 6 at 14:07
Yes, let the space be $L^2$. By complete family I mean encyclopediaofmath.org/index.php/Complete_system_of_functions
– Asatur Khurshudyan
Aug 6 at 14:07
1
1
So you work in $L^2([0,T])$, but the $phi_m$ must be $L^infty([0,T]) cap L^2([0,T])$.
– Paul Frost
Aug 6 at 14:28
So you work in $L^2([0,T])$, but the $phi_m$ must be $L^infty([0,T]) cap L^2([0,T])$.
– Paul Frost
Aug 6 at 14:28
1
1
Weirstrass $M$-test is an if theorem. You can have a series $sum f_n$ that converges absolutely uniformly but $sumsup|f_n|=infty$.
– Julián Aguirre
Aug 6 at 14:50
Weirstrass $M$-test is an if theorem. You can have a series $sum f_n$ that converges absolutely uniformly but $sumsup|f_n|=infty$.
– Julián Aguirre
Aug 6 at 14:50
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
According to the definition of complete system in the link provided, $phi_m$ must be an orthonormal system. This implies among other things that$int_0^T|phi_m|^2,dt=1$. Suppose that there is a constant $M>0$ such that
$$
|phi_m(t)|leq Mquadforall tin[0,T]quadtextandquad sum_m=1^inftyint_0^T|phi_m(t)|,dtleq M.
$$
This holds in particular if $|phi_m(t)|leq c_m$ with $sum_m=1^infty c_m<infty$. Then
$$
sum_m=1^infty|phi_m(t)|^2le Msum_m=1^infty|phi_m(t)|,quad0le tle T.
$$
Integrating we get $infty$ on the left hand side, and $M^2$ on the right hand side, a contradiction.
This shows that no such families exist.
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
In other words, there are complete families of $phi_m(t)$ that are bounded but their series is not absolutely convergent. That's clear. I am more interested in examples of $phi_m$ the series of which is absolutely convergent.
– Asatur Khurshudyan
Aug 7 at 0:41
What I showed is that such families do not exist.
– Julián Aguirre
Aug 7 at 10:39
My question could be too primitive for you, but I am just trying to understand. My understanding of your answer, is that there might be a family $phi_m in L^1[0, T]$ whose series is not absolutely convergent. I agree with that. But since not every family $phi_m$ with $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ satisfies $$ sum_m = 1^infty int_0^T |phi_m(t)| dt < infty, $$ I think we can't claim that such families do not exist at all.
– Asatur Khurshudyan
Aug 7 at 12:02
On the other hand, for every family satisfying $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ we have $$ sum_m = 1^infty |phi_m(t)| < infty. $$ Integrating $$ |phi_m(t)|^2 leq c_m^2, $$ over $t in [0, T]$, we obtain that if, in addition, $$ c_m^2 geq frac1T, $$ then $phi_m$ can be orthonormal.
– Asatur Khurshudyan
Aug 7 at 12:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
According to the definition of complete system in the link provided, $phi_m$ must be an orthonormal system. This implies among other things that$int_0^T|phi_m|^2,dt=1$. Suppose that there is a constant $M>0$ such that
$$
|phi_m(t)|leq Mquadforall tin[0,T]quadtextandquad sum_m=1^inftyint_0^T|phi_m(t)|,dtleq M.
$$
This holds in particular if $|phi_m(t)|leq c_m$ with $sum_m=1^infty c_m<infty$. Then
$$
sum_m=1^infty|phi_m(t)|^2le Msum_m=1^infty|phi_m(t)|,quad0le tle T.
$$
Integrating we get $infty$ on the left hand side, and $M^2$ on the right hand side, a contradiction.
This shows that no such families exist.
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
In other words, there are complete families of $phi_m(t)$ that are bounded but their series is not absolutely convergent. That's clear. I am more interested in examples of $phi_m$ the series of which is absolutely convergent.
– Asatur Khurshudyan
Aug 7 at 0:41
What I showed is that such families do not exist.
– Julián Aguirre
Aug 7 at 10:39
My question could be too primitive for you, but I am just trying to understand. My understanding of your answer, is that there might be a family $phi_m in L^1[0, T]$ whose series is not absolutely convergent. I agree with that. But since not every family $phi_m$ with $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ satisfies $$ sum_m = 1^infty int_0^T |phi_m(t)| dt < infty, $$ I think we can't claim that such families do not exist at all.
– Asatur Khurshudyan
Aug 7 at 12:02
On the other hand, for every family satisfying $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ we have $$ sum_m = 1^infty |phi_m(t)| < infty. $$ Integrating $$ |phi_m(t)|^2 leq c_m^2, $$ over $t in [0, T]$, we obtain that if, in addition, $$ c_m^2 geq frac1T, $$ then $phi_m$ can be orthonormal.
