Prove or disprove that $W$ is a Banach space.
$begingroup$
True or false, and justify.
Let $W ={ P: textrm{polynomial of degree} le100}$ and $|| P || = sum|a_n|$.
Then $|| P ||$ is a norm and $W$ is a Banach space under this norm.
I proved that $|| P ||$ is a norm, but how can I prove or disprove that $W$ is a Banach space?
I know that the set of all polynomials is not a Banach space, but in finite case it will be same?
Thanks a lot.
functional-analysis banach-spaces normed-spaces
edited Dec 16 '18 at 17:20
Song
14.2k1634
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
$endgroup$
|
show 1 more comment
$begingroup$
True or false, and justify.
Let $W ={ P: textrm{polynomial of degree} le100}$ and $|| P || = sum|a_n|$.
Then $|| P ||$ is a norm and $W$ is a Banach space under this norm.
I proved that $|| P ||$ is a norm, but how can I prove or disprove that $W$ is a Banach space?
I know that the set of all polynomials is not a Banach space, but in finite case it will be same?
Thanks a lot.
functional-analysis banach-spaces normed-spaces
edited Dec 16 '18 at 17:20
Song
14.2k1634
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
$endgroup$
1
$begingroup$
It's not the same in the finite case. It is left to show that a Cauchy sequence of polynomials of degree $<100$ is a polynomial of degree $<100$. Note that if $P=sum a_n x^n$ and $Q=sum b_n x^n$ then $|P-Q| = sum |a_n-b_n|$. Hence a Cauchy sequence of polynomials is somehow 100 Cauchy sequences of real/complex numbers (the coefficients).
$endgroup$
– Yanko
Dec 16 '18 at 13:35
1
$begingroup$
Note: in your formulation you left out the important connection between $P$ and $a_n$.
$endgroup$
– GEdgar
Dec 16 '18 at 14:01
1
$begingroup$
Any finite-dimensional normed space is a Banach space.
$endgroup$
– Song
Dec 16 '18 at 17:19
$begingroup$
I think it pertains that the tag of the question be moved from measure-theory to banach-spaces or functional-analysis.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:19
$begingroup$
@Song that is if you are considering it over the fields $mathbb C$ or $mathbb R$?
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:22
|
show 1 more comment
$begingroup$
True or false, and justify.
Let $W ={ P: textrm{polynomial of degree} le100}$ and $|| P || = sum|a_n|$.
Then $|| P ||$ is a norm and $W$ is a Banach space under this norm.
I proved that $|| P ||$ is a norm, but how can I prove or disprove that $W$ is a Banach space?
I know that the set of all polynomials is not a Banach space, but in finite case it will be same?
Thanks a lot.
functional-analysis banach-spaces normed-spaces
edited Dec 16 '18 at 17:20
Song
14.2k1634
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
$endgroup$
True or false, and justify.
Let $W ={ P: textrm{polynomial of degree} le100}$ and $|| P || = sum|a_n|$.
Then $|| P ||$ is a norm and $W$ is a Banach space under this norm.
I proved that $|| P ||$ is a norm, but how can I prove or disprove that $W$ is a Banach space?
I know that the set of all polynomials is not a Banach space, but in finite case it will be same?
Thanks a lot.
functional-analysis banach-spaces normed-spaces
functional-analysis banach-spaces normed-spaces
edited Dec 16 '18 at 17:20
Song
14.2k1634
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
edited Dec 16 '18 at 17:20
Song
14.2k1634
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
edited Dec 16 '18 at 17:20
Song
14.2k1634
edited Dec 16 '18 at 17:20
Song
14.2k1634
edited Dec 16 '18 at 17:20
Song
14.2k1634
14.2k1634
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
asked Dec 16 '18 at 13:33
Duaa HamzehDuaa Hamzeh
524
524
1
$begingroup$
It's not the same in the finite case. It is left to show that a Cauchy sequence of polynomials of degree $<100$ is a polynomial of degree $<100$. Note that if $P=sum a_n x^n$ and $Q=sum b_n x^n$ then $|P-Q| = sum |a_n-b_n|$. Hence a Cauchy sequence of polynomials is somehow 100 Cauchy sequences of real/complex numbers (the coefficients).
