Finding a function of two variables satisfying some condtions












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I am trying to find a function of two variables $f(x, y): [0, infty) times mathbb{R} to mathbb{R}$, satisfying the following conditions:



(i) $f(0, y) = 1$ for all $y in mathbb{R}$;



(ii) $f(x, y) > 0$ for all $x in (0, infty)$ if $y geq 0$;



(iii) $f(x, y) < 0$ for all $x in (0, infty)$ if $y < 0$;



Does there exist a function f(x, y) which satisfies the conditions (i)-(iii)?



Thank you very much for your support.






real-analysis functional-analysis numerical-methods







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$endgroup$








  • 3




    $begingroup$
    Your conditions $(i)$ and $(iii)$ are contradictory.
    $endgroup$
    – Song
    Dec 16 '18 at 6:47










  • $begingroup$
    I agree with @Song. Maybe in $(iii)$ you wanted to write $f(x,y) < 0$ for all $x in (0, infty)$ instead of $ [0, infty)$?
    $endgroup$
    – Hermione
    Dec 16 '18 at 12:04












  • $begingroup$
    @Hermione Thanks for your very good response. I have corrected this condition.
    $endgroup$
    – MichaelCarrick
    Dec 16 '18 at 15:32
















0












$begingroup$


I am trying to find a function of two variables $f(x, y): [0, infty) times mathbb{R} to mathbb{R}$, satisfying the following conditions:



(i) $f(0, y) = 1$ for all $y in mathbb{R}$;



(ii) $f(x, y) > 0$ for all $x in (0, infty)$ if $y geq 0$;



(iii) $f(x, y) < 0$ for all $x in (0, infty)$ if $y < 0$;



Does there exist a function f(x, y) which satisfies the conditions (i)-(iii)?



Thank you very much for your support.






real-analysis functional-analysis numerical-methods







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Your conditions $(i)$ and $(iii)$ are contradictory.
    $endgroup$
    – Song
    Dec 16 '18 at 6:47










  • $begingroup$
    I agree with @Song. Maybe in $(iii)$ you wanted to write $f(x,y) < 0$ for all $x in (0, infty)$ instead of $ [0, infty)$?
    $endgroup$
    – Hermione
    Dec 16 '18 at 12:04












  • $begingroup$
    @Hermione Thanks for your very good response. I have corrected this condition.
    $endgroup$
    – MichaelCarrick
    Dec 16 '18 at 15:32














0












0








0





$begingroup$


I am trying to find a function of two variables $f(x, y): [0, infty) times mathbb{R} to mathbb{R}$, satisfying the following conditions:



(i) $f(0, y) = 1$ for all $y in mathbb{R}$;



(ii) $f(x, y) > 0$ for all $x in (0, infty)$ if $y geq 0$;



(iii) $f(x, y) < 0$ for all $x in (0, infty)$ if $y < 0$;



Does there exist a function f(x, y) which satisfies the conditions (i)-(iii)?



Thank you very much for your support.






real-analysis functional-analysis numerical-methods







share|cite|improve this question











$endgroup$




I am trying to find a function of two variables $f(x, y): [0, infty) times mathbb{R} to mathbb{R}$, satisfying the following conditions:



(i) $f(0, y) = 1$ for all $y in mathbb{R}$;



(ii) $f(x, y) > 0$ for all $x in (0, infty)$ if $y geq 0$;



(iii) $f(x, y) < 0$ for all $x in (0, infty)$ if $y < 0$;



Does there exist a function f(x, y) which satisfies the conditions (i)-(iii)?



