Mixing
Definition of a smooth function on $[0,1]^k$
Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
asked Nov 21 '18 at 1:04
D_S
13.4k51551
add a comment |
Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
asked Nov 21 '18 at 1:04
D_S
13.4k51551
add a comment |
Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
asked Nov 21 '18 at 1:04
D_S
13.4k51551
Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
calculus differential-geometry smooth-manifolds
asked Nov 21 '18 at 1:04
D_S
13.4k51551
asked Nov 21 '18 at 1:04
D_S
13.4k51551
edited Nov 21 '18 at 1:20
asked Nov 21 '18 at 1:04
D_S
13.4k51551
asked Nov 21 '18 at 1:04
D_S
13.4k51551
asked Nov 21 '18 at 1:04
D_S
13.4k51551
13.4k51551
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
answered Nov 21 '18 at 1:18
rldias
2,9901522
add a comment |
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
answered Nov 21 '18 at 1:24
Dante Grevino
94319
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%2f3007119%2fdefinition-of-a-smooth-function-on-0-1k%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
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
answered Nov 21 '18 at 1:18
rldias
2,9901522
add a comment |
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
answered Nov 21 '18 at 1:18
rldias
2,9901522
add a comment |
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
answered Nov 21 '18 at 1:18
rldias
2,9901522
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
answered Nov 21 '18 at 1:18
rldias
2,9901522
answered Nov 21 '18 at 1:18
rldias
2,9901522
answered Nov 21 '18 at 1:18
rldias
2,9901522
answered Nov 21 '18 at 1:18
rldias
2,9901522
2,9901522
add a comment |
add a comment |
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
answered Nov 21 '18 at 1:24
Dante Grevino
94319
add a comment |
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
answered Nov 21 '18 at 1:24
Dante Grevino
94319
add a comment |
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
answered Nov 21 '18 at 1:24
Dante Grevino
94319
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
answered Nov 21 '18 at 1:24
Dante Grevino
94319
answered Nov 21 '18 at 1:24
Dante Grevino
94319
answered Nov 21 '18 at 1:24
Dante Grevino
94319
answered Nov 21 '18 at 1:24
Dante Grevino
94319
94319
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3007119%2fdefinition-of-a-smooth-function-on-0-1k%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
