Mixing
Embedding into dual space
Clash Royale CLAN TAG#URR8PPP
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Let $X$ be a Banach space and $X^*$ its dual
Does anyone have an example where $X$ does not embed isomorphically/isometrically into $X^*$.
Does $C([0,1])$ embed isomorphically/isometrically into its dual?
functional-analysis
asked Aug 5 at 23:15
Tom Chalmer
271212
add a comment |Â
up vote
0
down vote
favorite
Let $X$ be a Banach space and $X^*$ its dual
Does anyone have an example where $X$ does not embed isomorphically/isometrically into $X^*$.
Does $C([0,1])$ embed isomorphically/isometrically into its dual?
functional-analysis
asked Aug 5 at 23:15
Tom Chalmer
271212
You forgot to mention that $X$ is infinite dimensional.
– Kavi Rama Murthy
Aug 5 at 23:28
Yes, $X$ is infinite dimensional
– Tom Chalmer
Aug 6 at 0:12
1
Isn't $L^p$ a counterexample for suitable $p$?
– David C. Ullrich
Aug 6 at 3:31
I think you have to make clear what you mean by "embed isomorphically". To me, an isomorphism is bounded, injective and surjective. Is this what you mean? Why do you use "embed" then?
– amsmath
Aug 9 at 11:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a Banach space and $X^*$ its dual
Does anyone have an example where $X$ does not embed isomorphically/isometrically into $X^*$.
Does $C([0,1])$ embed isomorphically/isometrically into its dual?
functional-analysis
asked Aug 5 at 23:15
Tom Chalmer
271212
Let $X$ be a Banach space and $X^*$ its dual
Does anyone have an example where $X$ does not embed isomorphically/isometrically into $X^*$.
Does $C([0,1])$ embed isomorphically/isometrically into its dual?
functional-analysis
asked Aug 5 at 23:15
Tom Chalmer
271212
asked Aug 5 at 23:15
Tom Chalmer
271212
asked Aug 5 at 23:15
Tom Chalmer
271212
asked Aug 5 at 23:15
Tom Chalmer
271212
271212
You forgot to mention that $X$ is infinite dimensional.
– Kavi Rama Murthy
Aug 5 at 23:28
Yes, $X$ is infinite dimensional
– Tom Chalmer
Aug 6 at 0:12
1
Isn't $L^p$ a counterexample for suitable $p$?
– David C. Ullrich
Aug 6 at 3:31
I think you have to make clear what you mean by "embed isomorphically". To me, an isomorphism is bounded, injective and surjective. Is this what you mean? Why do you use "embed" then?
– amsmath
Aug 9 at 11:48
add a comment |Â
You forgot to mention that $X$ is infinite dimensional.
– Kavi Rama Murthy
Aug 5 at 23:28
Yes, $X$ is infinite dimensional
– Tom Chalmer
Aug 6 at 0:12
1
Isn't $L^p$ a counterexample for suitable $p$?
– David C. Ullrich
Aug 6 at 3:31
I think you have to make clear what you mean by "embed isomorphically". To me, an isomorphism is bounded, injective and surjective. Is this what you mean? Why do you use "embed" then?
– amsmath
Aug 9 at 11:48
You forgot to mention that $X$ is infinite dimensional.
– Kavi Rama Murthy
Aug 5 at 23:28
You forgot to mention that $X$ is infinite dimensional.
– Kavi Rama Murthy
Aug 5 at 23:28
Yes, $X$ is infinite dimensional
– Tom Chalmer
Aug 6 at 0:12
Yes, $X$ is infinite dimensional
– Tom Chalmer
Aug 6 at 0:12
1
1
Isn't $L^p$ a counterexample for suitable $p$?
– David C. Ullrich
Aug 6 at 3:31
Isn't $L^p$ a counterexample for suitable $p$?
– David C. Ullrich
Aug 6 at 3:31
I think you have to make clear what you mean by "embed isomorphically". To me, an isomorphism is bounded, injective and surjective. Is this what you mean? Why do you use "embed" then?
– amsmath
Aug 9 at 11:48
I think you have to make clear what you mean by "embed isomorphically". To me, an isomorphism is bounded, injective and surjective. Is this what you mean? Why do you use "embed" then?
– amsmath
Aug 9 at 11:48
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
If $X=l^p$ with $1<p<infty $ then $X^*$ does not contain an isomorphic copy of $C[0,1]$. To prove this we need two well known theorems from FA:
Theorem 1
(Pitt) and bounded linear map from $l^r$ to $l^s$ is compact if $1<s<r<infty$.
