If $f:mathbb{R}^ntimesmathbb{R}^ntomathbb{R}$ is a continuous and bounded function, is the map...












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I think is true but I didn't find a proof. Any idea is appreciated.






general-topology analysis







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closed as off-topic by user21820, José Carlos Santos, Did, saz, RRL Feb 5 at 23:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Did, saz, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    I think is true but I didn't find a proof. Any idea is appreciated.






    general-topology analysis







    share|cite|improve this question









    $endgroup$



    closed as off-topic by user21820, José Carlos Santos, Did, saz, RRL Feb 5 at 23:31


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Did, saz, RRL

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1





      $begingroup$


      I think is true but I didn't find a proof. Any idea is appreciated.






      general-topology analysis







      share|cite|improve this question









      $endgroup$




      I think is true but I didn't find a proof. Any idea is appreciated.






      general-topology analysis




      general-topology analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 21 at 11:32









      MarcoMarco

      331110




      331110




      closed as off-topic by user21820, José Carlos Santos, Did, saz, RRL Feb 5 at 23:31


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Did, saz, RRL

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by user21820, José Carlos Santos, Did, saz, RRL Feb 5 at 23:31


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Did, saz, RRL

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
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          1












          $begingroup$

          Let $f(x,y)=0$ for $x leq 0$, $e^{|y|} x$ for $0leq x leq e^{-|y|}$ and $1$ for $x >e^{-|y|}$. Then $f$ is continuous and $sup_y f(x,y) $ is $0$ for $xleq 0$, $1$ for $x > 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Ingix Thansk for pointing out.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:15



















          0












          $begingroup$

          Let $n=1$, $f(x,y)=arctan(xy)$ and , for $x in mathbb R$ let $g(x):= sup_yf(x,y).$



          If $x=0$, then $g(0)=0$ and for $x ne 0$ we have $g(x)= pi /2.$



          Conclusion ?






          share|cite|improve this answer









          $endgroup$




















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Let $f(x,y)=0$ for $x leq 0$, $e^{|y|} x$ for $0leq x leq e^{-|y|}$ and $1$ for $x >e^{-|y|}$. Then $f$ is continuous and $sup_y f(x,y) $ is $0$ for $xleq 0$, $1$ for $x > 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @Ingix Thansk for pointing out.
              $endgroup$
              – Kavi Rama Murthy
              Jan 21 at 23:15
















            1












            $begingroup$

            Let $f(x,y)=0$ for $x leq 0$, $e^{|y|} x$ for $0leq x leq e^{-|y|}$ and $1$ for $x >e^{-|y|}$. Then $f$ is continuous and $sup_y f(x,y) $ is $0$ for $xleq 0$, $1$ for $x > 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @Ingix Thansk for pointing out.
              $endgroup$
              – Kavi Rama Murthy
              Jan 21 at 23:15














            1












            1








            1





            $begingroup$

            Let $f(x,y)=0$ for $x leq 0$, $e^{|y|} x$ for $0leq x leq e^{-|y|}$ and $1$ for $x >e^{-|y|}$. Then $f$ is continuous and $sup_y f(x,y) $ is $0$ for $xleq 0$, $1$ for $x > 0$.






            share|cite|improve this answer











            $endgroup$



            Let $f(x,y)=0$ for $x leq 0$, $e^{|y|} x$ for $0leq x leq e^{-|y|}$ and $1$ for $x >e^{-|y|}$. Then $f$ is continuous and $sup_y f(x,y) $ is $0$ for $xleq 0$, $1$ for $x > 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 21 at 23:15

























            answered Jan 21 at 12:07









            Kavi Rama MurthyKavi Rama Murthy

            65.3k42766




            65.3k42766












            • $begingroup$
              @Ingix Thansk for pointing out.
              $endgroup$
              – Kavi Rama Murthy
              Jan 21 at 23:15


















            • $begingroup$
              @Ingix Thansk for pointing out.
              $endgroup$
              – Kavi Rama Murthy
              Jan 21 at 23:15
















            $begingroup$
            @Ingix Thansk for pointing out.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:15




            $begingroup$
            @Ingix Thansk for pointing out.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:15











            0












            $begingroup$

            Let $n=1$, $f(x,y)=arctan(xy)$ and , for $x in mathbb R$ let $g(x):= sup_yf(x,y).$



            If $x=0$, then $g(0)=0$ and for $x ne 0$ we have $g(x)= pi /2.$



            Conclusion ?






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let $n=1$, $f(x,y)=arctan(xy)$ and , for $x in mathbb R$ let $g(x):= sup_yf(x,y).$



              If $x=0$, then $g(0)=0$ and for $x ne 0$ we have $g(x)= pi /2.$



              Conclusion ?






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $n=1$, $f(x,y)=arctan(xy)$ and , for $x in mathbb R$ let $g(x):= sup_yf(x,y).$



                If $x=0$, then $g(0)=0$ and for $x ne 0$ we have $g(x)= pi /2.$



                Conclusion ?






                share|cite|improve this answer









                $endgroup$



                Let $n=1$, $f(x,y)=arctan(xy)$ and , for $x in mathbb R$ let $g(x):= sup_yf(x,y).$



                If $x=0$, then $g(0)=0$ and for $x ne 0$ we have $g(x)= pi /2.$



                Conclusion ?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 12:07









                FredFred

                47.7k1849




                47.7k1849















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