“Standard reference” for $C_c^infty(mathbb R)$ is dense in $C_c(mathbb R)$












2












$begingroup$


$C_c^infty(mathbb R)$ is dense in $C_c(mathbb R)$. This can be shown by mollification. This is a well-known, widely used fact. However, I wasn't able to find any book which I could point in a reference to. Is there any kind of "standard reference" with a readable proof?






functional-analysis reference-request smooth-functions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Which topology are you considering?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:31






  • 1




    $begingroup$
    @LorenzoQuarisa I would guess the sup-norm topology. But yes. the OP needs to tell us.
    $endgroup$
    – zhw.
    Jan 16 at 22:55












  • $begingroup$
    I don't know what you mean by mollification, but the proof I read in the Bourbaki series uses convolution by means of a $mathscr{C}^infty$ function with support in a tiny interval.
    $endgroup$
    – Will M.
    Jan 17 at 6:29












  • $begingroup$
    @LorenzoQuarisa I consider $C_c(mathbb R)$ s being equipped with the sup-norm.
    $endgroup$
    – 0xbadf00d
    Jan 17 at 9:47
















2












$begingroup$


$C_c^infty(mathbb R)$ is dense in $C_c(mathbb R)$. This can be shown by mollification. This is a well-known, widely used fact. However, I wasn't able to find any book which I could point in a reference to. Is there any kind of "standard reference" with a readable proof?






functional-analysis reference-request smooth-functions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Which topology are you considering?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:31






  • 1




    $begingroup$
    @LorenzoQuarisa I would guess the sup-norm topology. But yes. the OP needs to tell us.
    $endgroup$
    – zhw.
    Jan 16 at 22:55












  • $begingroup$
    I don't know what you mean by mollification, but the proof I read in the Bourbaki series uses convolution by means of a $mathscr{C}^infty$ function with support in a tiny interval.
    $endgroup$
    – Will M.
    Jan 17 at 6:29












  • $begingroup$
    @LorenzoQuarisa I consider $C_c(mathbb R)$ s being equipped with the sup-norm.
    $endgroup$
    – 0xbadf00d
    Jan 17 at 9:47














2












2








2





$begingroup$


$C_c^infty(mathbb R)$ is dense in $C_c(mathbb R)$. This can be shown by mollification. This is a well-known, widely used fact. However, I wasn't able to find any book which I could point in a reference to. Is there any kind of "standard reference" with a readable proof?






functional-analysis reference-request smooth-functions







share|cite|improve this question









$endgroup$




$C_c^infty(mathbb R)$ is dense in $C_c(mathbb R)$. This can be shown by mollification. This is a well-known, widely used fact. However, I wasn't able to find any book which I could point in a reference to. Is there any kind of "standard reference" with a readable proof?






functional-analysis reference-request smooth-functions




functional-analysis reference-request smooth-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 22:23









0xbadf00d0xbadf00d

1,90441531




1,90441531








  • 1




    $begingroup$
    Which topology are you considering?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:31






  • 1




    $begingroup$
    @LorenzoQuarisa I would guess the sup-norm topology. But yes. the OP needs to tell us.
    $endgroup$
    – zhw.
    Jan 16 at 22:55












  • $begingroup$
    I don't know what you mean by mollification, but the proof I read in the Bourbaki series uses convolution by means of a $mathscr{C}^infty$ function with support in a tiny interval.
    $endgroup$
    – Will M.
    Jan 17 at 6:29












  • $begingroup$
    @LorenzoQuarisa I consider $C_c(mathbb R)$ s being equipped with the sup-norm.
    $endgroup$
    – 0xbadf00d
    Jan 17 at 9:47














  • 1




    $begingroup$
    Which topology are you considering?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:31






  • 1




    $begingroup$
    @LorenzoQuarisa I would guess the sup-norm topology. But yes. the OP needs to tell us.
    $endgroup$
    – zhw.
    Jan 16 at 22:55












  • $begingroup$
    I don't know what you mean by mollification, but the proof I read in the Bourbaki series uses convolution by means of a $mathscr{C}^infty$ function with support in a tiny interval.
    $endgroup$
    – Will M.
    Jan 17 at 6:29












  • $begingroup$
    @LorenzoQuarisa I consider $C_c(mathbb R)$ s being equipped with the sup-norm.
    $endgroup$
    – 0xbadf00d
    Jan 17 at 9:47








1




1




$begingroup$
Which topology are you considering?
$endgroup$
– Lorenzo Quarisa
Jan 16 at 22:31




$begingroup$
Which topology are you considering?
$endgroup$
– Lorenzo Quarisa
Jan 16 at 22:31




1




1




$begingroup$
@LorenzoQuarisa I would guess the sup-norm topology. But yes. the OP needs to tell us.
$endgroup$
– zhw.
Jan 16 at 22:55






$begingroup$
@LorenzoQuarisa I would guess the sup-norm topology. But yes. the OP needs to tell us.
$endgroup$
– zhw.
Jan 16 at 22:55














$begingroup$
I don't know what you mean by mollification, but the proof I read in the Bourbaki series uses convolution by means of a $mathscr{C}^infty$ function with support in a tiny interval.
$endgroup$
– Will M.
Jan 17 at 6:29






$begingroup$
I don't know what you mean by mollification, but the proof I read in the Bourbaki series uses convolution by means of a $mathscr{C}^infty$ function with support in a tiny interval.
$endgroup$
– Will M.
Jan 17 at 6:29














$begingroup$
@LorenzoQuarisa I consider $C_c(mathbb R)$ s being equipped with the sup-norm.
$endgroup$
– 0xbadf00d
Jan 17 at 9:47




$begingroup$
@LorenzoQuarisa I consider $C_c(mathbb R)$ s being equipped with the sup-norm.
$endgroup$
– 0xbadf00d
Jan 17 at 9:47










2 Answers
2






active

oldest

votes


















2












$begingroup$

(I am assuming you are equipping $C_c^{infty}(mathbb R)$ and $C^{infty}(mathbb R)$ with the sup-norm). One can use the Stone-Weierstrass Theorem for locally compact Hausdorff spaces to show the result (for references to the Stone-Weierstrass Theorem, see Willard's General Topology Section 44 or Folland Chapter 4). In fact, the Stone-Weierstrass Theorem yields a stronger result: $C_c^{infty}(mathbb R^n)$ is dense in $C_0(mathbb R^n)$ when both spaces are given the topology of uniform convergence. The sum, product, scalar multiple, and complex conjugate of smooth compactly supported functions is easily verified to also be smooth and compactly supported.