$endgroup$
– Yanko
Dec 16 '18 at 13:35
1
$begingroup$
Note: in your formulation you left out the important connection between $P$ and $a_n$.
$endgroup$
– GEdgar
Dec 16 '18 at 14:01
1
$begingroup$
Any finite-dimensional normed space is a Banach space.
$endgroup$
– Song
Dec 16 '18 at 17:19
$begingroup$
I think it pertains that the tag of the question be moved from measure-theory to banach-spaces or functional-analysis.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:19
$begingroup$
@Song that is if you are considering it over the fields $mathbb C$ or $mathbb R$?
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:22
|
show 1 more comment
1
$begingroup$
It's not the same in the finite case. It is left to show that a Cauchy sequence of polynomials of degree $<100$ is a polynomial of degree $<100$. Note that if $P=sum a_n x^n$ and $Q=sum b_n x^n$ then $|P-Q| = sum |a_n-b_n|$. Hence a Cauchy sequence of polynomials is somehow 100 Cauchy sequences of real/complex numbers (the coefficients).
$endgroup$
– Yanko
Dec 16 '18 at 13:35
1
$begingroup$
Note: in your formulation you left out the important connection between $P$ and $a_n$.
$endgroup$
– GEdgar
Dec 16 '18 at 14:01
1
$begingroup$
Any finite-dimensional normed space is a Banach space.
$endgroup$
– Song
Dec 16 '18 at 17:19
$begingroup$
I think it pertains that the tag of the question be moved from measure-theory to banach-spaces or functional-analysis.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:19
$begingroup$
@Song that is if you are considering it over the fields $mathbb C$ or $mathbb R$?
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:22
1
1
$begingroup$
It's not the same in the finite case. It is left to show that a Cauchy sequence of polynomials of degree $<100$ is a polynomial of degree $<100$. Note that if $P=sum a_n x^n$ and $Q=sum b_n x^n$ then $|P-Q| = sum |a_n-b_n|$. Hence a Cauchy sequence of polynomials is somehow 100 Cauchy sequences of real/complex numbers (the coefficients).
$endgroup$
– Yanko
Dec 16 '18 at 13:35
$begingroup$
It's not the same in the finite case. It is left to show that a Cauchy sequence of polynomials of degree $<100$ is a polynomial of degree $<100$. Note that if $P=sum a_n x^n$ and $Q=sum b_n x^n$ then $|P-Q| = sum |a_n-b_n|$. Hence a Cauchy sequence of polynomials is somehow 100 Cauchy sequences of real/complex numbers (the coefficients).
$endgroup$
– Yanko
Dec 16 '18 at 13:35
1
1
$begingroup$
Note: in your formulation you left out the important connection between $P$ and $a_n$.
$endgroup$
– GEdgar
Dec 16 '18 at 14:01
$begingroup$
Note: in your formulation you left out the important connection between $P$ and $a_n$.
$endgroup$
– GEdgar
Dec 16 '18 at 14:01
1
1
$begingroup$
Any finite-dimensional normed space is a Banach space.
$endgroup$
– Song
Dec 16 '18 at 17:19
$begingroup$
Any finite-dimensional normed space is a Banach space.
$endgroup$
– Song
Dec 16 '18 at 17:19
$begingroup$
I think it pertains that the tag of the question be moved from measure-theory to banach-spaces or functional-analysis.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:19
$begingroup$
I think it pertains that the tag of the question be moved from measure-theory to banach-spaces or functional-analysis.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:19
$begingroup$
@Song that is if you are considering it over the fields $mathbb C$ or $mathbb R$?
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:22
$begingroup$
@Song that is if you are considering it over the fields $mathbb C$ or $mathbb R$?
