Continuous functions on unit discs can be extended to whole plane












3












$begingroup$


Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.




  1. Given continuous function $g : B to mathbb{R}$, there is a continuous function $f : mathbb{R} ^2 to mathbb{R}$ such that $f=g$ on $B$.

  2. Given continuous function $u : D to mathbb{R}$, there is a continuous function $v : mathbb{R} ^2 to mathbb{R}$ such that $ v=u $ on $D$.


I think the first one is doable. What I think is if we define $f$ as $ f(re^{iz}) = r*f(e^{iz})$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?






general-topology algebraic-topology continuity metric-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:51










  • $begingroup$
    You're on the right track. You could also simply let $f(re^{iz})=f(e^{iz})$ for $rgt1$. You could even let $f(re^{iz})=f(e^{iz})/r$. But whatever you do, you need to prove that your extended function is continuous.
    $endgroup$
    – Barry Cipra
    Dec 25 '18 at 16:52


















3












$begingroup$


Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.




  1. Given continuous function $g : B to mathbb{R}$, there is a continuous function $f : mathbb{R} ^2 to mathbb{R}$ such that $f=g$ on $B$.

  2. Given continuous function $u : D to mathbb{R}$, there is a continuous function $v : mathbb{R} ^2 to mathbb{R}$ such that $ v=u $ on $D$.


I think the first one is doable. What I think is if we define $f$ as $ f(re^{iz}) = r*f(e^{iz})$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?






general-topology algebraic-topology continuity metric-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:51










  • $begingroup$
    You're on the right track. You could also simply let $f(re^{iz})=f(e^{iz})$ for $rgt1$. You could even let $f(re^{iz})=f(e^{iz})/r$. But whatever you do, you need to prove that your extended function is continuous.
    $endgroup$
    – Barry Cipra
    Dec 25 '18 at 16:52
















3












3








3





$begingroup$


Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.




  1. Given continuous function $g : B to mathbb{R}$, there is a continuous function $f : mathbb{R} ^2 to mathbb{R}$ such that $f=g$ on $B$.

  2. Given continuous function $u : D to mathbb{R}$, there is a continuous function $v : mathbb{R} ^2 to mathbb{R}$ such that $ v=u $ on $D$.


I think the first one is doable. What I think is if we define $f$ as $ f(re^{iz}) = r*f(e^{iz})$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?






general-topology algebraic-topology continuity metric-spaces







share|cite|improve this question











$endgroup$




Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.




  1. Given continuous function $g : B to mathbb{R}$, there is a continuous function $f : mathbb{R} ^2 to mathbb{R}$ such that $f=g$ on $B$.

  2. Given continuous function $u : D to mathbb{R}$, there is a continuous function $v : mathbb{R} ^2 to mathbb{R}$ such that $ v=u $ on $D$.


I think the first one is doable. What I think is if we define $f$ as $ f(re^{iz}) = r*f(e^{iz})$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?






general-topology algebraic-topology continuity metric-spaces




general-topology algebraic-topology continuity metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 16:49









mathcounterexamples.net

26.7k22157




26.7k22157










asked Dec 25 '18 at 16:45









ChakSayantanChakSayantan

1426




1426












  • $begingroup$
    For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:51










  • $begingroup$
    You're on the right track. You could also simply let $f(re^{iz})=f(e^{iz})$ for $rgt1$. You could even let $f(re^{iz})=f(e^{iz})/r$. But whatever you do, you need to prove that your extended function is continuous.
    $endgroup$
    – Barry Cipra
    Dec 25 '18 at 16:52




















  • $begingroup$
    For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:51










  • $begingroup$
    You're on the right track. You could also simply let $f(re^{iz})=f(e^{iz})$ for $rgt1$. You could even let $f(re^{iz})=f(e^{iz})/r$. But whatever you do, you need to prove that your extended function is continuous.
    $endgroup$
    – Barry Cipra
    Dec 25 '18 at 16:52


















$begingroup$
For the second part think of a function that goes to $infty$ as you get close to the border of the disk
$endgroup$
– Alessandro Codenotti
Dec 25 '18 at 16:51




$begingroup$
For the second part think of a function that goes to $infty$ as you get close to the border of the disk
$endgroup$
– Alessandro Codenotti
Dec 25 '18 at 16:51












$begingroup$
You're on the right track. You could also simply let $f(re^{iz})=f(e^{iz})$ for $rgt1$. You could even let $f(re^{iz})=f(e^{iz})/r$. But whatever you do, you need to prove that your extended function is continuous.
$endgroup$
– Barry Cipra
Dec 25 '18 at 16:52






$begingroup$
You're on the right track. You could also simply let $f(re^{iz})=f(e^{iz})$ for $rgt1$. You could even let $f(re^{iz})=f(e^{iz})/r$. But whatever you do, you need to prove that your extended function is continuous.
$endgroup$
– Barry Cipra
Dec 25 '18 at 16:52












4 Answers
4






active

oldest

votes


















5












$begingroup$


  1. Is indeed true.


  2. Is false. Consider $u : (x,y) mapsto frac{1}{1-sqrt{x^2+y^2}}$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.







share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbb{R}$ being continuous determines the boundary behavior of $f|_{partial D}$. Indeed, radially extending as you propose should work fine.



    As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance



    $$ f(x)=sinleft(frac{1}{x-1}right)$$
    defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
    $$g(x,y)=sinleft(frac{1}{sqrt{x^2+y^2}-1}right)$$
    on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbb{R}^2$.






