Let $A = { x = { x_k } in l^2 vert x_2 = 2x_1 }$, show that It is a closed subspace of $l^2$
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Let $A = { x = { x_k } in l^2 vert x_2 = 2x_1 }$, I want to show that it is a closed subspace of $l^2$.
Now this is what I have done, but I am not sure if it is right.
Checking that is is a subspace is easy: I just pick $alpha,beta in mathbb{R}$ and $x,y in A$ and check that $alpha x + beta y in A$.
Now comes the hard part. I know that if $X subset Y$ is a subspace of Y complete, then being closed is equivalent to being complete.
I then verify it is complete and finish:
Let $x_n = {x_{n,k}}_{n=1}^infty$ be a Caychy sequence of A.
Then $forall epsilon > 0$, $exists N_epsilon$ such that $forall n, m > N_epsilon$
$$ vert x_{n,k} - x_{m,k} vert leq Vert x_n - x_m Vert_2 < epsilon $$
This tells me that: $exists lim_{nto infty} x_{n,j} = x_j$ which is my candiate limit. (I know it exists because I have a Cauchy sequence in $ mathbb{R}$ which is a complete space.
I know prove that it is indeed the limit by calculating:
$Vert x_n - x Vert_2^2$.
$$ Vert x_n - x Vert_2^2 = sum_{j=1}^{infty} vert x_{n,j} - x_j vert^2 = $$
$$ = sum_{j=1}^{infty} vert x_{n,j} - lim_{kto infty}x_{k,j} vert^2 = $$
$$ = lim_{kto infty} sum_{j=1}^{M} vert x_{n,j} - x_{k,j} vert^2 leq epsilon^2$$
I picked M finite so that I could bring out the limit, then I let it go to infinity to find that $Vert x_n - x Vert_2 < epsilon$
I have some doubts because I did not use the specific definition of A to find such limit. Is there a problem in my proof?
functional-analysis
asked Nov 27 at 17:57
qcc101
458113
add a comment |
up vote
0
down vote
favorite
Let $A = { x = { x_k } in l^2 vert x_2 = 2x_1 }$, I want to show that it is a closed subspace of $l^2$.
Now this is what I have done, but I am not sure if it is right.
Checking that is is a subspace is easy: I just pick $alpha,beta in mathbb{R}$ and $x,y in A$ and check that $alpha x + beta y in A$.
Now comes the hard part. I know that if $X subset Y$ is a subspace of Y complete, then being closed is equivalent to being complete.
I then verify it is complete and finish:
Let $x_n = {x_{n,k}}_{n=1}^infty$ be a Caychy sequence of A.
Then $forall epsilon > 0$, $exists N_epsilon$ such that $forall n, m > N_epsilon$
$$ vert x_{n,k} - x_{m,k} vert leq Vert x_n - x_m Vert_2 < epsilon $$
This tells me that: $exists lim_{nto infty} x_{n,j} = x_j$ which is my candiate limit. (I know it exists because I have a Cauchy sequence in $ mathbb{R}$ which is a complete space.
I know prove that it is indeed the limit by calculating:
$Vert x_n - x Vert_2^2$.
$$ Vert x_n - x Vert_2^2 = sum_{j=1}^{infty} vert x_{n,j} - x_j vert^2 = $$
$$ = sum_{j=1}^{infty} vert x_{n,j} - lim_{kto infty}x_{k,j} vert^2 = $$
$$ = lim_{kto infty} sum_{j=1}^{M} vert x_{n,j} - x_{k,j} vert^2 leq epsilon^2$$
I picked M finite so that I could bring out the limit, then I let it go to infinity to find that $Vert x_n - x Vert_2 < epsilon$
I have some doubts because I did not use the specific definition of A to find such limit. Is there a problem in my proof?
functional-analysis
asked Nov 27 at 17:57
qcc101
458113
Consider the linear functional $f$ from the space to the field, mapping $x$ to $x_2-2x_1$. It is continuous, and thus the preimage of the closed set ${0}$ is also closed.
– dan_fulea
Nov 27 at 18:00
@dan_fulea how do you know it is continuous
– mathworker21
Nov 27 at 18:01
... boundedness, $|f(x)-f(y)|=|(x_2-y_2)-2(x_1-y_1)|le|x_2-y_2|+2|x_1-y_1|le |x-y|+2|x-y|=3|x-y|$.
