$iint_D f = 0$ implies $f(p)=0$ for all $p$.
If $D$ is open, and if $f$ is continuous, bounded, and obeys $f(p) ge 0$ for all $p in D$, then $iint_D f = 0$ implies $f(p)=0$ for all $p$.
The hint in the back of my book says that
There's a neighborhood where $f(p) ge delta$.
From the definition in my book, it states that
The double integral $iint_R f$ exists and has value $v$ if and only if for any $epsilon > 0$ there is a $delta > 0$ such that
$|S(N,f,{p_{ij}})-v| < delta$.
There's another theorem in the book that states that
If $f(p) ge 0$ for all $pin D$, $iint_D f ge 0$.
So far, from the information I've been given, that would mean that I could use the contrapositive to prove that if $f(p) ne 0$ for all $p$, then $iint_D f ne 0$.
Since $f(p) ne 0$ I could use cases.
Case 1: If $f(p) > 0$, then I could use the aforementioned theorem to show that $iint_D f > 0$ which is not equal to $0$.
Case 2: If $f(p) < 0$, then $-f(p)=g(p)>0$, so that $iint_D g(p) > 0$. Then $iint_D g(p) > 0$ is equivalent to -$iint_D f(p) > 0$ which is not equal to $0$.
I would like to know if I'm approaching this proof the correct way? Also, would it be difficult to instead try a direct proof using the hint and the definition that I stated above?
calculus analysis multivariable-calculus
asked Nov 27 '18 at 20:20
K.M
681312
|
show 3 more comments
If $D$ is open, and if $f$ is continuous, bounded, and obeys $f(p) ge 0$ for all $p in D$, then $iint_D f = 0$ implies $f(p)=0$ for all $p$.
The hint in the back of my book says that
There's a neighborhood where $f(p) ge delta$.
From the definition in my book, it states that
The double integral $iint_R f$ exists and has value $v$ if and only if for any $epsilon > 0$ there is a $delta > 0$ such that
$|S(N,f,{p_{ij}})-v| < delta$.
There's another theorem in the book that states that
If $f(p) ge 0$ for all $pin D$, $iint_D f ge 0$.
So far, from the information I've been given, that would mean that I could use the contrapositive to prove that if $f(p) ne 0$ for all $p$, then $iint_D f ne 0$.
Since $f(p) ne 0$ I could use cases.
Case 1: If $f(p) > 0$, then I could use the aforementioned theorem to show that $iint_D f > 0$ which is not equal to $0$.
Case 2: If $f(p) < 0$, then $-f(p)=g(p)>0$, so that $iint_D g(p) > 0$. Then $iint_D g(p) > 0$ is equivalent to -$iint_D f(p) > 0$ which is not equal to $0$.
I would like to know if I'm approaching this proof the correct way? Also, would it be difficult to instead try a direct proof using the hint and the definition that I stated above?
calculus analysis multivariable-calculus
asked Nov 27 '18 at 20:20
K.M
681312
1
There is sort of a way to attack this if you break $D$ into two sets: places where $f(p) geq delta$ and places where it's not.
– Sean Roberson
Nov 27 '18 at 20:26
So that integral is improper? Do your theorems in the text permit that?
– Randall
Nov 27 '18 at 20:35
The integral is defined on $D$. Also, the theorem stated above requires that $D$ is bounded. Although, the definition says that the double integral exists for the conditions listed above for $R$, which the book states is a rectangle, so I think that that means that the integral must be bounded.
– K.M
Nov 27 '18 at 20:41
@SeanRoberson: is the $delta$ referring to the $delta$ used in $|S(N,f,{p_{ij}})| < delta$?
– K.M
Nov 27 '18 at 20:44
You do not need Case 2 as $f(x)ge 0$ for all $xin D$.
– A.Γ.