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{N}{mathbb N}$
$newcommand{C}{mathbb C}$
$newcommand{R}{mathbb R}$
$newcommand{e}{epsilon}$
As it has been pointed out in the comments you must specify over which ring of polynomials you are working. I suppose that you are working on $mathbb C[x]$ or $mathbb R[x]$ or over any other ring such that it is a Banach space. As you have proved that ||P|| is a norm we must see whether $W$ is a Banach space or not. Let's prove that $(W,||.||)$ is complete. Observe that as all the polynomials have the same maximum degree it means that the difference of them have the same maximum degree, that is $p_n -p_m in W$. For that matter let $(p_n)_{nin N}$ be a Cauchy sequence in $W$. This means that for a certain $n_0 in N$ for all $n,mgeq n_0$ we have that $$||p_m-p_n||<e.$$
This in turn can be rewritten as
begin{equation}sum_{i=0}^{100} |a_{ni} -{a_{mi}}| < e end{equation}
where $(a_{ni}), (a_{mi})$ are the coefficients of the polynomials $p_n$ and $p_m$ respectively. This means we have $101$ (what is important here is that it is a finite number) Cauchy sequences in $C$ or in $R$, and as they are complete they converge. That is a for a certain $n_1 in N$ we have that for all $ngeq n_1$
$$|a_{ni}-a_i|<dfrac{e}{101}$$
which combined with what we have seen before implies that for all $ngeq n_1$
begin{equation} sum_{i=0}^{100} |a_{ni} -{a_{i}}| < e. end{equation}
Let's name $p = sum_{i=0}^{100} a_i $. What we have seen before implies that $p_n to p$. As $p in W$ we have seen that W is a banach space.
This exercise illustrates that in this case the problem can be reduced to the study of the coefficients of the polynomial while the degree as long as it's bounded is not a problem. Remember that the set of the polynomials is not complete because the degree is not bounded and that helps you construct a Cauchy sequence that does not converge to a polynomial but to $e^x$, for example.
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
$endgroup$
$begingroup$
If something is not clearly written or needs more explaining please ask me.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:20
$begingroup$
It's very nice and clear answer.. Grateful 😍
$endgroup$
– Duaa Hamzeh
Dec 17 '18 at 3:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$newcommand{N}{mathbb N}$
$newcommand{C}{mathbb C}$
$newcommand{R}{mathbb R}$
$newcommand{e}{epsilon}$
As it has been pointed out in the comments you must specify over which ring of polynomials you are working. I suppose that you are working on $mathbb C[x]$ or $mathbb R[x]$ or over any other ring such that it is a Banach space. As you have proved that ||P|| is a norm we must see whether $W$ is a Banach space or not. Let's prove that $(W,||.||)$ is complete. Observe that as all the polynomials have the same maximum degree it means that the difference of them have the same maximum degree, that is $p_n -p_m in W$. For that matter let $(p_n)_{nin N}$ be a Cauchy sequence in $W$. This means that for a certain $n_0 in N$ for all $n,mgeq n_0$ we have that $$||p_m-p_n||<e.$$
This in turn can be rewritten as
begin{equation}sum_{i=0}^{100} |a_{ni} -{a_{mi}}| < e end{equation}
where $(a_{ni}), (a_{mi})$ are the coefficients of the polynomials $p_n$ and $p_m$ respectively. This means we have $101$ (what is important here is that it is a finite number) Cauchy sequences in $C$ or in $R$, and as they are complete they converge. That is a for a certain $n_1 in N$ we have that for all $ngeq n_1$
$$|a_{ni}-a_i|<dfrac{e}{101}$$
which combined with what we have seen before implies that for all $ngeq n_1$
begin{equation} sum_{i=0}^{100} |a_{ni} -{a_{i}}| < e. end{equation}
Let's name $p = sum_{i=0}^{100} a_i $. What we have seen before implies that $p_n to p$. As $p in W$ we have seen that W is a banach space.
This exercise illustrates that in this case the problem can be reduced to the study of the coefficients of the polynomial while the degree as long as it's bounded is not a problem. Remember that the set of the polynomials is not complete because the degree is not bounded and that helps you construct a Cauchy sequence that does not converge to a polynomial but to $e^x$, for example.
