Mixing
How to read $f(K)=bigcup_{kin K}langle f,krangle$ and what it means, mathematically
$begingroup$
Let $E$ be a real Banach space and let $K$ be a subset of $E$. Suppose for each $fin E^{*}$, the set
$$f(K)=bigcup_{kin K}langle f,krangle$$
is bounded in $Bbb{R}$. Then, $K$ is a bounded subset of $E.$
Question:
How to read $f(K)=bigcup_{kin K}langle f,krangle$ and what does it mean, mathematically?
functional-analysis proof-explanation definition
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
$endgroup$
add a comment |
$begingroup$
Let $E$ be a real Banach space and let $K$ be a subset of $E$. Suppose for each $fin E^{*}$, the set
$$f(K)=bigcup_{kin K}langle f,krangle$$
is bounded in $Bbb{R}$. Then, $K$ is a bounded subset of $E.$
Question:
How to read $f(K)=bigcup_{kin K}langle f,krangle$ and what does it mean, mathematically?
functional-analysis proof-explanation definition
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
$endgroup$
1
$begingroup$
I'm not a native English speaker but I think it's read something like this: $f$ of $K$ equals (pause) the union of $f$ dot $k$'s where the union is taken over all $k$'s in $K$. Also, mathematically, the equality you have written is kind of weird. I think it's better to write it as $f(K)=bigcup_{kin K}{langle f,krangle}$.
$endgroup$
– stressed out
Jan 16 at 20:40
add a comment |
$begingroup$
Let $E$ be a real Banach space and let $K$ be a subset of $E$. Suppose for each $fin E^{*}$, the set
$$f(K)=bigcup_{kin K}langle f,krangle$$
is bounded in $Bbb{R}$. Then, $K$ is a bounded subset of $E.$
Question:
How to read $f(K)=bigcup_{kin K}langle f,krangle$ and what does it mean, mathematically?
functional-analysis proof-explanation definition
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
$endgroup$
Let $E$ be a real Banach space and let $K$ be a subset of $E$. Suppose for each $fin E^{*}$, the set
$$f(K)=bigcup_{kin K}langle f,krangle$$
is bounded in $Bbb{R}$. Then, $K$ is a bounded subset of $E.$
Question:
How to read $f(K)=bigcup_{kin K}langle f,krangle$ and what does it mean, mathematically?
functional-analysis proof-explanation definition
functional-analysis proof-explanation definition
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
edited Jan 17 at 5:04
Omojola Micheal
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
asked Jan 16 at 20:36
Omojola MichealOmojola Micheal
1,889324
1,889324
1
$begingroup$
I'm not a native English speaker but I think it's read something like this: $f$ of $K$ equals (pause) the union of $f$ dot $k$'s where the union is taken over all $k$'s in $K$. Also, mathematically, the equality you have written is kind of weird. I think it's better to write it as $f(K)=bigcup_{kin K}{langle f,krangle}$.
$endgroup$
– stressed out
Jan 16 at 20:40
add a comment |
1
$begingroup$
I'm not a native English speaker but I think it's read something like this: $f$ of $K$ equals (pause) the union of $f$ dot $k$'s where the union is taken over all $k$'s in $K$. Also, mathematically, the equality you have written is kind of weird. I think it's better to write it as $f(K)=bigcup_{kin K}{langle f,krangle}$.
$endgroup$
– stressed out
Jan 16 at 20:40
1
1
$begingroup$
I'm not a native English speaker but I think it's read something like this: $f$ of $K$ equals (pause) the union of $f$ dot $k$'s where the union is taken over all $k$'s in $K$. Also, mathematically, the equality you have written is kind of weird. I think it's better to write it as $f(K)=bigcup_{kin K}{langle f,krangle}$.
$endgroup$
– stressed out
Jan 16 at 20:40
$begingroup$
I'm not a native English speaker but I think it's read something like this: $f$ of $K$ equals (pause) the union of $f$ dot $k$'s where the union is taken over all $k$'s in $K$. Also, mathematically, the equality you have written is kind of weird. I think it's better to write it as $f(K)=bigcup_{kin K}{langle f,krangle}$.
