How to determine Coercive functions












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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 $sqrt{x^2 + y^2} > K$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

For (c),
use
$e^x ge 1+x$,
so
$e^{x^2} ge 1+x^2$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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 $sqrt{x^2 + y^2} > K$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    0












    $begingroup$

    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 $sqrt{x^2 + y^2} > K$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    0












    0








    0





    $begingroup$

    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 $sqrt{x^2 + y^2} > K$.






    share|cite|improve this answer











    $endgroup$



    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 $sqrt{x^2 + y^2} > K$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 8 '15 at 18:49

























    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


















    • $begingroup$
      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











    0












    $begingroup$

    For (c),
    use
    $e^x ge 1+x$,
    so
    $e^{x^2} ge 1+x^2$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For (c),
      use
      $e^x ge 1+x$,
      so
      $e^{x^2} ge 1+x^2$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For (c),
        use
        $e^x ge 1+x$,
        so
        $e^{x^2} ge 1+x^2$.






        share|cite|improve this answer









        $endgroup$



        For (c),
        use
        $e^x ge 1+x$,
        so
        $e^{x^2} ge 1+x^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 8 '15 at 18:52









        marty cohenmarty cohen

        73.7k549128




        73.7k549128






























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