How to determine Coercive functions
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A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_Vert x Vert rightarrow infty f(x)=+ infty$.
I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.
$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^x^2+e^y^2+e^z^2$
To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?
optimization nonlinear-optimization
asked Sep 8 '15 at 16:43
clarkson
83111428
add a comment |Â
up vote
1
down vote
favorite
A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_Vert x Vert rightarrow infty f(x)=+ infty$.
I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.
$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^x^2+e^y^2+e^z^2$
To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?
optimization nonlinear-optimization
asked Sep 8 '15 at 16:43
clarkson
83111428
Yes, that is right.
– uniquesolution
Sep 8 '15 at 16:47
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
– MrYouMath
Sep 8 '15 at 16:54
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
– K. Miller
Sep 8 '15 at 17:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_Vert x Vert rightarrow infty f(x)=+ infty$.
I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.
$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^x^2+e^y^2+e^z^2$
To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?
optimization nonlinear-optimization
asked Sep 8 '15 at 16:43
clarkson
83111428
A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_Vert x Vert rightarrow infty f(x)=+ infty$.
I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.
$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^x^2+e^y^2+e^z^2$
To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?
optimization nonlinear-optimization
asked Sep 8 '15 at 16:43
clarkson
83111428
asked Sep 8 '15 at 16:43
clarkson
83111428
asked Sep 8 '15 at 16:43
clarkson
83111428
asked Sep 8 '15 at 16:43
clarkson
83111428
83111428
Yes, that is right.
– uniquesolution
Sep 8 '15 at 16:47
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
– MrYouMath
Sep 8 '15 at 16:54
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
– K. Miller
Sep 8 '15 at 17:06
add a comment |Â
Yes, that is right.
– uniquesolution
Sep 8 '15 at 16:47
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
– MrYouMath
Sep 8 '15 at 16:54
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
– K. Miller
Sep 8 '15 at 17:06
Yes, that is right.
– uniquesolution
Sep 8 '15 at 16:47
Yes, that is right.
– uniquesolution
Sep 8 '15 at 16:47
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
– MrYouMath
Sep 8 '15 at 16:54
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
– MrYouMath
Sep 8 '15 at 16:54
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
– K. Miller
Sep 8 '15 at 17:06
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
– K. Miller
Sep 8 '15 at 17:06
add a comment |Â
2 Answers
2
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0
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Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbfx) = |mathbfx|^2$. Now you can see that $f(mathbfx) to infty$ as $|mathbfx| to infty$.
Here is a hint for the second function. Use the inequality $-frac32(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrtx^2 + y^2 > K$.
answered Sep 8 '15 at 17:10
K. Miller
3,623612
What I don't understand is what is this $||x||$
– sam_rox
Sep 8 '15 at 17:14
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbfx = (x_1,ldots,x_n) in mathbbR^n$ the $2$-norm of $mathbfx$ is defined by $|mathbfx| = (x_1^2 + cdots + x_n^2)^1/2$. In two variables $x$ and $y$ you have $|(x,y)| = sqrtx^2 + y^2$.
– K. Miller
Sep 8 '15 at 17:20
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrtx^2 + y^2$. For $f(x,y,z)=e^x^2+e^y^2+e^z^2$ as there are three variables is the norm = $sqrtx^2+y^2+z^2$
– clarkson
Sep 8 '15 at 17:30
No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
– K. Miller
Sep 8 '15 at 17:35
@K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrtx^8+y^8$
– clarkson
Sep 8 '15 at 17:53
 |Â
show 1 more comment
up vote
0
down vote
For (c),
use
$e^x ge 1+x$,
so
$e^x^2 ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbfx) = |mathbfx|^2$. Now you can see that $f(mathbfx) to infty$ as $|mathbfx| to infty$.
Here is a hint for the second function. Use the inequality $-frac32(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrtx^2 + y^2 > K$.
answered Sep 8 '15 at 17:10
K. Miller
3,623612
What I don't understand is what is this $||x||$
– sam_rox
Sep 8 '15 at 17:14
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbfx = (x_1,ldots,x_n) in mathbbR^n$ the $2$-norm of $mathbfx$ is defined by $|mathbfx| = (x_1^2 + cdots + x_n^2)^1/2$. In two variables $x$ and $y$ you have $|(x,y)| = sqrtx^2 + y^2$.
– K. Miller
Sep 8 '15 at 17:20
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrtx^2 + y^2$. For $f(x,y,z)=e^x^2+e^y^2+e^z^2$ as there are three variables is the norm = $sqrtx^2+y^2+z^2$
– clarkson
Sep 8 '15 at 17:30
No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
– K. Miller
Sep 8 '15 at 17:35
@K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrtx^8+y^8$
– clarkson
Sep 8 '15 at 17:53
 |Â
show 1 more comment
up vote
0
down vote
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbfx) = |mathbfx|^2$. Now you can see that $f(mathbfx) to infty$ as $|mathbfx| to infty$.
Here is a hint for the second function. Use the inequality $-frac32(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrtx^2 + y^2 > K$.
answered Sep 8 '15 at 17:10
K. Miller
3,623612
What I don't understand is what is this $||x||$
– sam_rox
Sep 8 '15 at 17:14
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbfx = (x_1,ldots,x_n) in mathbbR^n$ the $2$-norm of $mathbfx$ is defined by $|mathbfx| = (x_1^2 + cdots + x_n^2)^1/2$. In two variables $x$ and $y$ you have $|(x,y)| = sqrtx^2 + y^2$.
