Extenstion of Intermediate Value Theorem.
$begingroup$
Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?
Thank you,
JF
real-analysis vector-spaces
asked Jun 2 '14 at 16:29
user154763user154763
61
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?
Thank you,
JF
real-analysis vector-spaces
asked Jun 2 '14 at 16:29
user154763user154763
61
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?
Thank you,
JF
real-analysis vector-spaces
asked Jun 2 '14 at 16:29
user154763user154763
61
$endgroup$
Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?
Thank you,
JF
real-analysis vector-spaces
real-analysis vector-spaces
asked Jun 2 '14 at 16:29
user154763user154763
61
asked Jun 2 '14 at 16:29
user154763user154763
61
asked Jun 2 '14 at 16:29
user154763user154763
61
asked Jun 2 '14 at 16:29
user154763user154763
61
asked Jun 2 '14 at 16:29
user154763user154763
61
61
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
$endgroup$
$begingroup$
If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
$endgroup$
– user154763
Jun 2 '14 at 18:25
$begingroup$
I see, still the concept holds, as the function can go around the point without having to go through it.
$endgroup$
– Ellya
Jun 2 '14 at 18:45
$begingroup$
How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
$endgroup$
– user154763
Jun 2 '14 at 19:13
add a comment |
$begingroup$
I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.
$d$ is a member of N.
P = $mathbb{R}^{d}$
Q is an $mathbb{R}^{d - 1}$ dimensional line in P.
A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…
When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:
The points may become equal and unequal any number of times without becoming equal to a member of Q.
The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
$endgroup$
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%2f818312%2fextenstion-of-intermediate-value-theorem%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 believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
$endgroup$
$begingroup$
If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
$endgroup$
– user154763
Jun 2 '14 at 18:25
$begingroup$
I see, still the concept holds, as the function can go around the point without having to go through it.
$endgroup$
– Ellya
Jun 2 '14 at 18:45
$begingroup$
How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
$endgroup$
– user154763
Jun 2 '14 at 19:13
add a comment |
$begingroup$
I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
$endgroup$
$begingroup$
If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
$endgroup$
– user154763
Jun 2 '14 at 18:25
$begingroup$
I see, still the concept holds, as the function can go around the point without having to go through it.
$endgroup$
– Ellya
Jun 2 '14 at 18:45
$begingroup$
How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
$endgroup$
– user154763
Jun 2 '14 at 19:13
add a comment |
$begingroup$
I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
$endgroup$
I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
answered Jun 2 '14 at 16:34
EllyaEllya
9,57211226
9,57211226
$begingroup$
If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
$endgroup$
– user154763
Jun 2 '14 at 18:25
$begingroup$
I see, still the concept holds, as the function can go around the point without having to go through it.
$endgroup$
– Ellya
Jun 2 '14 at 18:45
$begingroup$
How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
$endgroup$
– user154763
Jun 2 '14 at 19:13
add a comment |
$begingroup$
If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
$endgroup$
– user154763
Jun 2 '14 at 18:25
$begingroup$
I see, still the concept holds, as the function can go around the point without having to go through it.
$endgroup$
– Ellya
Jun 2 '14 at 18:45
$begingroup$
How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
$endgroup$
– user154763
Jun 2 '14 at 19:13
$begingroup$
If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
$endgroup$
– user154763
Jun 2 '14 at 18:25
$begingroup$
If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
$endgroup$
– user154763
Jun 2 '14 at 18:25
$begingroup$
I see, still the concept holds, as the function can go around the point without having to go through it.
$endgroup$
– Ellya
Jun 2 '14 at 18:45
$begingroup$
I see, still the concept holds, as the function can go around the point without having to go through it.
$endgroup$
– Ellya
Jun 2 '14 at 18:45
$begingroup$
How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
$endgroup$
– user154763
Jun 2 '14 at 19:13
$begingroup$
How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
$endgroup$
– user154763
Jun 2 '14 at 19:13
add a comment |
$begingroup$
I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.
$d$ is a member of N.
P = $mathbb{R}^{d}$
Q is an $mathbb{R}^{d - 1}$ dimensional line in P.
A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…
When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:
The points may become equal and unequal any number of times without becoming equal to a member of Q.
The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
$endgroup$
add a comment |
$begingroup$
I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.
$d$ is a member of N.
P = $mathbb{R}^{d}$
Q is an $mathbb{R}^{d - 1}$ dimensional line in P.
A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…
When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:
The points may become equal and unequal any number of times without becoming equal to a member of Q.
The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
$endgroup$
add a comment |
$begingroup$
I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.
$d$ is a member of N.
P = $mathbb{R}^{d}$
Q is an $mathbb{R}^{d - 1}$ dimensional line in P.
A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…
When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:
The points may become equal and unequal any number of times without becoming equal to a member of Q.
The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
$endgroup$
I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.
$d$ is a member of N.
P = $mathbb{R}^{d}$
Q is an $mathbb{R}^{d - 1}$ dimensional line in P.
A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…
When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:
The points may become equal and unequal any number of times without becoming equal to a member of Q.
The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
edited Dec 4 '18 at 13:56
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
answered Dec 4 '18 at 13:48
alan2herealan2here
504219
504219
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%2f818312%2fextenstion-of-intermediate-value-theorem%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
