Mixing
a reference for the isomorphism $ell_1(E')cong [c_0(E)]'$
$begingroup$
Let $E$ be a Banach space, and let $E'$ denote its topological dual.
Let us consider the spaces $ell_1(E')$ and $c_0(E)$ defined by
$ell_1(E')={(x_n^{'})_{n=1}^inftysubset E': sum_{n=1}^infty||x_n^{'}||<infty}$, and
$c_0(E)={(x_n)_{n=1}^inftysubset E: x_nlongrightarrow 0 {rm{in}} E}$.
Can anybody give a reference containing an elementary proof of the topological isomorphism $ell_1(E')cong [c_0(E)]'$ ?
Thanks in advance.
functional-analysis
asked Jan 9 at 21:15
serenusserenus
22416
$endgroup$
add a comment |
$begingroup$
Let $E$ be a Banach space, and let $E'$ denote its topological dual.
Let us consider the spaces $ell_1(E')$ and $c_0(E)$ defined by
$ell_1(E')={(x_n^{'})_{n=1}^inftysubset E': sum_{n=1}^infty||x_n^{'}||<infty}$, and
$c_0(E)={(x_n)_{n=1}^inftysubset E: x_nlongrightarrow 0 {rm{in}} E}$.
Can anybody give a reference containing an elementary proof of the topological isomorphism $ell_1(E')cong [c_0(E)]'$ ?
Thanks in advance.
functional-analysis
asked Jan 9 at 21:15
serenusserenus
22416
$endgroup$
add a comment |
$begingroup$
Let $E$ be a Banach space, and let $E'$ denote its topological dual.
Let us consider the spaces $ell_1(E')$ and $c_0(E)$ defined by
$ell_1(E')={(x_n^{'})_{n=1}^inftysubset E': sum_{n=1}^infty||x_n^{'}||<infty}$, and
$c_0(E)={(x_n)_{n=1}^inftysubset E: x_nlongrightarrow 0 {rm{in}} E}$.
Can anybody give a reference containing an elementary proof of the topological isomorphism $ell_1(E')cong [c_0(E)]'$ ?
Thanks in advance.
functional-analysis
asked Jan 9 at 21:15
serenusserenus
22416
$endgroup$
Let $E$ be a Banach space, and let $E'$ denote its topological dual.
Let us consider the spaces $ell_1(E')$ and $c_0(E)$ defined by
$ell_1(E')={(x_n^{'})_{n=1}^inftysubset E': sum_{n=1}^infty||x_n^{'}||<infty}$, and
$c_0(E)={(x_n)_{n=1}^inftysubset E: x_nlongrightarrow 0 {rm{in}} E}$.
Can anybody give a reference containing an elementary proof of the topological isomorphism $ell_1(E')cong [c_0(E)]'$ ?
Thanks in advance.
functional-analysis
functional-analysis
asked Jan 9 at 21:15
serenusserenus
22416
asked Jan 9 at 21:15
serenusserenus
22416
asked Jan 9 at 21:15
serenusserenus
22416
asked Jan 9 at 21:15
serenusserenus
22416
asked Jan 9 at 21:15
serenusserenus
22416
22416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With the same idea as when $E=mathbb C$, define $gamma:ell_1(E')to c_0(E)'$ by
$$tag1
gamma(x')x=sum_n x'_n(x_n), xin c_0(E).
$$
The conditions on $ell_1(E')$ and $c_0(E)$ guarantee that $gamma $ is well-defined (that is, the series makes sense). Moreover, $gamma $ is isometric; indeed, from $(1) $ we get $|gamma (x')|leqsum_n|x'_n|=|x'|$. Given $varepsilon>0$ and $minmathbb N $, for each $nleq m $ choose $x_nin E $ with $|x_n|=1$ and $x'_n (x_n)>|x'_n|-varepsilon/2^n $, and put $x_n=0$ for $n>m$; then $x={x_n} in c_0 (E) $ and $$gamma(x')x=sum_nx'_n (x_n)=sum_{n=1}^mx'_n (x_n)>-varepsilon+sum_{n=1}^m|x'_n|. $$ As we can do this for all $varepsilon $ and $m $, we get $|gamma (x')|geqsum_n|x'_n|=|x'|$, and thus $gamma$ is isometric.
