Mixing
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
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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
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add a comment |
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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 |
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$begingroup$
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