Weak cauchy operator sequence over hilbert spaces
$begingroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
$endgroup$
add a comment |
$begingroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
$endgroup$
add a comment |
$begingroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
$endgroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
399113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
$endgroup$
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%2f3037543%2fweak-cauchy-operator-sequence-over-hilbert-spaces%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
$begingroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
$endgroup$
add a comment |
$begingroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
$endgroup$
add a comment |
$begingroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
$endgroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
answered Dec 13 '18 at 3:32
SongSong
15.9k1739
15.9k1739
add a comment |
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%2f3037543%2fweak-cauchy-operator-sequence-over-hilbert-spaces%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