Thank you very much for your support.






real-analysis functional-analysis numerical-methods




real-analysis functional-analysis numerical-methods






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 15:33







MichaelCarrick

















asked Dec 16 '18 at 6:34









MichaelCarrickMichaelCarrick

1097




1097








  • 3




    $begingroup$
    Your conditions $(i)$ and $(iii)$ are contradictory.
    $endgroup$
    – Song
    Dec 16 '18 at 6:47










  • $begingroup$
    I agree with @Song. Maybe in $(iii)$ you wanted to write $f(x,y) < 0$ for all $x in (0, infty)$ instead of $ [0, infty)$?
    $endgroup$
    – Hermione
    Dec 16 '18 at 12:04












  • $begingroup$
    @Hermione Thanks for your very good response. I have corrected this condition.
    $endgroup$
    – MichaelCarrick
    Dec 16 '18 at 15:32














  • 3




    $begingroup$
    Your conditions $(i)$ and $(iii)$ are contradictory.
    $endgroup$
    – Song
    Dec 16 '18 at 6:47










  • $begingroup$
    I agree with @Song. Maybe in $(iii)$ you wanted to write $f(x,y) < 0$ for all $x in (0, infty)$ instead of $ [0, infty)$?
    $endgroup$
    – Hermione
    Dec 16 '18 at 12:04












  • $begingroup$
    @Hermione Thanks for your very good response. I have corrected this condition.
    $endgroup$
    – MichaelCarrick
    Dec 16 '18 at 15:32








3




3




$begingroup$
Your conditions $(i)$ and $(iii)$ are contradictory.
$endgroup$
– Song
Dec 16 '18 at 6:47




$begingroup$
Your conditions $(i)$ and $(iii)$ are contradictory.
$endgroup$
– Song
Dec 16 '18 at 6:47












$begingroup$
I agree with @Song. Maybe in $(iii)$ you wanted to write $f(x,y) < 0$ for all $x in (0, infty)$ instead of $ [0, infty)$?
$endgroup$
– Hermione
Dec 16 '18 at 12:04






$begingroup$
I agree with @Song. Maybe in $(iii)$ you wanted to write $f(x,y) < 0$ for all $x in (0, infty)$ instead of $ [0, infty)$?
$endgroup$
– Hermione
Dec 16 '18 at 12:04














$begingroup$
@Hermione Thanks for your very good response. I have corrected this condition.
$endgroup$
– MichaelCarrick
Dec 16 '18 at 15:32




$begingroup$
@Hermione Thanks for your very good response. I have corrected this condition.
$endgroup$
– MichaelCarrick
Dec 16 '18 at 15:32










2 Answers
2






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$begingroup$

You can consider the function $f$ defined as
$$ f (x, y) = 1 quad text{if } x=0 text{ for all } y in mathbb{R}$$
$$ f (x, y) = 1 quad text{if } x>0 text{ and } y geq 0$$
$$ f (x, y) = -1 quad text{if } x>0 text{ and } y < 0$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $${1over2}(2operatorname{heavi}(x)-operatorname{krondelta}(x))(2operatorname{heavi}(y)+operatorname{krondelta}(y)-1)+operatorname{krondelta}(x) ,$$where $$operatorname{heavi}$$is the Heaviside function, &$$operatorname{krondelta}$$the Kronecker $delta$ function.



    This could be expressed as a limit of continuous functions: $$lim_{atoinfty}{1over2}(tanh(ax)-exp(-(ax)^2)+1)(tanh(ay)+exp(-(ay)^2)+exp(-(ax)^2) .$$or$$lim_{atoinfty}{1over2}(tanh(ax)-operatorname{sech}(ax)+1)(tanh(ay)+operatorname{sech}(ay))+operatorname{sech}(ax) .$$






    share|cite|improve this answer











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      2 Answers
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      2 Answers
      2






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      0












      $begingroup$

      You can consider the function $f$ defined as
      $$ f (x, y) = 1 quad text{if } x=0 text{ for all } y in mathbb{R}$$
      $$ f (x, y) = 1 quad text{if } x>0 text{ and } y geq 0$$
      $$ f (x, y) = -1 quad text{if } x>0 text{ and } y < 0$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        You can consider the function $f$ defined as
        $$ f (x, y) = 1 quad text{if } x=0 text{ for all } y in mathbb{R}$$
        $$ f (x, y) = 1 quad text{if } x>0 text{ and } y geq 0$$
        $$ f (x, y) = -1 quad text{if } x>0 text{ and } y < 0$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          You can consider the function $f$ defined as
          $$ f (x, y) = 1 quad text{if } x=0 text{ for all } y in mathbb{R}$$
          $$ f (x, y) = 1 quad text{if } x>0 text{ and } y geq 0$$
          $$ f (x, y) = -1 quad text{if } x>0 text{ and } y < 0$$