– dan_fulea
Nov 27 at 18:04
@dan_fulea I do not see how this changes my proof, could you elaborate a bit?
– qcc101
Nov 27 at 18:23
$f(x)=x_2-2x_1$ is a continuous linear functional. So the inverse image of the closed set ${ 0 }$ is a closed subset of $ell^2$.
– DisintegratingByParts
Nov 28 at 18:32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A = { x = { x_k } in l^2 vert x_2 = 2x_1 }$, I want to show that it is a closed subspace of $l^2$.
Now this is what I have done, but I am not sure if it is right.
Checking that is is a subspace is easy: I just pick $alpha,beta in mathbb{R}$ and $x,y in A$ and check that $alpha x + beta y in A$.
Now comes the hard part. I know that if $X subset Y$ is a subspace of Y complete, then being closed is equivalent to being complete.
I then verify it is complete and finish:
Let $x_n = {x_{n,k}}_{n=1}^infty$ be a Caychy sequence of A.
Then $forall epsilon > 0$, $exists N_epsilon$ such that $forall n, m > N_epsilon$
$$ vert x_{n,k} - x_{m,k} vert leq Vert x_n - x_m Vert_2 < epsilon $$
This tells me that: $exists lim_{nto infty} x_{n,j} = x_j$ which is my candiate limit. (I know it exists because I have a Cauchy sequence in $ mathbb{R}$ which is a complete space.
I know prove that it is indeed the limit by calculating:
$Vert x_n - x Vert_2^2$.
$$ Vert x_n - x Vert_2^2 = sum_{j=1}^{infty} vert x_{n,j} - x_j vert^2 = $$
$$ = sum_{j=1}^{infty} vert x_{n,j} - lim_{kto infty}x_{k,j} vert^2 = $$
$$ = lim_{kto infty} sum_{j=1}^{M} vert x_{n,j} - x_{k,j} vert^2 leq epsilon^2$$
I picked M finite so that I could bring out the limit, then I let it go to infinity to find that $Vert x_n - x Vert_2 < epsilon$
I have some doubts because I did not use the specific definition of A to find such limit. Is there a problem in my proof?
functional-analysis
asked Nov 27 at 17:57
qcc101
458113
Let $A = { x = { x_k } in l^2 vert x_2 = 2x_1 }$, I want to show that it is a closed subspace of $l^2$.
Now this is what I have done, but I am not sure if it is right.
Checking that is is a subspace is easy: I just pick $alpha,beta in mathbb{R}$ and $x,y in A$ and check that $alpha x + beta y in A$.
Now comes the hard part. I know that if $X subset Y$ is a subspace of Y complete, then being closed is equivalent to being complete.
I then verify it is complete and finish:
Let $x_n = {x_{n,k}}_{n=1}^infty$ be a Caychy sequence of A.
Then $forall epsilon > 0$, $exists N_epsilon$ such that $forall n, m > N_epsilon$
$$ vert x_{n,k} - x_{m,k} vert leq Vert x_n - x_m Vert_2 < epsilon $$
This tells me that: $exists lim_{nto infty} x_{n,j} = x_j$ which is my candiate limit. (I know it exists because I have a Cauchy sequence in $ mathbb{R}$ which is a complete space.
I know prove that it is indeed the limit by calculating:
$Vert x_n - x Vert_2^2$.
$$ Vert x_n - x Vert_2^2 = sum_{j=1}^{infty} vert x_{n,j} - x_j vert^2 = $$
$$ = sum_{j=1}^{infty} vert x_{n,j} - lim_{kto infty}x_{k,j} vert^2 = $$
$$ = lim_{kto infty} sum_{j=1}^{M} vert x_{n,j} - x_{k,j} vert^2 leq epsilon^2$$
I picked M finite so that I could bring out the limit, then I let it go to infinity to find that $Vert x_n - x Vert_2 < epsilon$
I have some doubts because I did not use the specific definition of A to find such limit. Is there a problem in my proof?
functional-analysis
functional-analysis
asked Nov 27 at 17:57
qcc101
458113
asked Nov 27 at 17:57
qcc101
458113
asked Nov 27 at 17:57
qcc101
458113
asked Nov 27 at 17:57
qcc101
458113
asked Nov 27 at 17:57
qcc101
458113
458113
Consider the linear functional $f$ from the space to the field, mapping $x$ to $x_2-2x_1$. It is continuous, and thus the preimage of the closed set ${0}$ is also closed.