It is trivial that $gamma$ is linear. If $gamma(x)=0$, evaluating on sequences that have a single nonzero element we get that $x'_n=0$ for all $n$, so $x'=0$ and $gamma$ is injective. Now let $phiin c_0(E)'$. Any element in $c_0(E)$ is of the form $x=sum_n x_n,e_n$, where $x_nin E$ and $e_nin c_0(mathbb N)$ is the sequence with the $n^{rm th}$ entry equal to 1 and zeroes elsewhere. From $x_nto0$, we get that the series for $x$ converges (since the norm is the supremum norm). As $phi$ is continuous, $$phi(x)=sum_n phi(x_ne_n)=sum_nphi_n(x_n),$$ where $phi_nin E'$ is defined by $phi_n(x)=phi(xe_n)$ (by $xe_n$ we mean the sequence with $x$ in the $n^{rm th}$ position and zeroes elsewhere). Thus $phi=gamma({phi_n})$, if we show that ${phi_n}inell_1(E')$.
Fix $varepsilon>0$, $minmathbb N$. For each $nleq m$, choose $x_nin E$ with $|x_n|=1$, $phi_n(x_n)>|phi_n|-varepsilon/2^n$, and $x_n=0$ if $n>m$. Then $x={x_n}in c_0(E)$, and
$$
|phi|geq|phi(x)|=sum_nphi_n(x_n)=sum_{n=1}^mphi_n(x_n)geq-varepsilon+sum_{n=1}^m |phi_n|.
$$
As we can do this for all $varepsilon>0$ and all $minmathbb N$, we get that $sum_n|phi_n|leq|phi|$ and so ${phi_n}inell_1(E')$. So $gamma$ is surjective.
By the Inverse Mapping Theorem, $gamma$ is an isomorphism.
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
Thank for the very nice proof. Just one more question. It is easy to see that $||gamma(x')||leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also?
$endgroup$
– serenus
Jan 10 at 7:53
$begingroup$
Done. You'll notice that the argument is familiar.
$endgroup$
– Martin Argerami
Jan 10 at 10:06
$begingroup$
In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $gamma (x')x$, isn't it?
$endgroup$
– serenus
Jan 10 at 10:33
$begingroup$
Yes. $ $
$endgroup$
– Martin Argerami
Jan 10 at 15:35
$begingroup$
Thanks for the proof.
$endgroup$
– serenus
Jan 10 at 15:43
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067955%2fa-reference-for-the-isomorphism-ell-1e-cong-c-0e%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With the same idea as when $E=mathbb C$, define $gamma:ell_1(E')to c_0(E)'$ by
$$tag1
gamma(x')x=sum_n x'_n(x_n), xin c_0(E).
$$
The conditions on $ell_1(E')$ and $c_0(E)$ guarantee that $gamma $ is well-defined (that is, the series makes sense). Moreover, $gamma $ is isometric; indeed, from $(1) $ we get $|gamma (x')|leqsum_n|x'_n|=|x'|$. Given $varepsilon>0$ and $minmathbb N $, for each $nleq m $ choose $x_nin E $ with $|x_n|=1$ and $x'_n (x_n)>|x'_n|-varepsilon/2^n $, and put $x_n=0$ for $n>m$; then $x={x_n} in c_0 (E) $ and $$gamma(x')x=sum_nx'_n (x_n)=sum_{n=1}^mx'_n (x_n)>-varepsilon+sum_{n=1}^m|x'_n|. $$ As we can do this for all $varepsilon $ and $m $, we get $|gamma (x')|geqsum_n|x'_n|=|x'|$, and thus $gamma$ is isometric.
It is trivial that $gamma$ is linear. If $gamma(x)=0$, evaluating on sequences that have a single nonzero element we get that $x'_n=0$ for all $n$, so $x'=0$ and $gamma$ is injective. Now let $phiin c_0(E)'$. Any element in $c_0(E)$ is of the form $x=sum_n x_n,e_n$, where $x_nin E$ and $e_nin c_0(mathbb N)$ is the sequence with the $n^{rm th}$ entry equal to 1 and zeroes elsewhere. From $x_nto0$, we get that the series for $x$ converges (since the norm is the supremum norm). As $phi$ is continuous, $$phi(x)=sum_n phi(x_ne_n)=sum_nphi_n(x_n),$$ where $phi_nin E'$ is defined by $phi_n(x)=phi(xe_n)$ (by $xe_n$ we mean the sequence with $x$ in the $n^{rm th}$ position and zeroes elsewhere). Thus $phi=gamma({phi_n})$, if we show that ${phi_n}inell_1(E')$.