– K. Miller
Sep 8 '15 at 17:20
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrtx^2 + y^2$. For $f(x,y,z)=e^x^2+e^y^2+e^z^2$ as there are three variables is the norm = $sqrtx^2+y^2+z^2$
– clarkson
Sep 8 '15 at 17:30
No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
– K. Miller
Sep 8 '15 at 17:35
@K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrtx^8+y^8$
– clarkson
Sep 8 '15 at 17:53
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbfx) = |mathbfx|^2$. Now you can see that $f(mathbfx) to infty$ as $|mathbfx| to infty$.
Here is a hint for the second function. Use the inequality $-frac32(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrtx^2 + y^2 > K$.
answered Sep 8 '15 at 17:10
K. Miller
3,623612
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbfx) = |mathbfx|^2$. Now you can see that $f(mathbfx) to infty$ as $|mathbfx| to infty$.
Here is a hint for the second function. Use the inequality $-frac32(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrtx^2 + y^2 > K$.
answered Sep 8 '15 at 17:10
K. Miller
3,623612
edited Sep 8 '15 at 18:49
answered Sep 8 '15 at 17:10
K. Miller
3,623612
answered Sep 8 '15 at 17:10
K. Miller
3,623612
answered Sep 8 '15 at 17:10
K. Miller
3,623612
3,623612
What I don't understand is what is this $||x||$
– sam_rox
Sep 8 '15 at 17:14
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbfx = (x_1,ldots,x_n) in mathbbR^n$ the $2$-norm of $mathbfx$ is defined by $|mathbfx| = (x_1^2 + cdots + x_n^2)^1/2$. In two variables $x$ and $y$ you have $|(x,y)| = sqrtx^2 + y^2$.
– K. Miller
Sep 8 '15 at 17:20
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrtx^2 + y^2$. For $f(x,y,z)=e^x^2+e^y^2+e^z^2$ as there are three variables is the norm = $sqrtx^2+y^2+z^2$
– clarkson
Sep 8 '15 at 17:30
No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
– K. Miller
Sep 8 '15 at 17:35
@K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrtx^8+y^8$
– clarkson
Sep 8 '15 at 17:53
 |Â
show 1 more comment
What I don't understand is what is this $||x||$
– sam_rox
Sep 8 '15 at 17:14
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbfx = (x_1,ldots,x_n) in mathbbR^n$ the $2$-norm of $mathbfx$ is defined by $|mathbfx| = (x_1^2 + cdots + x_n^2)^1/2$. In two variables $x$ and $y$ you have $|(x,y)| = sqrtx^2 + y^2$.
– K. Miller
Sep 8 '15 at 17:20
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrtx^2 + y^2$. For $f(x,y,z)=e^x^2+e^y^2+e^z^2$ as there are three variables is the norm = $sqrtx^2+y^2+z^2$
– clarkson
Sep 8 '15 at 17:30
No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
– K. Miller
Sep 8 '15 at 17:35
@K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrtx^8+y^8$
– clarkson
Sep 8 '15 at 17:53
What I don't understand is what is this $||x||$
– sam_rox
Sep 8 '15 at 17:14
What I don't understand is what is this $||x||$
– sam_rox
Sep 8 '15 at 17:14
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbfx = (x_1,ldots,x_n) in mathbbR^n$ the $2$-norm of $mathbfx$ is defined by $|mathbfx| = (x_1^2 + cdots + x_n^2)^1/2$. In two variables $x$ and $y$ you have $|(x,y)| = sqrtx^2 + y^2$.
– K. Miller
Sep 8 '15 at 17:20
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbfx = (x_1,ldots,x_n) in mathbbR^n$ the $2$-norm of $mathbfx$ is defined by $|mathbfx| = (x_1^2 + cdots + x_n^2)^1/2$. In two variables $x$ and $y$ you have $|(x,y)| = sqrtx^2 + y^2$.
– K. Miller
Sep 8 '15 at 17:20
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrtx^2 + y^2$. For $f(x,y,z)=e^x^2+e^y^2+e^z^2$ as there are three variables is the norm = $sqrtx^2+y^2+z^2$
– clarkson
Sep 8 '15 at 17:30
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrtx^2 + y^2$. For $f(x,y,z)=e^x^2+e^y^2+e^z^2$ as there are three variables is the norm = $sqrtx^2+y^2+z^2$
– clarkson
Sep 8 '15 at 17:30
No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
– K. Miller
Sep 8 '15 at 17:35
No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
– K. Miller
Sep 8 '15 at 17:35
@K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrtx^8+y^8$
– clarkson
Sep 8 '15 at 17:53
@K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrtx^8+y^8$
– clarkson
Sep 8 '15 at 17:53
 |Â
show 1 more comment
up vote
0
down vote
For (c),
use
$e^x ge 1+x$,
so
$e^x^2 ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
add a comment |Â
up vote
0
down vote
For (c),
use
$e^x ge 1+x$,
so
$e^x^2 ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For (c),
use
$e^x ge 1+x$,
so
$e^x^2 ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
For (c),
use
$e^x ge 1+x$,
so
$e^x^2 ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
answered Sep 8 '15 at 18:52
marty cohen
69.8k446122
69.8k446122
add a comment |Â
add a comment |Â
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Yes, that is right.
– uniquesolution
Sep 8 '15 at 16:47
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
– MrYouMath
Sep 8 '15 at 16:54
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
– K. Miller
Sep 8 '15 at 17:06