Nov 27 '18 at 20:56
|
show 3 more comments
If $D$ is open, and if $f$ is continuous, bounded, and obeys $f(p) ge 0$ for all $p in D$, then $iint_D f = 0$ implies $f(p)=0$ for all $p$.
The hint in the back of my book says that
There's a neighborhood where $f(p) ge delta$.
From the definition in my book, it states that
The double integral $iint_R f$ exists and has value $v$ if and only if for any $epsilon > 0$ there is a $delta > 0$ such that
$|S(N,f,{p_{ij}})-v| < delta$.
There's another theorem in the book that states that
If $f(p) ge 0$ for all $pin D$, $iint_D f ge 0$.
So far, from the information I've been given, that would mean that I could use the contrapositive to prove that if $f(p) ne 0$ for all $p$, then $iint_D f ne 0$.
Since $f(p) ne 0$ I could use cases.
Case 1: If $f(p) > 0$, then I could use the aforementioned theorem to show that $iint_D f > 0$ which is not equal to $0$.
Case 2: If $f(p) < 0$, then $-f(p)=g(p)>0$, so that $iint_D g(p) > 0$. Then $iint_D g(p) > 0$ is equivalent to -$iint_D f(p) > 0$ which is not equal to $0$.
I would like to know if I'm approaching this proof the correct way? Also, would it be difficult to instead try a direct proof using the hint and the definition that I stated above?
calculus analysis multivariable-calculus
asked Nov 27 '18 at 20:20
K.M
681312
If $D$ is open, and if $f$ is continuous, bounded, and obeys $f(p) ge 0$ for all $p in D$, then $iint_D f = 0$ implies $f(p)=0$ for all $p$.
The hint in the back of my book says that
There's a neighborhood where $f(p) ge delta$.
From the definition in my book, it states that
The double integral $iint_R f$ exists and has value $v$ if and only if for any $epsilon > 0$ there is a $delta > 0$ such that
$|S(N,f,{p_{ij}})-v| < delta$.
There's another theorem in the book that states that
If $f(p) ge 0$ for all $pin D$, $iint_D f ge 0$.
So far, from the information I've been given, that would mean that I could use the contrapositive to prove that if $f(p) ne 0$ for all $p$, then $iint_D f ne 0$.
Since $f(p) ne 0$ I could use cases.
Case 1: If $f(p) > 0$, then I could use the aforementioned theorem to show that $iint_D f > 0$ which is not equal to $0$.
Case 2: If $f(p) < 0$, then $-f(p)=g(p)>0$, so that $iint_D g(p) > 0$. Then $iint_D g(p) > 0$ is equivalent to -$iint_D f(p) > 0$ which is not equal to $0$.
I would like to know if I'm approaching this proof the correct way? Also, would it be difficult to instead try a direct proof using the hint and the definition that I stated above?
calculus analysis multivariable-calculus
calculus analysis multivariable-calculus
asked Nov 27 '18 at 20:20
K.M
681312
asked Nov 27 '18 at 20:20
K.M
681312
edited Nov 27 '18 at 20:46
asked Nov 27 '18 at 20:20
K.M
681312
asked Nov 27 '18 at 20:20
K.M
681312
asked Nov 27 '18 at 20:20
K.M
681312
681312
1
There is sort of a way to attack this if you break $D$ into two sets: places where $f(p) geq delta$ and places where it's not.
– Sean Roberson
Nov 27 '18 at 20:26
So that integral is improper? Do your theorems in the text permit that?
– Randall
Nov 27 '18 at 20:35
The integral is defined on $D$. Also, the theorem stated above requires that $D$ is bounded. Although, the definition says that the double integral exists for the conditions listed above for $R$, which the book states is a rectangle, so I think that that means that the integral must be bounded.
– K.M
Nov 27 '18 at 20:41
@SeanRoberson: is the $delta$ referring to the $delta$ used in $|S(N,f,{p_{ij}})| < delta$?
– K.M
Nov 27 '18 at 20:44
You do not need Case 2 as $f(x)ge 0$ for all $xin D$.