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
$endgroup$
$begingroup$
If something is not clearly written or needs more explaining please ask me.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:20
$begingroup$
It's very nice and clear answer.. Grateful 😍
$endgroup$
– Duaa Hamzeh
Dec 17 '18 at 3:40
add a comment |
$begingroup$
$newcommand{N}{mathbb N}$
$newcommand{C}{mathbb C}$
$newcommand{R}{mathbb R}$
$newcommand{e}{epsilon}$
As it has been pointed out in the comments you must specify over which ring of polynomials you are working. I suppose that you are working on $mathbb C[x]$ or $mathbb R[x]$ or over any other ring such that it is a Banach space. As you have proved that ||P|| is a norm we must see whether $W$ is a Banach space or not. Let's prove that $(W,||.||)$ is complete. Observe that as all the polynomials have the same maximum degree it means that the difference of them have the same maximum degree, that is $p_n -p_m in W$. For that matter let $(p_n)_{nin N}$ be a Cauchy sequence in $W$. This means that for a certain $n_0 in N$ for all $n,mgeq n_0$ we have that $$||p_m-p_n||<e.$$
This in turn can be rewritten as
begin{equation}sum_{i=0}^{100} |a_{ni} -{a_{mi}}| < e end{equation}
where $(a_{ni}), (a_{mi})$ are the coefficients of the polynomials $p_n$ and $p_m$ respectively. This means we have $101$ (what is important here is that it is a finite number) Cauchy sequences in $C$ or in $R$, and as they are complete they converge. That is a for a certain $n_1 in N$ we have that for all $ngeq n_1$
$$|a_{ni}-a_i|<dfrac{e}{101}$$
which combined with what we have seen before implies that for all $ngeq n_1$
begin{equation} sum_{i=0}^{100} |a_{ni} -{a_{i}}| < e. end{equation}
Let's name $p = sum_{i=0}^{100} a_i $. What we have seen before implies that $p_n to p$. As $p in W$ we have seen that W is a banach space.
This exercise illustrates that in this case the problem can be reduced to the study of the coefficients of the polynomial while the degree as long as it's bounded is not a problem. Remember that the set of the polynomials is not complete because the degree is not bounded and that helps you construct a Cauchy sequence that does not converge to a polynomial but to $e^x$, for example.
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
$endgroup$
$begingroup$
If something is not clearly written or needs more explaining please ask me.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:20
$begingroup$
It's very nice and clear answer.. Grateful 😍
$endgroup$
– Duaa Hamzeh
Dec 17 '18 at 3:40
add a comment |
$begingroup$
$newcommand{N}{mathbb N}$
$newcommand{C}{mathbb C}$
$newcommand{R}{mathbb R}$
$newcommand{e}{epsilon}$
As it has been pointed out in the comments you must specify over which ring of polynomials you are working. I suppose that you are working on $mathbb C[x]$ or $mathbb R[x]$ or over any other ring such that it is a Banach space. As you have proved that ||P|| is a norm we must see whether $W$ is a Banach space or not. Let's prove that $(W,||.||)$ is complete. Observe that as all the polynomials have the same maximum degree it means that the difference of them have the same maximum degree, that is $p_n -p_m in W$. For that matter let $(p_n)_{nin N}$ be a Cauchy sequence in $W$. This means that for a certain $n_0 in N$ for all $n,mgeq n_0$ we have that $$||p_m-p_n||<e.$$
This in turn can be rewritten as
begin{equation}sum_{i=0}^{100} |a_{ni} -{a_{mi}}| < e end{equation}
where $(a_{ni}), (a_{mi})$ are the coefficients of the polynomials $p_n$ and $p_m$ respectively. This means we have $101$ (what is important here is that it is a finite number) Cauchy sequences in $C$ or in $R$, and as they are complete they converge. That is a for a certain $n_1 in N$ we have that for all $ngeq n_1$
$$|a_{ni}-a_i|<dfrac{e}{101}$$
which combined with what we have seen before implies that for all $ngeq n_1$
begin{equation} sum_{i=0}^{100} |a_{ni} -{a_{i}}| < e. end{equation}
Let's name $p = sum_{i=0}^{100} a_i $. What we have seen before implies that $p_n to p$. As $p in W$ we have seen that W is a banach space.
This exercise illustrates that in this case the problem can be reduced to the study of the coefficients of the polynomial while the degree as long as it's bounded is not a problem. Remember that the set of the polynomials is not complete because the degree is not bounded and that helps you construct a Cauchy sequence that does not converge to a polynomial but to $e^x$, for example.
