Looking for a certain function with compact support












0












$begingroup$


can someone help me with the following doubt?



I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$



How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?



I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.



Thanks.






real-analysis functional-analysis analysis







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    can someone help me with the following doubt?



    I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$



    How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?



    I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.



    Thanks.






    real-analysis functional-analysis analysis







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      can someone help me with the following doubt?



      I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$



      How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?



      I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.



      Thanks.






      real-analysis functional-analysis analysis







      share|cite|improve this question











      $endgroup$




      can someone help me with the following doubt?



      I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$



      How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?



      I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.



      Thanks.






      real-analysis functional-analysis analysis




      real-analysis functional-analysis analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 11:34









      Julián Aguirre

      69.4k24096




      69.4k24096










      asked Dec 26 '18 at 9:06









      mathlifemathlife

      609




      609






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Nothing will work.
          $$
          lim_{stoinfty}f_s(x)=f(0)
          $$
          is a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            And $f_s(x)=f(frac{x}{s})|x|^2$?
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:53










          • $begingroup$
            The limit will be $f(0),|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:54










          • $begingroup$
            But un muy cuestión $f(0)=1$
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:57










          • $begingroup$
            Then $f(0),|x|^2=|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:58










          • $begingroup$
            In that definition $f_s(x)$ has also compact support and is infinite derivable?
            $endgroup$
            – mathlife
            Dec 26 '18 at 14:02











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          Nothing will work.
          $$
          lim_{stoinfty}f_s(x)=f(0)
          $$
          is a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            And $f_s(x)=f(frac{x}{s})|x|^2$?
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:53










          • $begingroup$
            The limit will be $f(0),|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:54










          • $begingroup$
            But un muy cuestión $f(0)=1$
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:57










          • $begingroup$
            Then $f(0),|x|^2=|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:58










          • $begingroup$
            In that definition $f_s(x)$ has also compact support and is infinite derivable?
            $endgroup$
            – mathlife
            Dec 26 '18 at 14:02
















          1












          $begingroup$

          Nothing will work.
          $$
          lim_{stoinfty}f_s(x)=f(0)
          $$
          is a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            And $f_s(x)=f(frac{x}{s})|x|^2$?
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:53










          • $begingroup$
            The limit will be $f(0),|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:54










          • $begingroup$
            But un muy cuestión $f(0)=1$
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:57










          • $begingroup$
            Then $f(0),|x|^2=|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:58










          • $begingroup$
            In that definition $f_s(x)$ has also compact support and is infinite derivable?
            $endgroup$
            – mathlife
            Dec 26 '18 at 14:02














          1












          1








          1





          $begingroup$

          Nothing will work.
          $$
          lim_{stoinfty}f_s(x)=f(0)
          $$
          is a constant.






          share|cite|improve this answer











          $endgroup$



          Nothing will work.
          $$
          lim_{stoinfty}f_s(x)=f(0)
          $$
          is a constant.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 26 '18 at 13:56

























          answered Dec 26 '18 at 11:35









          Julián AguirreJulián Aguirre

          69.4k24096




          69.4k24096












          • $begingroup$
            And $f_s(x)=f(frac{x}{s})|x|^2$?
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:53










          • $begingroup$
            The limit will be $f(0),|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:54










          • $begingroup$
            But un muy cuestión $f(0)=1$
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:57










          • $begingroup$
            Then $f(0),|x|^2=|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:58










          • $begingroup$
            In that definition $f_s(x)$ has also compact support and is infinite derivable?
            $endgroup$
            – mathlife
            Dec 26 '18 at 14:02


















          • $begingroup$
            And $f_s(x)=f(frac{x}{s})|x|^2$?
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:53










          • $begingroup$
            The limit will be $f(0),|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:54










          • $begingroup$
            But un muy cuestión $f(0)=1$
            $endgroup$
            – mathlife
            Dec 26 '18 at 13:57










          • $begingroup$
            Then $f(0),|x|^2=|x|^2$.
            $endgroup$
            – Julián Aguirre
            Dec 26 '18 at 13:58










          • $begingroup$
            In that definition $f_s(x)$ has also compact support and is infinite derivable?
            $endgroup$
            – mathlife
            Dec 26 '18 at 14:02
















          $begingroup$
          And $f_s(x)=f(frac{x}{s})|x|^2$?
          $endgroup$
          – mathlife
          Dec 26 '18 at 13:53




          $begingroup$
          And $f_s(x)=f(frac{x}{s})|x|^2$?
          $endgroup$
          – mathlife
          Dec 26 '18 at 13:53












          $begingroup$
          The limit will be $f(0),|x|^2$.
          $endgroup$
          – Julián Aguirre
          Dec 26 '18 at 13:54




          $begingroup$
          The limit will be $f(0),|x|^2$.
          $endgroup$
          – Julián Aguirre
          Dec 26 '18 at 13:54












          $begingroup$
          But un muy cuestión $f(0)=1$
          $endgroup$
          – mathlife
          Dec 26 '18 at 13:57




          $begingroup$
          But un muy cuestión $f(0)=1$
          $endgroup$
          – mathlife
          Dec 26 '18 at 13:57












          $begingroup$
          Then $f(0),|x|^2=|x|^2$.
          $endgroup$
          – Julián Aguirre
          Dec 26 '18 at 13:58




          $begingroup$
          Then $f(0),|x|^2=|x|^2$.
          $endgroup$
          – Julián Aguirre
          Dec 26 '18 at 13:58












          $begingroup$
          In that definition $f_s(x)$ has also compact support and is infinite derivable?
          $endgroup$
          – mathlife
          Dec 26 '18 at 14:02




          $begingroup$
          In that definition $f_s(x)$ has also compact support and is infinite derivable?
          $endgroup$
          – mathlife
          Dec 26 '18 at 14:02


















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