Theorem 2
$C[0,1]$ is universal for separable Banach spaces (in the sense any separable Banach space can be embedded isometrically in $C[0,1]$). Ref.: Geometric FA by Holmes.
Now let $q>p^*$ where $p^*$ is the conjugate of $p$. Suppose there is an isomorphism $T: C[0,1] to X^*=l^p^*$. Let $S:l^q to C[0,1]$ be an isometric isomorphism (into). Then $Tcirc S$ is compact by Pitt's Theorem. But this map is also an isomorphism. It is easy to see that there cannot be an isomorphism on an infinite dimensional space which is compact.
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
For the last part note that $|T(S(x))| geq C |x|$ for some $c$. If $x_n$ is any sequence in the unit ball of $l^q$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $x_n$ is Cauchy. This makes the unit ball of $l^q$ compact, a contradictioin.
– Kavi Rama Murthy
Aug 6 at 4:57
Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$
– Tom Chalmer
Aug 6 at 6:18
$X$ does not embed isomorphically in $X^*$ if $X=l^p$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual.
– Kavi Rama Murthy
Aug 6 at 7:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $X=l^p$ with $1<p<infty $ then $X^*$ does not contain an isomorphic copy of $C[0,1]$. To prove this we need two well known theorems from FA:
Theorem 1
(Pitt) and bounded linear map from $l^r$ to $l^s$ is compact if $1<s<r<infty$.
Theorem 2
$C[0,1]$ is universal for separable Banach spaces (in the sense any separable Banach space can be embedded isometrically in $C[0,1]$). Ref.: Geometric FA by Holmes.
Now let $q>p^*$ where $p^*$ is the conjugate of $p$. Suppose there is an isomorphism $T: C[0,1] to X^*=l^p^*$. Let $S:l^q to C[0,1]$ be an isometric isomorphism (into). Then $Tcirc S$ is compact by Pitt's Theorem. But this map is also an isomorphism. It is easy to see that there cannot be an isomorphism on an infinite dimensional space which is compact.
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
For the last part note that $|T(S(x))| geq C |x|$ for some $c$. If $x_n$ is any sequence in the unit ball of $l^q$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $x_n$ is Cauchy. This makes the unit ball of $l^q$ compact, a contradictioin.
– Kavi Rama Murthy
Aug 6 at 4:57
Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$
– Tom Chalmer
Aug 6 at 6:18
$X$ does not embed isomorphically in $X^*$ if $X=l^p$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual.
– Kavi Rama Murthy
Aug 6 at 7:31
add a comment |Â
up vote
2
down vote
If $X=l^p$ with $1<p<infty $ then $X^*$ does not contain an isomorphic copy of $C[0,1]$. To prove this we need two well known theorems from FA:
Theorem 1
(Pitt) and bounded linear map from $l^r$ to $l^s$ is compact if $1<s<r<infty$.
Theorem 2
$C[0,1]$ is universal for separable Banach spaces (in the sense any separable Banach space can be embedded isometrically in $C[0,1]$). Ref.: Geometric FA by Holmes.
Now let $q>p^*$ where $p^*$ is the conjugate of $p$. Suppose there is an isomorphism $T: C[0,1] to X^*=l^p^*$. Let $S:l^q to C[0,1]$ be an isometric isomorphism (into). Then $Tcirc S$ is compact by Pitt's Theorem. But this map is also an isomorphism. It is easy to see that there cannot be an isomorphism on an infinite dimensional space which is compact.
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
For the last part note that $|T(S(x))| geq C |x|$ for some $c$. If $x_n$ is any sequence in the unit ball of $l^q$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $x_n$ is Cauchy. This makes the unit ball of $l^q$ compact, a contradictioin.
– Kavi Rama Murthy
Aug 6 at 4:57
Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$
– Tom Chalmer
Aug 6 at 6:18
$X$ does not embed isomorphically in $X^*$ if $X=l^p$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual.
– Kavi Rama Murthy
Aug 6 at 7:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $X=l^p$ with $1<p<infty $ then $X^*$ does not contain an isomorphic copy of $C[0,1]$. To prove this we need two well known theorems from FA:
Theorem 1
(Pitt) and bounded linear map from $l^r$ to $l^s$ is compact if $1<s<r<infty$.
Theorem 2
$C[0,1]$ is universal for separable Banach spaces (in the sense any separable Banach space can be embedded isometrically in $C[0,1]$). Ref.: Geometric FA by Holmes.