– A.Γ.
Nov 27 '18 at 20:56
|
show 3 more comments
1
There is sort of a way to attack this if you break $D$ into two sets: places where $f(p) geq delta$ and places where it's not.
– Sean Roberson
Nov 27 '18 at 20:26
So that integral is improper? Do your theorems in the text permit that?
– Randall
Nov 27 '18 at 20:35
The integral is defined on $D$. Also, the theorem stated above requires that $D$ is bounded. Although, the definition says that the double integral exists for the conditions listed above for $R$, which the book states is a rectangle, so I think that that means that the integral must be bounded.
– K.M
Nov 27 '18 at 20:41
@SeanRoberson: is the $delta$ referring to the $delta$ used in $|S(N,f,{p_{ij}})| < delta$?
– K.M
Nov 27 '18 at 20:44
You do not need Case 2 as $f(x)ge 0$ for all $xin D$.
– A.Γ.
Nov 27 '18 at 20:56
1
1
There is sort of a way to attack this if you break $D$ into two sets: places where $f(p) geq delta$ and places where it's not.
– Sean Roberson
Nov 27 '18 at 20:26
There is sort of a way to attack this if you break $D$ into two sets: places where $f(p) geq delta$ and places where it's not.
– Sean Roberson
Nov 27 '18 at 20:26
So that integral is improper? Do your theorems in the text permit that?
– Randall
Nov 27 '18 at 20:35
So that integral is improper? Do your theorems in the text permit that?
– Randall
Nov 27 '18 at 20:35
The integral is defined on $D$. Also, the theorem stated above requires that $D$ is bounded. Although, the definition says that the double integral exists for the conditions listed above for $R$, which the book states is a rectangle, so I think that that means that the integral must be bounded.
– K.M
Nov 27 '18 at 20:41
The integral is defined on $D$. Also, the theorem stated above requires that $D$ is bounded. Although, the definition says that the double integral exists for the conditions listed above for $R$, which the book states is a rectangle, so I think that that means that the integral must be bounded.
– K.M
Nov 27 '18 at 20:41
@SeanRoberson: is the $delta$ referring to the $delta$ used in $|S(N,f,{p_{ij}})| < delta$?
– K.M
Nov 27 '18 at 20:44
@SeanRoberson: is the $delta$ referring to the $delta$ used in $|S(N,f,{p_{ij}})| < delta$?
– K.M
Nov 27 '18 at 20:44
You do not need Case 2 as $f(x)ge 0$ for all $xin D$.
– A.Γ.
Nov 27 '18 at 20:56
You do not need Case 2 as $f(x)ge 0$ for all $xin D$.
– A.Γ.
Nov 27 '18 at 20:56
|
show 3 more comments
2 Answers
2
active
oldest
votes
Extended hint.
Assume $exists pin D$ such that $f(p)>0$. To prove $iint_D f>0$ follow the steps:
- Prove that there exists a neighborhood $N$ where $f(x)gedelta=frac{f(p)}{2}$. One can even choose $N$ as a disc around $p$.
- Estimate $iint_N fge ?$.
- Compare $iint_D f=iint_N f+iint_{Dsetminus N}f$ and $iint_N f$.
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
For the first part, since $f$ is continuous, and we've assumed that $f(p)>0$, does that mean that it would be better to set $epsilon = frac{f(p)}{2}$, so that we can choose a $delta$ about $p$, simply by convention?
– K.M
Nov 27 '18 at 22:12
@K.M No problem to call it $epsilon$, name does not really matter. I just followed the notation in the book hint that $f(x)gedelta$.
– A.Γ.
Nov 28 '18 at 7:47
add a comment |
Using the hint from your book, assume, toward a contradiction, that $f(p)=r >0$ for some $pin D$. Then continuity of $f$ at $p$ implies that there is a square $[a,b]times [c,d]$ containing $p$ such that $f(x)>r/2$ whenever $xin [a,b]times [c,d].$ (Drawing a picture and picking a suitable $epsilon$ for the standard $epsilon-delta$ argument will help here).
