Mixing
Show that there is a unique extension $overline{T}in(l^2)'$
$begingroup$
Consider the subspace $Y={{x_k}_kin l^2:x_k=0 $ whenever $k$ is odd $}subset l^2$ and a linear functional $Tin Y', Tne 0.$ Show to there is a unique linear extension $overline{T}in(l^2)'$ of $T$ with $||overline{T}||_{(l^2)'}=||T||_{Y'}.$
I know what to do if I have a defined functional $T:Wto mathbb{R}, {x_k}_kmapsto 2017x_k$, with $W={{x_k}_kin l^1, x_k=0, forall kinmathbb{N}, kne 2017}$. In this case I would calculate the dual norm of $T$ and then I would find $overline{T}$ extension of $T$ with $overline{T}({x_k})=sum_1^infty x_ky_k$ with ${y_k}in l^{infty}$.
How I can solve this problem without knowing the definition of the functional $T$?
real-analysis functional-analysis
asked Feb 14 '18 at 10:30
james wattjames watt
373110
$endgroup$
add a comment |
$begingroup$
Consider the subspace $Y={{x_k}_kin l^2:x_k=0 $ whenever $k$ is odd $}subset l^2$ and a linear functional $Tin Y', Tne 0.$ Show to there is a unique linear extension $overline{T}in(l^2)'$ of $T$ with $||overline{T}||_{(l^2)'}=||T||_{Y'}.$
I know what to do if I have a defined functional $T:Wto mathbb{R}, {x_k}_kmapsto 2017x_k$, with $W={{x_k}_kin l^1, x_k=0, forall kinmathbb{N}, kne 2017}$. In this case I would calculate the dual norm of $T$ and then I would find $overline{T}$ extension of $T$ with $overline{T}({x_k})=sum_1^infty x_ky_k$ with ${y_k}in l^{infty}$.
How I can solve this problem without knowing the definition of the functional $T$?
real-analysis functional-analysis
asked Feb 14 '18 at 10:30
james wattjames watt
373110
$endgroup$
add a comment |
$begingroup$
Consider the subspace $Y={{x_k}_kin l^2:x_k=0 $ whenever $k$ is odd $}subset l^2$ and a linear functional $Tin Y', Tne 0.$ Show to there is a unique linear extension $overline{T}in(l^2)'$ of $T$ with $||overline{T}||_{(l^2)'}=||T||_{Y'}.$
I know what to do if I have a defined functional $T:Wto mathbb{R}, {x_k}_kmapsto 2017x_k$, with $W={{x_k}_kin l^1, x_k=0, forall kinmathbb{N}, kne 2017}$. In this case I would calculate the dual norm of $T$ and then I would find $overline{T}$ extension of $T$ with $overline{T}({x_k})=sum_1^infty x_ky_k$ with ${y_k}in l^{infty}$.
How I can solve this problem without knowing the definition of the functional $T$?
real-analysis functional-analysis
asked Feb 14 '18 at 10:30
james wattjames watt
373110
$endgroup$
Consider the subspace $Y={{x_k}_kin l^2:x_k=0 $ whenever $k$ is odd $}subset l^2$ and a linear functional $Tin Y', Tne 0.$ Show to there is a unique linear extension $overline{T}in(l^2)'$ of $T$ with $||overline{T}||_{(l^2)'}=||T||_{Y'}.$
I know what to do if I have a defined functional $T:Wto mathbb{R}, {x_k}_kmapsto 2017x_k$, with $W={{x_k}_kin l^1, x_k=0, forall kinmathbb{N}, kne 2017}$. In this case I would calculate the dual norm of $T$ and then I would find $overline{T}$ extension of $T$ with $overline{T}({x_k})=sum_1^infty x_ky_k$ with ${y_k}in l^{infty}$.
How I can solve this problem without knowing the definition of the functional $T$?
real-analysis functional-analysis
real-analysis functional-analysis
asked Feb 14 '18 at 10:30
james wattjames watt
373110
asked Feb 14 '18 at 10:30
james wattjames watt
373110
asked Feb 14 '18 at 10:30
james wattjames watt
373110
asked Feb 14 '18 at 10:30
james wattjames watt
373110
asked Feb 14 '18 at 10:30
james wattjames watt
373110
373110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The existence of a extension is guaranteed by the Hahn-Banach-Theorem, but in general this extension is not unique (depending on the Banach space). Here, you have a Hilbert space and in this case you don't need the Hahn-Banach-Theorem and we also get uniqueness of the extension.