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Your approach is good. A concrete example for the second part is $f(z)=frac{1}{|z|-1}$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
        $endgroup$
        – ChakSayantan
        Dec 25 '18 at 16:59










      • $begingroup$
        Good point @ChakSayantan I will update my answer
        $endgroup$
        – Sorin Tirc
        Dec 25 '18 at 17:00



















      2












      $begingroup$

      Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.



      For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$


        1. Is indeed true.


        2. Is false. Consider $u : (x,y) mapsto frac{1}{1-sqrt{x^2+y^2}}$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.







        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$


          1. Is indeed true.


          2. Is false. Consider $u : (x,y) mapsto frac{1}{1-sqrt{x^2+y^2}}$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.







          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$


            1. Is indeed true.


            2. Is false. Consider $u : (x,y) mapsto frac{1}{1-sqrt{x^2+y^2}}$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.







            share|cite|improve this answer









            $endgroup$




            1. Is indeed true.


            2. Is false. Consider $u : (x,y) mapsto frac{1}{1-sqrt{x^2+y^2}}$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 25 '18 at 16:52









            mathcounterexamples.netmathcounterexamples.net

            26.7k22157




            26.7k22157























                3












                $begingroup$

                As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbb{R}$ being continuous determines the boundary behavior of $f|_{partial D}$. Indeed, radially extending as you propose should work fine.



                As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance



                $$ f(x)=sinleft(frac{1}{x-1}right)$$
                defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
                $$g(x,y)=sinleft(frac{1}{sqrt{x^2+y^2}-1}right)$$
                on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbb{R}^2$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbb{R}$ being continuous determines the boundary behavior of $f|_{partial D}$. Indeed, radially extending as you propose should work fine.



                  As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance



                  $$ f(x)=sinleft(frac{1}{x-1}right)$$
                  defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
                  $$g(x,y)=sinleft(frac{1}{sqrt{x^2+y^2}-1}right)$$
                  on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbb{R}^2$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbb{R}$ being continuous determines the boundary behavior of $f|_{partial D}$. Indeed, radially extending as you propose should work fine.



                    As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance



                    $$ f(x)=sinleft(frac{1}{x-1}right)$$
                    defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
                    $$g(x,y)=sinleft(frac{1}{sqrt{x^2+y^2}-1}right)$$
                    on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbb{R}^2$.






                    share|cite|improve this answer









                    $endgroup$



                    As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbb{R}$ being continuous determines the boundary behavior of $f|_{partial D}$. Indeed, radially extending as you propose should work fine.



                    As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance



                    $$ f(x)=sinleft(frac{1}{x-1}right)$$
                    defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
                    $$g(x,y)=sinleft(frac{1}{sqrt{x^2+y^2}-1}right)$$
                    on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbb{R}^2$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 25 '18 at 16:52









                    Antonios-Alexandros RobotisAntonios-Alexandros Robotis

                    10.3k41641




                    10.3k41641























                        3












                        $begingroup$

                        Your approach is good. A concrete example for the second part is $f(z)=frac{1}{|z|-1}$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
                          $endgroup$
                          – ChakSayantan
                          Dec 25 '18 at 16:59










                        • $begingroup$
                          Good point @ChakSayantan I will update my answer
                          $endgroup$
                          – Sorin Tirc
                          Dec 25 '18 at 17:00
















                        3












                        $begingroup$

                        Your approach is good. A concrete example for the second part is $f(z)=frac{1}{|z|-1}$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
                          $endgroup$
                          – ChakSayantan
                          Dec 25 '18 at 16:59










                        • $begingroup$
                          Good point @ChakSayantan I will update my answer
                          $endgroup$
                          – Sorin Tirc
                          Dec 25 '18 at 17:00














                        3












                        3








                        3





                        $begingroup$

                        Your approach is good. A concrete example for the second part is $f(z)=frac{1}{|z|-1}$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane






                        share|cite|improve this answer











                        $endgroup$



                        Your approach is good. A concrete example for the second part is $f(z)=frac{1}{|z|-1}$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Dec 25 '18 at 17:00

























                        answered Dec 25 '18 at 16:50









                        Sorin TircSorin Tirc

                        1,755213




                        1,755213








                        • 1




                          $begingroup$
                          f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
                          $endgroup$
                          – ChakSayantan
                          Dec 25 '18 at 16:59










                        • $begingroup$
                          Good point @ChakSayantan I will update my answer
                          $endgroup$
                          – Sorin Tirc
                          Dec 25 '18 at 17:00














                        • 1




                          $begingroup$
                          f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
                          $endgroup$
                          – ChakSayantan
                          Dec 25 '18 at 16:59










                        • $begingroup$
                          Good point @ChakSayantan I will update my answer
                          $endgroup$
                          – Sorin Tirc
                          Dec 25 '18 at 17:00








                        1




                        1




                        $begingroup$
                        f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
                        $endgroup$
                        – ChakSayantan
                        Dec 25 '18 at 16:59




                        $begingroup$
                        f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
                        $endgroup$
                        – ChakSayantan
                        Dec 25 '18 at 16:59












                        $begingroup$
                        Good point @ChakSayantan I will update my answer
                        $endgroup$
                        – Sorin Tirc
                        Dec 25 '18 at 17:00




                        $begingroup$
                        Good point @ChakSayantan I will update my answer
                        $endgroup$
                        – Sorin Tirc
                        Dec 25 '18 at 17:00











                        2












                        $begingroup$

                        Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.



                        For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.



                          For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.



                            For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.






                            share|cite|improve this answer









                            $endgroup$



                            Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.



                            For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 25 '18 at 16:51









                            Matt SamuelMatt Samuel

                            38.4k63768




                            38.4k63768






























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                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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