But then, using an elementary property of the Riemann integral, we can write
$int_D fge int_{[a,b]times [c,d]}f>frac{r}{2}(b-a)(d-c)>0$ and we have our contradiction.
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
I was wondering if you meant that $f(p) = r < 0$ for some $p in D$ and that continuity of $f$ at $p$ implies that there is a square containing '$p$ such that $f(x) < r/2$? Also, should the square be an open set?
– K.M
Nov 27 '18 at 23:35
1
You know $fge0$ and you want to prove it is never strictly positive; so you are assuming it is positive somewhere, and getting a contradiction. The original square you get using a $delta-epsilon$ argument will be an open set, but you can then always get a CLOSED square inside it, so the conclusion still holds on the closed square. That's why I said to draw a picture!
– Matematleta
Nov 27 '18 at 23:40
Using $r = f(p_0)$, I can set $epsilon = frac{r}{2}$ and so I see that by definition of continuity of $f$, we have that $|f(x)-f(p)|< epsilon$ whenever $x in [a,b]$ x $[c,b]$ implies that $-epsilon< f(x) - f(p) < epsilon$, so that $f(x)>f(p)-epsilon = r - frac{r}{2} = frac{r}{2}$. Although, I would like to know as an aside if you could possibly explain how I could incorporate $delta$ into an $epsilon-delta$ argument?
– K.M
Nov 28 '18 at 0:19
You almost have it: set $p=(p_1,p_2)$. With your choice of $epsilon$ you can find a $delta>0$ (the actual value is not important) such that $xin (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})Rightarrow f(x)in (f(p)-epsilon,f(p)-epsilon)$. To conclude, take any closed rectangle inside $ (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})$ and follow the proof I gave.
– Matematleta
Nov 28 '18 at 1:23
add a comment |
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2 Answers
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2 Answers
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oldest
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oldest
votes
Extended hint.
Assume $exists pin D$ such that $f(p)>0$. To prove $iint_D f>0$ follow the steps:
- Prove that there exists a neighborhood $N$ where $f(x)gedelta=frac{f(p)}{2}$. One can even choose $N$ as a disc around $p$.
- Estimate $iint_N fge ?$.
- Compare $iint_D f=iint_N f+iint_{Dsetminus N}f$ and $iint_N f$.
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
For the first part, since $f$ is continuous, and we've assumed that $f(p)>0$, does that mean that it would be better to set $epsilon = frac{f(p)}{2}$, so that we can choose a $delta$ about $p$, simply by convention?
– K.M
Nov 27 '18 at 22:12
@K.M No problem to call it $epsilon$, name does not really matter. I just followed the notation in the book hint that $f(x)gedelta$.
– A.Γ.
Nov 28 '18 at 7:47
add a comment |
Extended hint.
Assume $exists pin D$ such that $f(p)>0$. To prove $iint_D f>0$ follow the steps:
- Prove that there exists a neighborhood $N$ where $f(x)gedelta=frac{f(p)}{2}$. One can even choose $N$ as a disc around $p$.
- Estimate $iint_N fge ?$.
- Compare $iint_D f=iint_N f+iint_{Dsetminus N}f$ and $iint_N f$.
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
For the first part, since $f$ is continuous, and we've assumed that $f(p)>0$, does that mean that it would be better to set $epsilon = frac{f(p)}{2}$, so that we can choose a $delta$ about $p$, simply by convention?
– K.M
Nov 27 '18 at 22:12
@K.M No problem to call it $epsilon$, name does not really matter. I just followed the notation in the book hint that $f(x)gedelta$.
– A.Γ.
Nov 28 '18 at 7:47
add a comment |
Extended hint.