Fix $varepsilon>0$, $minmathbb N$. For each $nleq m$, choose $x_nin E$ with $|x_n|=1$, $phi_n(x_n)>|phi_n|-varepsilon/2^n$, and $x_n=0$ if $n>m$. Then $x={x_n}in c_0(E)$, and
$$
|phi|geq|phi(x)|=sum_nphi_n(x_n)=sum_{n=1}^mphi_n(x_n)geq-varepsilon+sum_{n=1}^m |phi_n|.
$$
As we can do this for all $varepsilon>0$ and all $minmathbb N$, we get that $sum_n|phi_n|leq|phi|$ and so ${phi_n}inell_1(E')$. So $gamma$ is surjective.
By the Inverse Mapping Theorem, $gamma$ is an isomorphism.
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
Thank for the very nice proof. Just one more question. It is easy to see that $||gamma(x')||leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also?
$endgroup$
– serenus
Jan 10 at 7:53
$begingroup$
Done. You'll notice that the argument is familiar.
$endgroup$
– Martin Argerami
Jan 10 at 10:06
$begingroup$
In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $gamma (x')x$, isn't it?
$endgroup$
– serenus
Jan 10 at 10:33
$begingroup$
Yes. $ $
$endgroup$
– Martin Argerami
Jan 10 at 15:35
$begingroup$
Thanks for the proof.
$endgroup$
– serenus
Jan 10 at 15:43
add a comment |
$begingroup$
With the same idea as when $E=mathbb C$, define $gamma:ell_1(E')to c_0(E)'$ by
$$tag1
gamma(x')x=sum_n x'_n(x_n), xin c_0(E).
$$
The conditions on $ell_1(E')$ and $c_0(E)$ guarantee that $gamma $ is well-defined (that is, the series makes sense). Moreover, $gamma $ is isometric; indeed, from $(1) $ we get $|gamma (x')|leqsum_n|x'_n|=|x'|$. Given $varepsilon>0$ and $minmathbb N $, for each $nleq m $ choose $x_nin E $ with $|x_n|=1$ and $x'_n (x_n)>|x'_n|-varepsilon/2^n $, and put $x_n=0$ for $n>m$; then $x={x_n} in c_0 (E) $ and $$gamma(x')x=sum_nx'_n (x_n)=sum_{n=1}^mx'_n (x_n)>-varepsilon+sum_{n=1}^m|x'_n|. $$ As we can do this for all $varepsilon $ and $m $, we get $|gamma (x')|geqsum_n|x'_n|=|x'|$, and thus $gamma$ is isometric.
It is trivial that $gamma$ is linear. If $gamma(x)=0$, evaluating on sequences that have a single nonzero element we get that $x'_n=0$ for all $n$, so $x'=0$ and $gamma$ is injective. Now let $phiin c_0(E)'$. Any element in $c_0(E)$ is of the form $x=sum_n x_n,e_n$, where $x_nin E$ and $e_nin c_0(mathbb N)$ is the sequence with the $n^{rm th}$ entry equal to 1 and zeroes elsewhere. From $x_nto0$, we get that the series for $x$ converges (since the norm is the supremum norm). As $phi$ is continuous, $$phi(x)=sum_n phi(x_ne_n)=sum_nphi_n(x_n),$$ where $phi_nin E'$ is defined by $phi_n(x)=phi(xe_n)$ (by $xe_n$ we mean the sequence with $x$ in the $n^{rm th}$ position and zeroes elsewhere). Thus $phi=gamma({phi_n})$, if we show that ${phi_n}inell_1(E')$.
Fix $varepsilon>0$, $minmathbb N$. For each $nleq m$, choose $x_nin E$ with $|x_n|=1$, $phi_n(x_n)>|phi_n|-varepsilon/2^n$, and $x_n=0$ if $n>m$. Then $x={x_n}in c_0(E)$, and
$$
|phi|geq|phi(x)|=sum_nphi_n(x_n)=sum_{n=1}^mphi_n(x_n)geq-varepsilon+sum_{n=1}^m |phi_n|.
$$
As we can do this for all $varepsilon>0$ and all $minmathbb N$, we get that $sum_n|phi_n|leq|phi|$ and so ${phi_n}inell_1(E')$. So $gamma$ is surjective.
By the Inverse Mapping Theorem, $gamma$ is an isomorphism.
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
Thank for the very nice proof. Just one more question. It is easy to see that $||gamma(x')||leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also?
$endgroup$
– serenus
Jan 10 at 7:53
$begingroup$
Done. You'll notice that the argument is familiar.
