Mixing
Is a Schauder Basis a minimal total set?
1
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Given that the span of Schauder Basis - $Ssubset V$ is $V$ itself, this implies that it is a total set. My teacher said that it was minimal as well but did not go on to prove it. I suspect it might be true given the uniqueness of representations of vectors in the "span" of a Schauder basis but I can't seem to show it.
linear-algebra functional-analysis banach-spaces
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
$endgroup$
|
show 4 more comments
1
$begingroup$
Given that the span of Schauder Basis - $Ssubset V$ is $V$ itself, this implies that it is a total set. My teacher said that it was minimal as well but did not go on to prove it. I suspect it might be true given the uniqueness of representations of vectors in the "span" of a Schauder basis but I can't seem to show it.
linear-algebra functional-analysis banach-spaces
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
$endgroup$
$begingroup$
If $(s_n)$ is the basis, for fixed $m$, $s_m^*(x)=0$ whenever $x$ is an element of $U={rm span}{ s_i : ine m}$. Then $s_m (x)=0$ for any $x$ in the closure of $U$ by continuity.
$endgroup$
– David Mitra
Feb 1 at 19:22
$begingroup$
... I meant to write "$s_m^*(x)$" in the last sentence.
$endgroup$
– David Mitra
Feb 1 at 19:30
$begingroup$
I'm sorry what is $s*_m(x)$?
$endgroup$
– Jhon Doe
Feb 2 at 2:34
$begingroup$
The coefficient functional for $s_m$. $s_m^*(sumalpha_i s_i) =alpha_m$. We have $s_j^*(s_i)=0$ if $ine j$ and $s_j^*(s_j)=1$ for all $j$. One can show the coefficient functionals are continuous.
$endgroup$
– David Mitra
Feb 2 at 3:41
$begingroup$
Im sorry but im kinda slow so ive got a couple of questions. To show that $S$ is a minimal total set we have to show that if $F$ is a total set it is not a subset of $S$. I dont see how the coefficient functional describes that. Secondly, whats the domain of the function is it V?
$endgroup$
– Jhon Doe
Feb 2 at 13:24
|
show 4 more comments
1
1
1
$begingroup$
Given that the span of Schauder Basis - $Ssubset V$ is $V$ itself, this implies that it is a total set. My teacher said that it was minimal as well but did not go on to prove it. I suspect it might be true given the uniqueness of representations of vectors in the "span" of a Schauder basis but I can't seem to show it.
linear-algebra functional-analysis banach-spaces
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
$endgroup$
Given that the span of Schauder Basis - $Ssubset V$ is $V$ itself, this implies that it is a total set. My teacher said that it was minimal as well but did not go on to prove it. I suspect it might be true given the uniqueness of representations of vectors in the "span" of a Schauder basis but I can't seem to show it.
linear-algebra functional-analysis banach-spaces
linear-algebra functional-analysis banach-spaces
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
edited Feb 2 at 14:47
Jhon Doe
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
asked Feb 1 at 13:47
Jhon DoeJhon Doe
695414
695414
$begingroup$
If $(s_n)$ is the basis, for fixed $m$, $s_m^*(x)=0$ whenever $x$ is an element of $U={rm span}{ s_i : ine m}$. Then $s_m (x)=0$ for any $x$ in the closure of $U$ by continuity.
$endgroup$
– David Mitra
Feb 1 at 19:22
$begingroup$
... I meant to write "$s_m^*(x)$" in the last sentence.
$endgroup$
– David Mitra
Feb 1 at 19:30
$begingroup$
I'm sorry what is $s*_m(x)$?
$endgroup$
– Jhon Doe
Feb 2 at 2:34
$begingroup$
The coefficient functional for $s_m$. $s_m^*(sumalpha_i s_i) =alpha_m$. We have $s_j^*(s_i)=0$ if $ine j$ and $s_j^*(s_j)=1$ for all $j$. One can show the coefficient functionals are continuous.
$endgroup$
– David Mitra
Feb 2 at 3:41
$begingroup$
Im sorry but im kinda slow so ive got a couple of questions. To show that $S$ is a minimal total set we have to show that if $F$ is a total set it is not a subset of $S$. I dont see how the coefficient functional describes that. Secondly, whats the domain of the function is it V?
$endgroup$
– Jhon Doe
Feb 2 at 13:24
|
show 4 more comments
$begingroup$
If $(s_n)$ is the basis, for fixed $m$, $s_m^*(x)=0$ whenever $x$ is an element of $U={rm span}{ s_i : ine m}$. Then $s_m (x)=0$ for any $x$ in the closure of $U$ by continuity.
$endgroup$
– David Mitra
Feb 1 at 19:22
$begingroup$
... I meant to write "$s_m^*(x)$" in the last sentence.
