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Let $\Omega \subset \Bbb R ^2$ be bounded and closed, and let $g : \Omega \to \Bbb [0, \infty)$ be continuous. Let $g(x_0,y_0)=\max \limits_{\Omega} g(x,y)$. Show that:

$\exists \rho_{x_0},\rho_{y_0}>0$, $\forall(x,y)\in \Omega,|x-x_0|<\rho_{x_0},|y-y_0|<\rho_{y_0}$, then $g(x,y_0)\geq g(x,y)$.

In fact, I think the $\Omega$ is bounded and closed is not necessary. All of the above is my guess, I really don't know whether it is right. Thanks for any answer or advice.

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  • $\begingroup$ The assumption of $\Omega$ being closed and bounded (i.e. compact) is essential: without it, there is no guarantee that $g$ is bounded or, if it is, there is no guarantee that it reaches its upper bound. $\endgroup$
    – Alex M.
    Sep 25 '15 at 9:57
  • $\begingroup$ @AlexM. If the upper bounde can be reached ,I think closed and bounded is needless $\endgroup$
    – lanse2pty
    Sep 26 '15 at 0:37
  • $\begingroup$ If it can be reached, then yes, you are right. $\endgroup$
    – Alex M.
    Sep 26 '15 at 9:35
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It is not true. The strict inequality is trivially false by taking $x=x_0$, $y=y_0$. The non-strict one is not true either. Consider $g(x,y)=-(x-y)^2$ and $(x_0,y_0)=(0,0)$. Then $$ g(x,y_0)=-x^2<g(x,x)=0 $$ for all $x\ne x_0$.

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  • $\begingroup$ I have edited it , Now ,is it right? $\endgroup$
    – lanse2pty
    Sep 26 '15 at 0:41
  • $\begingroup$ @lanse2pty Read my answer, it has a counterexample for non-strict inequality. $\endgroup$
    – A.Γ.
    Sep 26 '15 at 8:17
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The first condition $$g(x_0,y_0)=\max \limits_{\Omega} g(x,y)$$ means that the function has a global maximum at $(x_0,y_0)$. But that alone does not imply it is a strict maximum – a constant function $g=\mathrm{const.}$ will have a global maximum everywhere, so for any two points $g$ values at those points are equal (a strict inequality doesn't hold).

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  • $\begingroup$ yes, I have edited it . $\endgroup$
    – lanse2pty
    Sep 26 '15 at 0:41

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