$endgroup$
– Martin Argerami
Jan 10 at 10:06
$begingroup$
In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $gamma (x')x$, isn't it?
$endgroup$
– serenus
Jan 10 at 10:33
$begingroup$
Yes. $ $
$endgroup$
– Martin Argerami
Jan 10 at 15:35
$begingroup$
Thanks for the proof.
$endgroup$
– serenus
Jan 10 at 15:43
add a comment |
$begingroup$
With the same idea as when $E=mathbb C$, define $gamma:ell_1(E')to c_0(E)'$ by
$$tag1
gamma(x')x=sum_n x'_n(x_n), xin c_0(E).
$$
The conditions on $ell_1(E')$ and $c_0(E)$ guarantee that $gamma $ is well-defined (that is, the series makes sense). Moreover, $gamma $ is isometric; indeed, from $(1) $ we get $|gamma (x')|leqsum_n|x'_n|=|x'|$. Given $varepsilon>0$ and $minmathbb N $, for each $nleq m $ choose $x_nin E $ with $|x_n|=1$ and $x'_n (x_n)>|x'_n|-varepsilon/2^n $, and put $x_n=0$ for $n>m$; then $x={x_n} in c_0 (E) $ and $$gamma(x')x=sum_nx'_n (x_n)=sum_{n=1}^mx'_n (x_n)>-varepsilon+sum_{n=1}^m|x'_n|. $$ As we can do this for all $varepsilon $ and $m $, we get $|gamma (x')|geqsum_n|x'_n|=|x'|$, and thus $gamma$ is isometric.
It is trivial that $gamma$ is linear. If $gamma(x)=0$, evaluating on sequences that have a single nonzero element we get that $x'_n=0$ for all $n$, so $x'=0$ and $gamma$ is injective. Now let $phiin c_0(E)'$. Any element in $c_0(E)$ is of the form $x=sum_n x_n,e_n$, where $x_nin E$ and $e_nin c_0(mathbb N)$ is the sequence with the $n^{rm th}$ entry equal to 1 and zeroes elsewhere. From $x_nto0$, we get that the series for $x$ converges (since the norm is the supremum norm). As $phi$ is continuous, $$phi(x)=sum_n phi(x_ne_n)=sum_nphi_n(x_n),$$ where $phi_nin E'$ is defined by $phi_n(x)=phi(xe_n)$ (by $xe_n$ we mean the sequence with $x$ in the $n^{rm th}$ position and zeroes elsewhere). Thus $phi=gamma({phi_n})$, if we show that ${phi_n}inell_1(E')$.
Fix $varepsilon>0$, $minmathbb N$. For each $nleq m$, choose $x_nin E$ with $|x_n|=1$, $phi_n(x_n)>|phi_n|-varepsilon/2^n$, and $x_n=0$ if $n>m$. Then $x={x_n}in c_0(E)$, and
$$
|phi|geq|phi(x)|=sum_nphi_n(x_n)=sum_{n=1}^mphi_n(x_n)geq-varepsilon+sum_{n=1}^m |phi_n|.
$$
As we can do this for all $varepsilon>0$ and all $minmathbb N$, we get that $sum_n|phi_n|leq|phi|$ and so ${phi_n}inell_1(E')$. So $gamma$ is surjective.
By the Inverse Mapping Theorem, $gamma$ is an isomorphism.
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
With the same idea as when $E=mathbb C$, define $gamma:ell_1(E')to c_0(E)'$ by
$$tag1
gamma(x')x=sum_n x'_n(x_n), xin c_0(E).
$$
The conditions on $ell_1(E')$ and $c_0(E)$ guarantee that $gamma $ is well-defined (that is, the series makes sense). Moreover, $gamma $ is isometric; indeed, from $(1) $ we get $|gamma (x')|leqsum_n|x'_n|=|x'|$. Given $varepsilon>0$ and $minmathbb N $, for each $nleq m $ choose $x_nin E $ with $|x_n|=1$ and $x'_n (x_n)>|x'_n|-varepsilon/2^n $, and put $x_n=0$ for $n>m$; then $x={x_n} in c_0 (E) $ and $$gamma(x')x=sum_nx'_n (x_n)=sum_{n=1}^mx'_n (x_n)>-varepsilon+sum_{n=1}^m|x'_n|. $$ As we can do this for all $varepsilon $ and $m $, we get $|gamma (x')|geqsum_n|x'_n|=|x'|$, and thus $gamma$ is isometric.