$endgroup$
– David Mitra
Feb 1 at 19:30
$begingroup$
I'm sorry what is $s*_m(x)$?
$endgroup$
– Jhon Doe
Feb 2 at 2:34
$begingroup$
The coefficient functional for $s_m$. $s_m^*(sumalpha_i s_i) =alpha_m$. We have $s_j^*(s_i)=0$ if $ine j$ and $s_j^*(s_j)=1$ for all $j$. One can show the coefficient functionals are continuous.
$endgroup$
– David Mitra
Feb 2 at 3:41
$begingroup$
Im sorry but im kinda slow so ive got a couple of questions. To show that $S$ is a minimal total set we have to show that if $F$ is a total set it is not a subset of $S$. I dont see how the coefficient functional describes that. Secondly, whats the domain of the function is it V?
$endgroup$
– Jhon Doe
Feb 2 at 13:24
$begingroup$
If $(s_n)$ is the basis, for fixed $m$, $s_m^*(x)=0$ whenever $x$ is an element of $U={rm span}{ s_i : ine m}$. Then $s_m (x)=0$ for any $x$ in the closure of $U$ by continuity.
$endgroup$
– David Mitra
Feb 1 at 19:22
$begingroup$
If $(s_n)$ is the basis, for fixed $m$, $s_m^*(x)=0$ whenever $x$ is an element of $U={rm span}{ s_i : ine m}$. Then $s_m (x)=0$ for any $x$ in the closure of $U$ by continuity.
$endgroup$
– David Mitra
Feb 1 at 19:22
$begingroup$
... I meant to write "$s_m^*(x)$" in the last sentence.
$endgroup$
– David Mitra
Feb 1 at 19:30
$begingroup$
... I meant to write "$s_m^*(x)$" in the last sentence.
$endgroup$
– David Mitra
Feb 1 at 19:30
$begingroup$
I'm sorry what is $s*_m(x)$?
$endgroup$
– Jhon Doe
Feb 2 at 2:34
$begingroup$
I'm sorry what is $s*_m(x)$?
$endgroup$
– Jhon Doe
Feb 2 at 2:34
$begingroup$
The coefficient functional for $s_m$. $s_m^*(sumalpha_i s_i) =alpha_m$. We have $s_j^*(s_i)=0$ if $ine j$ and $s_j^*(s_j)=1$ for all $j$. One can show the coefficient functionals are continuous.
$endgroup$
– David Mitra
Feb 2 at 3:41
$begingroup$
The coefficient functional for $s_m$. $s_m^*(sumalpha_i s_i) =alpha_m$. We have $s_j^*(s_i)=0$ if $ine j$ and $s_j^*(s_j)=1$ for all $j$. One can show the coefficient functionals are continuous.
$endgroup$
– David Mitra
Feb 2 at 3:41
$begingroup$
Im sorry but im kinda slow so ive got a couple of questions. To show that $S$ is a minimal total set we have to show that if $F$ is a total set it is not a subset of $S$. I dont see how the coefficient functional describes that. Secondly, whats the domain of the function is it V?
$endgroup$
– Jhon Doe
Feb 2 at 13:24
$begingroup$
Im sorry but im kinda slow so ive got a couple of questions. To show that $S$ is a minimal total set we have to show that if $F$ is a total set it is not a subset of $S$. I dont see how the coefficient functional describes that. Secondly, whats the domain of the function is it V?
$endgroup$
– Jhon Doe
Feb 2 at 13:24
|
show 4 more comments
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$begingroup$
If $(s_n)$ is the basis, for fixed $m$, $s_m^*(x)=0$ whenever $x$ is an element of $U={rm span}{ s_i : ine m}$. Then $s_m (x)=0$ for any $x$ in the closure of $U$ by continuity.
$endgroup$
– David Mitra
Feb 1 at 19:22
$begingroup$
... I meant to write "$s_m^*(x)$" in the last sentence.
$endgroup$
– David Mitra
Feb 1 at 19:30
$begingroup$
I'm sorry what is $s*_m(x)$?
$endgroup$
– Jhon Doe
Feb 2 at 2:34
$begingroup$
The coefficient functional for $s_m$. $s_m^*(sumalpha_i s_i) =alpha_m$. We have $s_j^*(s_i)=0$ if $ine j$ and $s_j^*(s_j)=1$ for all $j$. One can show the coefficient functionals are continuous.
$endgroup$
– David Mitra
Feb 2 at 3:41
$begingroup$
Im sorry but im kinda slow so ive got a couple of questions. To show that $S$ is a minimal total set we have to show that if $F$ is a total set it is not a subset of $S$. I dont see how the coefficient functional describes that. Secondly, whats the domain of the function is it V?
$endgroup$
– Jhon Doe
Feb 2 at 13:24