Is it true that $( forall S int_{S} A dS = 0 )implies A equiv 0 $
$begingroup$
Is it true that
$( forall S int_{S} A dS = 0 )implies A equiv 0 $
where $S$ is a surface and $A$ is some function which takes values on $S$?
Is there a requirement on the smoothness of $S$? (In my case it is the boundary of a region in which some function $u$ satisfies Laplace's equation.)
Does this result generalise to
$( forall R int_{R} A dR = 0) implies A equiv 0 $?
For $R$ $subset mathbb{R}^n$ with $dim(R) = k$ - some specified type of region where the integral is defined, e.g. a surface, a volume, etc.?
I have a feeling that the first is true, but I can't manage to prove it.
Given the counterexample in the answers, let's impose the restriction that A, R are continuous.
calculus integration
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
$endgroup$
add a comment |
$begingroup$
Is it true that
$( forall S int_{S} A dS = 0 )implies A equiv 0 $
where $S$ is a surface and $A$ is some function which takes values on $S$?
Is there a requirement on the smoothness of $S$? (In my case it is the boundary of a region in which some function $u$ satisfies Laplace's equation.)
Does this result generalise to
$( forall R int_{R} A dR = 0) implies A equiv 0 $?
For $R$ $subset mathbb{R}^n$ with $dim(R) = k$ - some specified type of region where the integral is defined, e.g. a surface, a volume, etc.?
I have a feeling that the first is true, but I can't manage to prove it.
Given the counterexample in the answers, let's impose the restriction that A, R are continuous.
calculus integration
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
$endgroup$
$begingroup$
Do you at least have that $A$ is continuous?
$endgroup$
– Cocopuffs
Mar 27 '13 at 12:38
$begingroup$
Erm. Well, in my problem, yes, but I am interested in how this works in the general case, so if there is an answer which is restricted to $A in C^{n}[R]$ or $A in C^{infty}[R]$ then I would very much like to know. @Cocopuffs.
$endgroup$
– user27182
Mar 27 '13 at 12:42
add a comment |
$begingroup$
Is it true that
$( forall S int_{S} A dS = 0 )implies A equiv 0 $
where $S$ is a surface and $A$ is some function which takes values on $S$?
Is there a requirement on the smoothness of $S$? (In my case it is the boundary of a region in which some function $u$ satisfies Laplace's equation.)
Does this result generalise to
$( forall R int_{R} A dR = 0) implies A equiv 0 $?
For $R$ $subset mathbb{R}^n$ with $dim(R) = k$ - some specified type of region where the integral is defined, e.g. a surface, a volume, etc.?
I have a feeling that the first is true, but I can't manage to prove it.
Given the counterexample in the answers, let's impose the restriction that A, R are continuous.
calculus integration
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
$endgroup$
Is it true that
$( forall S int_{S} A dS = 0 )implies A equiv 0 $
where $S$ is a surface and $A$ is some function which takes values on $S$?
Is there a requirement on the smoothness of $S$? (In my case it is the boundary of a region in which some function $u$ satisfies Laplace's equation.)
Does this result generalise to
$( forall R int_{R} A dR = 0) implies A equiv 0 $?
For $R$ $subset mathbb{R}^n$ with $dim(R) = k$ - some specified type of region where the integral is defined, e.g. a surface, a volume, etc.?
I have a feeling that the first is true, but I can't manage to prove it.
Given the counterexample in the answers, let's impose the restriction that A, R are continuous.
calculus integration
calculus integration
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
edited Mar 27 '13 at 12:49
user27182
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
asked Mar 27 '13 at 12:23
user27182user27182
1,032627
1,032627
$begingroup$
Do you at least have that $A$ is continuous?
$endgroup$
– Cocopuffs
Mar 27 '13 at 12:38
$begingroup$
Erm. Well, in my problem, yes, but I am interested in how this works in the general case, so if there is an answer which is restricted to $A in C^{n}[R]$ or $A in C^{infty}[R]$ then I would very much like to know. @Cocopuffs.
$endgroup$
– user27182
Mar 27 '13 at 12:42
add a comment |
$begingroup$
Do you at least have that $A$ is continuous?
$endgroup$
– Cocopuffs
Mar 27 '13 at 12:38
$begingroup$
Erm. Well, in my problem, yes, but I am interested in how this works in the general case, so if there is an answer which is restricted to $A in C^{n}[R]$ or $A in C^{infty}[R]$ then I would very much like to know. @Cocopuffs.
$endgroup$
– user27182
Mar 27 '13 at 12:42
$begingroup$
Do you at least have that $A$ is continuous?
$endgroup$
– Cocopuffs
Mar 27 '13 at 12:38
$begingroup$
Do you at least have that $A$ is continuous?
$endgroup$
– Cocopuffs
Mar 27 '13 at 12:38
$begingroup$
Erm. Well, in my problem, yes, but I am interested in how this works in the general case, so if there is an answer which is restricted to $A in C^{n}[R]$ or $A in C^{infty}[R]$ then I would very much like to know. @Cocopuffs.