Assume $exists pin D$ such that $f(p)>0$. To prove $iint_D f>0$ follow the steps:
- Prove that there exists a neighborhood $N$ where $f(x)gedelta=frac{f(p)}{2}$. One can even choose $N$ as a disc around $p$.
- Estimate $iint_N fge ?$.
- Compare $iint_D f=iint_N f+iint_{Dsetminus N}f$ and $iint_N f$.
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
Extended hint.
Assume $exists pin D$ such that $f(p)>0$. To prove $iint_D f>0$ follow the steps:
- Prove that there exists a neighborhood $N$ where $f(x)gedelta=frac{f(p)}{2}$. One can even choose $N$ as a disc around $p$.
- Estimate $iint_N fge ?$.
- Compare $iint_D f=iint_N f+iint_{Dsetminus N}f$ and $iint_N f$.
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
edited Nov 27 '18 at 21:28
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
answered Nov 27 '18 at 21:22
A.Γ.
22k32455
22k32455
For the first part, since $f$ is continuous, and we've assumed that $f(p)>0$, does that mean that it would be better to set $epsilon = frac{f(p)}{2}$, so that we can choose a $delta$ about $p$, simply by convention?
– K.M
Nov 27 '18 at 22:12
@K.M No problem to call it $epsilon$, name does not really matter. I just followed the notation in the book hint that $f(x)gedelta$.
– A.Γ.
Nov 28 '18 at 7:47
add a comment |
For the first part, since $f$ is continuous, and we've assumed that $f(p)>0$, does that mean that it would be better to set $epsilon = frac{f(p)}{2}$, so that we can choose a $delta$ about $p$, simply by convention?
– K.M
Nov 27 '18 at 22:12
@K.M No problem to call it $epsilon$, name does not really matter. I just followed the notation in the book hint that $f(x)gedelta$.
– A.Γ.
Nov 28 '18 at 7:47
For the first part, since $f$ is continuous, and we've assumed that $f(p)>0$, does that mean that it would be better to set $epsilon = frac{f(p)}{2}$, so that we can choose a $delta$ about $p$, simply by convention?
– K.M
Nov 27 '18 at 22:12
For the first part, since $f$ is continuous, and we've assumed that $f(p)>0$, does that mean that it would be better to set $epsilon = frac{f(p)}{2}$, so that we can choose a $delta$ about $p$, simply by convention?
– K.M
Nov 27 '18 at 22:12
@K.M No problem to call it $epsilon$, name does not really matter. I just followed the notation in the book hint that $f(x)gedelta$.
– A.Γ.
Nov 28 '18 at 7:47
@K.M No problem to call it $epsilon$, name does not really matter. I just followed the notation in the book hint that $f(x)gedelta$.
– A.Γ.
Nov 28 '18 at 7:47
add a comment |
Using the hint from your book, assume, toward a contradiction, that $f(p)=r >0$ for some $pin D$. Then continuity of $f$ at $p$ implies that there is a square $[a,b]times [c,d]$ containing $p$ such that $f(x)>r/2$ whenever $xin [a,b]times [c,d].$ (Drawing a picture and picking a suitable $epsilon$ for the standard $epsilon-delta$ argument will help here).
But then, using an elementary property of the Riemann integral, we can write
$int_D fge int_{[a,b]times [c,d]}f>frac{r}{2}(b-a)(d-c)>0$ and we have our contradiction.
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
I was wondering if you meant that $f(p) = r < 0$ for some $p in D$ and that continuity of $f$ at $p$ implies that there is a square containing '$p$ such that $f(x) < r/2$? Also, should the square be an open set?
– K.M
Nov 27 '18 at 23:35
1
You know $fge0$ and you want to prove it is never strictly positive; so you are assuming it is positive somewhere, and getting a contradiction. The original square you get using a $delta-epsilon$ argument will be an open set, but you can then always get a CLOSED square inside it, so the conclusion still holds on the closed square. That's why I said to draw a picture!