Note that $Y$ is closed subspace and thus also a Hilbert-Space. The Riesz representation theorem implies that $T(x) = langle x,y rangle$ for a unique $y in Y$ and $|T|=|y|$. This gives the unique extension $overline{T}(x) = langle x, y rangle$.
Let us prove that the extension is unique: Any other extension $L colon H rightarrow mathbb{R}$ has a representation $L(x) = langle x, l rangle$ with $l in H$ and $|l|=|y|$. Decompose $$l = y' + u$$ with $y' in Y$ and $u in Y^perp$. Note that we used the geometry of Hilbert spaces in form of the existence of an orthogonal projection. Then we have for $x in Y$ that $$langle x, y rangle = overline{T}(x) = T(x) = L(x) = langle x, y' rangle.$$
Thus $langle x, y-y' rangle =0$. Taking $x=y-y'$ we get that $y=y'$. Now we have
$$|y|^2 = |l|^2 = |y'|^2 + |u|^2 = |y|^2 +|u|^2$$
and thus $u=0$. This proves $l=y$.
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
$endgroup$
$begingroup$
$W$ is a Hilbert-Space too? So in general by Riesz representation theorem there always exists an extension $overline{T}$ caused by the subspace is a Hilbert-Space?
$endgroup$
– james watt
Feb 15 '18 at 9:28
1
$begingroup$
In the first case, $Y$ is closed and the metric is induced by an inner product. Thus $Y$ is a Hilbert space. In general, the extension is not unique and this question depends on the geometry of the corresponding Banach space $X$. (E.g. if $X'$ is strict convex, then every extension is unique.) In the case of $l^1$ the extension is not unique. If $T(x) = 2017 x_{2017}$ on $W$, then you can take any $(y_k)_{k in mathbb{N}} in l^infty$ with $y_{2017} = 2017$ and $|y_i| leq 2017$ for $i neq k$.
$endgroup$
– p4sch
Feb 15 '18 at 9:52
1
$begingroup$
We have simply $|overline{T}| = |y| = |T|$ again by the Riesz representation theorem. If $L$ is another extension, i.e. the restriction to $Y$ is $T$ and $|L| = |T|$, then there exists an $l in l^2$ with $L(x) = langle x,l rangle$ and $|l| = |T|$. Now, we have $0=overline{T}(u)-L(u) = langle y-l,u rangle$ for any $u in Y$, i.e. $y-l in Y^{perp}$. That implies $|overline{T}|=|L|=|l| =|(l-y)+y|= |y|+|l-y| = |overline{T}| + |overline{T}-L|$. Thus $|overline{T}-L|=0$.
$endgroup$
– p4sch
Feb 15 '18 at 10:29
1
$begingroup$
Yes, of course! We didn't use any special property of $l^2$, i.e. we can just replace $^l2$ by a Hilbert space, say $H$, and $Y$ is now just a closed subspace.
$endgroup$
– p4sch
Feb 20 '18 at 20:00
1
$begingroup$
What you meant to say in your last comment is that we can identify the DUAL of $l^p$ with $l^q$.
$endgroup$
– DanielWainfleet
Jan 24 at 18:36
|
show 4 more comments
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The existence of a extension is guaranteed by the Hahn-Banach-Theorem, but in general this extension is not unique (depending on the Banach space). Here, you have a Hilbert space and in this case you don't need the Hahn-Banach-Theorem and we also get uniqueness of the extension.
Note that $Y$ is closed subspace and thus also a Hilbert-Space. The Riesz representation theorem implies that $T(x) = langle x,y rangle$ for a unique $y in Y$ and $|T|=|y|$. This gives the unique extension $overline{T}(x) = langle x, y rangle$.
