Completion of inner product space
$begingroup$
Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.
I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.
Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?
linear-algebra functional-analysis inner-product-space complete-spaces
asked Jan 23 at 20:14
JamesJames
916318
$endgroup$
add a comment |
$begingroup$
Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.
I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.
Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?
linear-algebra functional-analysis inner-product-space complete-spaces
asked Jan 23 at 20:14
JamesJames
916318
$endgroup$
add a comment |
$begingroup$
Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.
I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.
Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?
linear-algebra functional-analysis inner-product-space complete-spaces
asked Jan 23 at 20:14
JamesJames
916318
$endgroup$
Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.
I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.
Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?
linear-algebra functional-analysis inner-product-space complete-spaces
linear-algebra functional-analysis inner-product-space complete-spaces
asked Jan 23 at 20:14
JamesJames
916318
asked Jan 23 at 20:14
JamesJames
916318
asked Jan 23 at 20:14
JamesJames
916318
asked Jan 23 at 20:14
JamesJames
916318
asked Jan 23 at 20:14
JamesJames
916318
916318
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
But what I want to show that $Phi_X$ is a completion.
$endgroup$
– James
Jan 23 at 20:29
$begingroup$
Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
$endgroup$
– Martin Argerami
Jan 23 at 20:32
$begingroup$
Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
$endgroup$
– James
Jan 23 at 20:33
$begingroup$
Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
$endgroup$
– Martin Argerami
Jan 23 at 20:44
1
$begingroup$
Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
$endgroup$
– Martin Argerami
Jan 23 at 23:05
|
show 3 more comments
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1 Answer
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1 Answer
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active
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votes
$begingroup$
The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
But what I want to show that $Phi_X$ is a completion.
$endgroup$
– James
Jan 23 at 20:29
$begingroup$
Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
$endgroup$
– Martin Argerami
Jan 23 at 20:32
$begingroup$
Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
$endgroup$
– James
Jan 23 at 20:33
$begingroup$
Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
$endgroup$
– Martin Argerami
Jan 23 at 20:44
1
$begingroup$
Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
$endgroup$
– Martin Argerami
Jan 23 at 23:05
|
show 3 more comments
$begingroup$
The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
But what I want to show that $Phi_X$ is a completion.
$endgroup$
– James
Jan 23 at 20:29
$begingroup$
Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
$endgroup$
– Martin Argerami
Jan 23 at 20:32
$begingroup$
Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
$endgroup$
– James
Jan 23 at 20:33
$begingroup$
Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
$endgroup$
– Martin Argerami
Jan 23 at 20:44
1
$begingroup$
Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
$endgroup$
– Martin Argerami
Jan 23 at 23:05
|
show 3 more comments
$begingroup$
The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 23 at 20:27
Martin ArgeramiMartin Argerami
128k1183183
128k1183183
$begingroup$
But what I want to show that $Phi_X$ is a completion.
$endgroup$
– James
Jan 23 at 20:29
$begingroup$
Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
$endgroup$
– Martin Argerami
Jan 23 at 20:32
$begingroup$
Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
$endgroup$
– James
Jan 23 at 20:33
$begingroup$
Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
$endgroup$
– Martin Argerami
Jan 23 at 20:44
1
$begingroup$
Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
$endgroup$
– Martin Argerami
Jan 23 at 23:05
|
show 3 more comments
$begingroup$
But what I want to show that $Phi_X$ is a completion.
$endgroup$
– James
Jan 23 at 20:29
$begingroup$
Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
$endgroup$
– Martin Argerami
Jan 23 at 20:32
$begingroup$
Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
$endgroup$
– James
Jan 23 at 20:33
$begingroup$
Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
$endgroup$
– Martin Argerami
Jan 23 at 20:44
1
$begingroup$
Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
$endgroup$
– Martin Argerami
Jan 23 at 23:05
$begingroup$
But what I want to show that $Phi_X$ is a completion.
$endgroup$
– James
Jan 23 at 20:29
$begingroup$
But what I want to show that $Phi_X$ is a completion.
$endgroup$
– James
Jan 23 at 20:29
$begingroup$
Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
$endgroup$
– Martin Argerami
Jan 23 at 20:32
$begingroup$
Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
$endgroup$
– Martin Argerami
Jan 23 at 20:32
$begingroup$
Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
$endgroup$
– James
Jan 23 at 20:33
$begingroup$
Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
$endgroup$
– James
Jan 23 at 20:33
$begingroup$
Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
$endgroup$
– Martin Argerami
Jan 23 at 20:44
$begingroup$
Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
$endgroup$
– Martin Argerami
Jan 23 at 20:44
1
1
$begingroup$
Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
$endgroup$
– Martin Argerami
Jan 23 at 23:05
$begingroup$
Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
$endgroup$
– Martin Argerami
Jan 23 at 23:05
|
show 3 more comments
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