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
$endgroup$
$newcommand{N}{mathbb N}$
$newcommand{C}{mathbb C}$
$newcommand{R}{mathbb R}$
$newcommand{e}{epsilon}$
As it has been pointed out in the comments you must specify over which ring of polynomials you are working. I suppose that you are working on $mathbb C[x]$ or $mathbb R[x]$ or over any other ring such that it is a Banach space. As you have proved that ||P|| is a norm we must see whether $W$ is a Banach space or not. Let's prove that $(W,||.||)$ is complete. Observe that as all the polynomials have the same maximum degree it means that the difference of them have the same maximum degree, that is $p_n -p_m in W$. For that matter let $(p_n)_{nin N}$ be a Cauchy sequence in $W$. This means that for a certain $n_0 in N$ for all $n,mgeq n_0$ we have that $$||p_m-p_n||<e.$$
This in turn can be rewritten as
begin{equation}sum_{i=0}^{100} |a_{ni} -{a_{mi}}| < e end{equation}
where $(a_{ni}), (a_{mi})$ are the coefficients of the polynomials $p_n$ and $p_m$ respectively. This means we have $101$ (what is important here is that it is a finite number) Cauchy sequences in $C$ or in $R$, and as they are complete they converge. That is a for a certain $n_1 in N$ we have that for all $ngeq n_1$
$$|a_{ni}-a_i|<dfrac{e}{101}$$
which combined with what we have seen before implies that for all $ngeq n_1$
begin{equation} sum_{i=0}^{100} |a_{ni} -{a_{i}}| < e. end{equation}
Let's name $p = sum_{i=0}^{100} a_i $. What we have seen before implies that $p_n to p$. As $p in W$ we have seen that W is a banach space.
This exercise illustrates that in this case the problem can be reduced to the study of the coefficients of the polynomial while the degree as long as it's bounded is not a problem. Remember that the set of the polynomials is not complete because the degree is not bounded and that helps you construct a Cauchy sequence that does not converge to a polynomial but to $e^x$, for example.
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
answered Dec 16 '18 at 17:16
Leo LerenaLeo Lerena
445511
445511
$begingroup$
If something is not clearly written or needs more explaining please ask me.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:20
$begingroup$
It's very nice and clear answer.. Grateful 😍
$endgroup$
– Duaa Hamzeh
Dec 17 '18 at 3:40
add a comment |
$begingroup$
If something is not clearly written or needs more explaining please ask me.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:20
$begingroup$
It's very nice and clear answer.. Grateful 😍
$endgroup$
– Duaa Hamzeh
Dec 17 '18 at 3:40
$begingroup$
If something is not clearly written or needs more explaining please ask me.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:20
$begingroup$
If something is not clearly written or needs more explaining please ask me.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:20
$begingroup$
It's very nice and clear answer.. Grateful 😍
$endgroup$
– Duaa Hamzeh
Dec 17 '18 at 3:40
$begingroup$
It's very nice and clear answer.. Grateful 😍
$endgroup$
– Duaa Hamzeh
Dec 17 '18 at 3:40
add a comment |
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$begingroup$
It's not the same in the finite case. It is left to show that a Cauchy sequence of polynomials of degree $<100$ is a polynomial of degree $<100$. Note that if $P=sum a_n x^n$ and $Q=sum b_n x^n$ then $|P-Q| = sum |a_n-b_n|$. Hence a Cauchy sequence of polynomials is somehow 100 Cauchy sequences of real/complex numbers (the coefficients).
$endgroup$
– Yanko
Dec 16 '18 at 13:35
1
$begingroup$
Note: in your formulation you left out the important connection between $P$ and $a_n$.
$endgroup$
– GEdgar
Dec 16 '18 at 14:01
1
$begingroup$
Any finite-dimensional normed space is a Banach space.
$endgroup$
– Song
Dec 16 '18 at 17:19
$begingroup$
I think it pertains that the tag of the question be moved from measure-theory to banach-spaces or functional-analysis.
$endgroup$
– Leo Lerena
Dec 16 '18 at 17:19
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@Song that is if you are considering it over the fields $mathbb C$ or $mathbb R$?
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– Leo Lerena
Dec 16 '18 at 17:22