Let us prove that the extension is unique: Any other extension $L colon H rightarrow mathbb{R}$ has a representation $L(x) = langle x, l rangle$ with $l in H$ and $|l|=|y|$. Decompose $$l = y' + u$$ with $y' in Y$ and $u in Y^perp$. Note that we used the geometry of Hilbert spaces in form of the existence of an orthogonal projection. Then we have for $x in Y$ that $$langle x, y rangle = overline{T}(x) = T(x) = L(x) = langle x, y' rangle.$$
Thus $langle x, y-y' rangle =0$. Taking $x=y-y'$ we get that $y=y'$. Now we have
$$|y|^2 = |l|^2 = |y'|^2 + |u|^2 = |y|^2 +|u|^2$$
and thus $u=0$. This proves $l=y$.
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
$endgroup$
$begingroup$
$W$ is a Hilbert-Space too? So in general by Riesz representation theorem there always exists an extension $overline{T}$ caused by the subspace is a Hilbert-Space?
$endgroup$
– james watt
Feb 15 '18 at 9:28
1
$begingroup$
In the first case, $Y$ is closed and the metric is induced by an inner product. Thus $Y$ is a Hilbert space. In general, the extension is not unique and this question depends on the geometry of the corresponding Banach space $X$. (E.g. if $X'$ is strict convex, then every extension is unique.) In the case of $l^1$ the extension is not unique. If $T(x) = 2017 x_{2017}$ on $W$, then you can take any $(y_k)_{k in mathbb{N}} in l^infty$ with $y_{2017} = 2017$ and $|y_i| leq 2017$ for $i neq k$.
$endgroup$
– p4sch
Feb 15 '18 at 9:52
1
$begingroup$
We have simply $|overline{T}| = |y| = |T|$ again by the Riesz representation theorem. If $L$ is another extension, i.e. the restriction to $Y$ is $T$ and $|L| = |T|$, then there exists an $l in l^2$ with $L(x) = langle x,l rangle$ and $|l| = |T|$. Now, we have $0=overline{T}(u)-L(u) = langle y-l,u rangle$ for any $u in Y$, i.e. $y-l in Y^{perp}$. That implies $|overline{T}|=|L|=|l| =|(l-y)+y|= |y|+|l-y| = |overline{T}| + |overline{T}-L|$. Thus $|overline{T}-L|=0$.
$endgroup$
– p4sch
Feb 15 '18 at 10:29
1
$begingroup$
Yes, of course! We didn't use any special property of $l^2$, i.e. we can just replace $^l2$ by a Hilbert space, say $H$, and $Y$ is now just a closed subspace.
$endgroup$
– p4sch
Feb 20 '18 at 20:00
1
$begingroup$
What you meant to say in your last comment is that we can identify the DUAL of $l^p$ with $l^q$.
$endgroup$
– DanielWainfleet
Jan 24 at 18:36
|
show 4 more comments
$begingroup$
The existence of a extension is guaranteed by the Hahn-Banach-Theorem, but in general this extension is not unique (depending on the Banach space). Here, you have a Hilbert space and in this case you don't need the Hahn-Banach-Theorem and we also get uniqueness of the extension.
Note that $Y$ is closed subspace and thus also a Hilbert-Space. The Riesz representation theorem implies that $T(x) = langle x,y rangle$ for a unique $y in Y$ and $|T|=|y|$. This gives the unique extension $overline{T}(x) = langle x, y rangle$.
Let us prove that the extension is unique: Any other extension $L colon H rightarrow mathbb{R}$ has a representation $L(x) = langle x, l rangle$ with $l in H$ and $|l|=|y|$. Decompose $$l = y' + u$$ with $y' in Y$ and $u in Y^perp$. Note that we used the geometry of Hilbert spaces in form of the existence of an orthogonal projection. Then we have for $x in Y$ that $$langle x, y rangle = overline{T}(x) = T(x) = L(x) = langle x, y' rangle.$$
Thus $langle x, y-y' rangle =0$. Taking $x=y-y'$ we get that $y=y'$. Now we have
$$|y|^2 = |l|^2 = |y'|^2 + |u|^2 = |y|^2 +|u|^2$$
and thus $u=0$. This proves $l=y$.
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
$endgroup$
$begingroup$
$W$ is a Hilbert-Space too? So in general by Riesz representation theorem there always exists an extension $overline{T}$ caused by the subspace is a Hilbert-Space?