$endgroup$
– stressed out
Jan 16 at 20:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I can tell you what it means mathematically, but I prefer to leave your question about how you should read the symbols out loud to a native English speaker.
A functional $f$ is a function $f: E to mathbb{R}$ (sometimes $mathbb{C})$ that takes a vector in $E$ and outputs a real(complex) number. When you have a Hilbert space, every functional arises as $varphi: E to mathbb{R}$ where $varphi(cdot)=langle v,cdotrangle$ for some $v in E$. This follows from the Riesz representation theorem for Hilbert spaces.
Because of this, you can denote a functional $f$ using the bra-ket notation which is popular in physics and real analysis. Note that a Banach space does not necessarily have to be an inner product space. So, when we're dealing with a Banach space, $langle f, k rangle$ is just a notation that means the evaluation of $f$ at $k$. Your equality is now this:
$$f(K) = bigcup_{kin K} {f(k)} =bigcup_{k in K} {langle f,krangle}$$
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
$endgroup$
add a comment |
$begingroup$
Since $K$ is a Banach space $langle f,k rangle$ is the evaluation of $f$ at $k$ where $fin E^*$ is a functional on $E$ (i.e. $f(k)$). So $f(K)$ here is defined completely literally: it's the union of the evaluation of the functional $f$ at each point $kin K$.
The point of writing it this way though, is that the norm of $f$ is $|f| = sup_{|k|=1} |langle f,k rangle |$ so this is suggestive of how to go about proving boundedness.
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
$endgroup$
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%2f3076258%2fhow-to-read-fk-bigcup-k-in-k-langle-f-k-rangle-and-what-it-means-mathema%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I can tell you what it means mathematically, but I prefer to leave your question about how you should read the symbols out loud to a native English speaker.
A functional $f$ is a function $f: E to mathbb{R}$ (sometimes $mathbb{C})$ that takes a vector in $E$ and outputs a real(complex) number. When you have a Hilbert space, every functional arises as $varphi: E to mathbb{R}$ where $varphi(cdot)=langle v,cdotrangle$ for some $v in E$. This follows from the Riesz representation theorem for Hilbert spaces.
Because of this, you can denote a functional $f$ using the bra-ket notation which is popular in physics and real analysis. Note that a Banach space does not necessarily have to be an inner product space. So, when we're dealing with a Banach space, $langle f, k rangle$ is just a notation that means the evaluation of $f$ at $k$. Your equality is now this:
$$f(K) = bigcup_{kin K} {f(k)} =bigcup_{k in K} {langle f,krangle}$$
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
$endgroup$
add a comment |
$begingroup$
I can tell you what it means mathematically, but I prefer to leave your question about how you should read the symbols out loud to a native English speaker.
A functional $f$ is a function $f: E to mathbb{R}$ (sometimes $mathbb{C})$ that takes a vector in $E$ and outputs a real(complex) number. When you have a Hilbert space, every functional arises as $varphi: E to mathbb{R}$ where $varphi(cdot)=langle v,cdotrangle$ for some $v in E$. This follows from the Riesz representation theorem for Hilbert spaces.
Because of this, you can denote a functional $f$ using the bra-ket notation which is popular in physics and real analysis. Note that a Banach space does not necessarily have to be an inner product space. So, when we're dealing with a Banach space, $langle f, k rangle$ is just a notation that means the evaluation of $f$ at $k$. Your equality is now this:
$$f(K) = bigcup_{kin K} {f(k)} =bigcup_{k in K} {langle f,krangle}$$
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
$endgroup$
add a comment |
$begingroup$
I can tell you what it means mathematically, but I prefer to leave your question about how you should read the symbols out loud to a native English speaker.
A functional $f$ is a function $f: E to mathbb{R}$ (sometimes $mathbb{C})$ that takes a vector in $E$ and outputs a real(complex) number. When you have a Hilbert space, every functional arises as $varphi: E to mathbb{R}$ where $varphi(cdot)=langle v,cdotrangle$ for some $v in E$. This follows from the Riesz representation theorem for Hilbert spaces.