          share|cite|improve this answer









          $endgroup$



          You can consider the function $f$ defined as
          $$ f (x, y) = 1 quad text{if } x=0 text{ for all } y in mathbb{R}$$
          $$ f (x, y) = 1 quad text{if } x>0 text{ and } y geq 0$$
          $$ f (x, y) = -1 quad text{if } x>0 text{ and } y < 0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 16:12









          HermioneHermione

          19619




          19619























              1












              $begingroup$

              $${1over2}(2operatorname{heavi}(x)-operatorname{krondelta}(x))(2operatorname{heavi}(y)+operatorname{krondelta}(y)-1)+operatorname{krondelta}(x) ,$$where $$operatorname{heavi}$$is the Heaviside function, &$$operatorname{krondelta}$$the Kronecker $delta$ function.



              This could be expressed as a limit of continuous functions: $$lim_{atoinfty}{1over2}(tanh(ax)-exp(-(ax)^2)+1)(tanh(ay)+exp(-(ay)^2)+exp(-(ax)^2) .$$or$$lim_{atoinfty}{1over2}(tanh(ax)-operatorname{sech}(ax)+1)(tanh(ay)+operatorname{sech}(ay))+operatorname{sech}(ax) .$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                $${1over2}(2operatorname{heavi}(x)-operatorname{krondelta}(x))(2operatorname{heavi}(y)+operatorname{krondelta}(y)-1)+operatorname{krondelta}(x) ,$$where $$operatorname{heavi}$$is the Heaviside function, &$$operatorname{krondelta}$$the Kronecker $delta$ function.



                This could be expressed as a limit of continuous functions: $$lim_{atoinfty}{1over2}(tanh(ax)-exp(-(ax)^2)+1)(tanh(ay)+exp(-(ay)^2)+exp(-(ax)^2) .$$or$$lim_{atoinfty}{1over2}(tanh(ax)-operatorname{sech}(ax)+1)(tanh(ay)+operatorname{sech}(ay))+operatorname{sech}(ax) .$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $${1over2}(2operatorname{heavi}(x)-operatorname{krondelta}(x))(2operatorname{heavi}(y)+operatorname{krondelta}(y)-1)+operatorname{krondelta}(x) ,$$where $$operatorname{heavi}$$is the Heaviside function, &$$operatorname{krondelta}$$the Kronecker $delta$ function.



                  This could be expressed as a limit of continuous functions: $$lim_{atoinfty}{1over2}(tanh(ax)-exp(-(ax)^2)+1)(tanh(ay)+exp(-(ay)^2)+exp(-(ax)^2) .$$or$$lim_{atoinfty}{1over2}(tanh(ax)-operatorname{sech}(ax)+1)(tanh(ay)+operatorname{sech}(ay))+operatorname{sech}(ax) .$$






                  share|cite|improve this answer











                  $endgroup$



                  $${1over2}(2operatorname{heavi}(x)-operatorname{krondelta}(x))(2operatorname{heavi}(y)+operatorname{krondelta}(y)-1)+operatorname{krondelta}(x) ,$$where $$operatorname{heavi}$$is the Heaviside function, &$$operatorname{krondelta}$$the Kronecker $delta$ function.



                  This could be expressed as a limit of continuous functions: $$lim_{atoinfty}{1over2}(tanh(ax)-exp(-(ax)^2)+1)(tanh(ay)+exp(-(ay)^2)+exp(-(ax)^2) .$$or$$lim_{atoinfty}{1over2}(tanh(ax)-operatorname{sech}(ax)+1)(tanh(ay)+operatorname{sech}(ay))+operatorname{sech}(ax) .$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 17 '18 at 2:22

























                  answered Dec 16 '18 at 17:57









                  AmbretteOrriseyAmbretteOrrisey

                  54210




                  54210






























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