– Asatur Khurshudyan
Aug 7 at 12:07
add a comment |Â
up vote
0
down vote
According to the definition of complete system in the link provided, $phi_m$ must be an orthonormal system. This implies among other things that$int_0^T|phi_m|^2,dt=1$. Suppose that there is a constant $M>0$ such that
$$
|phi_m(t)|leq Mquadforall tin[0,T]quadtextandquad sum_m=1^inftyint_0^T|phi_m(t)|,dtleq M.
$$
This holds in particular if $|phi_m(t)|leq c_m$ with $sum_m=1^infty c_m<infty$. Then
$$
sum_m=1^infty|phi_m(t)|^2le Msum_m=1^infty|phi_m(t)|,quad0le tle T.
$$
Integrating we get $infty$ on the left hand side, and $M^2$ on the right hand side, a contradiction.
This shows that no such families exist.
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
In other words, there are complete families of $phi_m(t)$ that are bounded but their series is not absolutely convergent. That's clear. I am more interested in examples of $phi_m$ the series of which is absolutely convergent.
– Asatur Khurshudyan
Aug 7 at 0:41
What I showed is that such families do not exist.
– Julián Aguirre
Aug 7 at 10:39
My question could be too primitive for you, but I am just trying to understand. My understanding of your answer, is that there might be a family $phi_m in L^1[0, T]$ whose series is not absolutely convergent. I agree with that. But since not every family $phi_m$ with $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ satisfies $$ sum_m = 1^infty int_0^T |phi_m(t)| dt < infty, $$ I think we can't claim that such families do not exist at all.
– Asatur Khurshudyan
Aug 7 at 12:02
On the other hand, for every family satisfying $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ we have $$ sum_m = 1^infty |phi_m(t)| < infty. $$ Integrating $$ |phi_m(t)|^2 leq c_m^2, $$ over $t in [0, T]$, we obtain that if, in addition, $$ c_m^2 geq frac1T, $$ then $phi_m$ can be orthonormal.
– Asatur Khurshudyan
Aug 7 at 12:07
add a comment |Â
up vote
0
down vote
up vote
0
down vote
According to the definition of complete system in the link provided, $phi_m$ must be an orthonormal system. This implies among other things that$int_0^T|phi_m|^2,dt=1$. Suppose that there is a constant $M>0$ such that
$$
|phi_m(t)|leq Mquadforall tin[0,T]quadtextandquad sum_m=1^inftyint_0^T|phi_m(t)|,dtleq M.
$$
This holds in particular if $|phi_m(t)|leq c_m$ with $sum_m=1^infty c_m<infty$. Then
$$
sum_m=1^infty|phi_m(t)|^2le Msum_m=1^infty|phi_m(t)|,quad0le tle T.
$$
Integrating we get $infty$ on the left hand side, and $M^2$ on the right hand side, a contradiction.
This shows that no such families exist.
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
According to the definition of complete system in the link provided, $phi_m$ must be an orthonormal system. This implies among other things that$int_0^T|phi_m|^2,dt=1$. Suppose that there is a constant $M>0$ such that
$$
|phi_m(t)|leq Mquadforall tin[0,T]quadtextandquad sum_m=1^inftyint_0^T|phi_m(t)|,dtleq M.
$$
This holds in particular if $|phi_m(t)|leq c_m$ with $sum_m=1^infty c_m<infty$. Then
$$
sum_m=1^infty|phi_m(t)|^2le Msum_m=1^infty|phi_m(t)|,quad0le tle T.
$$
Integrating we get $infty$ on the left hand side, and $M^2$ on the right hand side, a contradiction.
This shows that no such families exist.
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
edited Aug 7 at 10:42
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
answered Aug 6 at 15:09
Julián Aguirre
64.8k23894
64.8k23894
In other words, there are complete families of $phi_m(t)$ that are bounded but their series is not absolutely convergent. That's clear. I am more interested in examples of $phi_m$ the series of which is absolutely convergent.
– Asatur Khurshudyan
Aug 7 at 0:41
What I showed is that such families do not exist.
– Julián Aguirre
Aug 7 at 10:39
My question could be too primitive for you, but I am just trying to understand. My understanding of your answer, is that there might be a family $phi_m in L^1[0, T]$ whose series is not absolutely convergent. I agree with that. But since not every family $phi_m$ with $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ satisfies $$ sum_m = 1^infty int_0^T |phi_m(t)| dt < infty, $$ I think we can't claim that such families do not exist at all.