– Matematleta
Nov 27 '18 at 23:40
Using $r = f(p_0)$, I can set $epsilon = frac{r}{2}$ and so I see that by definition of continuity of $f$, we have that $|f(x)-f(p)|< epsilon$ whenever $x in [a,b]$ x $[c,b]$ implies that $-epsilon< f(x) - f(p) < epsilon$, so that $f(x)>f(p)-epsilon = r - frac{r}{2} = frac{r}{2}$. Although, I would like to know as an aside if you could possibly explain how I could incorporate $delta$ into an $epsilon-delta$ argument?
– K.M
Nov 28 '18 at 0:19
You almost have it: set $p=(p_1,p_2)$. With your choice of $epsilon$ you can find a $delta>0$ (the actual value is not important) such that $xin (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})Rightarrow f(x)in (f(p)-epsilon,f(p)-epsilon)$. To conclude, take any closed rectangle inside $ (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})$ and follow the proof I gave.
– Matematleta
Nov 28 '18 at 1:23
add a comment |
Using the hint from your book, assume, toward a contradiction, that $f(p)=r >0$ for some $pin D$. Then continuity of $f$ at $p$ implies that there is a square $[a,b]times [c,d]$ containing $p$ such that $f(x)>r/2$ whenever $xin [a,b]times [c,d].$ (Drawing a picture and picking a suitable $epsilon$ for the standard $epsilon-delta$ argument will help here).
But then, using an elementary property of the Riemann integral, we can write
$int_D fge int_{[a,b]times [c,d]}f>frac{r}{2}(b-a)(d-c)>0$ and we have our contradiction.
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
I was wondering if you meant that $f(p) = r < 0$ for some $p in D$ and that continuity of $f$ at $p$ implies that there is a square containing '$p$ such that $f(x) < r/2$? Also, should the square be an open set?
– K.M
Nov 27 '18 at 23:35
1
You know $fge0$ and you want to prove it is never strictly positive; so you are assuming it is positive somewhere, and getting a contradiction. The original square you get using a $delta-epsilon$ argument will be an open set, but you can then always get a CLOSED square inside it, so the conclusion still holds on the closed square. That's why I said to draw a picture!
– Matematleta
Nov 27 '18 at 23:40
Using $r = f(p_0)$, I can set $epsilon = frac{r}{2}$ and so I see that by definition of continuity of $f$, we have that $|f(x)-f(p)|< epsilon$ whenever $x in [a,b]$ x $[c,b]$ implies that $-epsilon< f(x) - f(p) < epsilon$, so that $f(x)>f(p)-epsilon = r - frac{r}{2} = frac{r}{2}$. Although, I would like to know as an aside if you could possibly explain how I could incorporate $delta$ into an $epsilon-delta$ argument?
– K.M
Nov 28 '18 at 0:19
You almost have it: set $p=(p_1,p_2)$. With your choice of $epsilon$ you can find a $delta>0$ (the actual value is not important) such that $xin (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})Rightarrow f(x)in (f(p)-epsilon,f(p)-epsilon)$. To conclude, take any closed rectangle inside $ (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})$ and follow the proof I gave.
– Matematleta
Nov 28 '18 at 1:23
add a comment |
Using the hint from your book, assume, toward a contradiction, that $f(p)=r >0$ for some $pin D$. Then continuity of $f$ at $p$ implies that there is a square $[a,b]times [c,d]$ containing $p$ such that $f(x)>r/2$ whenever $xin [a,b]times [c,d].$ (Drawing a picture and picking a suitable $epsilon$ for the standard $epsilon-delta$ argument will help here).
But then, using an elementary property of the Riemann integral, we can write
$int_D fge int_{[a,b]times [c,d]}f>frac{r}{2}(b-a)(d-c)>0$ and we have our contradiction.