$endgroup$
– user27182
Mar 27 '13 at 12:42
$begingroup$
Erm. Well, in my problem, yes, but I am interested in how this works in the general case, so if there is an answer which is restricted to $A in C^{n}[R]$ or $A in C^{infty}[R]$ then I would very much like to know. @Cocopuffs.
$endgroup$
– user27182
Mar 27 '13 at 12:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Ross B. has demonstrated a counterexample if $A$ is discontinuous.
Suppose $A ne 0$ is continuous, and let $A(x_0) > 0$ WLOG. By continuity, there is a neighborhood of $x_0$ on which $A>0;$ form a surface $S'$ from this neighborhood.
Then $int_{S'} A dS' > int_{S'} min(A) dS' = min(A)text{vol}(S') > 0,$ a contradiction. Thus $A = 0.$
answered Jan 7 at 2:41
Display nameDisplay name
836314
$endgroup$
add a comment |
$begingroup$
If $A$ is not continuous, for your second answer a counterexample would be the function which is $1$ at $x=0$ and $0$ everywhere else.
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
$endgroup$
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%2f342725%2fis-it-true-that-forall-s-int-s-a-ds-0-implies-a-equiv-0%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
$begingroup$
Ross B. has demonstrated a counterexample if $A$ is discontinuous.
Suppose $A ne 0$ is continuous, and let $A(x_0) > 0$ WLOG. By continuity, there is a neighborhood of $x_0$ on which $A>0;$ form a surface $S'$ from this neighborhood.
Then $int_{S'} A dS' > int_{S'} min(A) dS' = min(A)text{vol}(S') > 0,$ a contradiction. Thus $A = 0.$
answered Jan 7 at 2:41
Display nameDisplay name
836314
$endgroup$
add a comment |
$begingroup$
Ross B. has demonstrated a counterexample if $A$ is discontinuous.
Suppose $A ne 0$ is continuous, and let $A(x_0) > 0$ WLOG. By continuity, there is a neighborhood of $x_0$ on which $A>0;$ form a surface $S'$ from this neighborhood.
Then $int_{S'} A dS' > int_{S'} min(A) dS' = min(A)text{vol}(S') > 0,$ a contradiction. Thus $A = 0.$
answered Jan 7 at 2:41
Display nameDisplay name
836314
$endgroup$
add a comment |
$begingroup$
Ross B. has demonstrated a counterexample if $A$ is discontinuous.
Suppose $A ne 0$ is continuous, and let $A(x_0) > 0$ WLOG. By continuity, there is a neighborhood of $x_0$ on which $A>0;$ form a surface $S'$ from this neighborhood.
Then $int_{S'} A dS' > int_{S'} min(A) dS' = min(A)text{vol}(S') > 0,$ a contradiction. Thus $A = 0.$
answered Jan 7 at 2:41
Display nameDisplay name
836314
$endgroup$
Ross B. has demonstrated a counterexample if $A$ is discontinuous.
Suppose $A ne 0$ is continuous, and let $A(x_0) > 0$ WLOG. By continuity, there is a neighborhood of $x_0$ on which $A>0;$ form a surface $S'$ from this neighborhood.
Then $int_{S'} A dS' > int_{S'} min(A) dS' = min(A)text{vol}(S') > 0,$ a contradiction. Thus $A = 0.$
answered Jan 7 at 2:41
Display nameDisplay name
836314
answered Jan 7 at 2:41
Display nameDisplay name
836314
answered Jan 7 at 2:41
Display nameDisplay name
836314
answered Jan 7 at 2:41
Display nameDisplay name
836314
836314
add a comment |
add a comment |
$begingroup$
If $A$ is not continuous, for your second answer a counterexample would be the function which is $1$ at $x=0$ and $0$ everywhere else.
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
$endgroup$
add a comment |
$begingroup$
If $A$ is not continuous, for your second answer a counterexample would be the function which is $1$ at $x=0$ and $0$ everywhere else.
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
$endgroup$
add a comment |
$begingroup$
If $A$ is not continuous, for your second answer a counterexample would be the function which is $1$ at $x=0$ and $0$ everywhere else.
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
$endgroup$
If $A$ is not continuous, for your second answer a counterexample would be the function which is $1$ at $x=0$ and $0$ everywhere else.
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
answered Mar 27 '13 at 12:47
Ross B.Ross B.
1,657516
1,657516
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%2f342725%2fis-it-true-that-forall-s-int-s-a-ds-0-implies-a-equiv-0%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
$begingroup$
Do you at least have that $A$ is continuous?
$endgroup$
– Cocopuffs
Mar 27 '13 at 12:38
$begingroup$
Erm. Well, in my problem, yes, but I am interested in how this works in the general case, so if there is an answer which is restricted to $A in C^{n}[R]$ or $A in C^{infty}[R]$ then I would very much like to know. @Cocopuffs.
$endgroup$
– user27182
Mar 27 '13 at 12:42