– Asatur Khurshudyan
Aug 7 at 12:02
On the other hand, for every family satisfying $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ we have $$ sum_m = 1^infty |phi_m(t)| < infty. $$ Integrating $$ |phi_m(t)|^2 leq c_m^2, $$ over $t in [0, T]$, we obtain that if, in addition, $$ c_m^2 geq frac1T, $$ then $phi_m$ can be orthonormal.
– Asatur Khurshudyan
Aug 7 at 12:07
add a comment |Â
In other words, there are complete families of $phi_m(t)$ that are bounded but their series is not absolutely convergent. That's clear. I am more interested in examples of $phi_m$ the series of which is absolutely convergent.
– Asatur Khurshudyan
Aug 7 at 0:41
What I showed is that such families do not exist.
– Julián Aguirre
Aug 7 at 10:39
My question could be too primitive for you, but I am just trying to understand. My understanding of your answer, is that there might be a family $phi_m in L^1[0, T]$ whose series is not absolutely convergent. I agree with that. But since not every family $phi_m$ with $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ satisfies $$ sum_m = 1^infty int_0^T |phi_m(t)| dt < infty, $$ I think we can't claim that such families do not exist at all.
– Asatur Khurshudyan
Aug 7 at 12:02
On the other hand, for every family satisfying $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ we have $$ sum_m = 1^infty |phi_m(t)| < infty. $$ Integrating $$ |phi_m(t)|^2 leq c_m^2, $$ over $t in [0, T]$, we obtain that if, in addition, $$ c_m^2 geq frac1T, $$ then $phi_m$ can be orthonormal.
– Asatur Khurshudyan
Aug 7 at 12:07
In other words, there are complete families of $phi_m(t)$ that are bounded but their series is not absolutely convergent. That's clear. I am more interested in examples of $phi_m$ the series of which is absolutely convergent.
– Asatur Khurshudyan
Aug 7 at 0:41
In other words, there are complete families of $phi_m(t)$ that are bounded but their series is not absolutely convergent. That's clear. I am more interested in examples of $phi_m$ the series of which is absolutely convergent.
– Asatur Khurshudyan
Aug 7 at 0:41
What I showed is that such families do not exist.
– Julián Aguirre
Aug 7 at 10:39
What I showed is that such families do not exist.
– Julián Aguirre
Aug 7 at 10:39
My question could be too primitive for you, but I am just trying to understand. My understanding of your answer, is that there might be a family $phi_m in L^1[0, T]$ whose series is not absolutely convergent. I agree with that. But since not every family $phi_m$ with $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ satisfies $$ sum_m = 1^infty int_0^T |phi_m(t)| dt < infty, $$ I think we can't claim that such families do not exist at all.
– Asatur Khurshudyan
Aug 7 at 12:02
My question could be too primitive for you, but I am just trying to understand. My understanding of your answer, is that there might be a family $phi_m in L^1[0, T]$ whose series is not absolutely convergent. I agree with that. But since not every family $phi_m$ with $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ satisfies $$ sum_m = 1^infty int_0^T |phi_m(t)| dt < infty, $$ I think we can't claim that such families do not exist at all.
– Asatur Khurshudyan
Aug 7 at 12:02
On the other hand, for every family satisfying $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ we have $$ sum_m = 1^infty |phi_m(t)| < infty. $$ Integrating $$ |phi_m(t)|^2 leq c_m^2, $$ over $t in [0, T]$, we obtain that if, in addition, $$ c_m^2 geq frac1T, $$ then $phi_m$ can be orthonormal.
– Asatur Khurshudyan
Aug 7 at 12:07
On the other hand, for every family satisfying $$ |phi_m(t)| leq c_m, ~~ sum_m = 1^infty c_m < infty, $$ we have $$ sum_m = 1^infty |phi_m(t)| < infty. $$ Integrating $$ |phi_m(t)|^2 leq c_m^2, $$ over $t in [0, T]$, we obtain that if, in addition, $$ c_m^2 geq frac1T, $$ then $phi_m$ can be orthonormal.
– Asatur Khurshudyan
Aug 7 at 12:07
add a comment |Â
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Do you work in the Hilbert space $L^2([0,T])$? What is the precise definition of a complete family?
– Paul Frost
Aug 6 at 13:44
Yes, let the space be $L^2$. By complete family I mean encyclopediaofmath.org/index.php/Complete_system_of_functions
– Asatur Khurshudyan
Aug 6 at 14:07
1
So you work in $L^2([0,T])$, but the $phi_m$ must be $L^infty([0,T]) cap L^2([0,T])$.
– Paul Frost
Aug 6 at 14:28
1
Weirstrass $M$-test is an if theorem. You can have a series $sum f_n$ that converges absolutely uniformly but $sumsup|f_n|=infty$.
– Julián Aguirre
Aug 6 at 14:50