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
Using the hint from your book, assume, toward a contradiction, that $f(p)=r >0$ for some $pin D$. Then continuity of $f$ at $p$ implies that there is a square $[a,b]times [c,d]$ containing $p$ such that $f(x)>r/2$ whenever $xin [a,b]times [c,d].$ (Drawing a picture and picking a suitable $epsilon$ for the standard $epsilon-delta$ argument will help here).
But then, using an elementary property of the Riemann integral, we can write
$int_D fge int_{[a,b]times [c,d]}f>frac{r}{2}(b-a)(d-c)>0$ and we have our contradiction.
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
answered Nov 27 '18 at 22:28
Matematleta
9,9522918
9,9522918
I was wondering if you meant that $f(p) = r < 0$ for some $p in D$ and that continuity of $f$ at $p$ implies that there is a square containing '$p$ such that $f(x) < r/2$? Also, should the square be an open set?
– K.M
Nov 27 '18 at 23:35
1
You know $fge0$ and you want to prove it is never strictly positive; so you are assuming it is positive somewhere, and getting a contradiction. The original square you get using a $delta-epsilon$ argument will be an open set, but you can then always get a CLOSED square inside it, so the conclusion still holds on the closed square. That's why I said to draw a picture!
– Matematleta
Nov 27 '18 at 23:40
Using $r = f(p_0)$, I can set $epsilon = frac{r}{2}$ and so I see that by definition of continuity of $f$, we have that $|f(x)-f(p)|< epsilon$ whenever $x in [a,b]$ x $[c,b]$ implies that $-epsilon< f(x) - f(p) < epsilon$, so that $f(x)>f(p)-epsilon = r - frac{r}{2} = frac{r}{2}$. Although, I would like to know as an aside if you could possibly explain how I could incorporate $delta$ into an $epsilon-delta$ argument?
– K.M
Nov 28 '18 at 0:19
You almost have it: set $p=(p_1,p_2)$. With your choice of $epsilon$ you can find a $delta>0$ (the actual value is not important) such that $xin (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})Rightarrow f(x)in (f(p)-epsilon,f(p)-epsilon)$. To conclude, take any closed rectangle inside $ (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})$ and follow the proof I gave.
– Matematleta
Nov 28 '18 at 1:23
add a comment |
I was wondering if you meant that $f(p) = r < 0$ for some $p in D$ and that continuity of $f$ at $p$ implies that there is a square containing '$p$ such that $f(x) < r/2$? Also, should the square be an open set?
– K.M
Nov 27 '18 at 23:35
1
You know $fge0$ and you want to prove it is never strictly positive; so you are assuming it is positive somewhere, and getting a contradiction. The original square you get using a $delta-epsilon$ argument will be an open set, but you can then always get a CLOSED square inside it, so the conclusion still holds on the closed square. That's why I said to draw a picture!
– Matematleta
Nov 27 '18 at 23:40
Using $r = f(p_0)$, I can set $epsilon = frac{r}{2}$ and so I see that by definition of continuity of $f$, we have that $|f(x)-f(p)|< epsilon$ whenever $x in [a,b]$ x $[c,b]$ implies that $-epsilon< f(x) - f(p) < epsilon$, so that $f(x)>f(p)-epsilon = r - frac{r}{2} = frac{r}{2}$. Although, I would like to know as an aside if you could possibly explain how I could incorporate $delta$ into an $epsilon-delta$ argument?
– K.M
Nov 28 '18 at 0:19
You almost have it: set $p=(p_1,p_2)$. With your choice of $epsilon$ you can find a $delta>0$ (the actual value is not important) such that $xin (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})Rightarrow f(x)in (f(p)-epsilon,f(p)-epsilon)$. To conclude, take any closed rectangle inside $ (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})$ and follow the proof I gave.
– Matematleta
Nov 28 '18 at 1:23
I was wondering if you meant that $f(p) = r < 0$ for some $p in D$ and that continuity of $f$ at $p$ implies that there is a square containing '$p$ such that $f(x) < r/2$? Also, should the square be an open set?