Because of this, you can denote a functional $f$ using the bra-ket notation which is popular in physics and real analysis. Note that a Banach space does not necessarily have to be an inner product space. So, when we're dealing with a Banach space, $langle f, k rangle$ is just a notation that means the evaluation of $f$ at $k$. Your equality is now this:
$$f(K) = bigcup_{kin K} {f(k)} =bigcup_{k in K} {langle f,krangle}$$
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
$endgroup$
I can tell you what it means mathematically, but I prefer to leave your question about how you should read the symbols out loud to a native English speaker.
A functional $f$ is a function $f: E to mathbb{R}$ (sometimes $mathbb{C})$ that takes a vector in $E$ and outputs a real(complex) number. When you have a Hilbert space, every functional arises as $varphi: E to mathbb{R}$ where $varphi(cdot)=langle v,cdotrangle$ for some $v in E$. This follows from the Riesz representation theorem for Hilbert spaces.
Because of this, you can denote a functional $f$ using the bra-ket notation which is popular in physics and real analysis. Note that a Banach space does not necessarily have to be an inner product space. So, when we're dealing with a Banach space, $langle f, k rangle$ is just a notation that means the evaluation of $f$ at $k$. Your equality is now this:
$$f(K) = bigcup_{kin K} {f(k)} =bigcup_{k in K} {langle f,krangle}$$
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
edited Jan 16 at 20:54
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
answered Jan 16 at 20:48
stressed outstressed out
5,5981638
5,5981638
add a comment |
add a comment |
$begingroup$
Since $K$ is a Banach space $langle f,k rangle$ is the evaluation of $f$ at $k$ where $fin E^*$ is a functional on $E$ (i.e. $f(k)$). So $f(K)$ here is defined completely literally: it's the union of the evaluation of the functional $f$ at each point $kin K$.
The point of writing it this way though, is that the norm of $f$ is $|f| = sup_{|k|=1} |langle f,k rangle |$ so this is suggestive of how to go about proving boundedness.
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
$endgroup$
add a comment |
$begingroup$
Since $K$ is a Banach space $langle f,k rangle$ is the evaluation of $f$ at $k$ where $fin E^*$ is a functional on $E$ (i.e. $f(k)$). So $f(K)$ here is defined completely literally: it's the union of the evaluation of the functional $f$ at each point $kin K$.
The point of writing it this way though, is that the norm of $f$ is $|f| = sup_{|k|=1} |langle f,k rangle |$ so this is suggestive of how to go about proving boundedness.
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
$endgroup$
add a comment |
$begingroup$
Since $K$ is a Banach space $langle f,k rangle$ is the evaluation of $f$ at $k$ where $fin E^*$ is a functional on $E$ (i.e. $f(k)$). So $f(K)$ here is defined completely literally: it's the union of the evaluation of the functional $f$ at each point $kin K$.
The point of writing it this way though, is that the norm of $f$ is $|f| = sup_{|k|=1} |langle f,k rangle |$ so this is suggestive of how to go about proving boundedness.
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
$endgroup$
Since $K$ is a Banach space $langle f,k rangle$ is the evaluation of $f$ at $k$ where $fin E^*$ is a functional on $E$ (i.e. $f(k)$). So $f(K)$ here is defined completely literally: it's the union of the evaluation of the functional $f$ at each point $kin K$.
The point of writing it this way though, is that the norm of $f$ is $|f| = sup_{|k|=1} |langle f,k rangle |$ so this is suggestive of how to go about proving boundedness.
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
answered Jan 16 at 20:47
postmortespostmortes
2,04131119
2,04131119
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%2f3076258%2fhow-to-read-fk-bigcup-k-in-k-langle-f-k-rangle-and-what-it-means-mathema%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
1
$begingroup$
I'm not a native English speaker but I think it's read something like this: $f$ of $K$ equals (pause) the union of $f$ dot $k$'s where the union is taken over all $k$'s in $K$. Also, mathematically, the equality you have written is kind of weird. I think it's better to write it as $f(K)=bigcup_{kin K}{langle f,krangle}$.
$endgroup$
– stressed out
Jan 16 at 20:40