Now let $q>p^*$ where $p^*$ is the conjugate of $p$. Suppose there is an isomorphism $T: C[0,1] to X^*=l^p^*$. Let $S:l^q to C[0,1]$ be an isometric isomorphism (into). Then $Tcirc S$ is compact by Pitt's Theorem. But this map is also an isomorphism. It is easy to see that there cannot be an isomorphism on an infinite dimensional space which is compact.
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
If $X=l^p$ with $1<p<infty $ then $X^*$ does not contain an isomorphic copy of $C[0,1]$. To prove this we need two well known theorems from FA:
Theorem 1
(Pitt) and bounded linear map from $l^r$ to $l^s$ is compact if $1<s<r<infty$.
Theorem 2
$C[0,1]$ is universal for separable Banach spaces (in the sense any separable Banach space can be embedded isometrically in $C[0,1]$). Ref.: Geometric FA by Holmes.
Now let $q>p^*$ where $p^*$ is the conjugate of $p$. Suppose there is an isomorphism $T: C[0,1] to X^*=l^p^*$. Let $S:l^q to C[0,1]$ be an isometric isomorphism (into). Then $Tcirc S$ is compact by Pitt's Theorem. But this map is also an isomorphism. It is easy to see that there cannot be an isomorphism on an infinite dimensional space which is compact.
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
edited Aug 6 at 4:59
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
answered Aug 6 at 4:51
Kavi Rama Murthy
21k2830
21k2830
For the last part note that $|T(S(x))| geq C |x|$ for some $c$. If $x_n$ is any sequence in the unit ball of $l^q$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $x_n$ is Cauchy. This makes the unit ball of $l^q$ compact, a contradictioin.
– Kavi Rama Murthy
Aug 6 at 4:57
Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$
– Tom Chalmer
Aug 6 at 6:18
$X$ does not embed isomorphically in $X^*$ if $X=l^p$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual.
– Kavi Rama Murthy
Aug 6 at 7:31
add a comment |Â
For the last part note that $|T(S(x))| geq C |x|$ for some $c$. If $x_n$ is any sequence in the unit ball of $l^q$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $x_n$ is Cauchy. This makes the unit ball of $l^q$ compact, a contradictioin.
– Kavi Rama Murthy
Aug 6 at 4:57
Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$
– Tom Chalmer
Aug 6 at 6:18
$X$ does not embed isomorphically in $X^*$ if $X=l^p$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual.
– Kavi Rama Murthy
Aug 6 at 7:31
For the last part note that $|T(S(x))| geq C |x|$ for some $c$. If $x_n$ is any sequence in the unit ball of $l^q$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $x_n$ is Cauchy. This makes the unit ball of $l^q$ compact, a contradictioin.
– Kavi Rama Murthy
Aug 6 at 4:57
For the last part note that $|T(S(x))| geq C |x|$ for some $c$. If $x_n$ is any sequence in the unit ball of $l^q$ then $T(S(x_n))$ has a convergent subsequence and the inequality implies that the corresponding subsequence of $x_n$ is Cauchy. This makes the unit ball of $l^q$ compact, a contradictioin.
– Kavi Rama Murthy
Aug 6 at 4:57
Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$
– Tom Chalmer
Aug 6 at 6:18
Hi Kavi. Thanks for your answer. I think my question was a little ambiguous, or I may not have understood your answer. It seems you showed there is a dual that does not contain an isomorphic copy of $C([0,1])$. I'm wondering if $C([0,1])$ embeds into its dual $C([0,1])^*$
– Tom Chalmer
Aug 6 at 6:18
$X$ does not embed isomorphically in $X^*$ if $X=l^p$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual.
– Kavi Rama Murthy
Aug 6 at 7:31
$X$ does not embed isomorphically in $X^*$ if $X=l^p$ with $p>2$ (by the Pitt's Theorem argument). I do not know if $C[0,1]$ embeds in its dual.
– Kavi Rama Murthy
Aug 6 at 7:31
add a comment |Â
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You forgot to mention that $X$ is infinite dimensional.
– Kavi Rama Murthy
Aug 5 at 23:28
Yes, $X$ is infinite dimensional
– Tom Chalmer
Aug 6 at 0:12
1
Isn't $L^p$ a counterexample for suitable $p$?
– David C. Ullrich
Aug 6 at 3:31
I think you have to make clear what you mean by "embed isomorphically". To me, an isomorphism is bounded, injective and surjective. Is this what you mean? Why do you use "embed" then?
– amsmath
Aug 9 at 11:48