$endgroup$
– james watt
Feb 15 '18 at 9:28
1
$begingroup$
In the first case, $Y$ is closed and the metric is induced by an inner product. Thus $Y$ is a Hilbert space. In general, the extension is not unique and this question depends on the geometry of the corresponding Banach space $X$. (E.g. if $X'$ is strict convex, then every extension is unique.) In the case of $l^1$ the extension is not unique. If $T(x) = 2017 x_{2017}$ on $W$, then you can take any $(y_k)_{k in mathbb{N}} in l^infty$ with $y_{2017} = 2017$ and $|y_i| leq 2017$ for $i neq k$.
$endgroup$
– p4sch
Feb 15 '18 at 9:52
1
$begingroup$
We have simply $|overline{T}| = |y| = |T|$ again by the Riesz representation theorem. If $L$ is another extension, i.e. the restriction to $Y$ is $T$ and $|L| = |T|$, then there exists an $l in l^2$ with $L(x) = langle x,l rangle$ and $|l| = |T|$. Now, we have $0=overline{T}(u)-L(u) = langle y-l,u rangle$ for any $u in Y$, i.e. $y-l in Y^{perp}$. That implies $|overline{T}|=|L|=|l| =|(l-y)+y|= |y|+|l-y| = |overline{T}| + |overline{T}-L|$. Thus $|overline{T}-L|=0$.
$endgroup$
– p4sch
Feb 15 '18 at 10:29
1
$begingroup$
Yes, of course! We didn't use any special property of $l^2$, i.e. we can just replace $^l2$ by a Hilbert space, say $H$, and $Y$ is now just a closed subspace.
$endgroup$
– p4sch
Feb 20 '18 at 20:00
1
$begingroup$
What you meant to say in your last comment is that we can identify the DUAL of $l^p$ with $l^q$.
$endgroup$
– DanielWainfleet
Jan 24 at 18:36
|
show 4 more comments
$begingroup$
The existence of a extension is guaranteed by the Hahn-Banach-Theorem, but in general this extension is not unique (depending on the Banach space). Here, you have a Hilbert space and in this case you don't need the Hahn-Banach-Theorem and we also get uniqueness of the extension.
Note that $Y$ is closed subspace and thus also a Hilbert-Space. The Riesz representation theorem implies that $T(x) = langle x,y rangle$ for a unique $y in Y$ and $|T|=|y|$. This gives the unique extension $overline{T}(x) = langle x, y rangle$.
Let us prove that the extension is unique: Any other extension $L colon H rightarrow mathbb{R}$ has a representation $L(x) = langle x, l rangle$ with $l in H$ and $|l|=|y|$. Decompose $$l = y' + u$$ with $y' in Y$ and $u in Y^perp$. Note that we used the geometry of Hilbert spaces in form of the existence of an orthogonal projection. Then we have for $x in Y$ that $$langle x, y rangle = overline{T}(x) = T(x) = L(x) = langle x, y' rangle.$$
Thus $langle x, y-y' rangle =0$. Taking $x=y-y'$ we get that $y=y'$. Now we have
$$|y|^2 = |l|^2 = |y'|^2 + |u|^2 = |y|^2 +|u|^2$$
and thus $u=0$. This proves $l=y$.
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
$endgroup$
The existence of a extension is guaranteed by the Hahn-Banach-Theorem, but in general this extension is not unique (depending on the Banach space). Here, you have a Hilbert space and in this case you don't need the Hahn-Banach-Theorem and we also get uniqueness of the extension.
Note that $Y$ is closed subspace and thus also a Hilbert-Space. The Riesz representation theorem implies that $T(x) = langle x,y rangle$ for a unique $y in Y$ and $|T|=|y|$. This gives the unique extension $overline{T}(x) = langle x, y rangle$.
Let us prove that the extension is unique: Any other extension $L colon H rightarrow mathbb{R}$ has a representation $L(x) = langle x, l rangle$ with $l in H$ and $|l|=|y|$. Decompose $$l = y' + u$$ with $y' in Y$ and $u in Y^perp$. Note that we used the geometry of Hilbert spaces in form of the existence of an orthogonal projection. Then we have for $x in Y$ that $$langle x, y rangle = overline{T}(x) = T(x) = L(x) = langle x, y' rangle.$$
Thus $langle x, y-y' rangle =0$. Taking $x=y-y'$ we get that $y=y'$. Now we have
$$|y|^2 = |l|^2 = |y'|^2 + |u|^2 = |y|^2 +|u|^2$$
and thus $u=0$. This proves $l=y$.