– K.M
Nov 27 '18 at 23:35
I was wondering if you meant that $f(p) = r < 0$ for some $p in D$ and that continuity of $f$ at $p$ implies that there is a square containing '$p$ such that $f(x) < r/2$? Also, should the square be an open set?
– K.M
Nov 27 '18 at 23:35
1
1
You know $fge0$ and you want to prove it is never strictly positive; so you are assuming it is positive somewhere, and getting a contradiction. The original square you get using a $delta-epsilon$ argument will be an open set, but you can then always get a CLOSED square inside it, so the conclusion still holds on the closed square. That's why I said to draw a picture!
– Matematleta
Nov 27 '18 at 23:40
You know $fge0$ and you want to prove it is never strictly positive; so you are assuming it is positive somewhere, and getting a contradiction. The original square you get using a $delta-epsilon$ argument will be an open set, but you can then always get a CLOSED square inside it, so the conclusion still holds on the closed square. That's why I said to draw a picture!
– Matematleta
Nov 27 '18 at 23:40
Using $r = f(p_0)$, I can set $epsilon = frac{r}{2}$ and so I see that by definition of continuity of $f$, we have that $|f(x)-f(p)|< epsilon$ whenever $x in [a,b]$ x $[c,b]$ implies that $-epsilon< f(x) - f(p) < epsilon$, so that $f(x)>f(p)-epsilon = r - frac{r}{2} = frac{r}{2}$. Although, I would like to know as an aside if you could possibly explain how I could incorporate $delta$ into an $epsilon-delta$ argument?
– K.M
Nov 28 '18 at 0:19
Using $r = f(p_0)$, I can set $epsilon = frac{r}{2}$ and so I see that by definition of continuity of $f$, we have that $|f(x)-f(p)|< epsilon$ whenever $x in [a,b]$ x $[c,b]$ implies that $-epsilon< f(x) - f(p) < epsilon$, so that $f(x)>f(p)-epsilon = r - frac{r}{2} = frac{r}{2}$. Although, I would like to know as an aside if you could possibly explain how I could incorporate $delta$ into an $epsilon-delta$ argument?
– K.M
Nov 28 '18 at 0:19
You almost have it: set $p=(p_1,p_2)$. With your choice of $epsilon$ you can find a $delta>0$ (the actual value is not important) such that $xin (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})Rightarrow f(x)in (f(p)-epsilon,f(p)-epsilon)$. To conclude, take any closed rectangle inside $ (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})$ and follow the proof I gave.
– Matematleta
Nov 28 '18 at 1:23
You almost have it: set $p=(p_1,p_2)$. With your choice of $epsilon$ you can find a $delta>0$ (the actual value is not important) such that $xin (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})Rightarrow f(x)in (f(p)-epsilon,f(p)-epsilon)$. To conclude, take any closed rectangle inside $ (p_1-sqrt {delta},p_1+sqrt {delta}times(p_2-sqrt {delta},p_2+sqrt {delta})$ and follow the proof I gave.
– Matematleta
Nov 28 '18 at 1:23
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There is sort of a way to attack this if you break $D$ into two sets: places where $f(p) geq delta$ and places where it's not.
– Sean Roberson
Nov 27 '18 at 20:26
So that integral is improper? Do your theorems in the text permit that?
– Randall
Nov 27 '18 at 20:35
The integral is defined on $D$. Also, the theorem stated above requires that $D$ is bounded. Although, the definition says that the double integral exists for the conditions listed above for $R$, which the book states is a rectangle, so I think that that means that the integral must be bounded.
– K.M
Nov 27 '18 at 20:41
@SeanRoberson: is the $delta$ referring to the $delta$ used in $|S(N,f,{p_{ij}})| < delta$?
– K.M
Nov 27 '18 at 20:44
You do not need Case 2 as $f(x)ge 0$ for all $xin D$.
– A.Γ.
Nov 27 '18 at 20:56