It is trivial that $gamma$ is linear. If $gamma(x)=0$, evaluating on sequences that have a single nonzero element we get that $x'_n=0$ for all $n$, so $x'=0$ and $gamma$ is injective. Now let $phiin c_0(E)'$. Any element in $c_0(E)$ is of the form $x=sum_n x_n,e_n$, where $x_nin E$ and $e_nin c_0(mathbb N)$ is the sequence with the $n^{rm th}$ entry equal to 1 and zeroes elsewhere. From $x_nto0$, we get that the series for $x$ converges (since the norm is the supremum norm). As $phi$ is continuous, $$phi(x)=sum_n phi(x_ne_n)=sum_nphi_n(x_n),$$ where $phi_nin E'$ is defined by $phi_n(x)=phi(xe_n)$ (by $xe_n$ we mean the sequence with $x$ in the $n^{rm th}$ position and zeroes elsewhere). Thus $phi=gamma({phi_n})$, if we show that ${phi_n}inell_1(E')$.
Fix $varepsilon>0$, $minmathbb N$. For each $nleq m$, choose $x_nin E$ with $|x_n|=1$, $phi_n(x_n)>|phi_n|-varepsilon/2^n$, and $x_n=0$ if $n>m$. Then $x={x_n}in c_0(E)$, and
$$
|phi|geq|phi(x)|=sum_nphi_n(x_n)=sum_{n=1}^mphi_n(x_n)geq-varepsilon+sum_{n=1}^m |phi_n|.
$$
As we can do this for all $varepsilon>0$ and all $minmathbb N$, we get that $sum_n|phi_n|leq|phi|$ and so ${phi_n}inell_1(E')$. So $gamma$ is surjective.
By the Inverse Mapping Theorem, $gamma$ is an isomorphism.
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
edited Jan 10 at 15:35
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 3:50
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
$begingroup$
Thank for the very nice proof. Just one more question. It is easy to see that $||gamma(x')||leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also?
$endgroup$
– serenus
Jan 10 at 7:53
$begingroup$
Done. You'll notice that the argument is familiar.
$endgroup$
– Martin Argerami
Jan 10 at 10:06
$begingroup$
In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $gamma (x')x$, isn't it?
$endgroup$
– serenus
Jan 10 at 10:33
$begingroup$
Yes. $ $
$endgroup$
– Martin Argerami
Jan 10 at 15:35
$begingroup$
Thanks for the proof.
$endgroup$
– serenus
Jan 10 at 15:43
add a comment |
$begingroup$
Thank for the very nice proof. Just one more question. It is easy to see that $||gamma(x')||leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also?
$endgroup$
– serenus
Jan 10 at 7:53
$begingroup$
Done. You'll notice that the argument is familiar.
$endgroup$
– Martin Argerami
Jan 10 at 10:06
$begingroup$
In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $gamma (x')x$, isn't it?
$endgroup$
– serenus
Jan 10 at 10:33
$begingroup$
Yes. $ $
$endgroup$
– Martin Argerami
Jan 10 at 15:35
$begingroup$
Thanks for the proof.
$endgroup$
– serenus
Jan 10 at 15:43
$begingroup$
Thank for the very nice proof. Just one more question. It is easy to see that $||gamma(x')||leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also?
$endgroup$
– serenus
Jan 10 at 7:53
$begingroup$
Thank for the very nice proof. Just one more question. It is easy to see that $||gamma(x')||leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also?
$endgroup$
– serenus
Jan 10 at 7:53
$begingroup$
Done. You'll notice that the argument is familiar.
$endgroup$
– Martin Argerami
Jan 10 at 10:06
$begingroup$
Done. You'll notice that the argument is familiar.
$endgroup$
– Martin Argerami
Jan 10 at 10:06
$begingroup$
In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $gamma (x')x$, isn't it?
$endgroup$
– serenus
Jan 10 at 10:33
$begingroup$
In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $gamma (x')x$, isn't it?
$endgroup$
– serenus
Jan 10 at 10:33
$begingroup$
Yes. $ $
$endgroup$
– Martin Argerami
Jan 10 at 15:35
$begingroup$
Yes. $ $
$endgroup$
– Martin Argerami
Jan 10 at 15:35
$begingroup$
Thanks for the proof.
$endgroup$
– serenus
Jan 10 at 15:43
$begingroup$
Thanks for the proof.
$endgroup$
– serenus
Jan 10 at 15:43
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067955%2fa-reference-for-the-isomorphism-ell-1e-cong-c-0e%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