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
edited Jan 24 at 11:16
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
answered Feb 14 '18 at 21:24
p4schp4sch
5,440318
5,440318
$begingroup$
$W$ is a Hilbert-Space too? So in general by Riesz representation theorem there always exists an extension $overline{T}$ caused by the subspace is a Hilbert-Space?
$endgroup$
– james watt
Feb 15 '18 at 9:28
1
$begingroup$
In the first case, $Y$ is closed and the metric is induced by an inner product. Thus $Y$ is a Hilbert space. In general, the extension is not unique and this question depends on the geometry of the corresponding Banach space $X$. (E.g. if $X'$ is strict convex, then every extension is unique.) In the case of $l^1$ the extension is not unique. If $T(x) = 2017 x_{2017}$ on $W$, then you can take any $(y_k)_{k in mathbb{N}} in l^infty$ with $y_{2017} = 2017$ and $|y_i| leq 2017$ for $i neq k$.
$endgroup$
– p4sch
Feb 15 '18 at 9:52
1
$begingroup$
We have simply $|overline{T}| = |y| = |T|$ again by the Riesz representation theorem. If $L$ is another extension, i.e. the restriction to $Y$ is $T$ and $|L| = |T|$, then there exists an $l in l^2$ with $L(x) = langle x,l rangle$ and $|l| = |T|$. Now, we have $0=overline{T}(u)-L(u) = langle y-l,u rangle$ for any $u in Y$, i.e. $y-l in Y^{perp}$. That implies $|overline{T}|=|L|=|l| =|(l-y)+y|= |y|+|l-y| = |overline{T}| + |overline{T}-L|$. Thus $|overline{T}-L|=0$.
$endgroup$
– p4sch
Feb 15 '18 at 10:29
1
$begingroup$
Yes, of course! We didn't use any special property of $l^2$, i.e. we can just replace $^l2$ by a Hilbert space, say $H$, and $Y$ is now just a closed subspace.
$endgroup$
– p4sch
Feb 20 '18 at 20:00
1
$begingroup$
What you meant to say in your last comment is that we can identify the DUAL of $l^p$ with $l^q$.
$endgroup$
– DanielWainfleet
Jan 24 at 18:36
|
show 4 more comments
$begingroup$
$W$ is a Hilbert-Space too? So in general by Riesz representation theorem there always exists an extension $overline{T}$ caused by the subspace is a Hilbert-Space?
$endgroup$
– james watt
Feb 15 '18 at 9:28
1
$begingroup$
In the first case, $Y$ is closed and the metric is induced by an inner product. Thus $Y$ is a Hilbert space. In general, the extension is not unique and this question depends on the geometry of the corresponding Banach space $X$. (E.g. if $X'$ is strict convex, then every extension is unique.) In the case of $l^1$ the extension is not unique. If $T(x) = 2017 x_{2017}$ on $W$, then you can take any $(y_k)_{k in mathbb{N}} in l^infty$ with $y_{2017} = 2017$ and $|y_i| leq 2017$ for $i neq k$.
$endgroup$
– p4sch
Feb 15 '18 at 9:52
1
$begingroup$
We have simply $|overline{T}| = |y| = |T|$ again by the Riesz representation theorem. If $L$ is another extension, i.e. the restriction to $Y$ is $T$ and $|L| = |T|$, then there exists an $l in l^2$ with $L(x) = langle x,l rangle$ and $|l| = |T|$. Now, we have $0=overline{T}(u)-L(u) = langle y-l,u rangle$ for any $u in Y$, i.e. $y-l in Y^{perp}$. That implies $|overline{T}|=|L|=|l| =|(l-y)+y|= |y|+|l-y| = |overline{T}| + |overline{T}-L|$. Thus $|overline{T}-L|=0$.
$endgroup$
– p4sch
Feb 15 '18 at 10:29
1
$begingroup$
Yes, of course! We didn't use any special property of $l^2$, i.e. we can just replace $^l2$ by a Hilbert space, say $H$, and $Y$ is now just a closed subspace.
