How to determine Coercive functions
$begingroup$
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
clarksonclarkson
87111533
$endgroup$
add a comment |
$begingroup$
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
clarksonclarkson
87111533
$endgroup$
$begingroup$
Yes, that is right.
$endgroup$
– uniquesolution
Sep 8 '15 at 16:47
$begingroup$
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
$endgroup$
– MrYouMath
Sep 8 '15 at 16:54
$begingroup$
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
$endgroup$
– K. Miller
Sep 8 '15 at 17:06
add a comment |
$begingroup$
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
clarksonclarkson
87111533
$endgroup$
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
optimization nonlinear-optimization
asked Sep 8 '15 at 16:43
clarksonclarkson
87111533
asked Sep 8 '15 at 16:43
clarksonclarkson
87111533
asked Sep 8 '15 at 16:43
clarksonclarkson
87111533
asked Sep 8 '15 at 16:43
clarksonclarkson
87111533
asked Sep 8 '15 at 16:43
clarksonclarkson
87111533
87111533
$begingroup$
Yes, that is right.
$endgroup$
– uniquesolution
Sep 8 '15 at 16:47
$begingroup$
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
$endgroup$
– MrYouMath
Sep 8 '15 at 16:54
$begingroup$
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
$endgroup$
– K. Miller
Sep 8 '15 at 17:06
add a comment |
$begingroup$
Yes, that is right.
$endgroup$
– uniquesolution
Sep 8 '15 at 16:47
$begingroup$
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
$endgroup$
– MrYouMath
Sep 8 '15 at 16:54
$begingroup$
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
$endgroup$
– K. Miller
Sep 8 '15 at 17:06
$begingroup$
Yes, that is right.
$endgroup$
– uniquesolution
Sep 8 '15 at 16:47
$begingroup$
Yes, that is right.
$endgroup$
– uniquesolution
Sep 8 '15 at 16:47
$begingroup$
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
$endgroup$
– MrYouMath
Sep 8 '15 at 16:54
$begingroup$
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
$endgroup$
– MrYouMath
Sep 8 '15 at 16:54
$begingroup$
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
$endgroup$
– K. Miller
Sep 8 '15 at 17:06
$begingroup$
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
$endgroup$
– K. Miller
Sep 8 '15 at 17:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.
Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ 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 $sqrt{x^2 + y^2} > K$.
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
$endgroup$
$begingroup$
What I don't understand is what is this $||x||$
$endgroup$
– sam_rox
Sep 8 '15 at 17:14
$begingroup$
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:20
$begingroup$
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
$endgroup$
– clarkson
Sep 8 '15 at 17:30
$begingroup$
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 $f(x,y)$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:35
$begingroup$
@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 $sqrt{x^8+y^8}$
$endgroup$
– clarkson
Sep 8 '15 at 17:53
|
show 1 more comment
$begingroup$
For (c),
use
$e^x ge 1+x$,
so
$e^{x^2} ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.
Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ 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 $sqrt{x^2 + y^2} > K$.
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
$endgroup$
$begingroup$
What I don't understand is what is this $||x||$
$endgroup$
– sam_rox
Sep 8 '15 at 17:14
$begingroup$
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:20
$begingroup$
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
$endgroup$
– clarkson
Sep 8 '15 at 17:30
$begingroup$
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 $f(x,y)$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:35
$begingroup$
@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 $sqrt{x^8+y^8}$
$endgroup$
– clarkson
Sep 8 '15 at 17:53
|
show 1 more comment
$begingroup$
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.
Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ 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 $sqrt{x^2 + y^2} > K$.
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
$endgroup$
$begingroup$
What I don't understand is what is this $||x||$
$endgroup$
– sam_rox
Sep 8 '15 at 17:14
$begingroup$
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:20
$begingroup$
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
$endgroup$
– clarkson
Sep 8 '15 at 17:30
$begingroup$
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 $f(x,y)$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:35
$begingroup$
@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 $sqrt{x^8+y^8}$
$endgroup$
– clarkson
Sep 8 '15 at 17:53
|
show 1 more comment
$begingroup$
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.
Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ 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 $sqrt{x^2 + y^2} > K$.
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
$endgroup$
Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.
Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ 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 $sqrt{x^2 + y^2} > K$.
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
edited Sep 8 '15 at 18:49
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
answered Sep 8 '15 at 17:10
K. MillerK. Miller
3,633612
3,633612
$begingroup$
What I don't understand is what is this $||x||$
$endgroup$
– sam_rox
Sep 8 '15 at 17:14
$begingroup$
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:20
$begingroup$
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
$endgroup$
– clarkson
Sep 8 '15 at 17:30
$begingroup$
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 $f(x,y)$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:35
$begingroup$
@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 $sqrt{x^8+y^8}$
$endgroup$
– clarkson
Sep 8 '15 at 17:53
|
show 1 more comment
$begingroup$
What I don't understand is what is this $||x||$
$endgroup$
– sam_rox
Sep 8 '15 at 17:14
$begingroup$
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:20
$begingroup$
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
$endgroup$
– clarkson
Sep 8 '15 at 17:30
$begingroup$
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 $f(x,y)$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:35
$begingroup$
@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 $sqrt{x^8+y^8}$
$endgroup$
– clarkson
Sep 8 '15 at 17:53
$begingroup$
What I don't understand is what is this $||x||$
$endgroup$
– sam_rox
Sep 8 '15 at 17:14
$begingroup$
What I don't understand is what is this $||x||$
$endgroup$
– sam_rox
Sep 8 '15 at 17:14
$begingroup$
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:20
$begingroup$
The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:20
$begingroup$
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
$endgroup$
– clarkson
Sep 8 '15 at 17:30
$begingroup$
@K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
$endgroup$
– clarkson
Sep 8 '15 at 17:30
$begingroup$
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 $f(x,y)$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:35
$begingroup$
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 $f(x,y)$.
$endgroup$
– K. Miller
Sep 8 '15 at 17:35
$begingroup$
@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 $sqrt{x^8+y^8}$
$endgroup$
– clarkson
Sep 8 '15 at 17:53
$begingroup$
@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 $sqrt{x^8+y^8}$
$endgroup$
– clarkson
Sep 8 '15 at 17:53
|
show 1 more comment
$begingroup$
For (c),
use
$e^x ge 1+x$,
so
$e^{x^2} ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
$endgroup$
add a comment |
$begingroup$
For (c),
use
$e^x ge 1+x$,
so
$e^{x^2} ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
$endgroup$
add a comment |
$begingroup$
For (c),
use
$e^x ge 1+x$,
so
$e^{x^2} ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
$endgroup$
For (c),
use
$e^x ge 1+x$,
so
$e^{x^2} ge 1+x^2$.
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
answered Sep 8 '15 at 18:52
marty cohenmarty cohen
73.7k549128
73.7k549128
add a comment |
add a comment |
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Yes, that is right.
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– uniquesolution
Sep 8 '15 at 16:47
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I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
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– MrYouMath
Sep 8 '15 at 16:54
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The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
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– K. Miller
Sep 8 '15 at 17:06