Continuous functions on unit discs can be extended to whole plane












2














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







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  • For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    – Alessandro Codenotti
    59 mins ago










  • 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.
    – Barry Cipra
    58 mins ago


















2














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
























  • For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    – Alessandro Codenotti
    59 mins ago










  • 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.
    – Barry Cipra
    58 mins ago
















2












2








2







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















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 1 hour ago









mathcounterexamples.net

23.9k21753




23.9k21753










asked 1 hour ago









ChakSayantan

1376




1376












  • For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    – Alessandro Codenotti
    59 mins ago










  • 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.
    – Barry Cipra
    58 mins ago




















  • For the second part think of a function that goes to $infty$ as you get close to the border of the disk
    – Alessandro Codenotti
    59 mins ago










  • 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.
    – Barry Cipra
    58 mins ago


















For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
59 mins ago




For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
59 mins ago












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.
– Barry Cipra
58 mins ago






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.
– Barry Cipra
58 mins ago












4 Answers
4






active

oldest

votes


















1















  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





























    1














    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





























      1














      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





























        1














        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























        • 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.
          – ChakSayantan
          51 mins ago










        • Good point @ChakSayantan I will update my answer
          – Sorin Tirc
          50 mins ago











        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1















        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


























          1















          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
























            1












            1








            1







            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













            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 58 mins ago









            mathcounterexamples.net

            23.9k21753




            23.9k21753























                1














                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


























                  1














                  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
























                    1












                    1








                    1






                    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












                    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 59 mins ago









                    Matt Samuel

                    36.9k63465




                    36.9k63465























                        1














                        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


























                          1














                          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
























                            1












                            1








                            1






                            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












                            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 58 mins ago









                            Antonios-Alexandros Robotis

                            9,14541640




                            9,14541640























                                1














                                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























                                • 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.
                                  – ChakSayantan
                                  51 mins ago










                                • Good point @ChakSayantan I will update my answer
                                  – Sorin Tirc
                                  50 mins ago
















                                1














                                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























                                • 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.
                                  – ChakSayantan
                                  51 mins ago










                                • Good point @ChakSayantan I will update my answer
                                  – Sorin Tirc
                                  50 mins ago














                                1












                                1








                                1






                                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














                                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 49 mins ago

























                                answered 1 hour ago









                                Sorin Tirc

                                98711




                                98711












                                • 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.
                                  – ChakSayantan
                                  51 mins ago










                                • Good point @ChakSayantan I will update my answer
                                  – Sorin Tirc
                                  50 mins ago


















                                • 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.
                                  – ChakSayantan
                                  51 mins ago










                                • Good point @ChakSayantan I will update my answer
                                  – Sorin Tirc
                                  50 mins ago
















                                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.
                                – ChakSayantan
                                51 mins ago




                                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.
                                – ChakSayantan
                                51 mins ago












                                Good point @ChakSayantan I will update my answer
                                – Sorin Tirc
                                50 mins ago




                                Good point @ChakSayantan I will update my answer
                                – Sorin Tirc
                                50 mins ago


















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