Check if function is coercive

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OK, but you need to stay in the domain of definition of $U$ which is $D=]0,+∞[×]0,+∞[$.

So you need $||(x,y)||\to+\infty$ only in $D$ for the norm you have chosen.

WARNING: It is different from $|x|\to+\infty\text{ AND }|y|\to+\infty$


For instance for $n\in\mathbb N^*$ and $n\to\infty$ we have $||(n,\frac 1{n^6})||\to+\infty$

$U(n,\frac 1{n^6})=6\ln(n)+\ln(\frac 1{n^6})=6\ln(n)-6\ln(n)=0$

Thus $U$ cannot be coercive.

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lala_12
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lala_12

Updated on September 28, 2020

Comments

  • lala_12
    lala_12 about 3 years

    I am trying to check if the function $U(x,y)=6\ln x+\ln y$ is coercive. I know, that i need to check if $\lim_{\vert (x,y) \vert \rightarrow \infty} U(x,y) = +\infty$, and so far I have

    $\lim_{(x,y) \rightarrow \infty} U(x,y) = +\infty$

    $\lim_{ -(x,y) \rightarrow \infty} U(x,y) = -\infty$

    Does this mean that the function is not coercive?

    Thank you.

    • zwim
      zwim about 6 years
      Why do you consider the case $(x,y)\to 0$ ?
    • lala_12
      lala_12 about 6 years
      @zwim that was a typo, I have corrected my question.
  • Fred
    Fred about 6 years
    Why the downvote ??????