$endgroup$
– p4sch
Feb 20 '18 at 20:00
1
$begingroup$
What you meant to say in your last comment is that we can identify the DUAL of $l^p$ with $l^q$.
$endgroup$
– DanielWainfleet
Jan 24 at 18:36
$begingroup$
$W$ is a Hilbert-Space too? So in general by Riesz representation theorem there always exists an extension $overline{T}$ caused by the subspace is a Hilbert-Space?
$endgroup$
– james watt
Feb 15 '18 at 9:28
$begingroup$
$W$ is a Hilbert-Space too? So in general by Riesz representation theorem there always exists an extension $overline{T}$ caused by the subspace is a Hilbert-Space?
$endgroup$
– james watt
Feb 15 '18 at 9:28
1
1
$begingroup$
In the first case, $Y$ is closed and the metric is induced by an inner product. Thus $Y$ is a Hilbert space. In general, the extension is not unique and this question depends on the geometry of the corresponding Banach space $X$. (E.g. if $X'$ is strict convex, then every extension is unique.) In the case of $l^1$ the extension is not unique. If $T(x) = 2017 x_{2017}$ on $W$, then you can take any $(y_k)_{k in mathbb{N}} in l^infty$ with $y_{2017} = 2017$ and $|y_i| leq 2017$ for $i neq k$.
$endgroup$
– p4sch
Feb 15 '18 at 9:52
$begingroup$
In the first case, $Y$ is closed and the metric is induced by an inner product. Thus $Y$ is a Hilbert space. In general, the extension is not unique and this question depends on the geometry of the corresponding Banach space $X$. (E.g. if $X'$ is strict convex, then every extension is unique.) In the case of $l^1$ the extension is not unique. If $T(x) = 2017 x_{2017}$ on $W$, then you can take any $(y_k)_{k in mathbb{N}} in l^infty$ with $y_{2017} = 2017$ and $|y_i| leq 2017$ for $i neq k$.
$endgroup$
– p4sch
Feb 15 '18 at 9:52
1
1
$begingroup$
We have simply $|overline{T}| = |y| = |T|$ again by the Riesz representation theorem. If $L$ is another extension, i.e. the restriction to $Y$ is $T$ and $|L| = |T|$, then there exists an $l in l^2$ with $L(x) = langle x,l rangle$ and $|l| = |T|$. Now, we have $0=overline{T}(u)-L(u) = langle y-l,u rangle$ for any $u in Y$, i.e. $y-l in Y^{perp}$. That implies $|overline{T}|=|L|=|l| =|(l-y)+y|= |y|+|l-y| = |overline{T}| + |overline{T}-L|$. Thus $|overline{T}-L|=0$.
$endgroup$
– p4sch
Feb 15 '18 at 10:29
$begingroup$
We have simply $|overline{T}| = |y| = |T|$ again by the Riesz representation theorem. If $L$ is another extension, i.e. the restriction to $Y$ is $T$ and $|L| = |T|$, then there exists an $l in l^2$ with $L(x) = langle x,l rangle$ and $|l| = |T|$. Now, we have $0=overline{T}(u)-L(u) = langle y-l,u rangle$ for any $u in Y$, i.e. $y-l in Y^{perp}$. That implies $|overline{T}|=|L|=|l| =|(l-y)+y|= |y|+|l-y| = |overline{T}| + |overline{T}-L|$. Thus $|overline{T}-L|=0$.
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– p4sch
Feb 15 '18 at 10:29
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1
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Yes, of course! We didn't use any special property of $l^2$, i.e. we can just replace $^l2$ by a Hilbert space, say $H$, and $Y$ is now just a closed subspace.
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– p4sch
Feb 20 '18 at 20:00
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Yes, of course! We didn't use any special property of $l^2$, i.e. we can just replace $^l2$ by a Hilbert space, say $H$, and $Y$ is now just a closed subspace.
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– p4sch
Feb 20 '18 at 20:00
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What you meant to say in your last comment is that we can identify the DUAL of $l^p$ with $l^q$.
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– DanielWainfleet
Jan 24 at 18:36
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What you meant to say in your last comment is that we can identify the DUAL of $l^p$ with $l^q$.
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– DanielWainfleet
Jan 24 at 18:36
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