– dan_fulea
Nov 27 at 18:00
@dan_fulea how do you know it is continuous
– mathworker21
Nov 27 at 18:01
... boundedness, $|f(x)-f(y)|=|(x_2-y_2)-2(x_1-y_1)|le|x_2-y_2|+2|x_1-y_1|le |x-y|+2|x-y|=3|x-y|$.
– dan_fulea
Nov 27 at 18:04
@dan_fulea I do not see how this changes my proof, could you elaborate a bit?
– qcc101
Nov 27 at 18:23
$f(x)=x_2-2x_1$ is a continuous linear functional. So the inverse image of the closed set ${ 0 }$ is a closed subset of $ell^2$.
– DisintegratingByParts
Nov 28 at 18:32
add a comment |
Consider the linear functional $f$ from the space to the field, mapping $x$ to $x_2-2x_1$. It is continuous, and thus the preimage of the closed set ${0}$ is also closed.
– dan_fulea
Nov 27 at 18:00
@dan_fulea how do you know it is continuous
– mathworker21
Nov 27 at 18:01
... boundedness, $|f(x)-f(y)|=|(x_2-y_2)-2(x_1-y_1)|le|x_2-y_2|+2|x_1-y_1|le |x-y|+2|x-y|=3|x-y|$.
– dan_fulea
Nov 27 at 18:04
@dan_fulea I do not see how this changes my proof, could you elaborate a bit?
– qcc101
Nov 27 at 18:23
$f(x)=x_2-2x_1$ is a continuous linear functional. So the inverse image of the closed set ${ 0 }$ is a closed subset of $ell^2$.
– DisintegratingByParts
Nov 28 at 18:32
Consider the linear functional $f$ from the space to the field, mapping $x$ to $x_2-2x_1$. It is continuous, and thus the preimage of the closed set ${0}$ is also closed.
– dan_fulea
Nov 27 at 18:00
Consider the linear functional $f$ from the space to the field, mapping $x$ to $x_2-2x_1$. It is continuous, and thus the preimage of the closed set ${0}$ is also closed.
– dan_fulea
Nov 27 at 18:00
@dan_fulea how do you know it is continuous
– mathworker21
Nov 27 at 18:01
@dan_fulea how do you know it is continuous
– mathworker21
Nov 27 at 18:01
... boundedness, $|f(x)-f(y)|=|(x_2-y_2)-2(x_1-y_1)|le|x_2-y_2|+2|x_1-y_1|le |x-y|+2|x-y|=3|x-y|$.
– dan_fulea
Nov 27 at 18:04
... boundedness, $|f(x)-f(y)|=|(x_2-y_2)-2(x_1-y_1)|le|x_2-y_2|+2|x_1-y_1|le |x-y|+2|x-y|=3|x-y|$.
– dan_fulea
Nov 27 at 18:04
@dan_fulea I do not see how this changes my proof, could you elaborate a bit?
– qcc101
Nov 27 at 18:23
@dan_fulea I do not see how this changes my proof, could you elaborate a bit?
– qcc101
Nov 27 at 18:23
$f(x)=x_2-2x_1$ is a continuous linear functional. So the inverse image of the closed set ${ 0 }$ is a closed subset of $ell^2$.
– DisintegratingByParts
Nov 28 at 18:32
$f(x)=x_2-2x_1$ is a continuous linear functional. So the inverse image of the closed set ${ 0 }$ is a closed subset of $ell^2$.
– DisintegratingByParts
Nov 28 at 18:32
add a comment |
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Consider the linear functional $f$ from the space to the field, mapping $x$ to $x_2-2x_1$. It is continuous, and thus the preimage of the closed set ${0}$ is also closed.
– dan_fulea
Nov 27 at 18:00
@dan_fulea how do you know it is continuous
– mathworker21
Nov 27 at 18:01
... boundedness, $|f(x)-f(y)|=|(x_2-y_2)-2(x_1-y_1)|le|x_2-y_2|+2|x_1-y_1|le |x-y|+2|x-y|=3|x-y|$.
– dan_fulea
Nov 27 at 18:04
@dan_fulea I do not see how this changes my proof, could you elaborate a bit?
– qcc101
Nov 27 at 18:23
$f(x)=x_2-2x_1$ is a continuous linear functional. So the inverse image of the closed set ${ 0 }$ is a closed subset of $ell^2$.
– DisintegratingByParts
Nov 28 at 18:32