The fact that $C_c^{infty}(mathbb R^n)$ separates points and vanishes nowhere follows from the following theorem in Folland:



Theorem (Folland, 8.18). Let $K subseteq mathbb R^n$ be nonempty and compact, and let $U$ be an open set with $U supseteq K$. Then, there is $f in C_c^{infty}(mathbb R^n) $ such that $0 leq f(x) leq 1 $ for all $x in mathbb R^n$, $f(K)={1}$, and $text{supp}(f) subseteq U$.



Theorem 8.17 in Folland also proves this in a different way using an approximation of the identity.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I don't have a reference at hand. But one can prove this without too much trouble using the Weierstrass approximation theorem.



    Suppose $fin C_c$ with support contained in $[a,b].$ Let $epsilon>0.$ Choose $a'<a$ and $b'>b.$ Then there exists a polynomial $p$ such that $|p-f|<epsilon$ on $[a',b'].$



    Now there exists $gin C^infty_c(mathbb R)$ with support in $[a',b']$ such that $0le gle 1$ everywhere, and $g=1$ on $[a,b].$ We then have $gpin C^infty_c(mathbb R),$ and $|gp -f|<epsilon$ everywhere. I'll leave the vefication of the last line to the reader; ask questions if you like.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      (I am assuming you are equipping $C_c^{infty}(mathbb R)$ and $C^{infty}(mathbb R)$ with the sup-norm). One can use the Stone-Weierstrass Theorem for locally compact Hausdorff spaces to show the result (for references to the Stone-Weierstrass Theorem, see Willard's General Topology Section 44 or Folland Chapter 4). In fact, the Stone-Weierstrass Theorem yields a stronger result: $C_c^{infty}(mathbb R^n)$ is dense in $C_0(mathbb R^n)$ when both spaces are given the topology of uniform convergence. The sum, product, scalar multiple, and complex conjugate of smooth compactly supported functions is easily verified to also be smooth and compactly supported.



      The fact that $C_c^{infty}(mathbb R^n)$ separates points and vanishes nowhere follows from the following theorem in Folland:



      Theorem (Folland, 8.18). Let $K subseteq mathbb R^n$ be nonempty and compact, and let $U$ be an open set with $U supseteq K$. Then, there is $f in C_c^{infty}(mathbb R^n) $ such that $0 leq f(x) leq 1 $ for all $x in mathbb R^n$, $f(K)={1}$, and $text{supp}(f) subseteq U$.



      Theorem 8.17 in Folland also proves this in a different way using an approximation of the identity.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        (I am assuming you are equipping $C_c^{infty}(mathbb R)$ and $C^{infty}(mathbb R)$ with the sup-norm). One can use the Stone-Weierstrass Theorem for locally compact Hausdorff spaces to show the result (for references to the Stone-Weierstrass Theorem, see Willard's General Topology Section 44 or Folland Chapter 4). In fact, the Stone-Weierstrass Theorem yields a stronger result: $C_c^{infty}(mathbb R^n)$ is dense in $C_0(mathbb R^n)$ when both spaces are given the topology of uniform convergence. The sum, product, scalar multiple, and complex conjugate of smooth compactly supported functions is easily verified to also be smooth and compactly supported.



        The fact that $C_c^{infty}(mathbb R^n)$ separates points and vanishes nowhere follows from the following theorem in Folland:



        Theorem (Folland, 8.18). Let $K subseteq mathbb R^n$ be nonempty and compact, and let $U$ be an open set with $U supseteq K$. Then, there is $f in C_c^{infty}(mathbb R^n) $ such that $0 leq f(x) leq 1 $ for all $x in mathbb R^n$, $f(K)={1}$, and $text{supp}(f) subseteq U$.



        Theorem 8.17 in Folland also proves this in a different way using an approximation of the identity.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          (I am assuming you are equipping $C_c^{infty}(mathbb R)$ and $C^{infty}(mathbb R)$ with the sup-norm). One can use the Stone-Weierstrass Theorem for locally compact Hausdorff spaces to show the result (for references to the Stone-Weierstrass Theorem, see Willard's General Topology Section 44 or Folland Chapter 4). In fact, the Stone-Weierstrass Theorem yields a stronger result: $C_c^{infty}(mathbb R^n)$ is dense in $C_0(mathbb R^n)$ when both spaces are given the topology of uniform convergence. The sum, product, scalar multiple, and complex conjugate of smooth compactly supported functions is easily verified to also be smooth and compactly supported.



          The fact that $C_c^{infty}(mathbb R^n)$ separates points and vanishes nowhere follows from the following theorem in Folland:



          Theorem (Folland, 8.18). Let $K subseteq mathbb R^n$ be nonempty and compact, and let $U$ be an open set with $U supseteq K$. Then, there is $f in C_c^{infty}(mathbb R^n) $ such that $0 leq f(x) leq 1 $ for all $x in mathbb R^n$, $f(K)={1}$, and $text{supp}(f) subseteq U$.



          Theorem 8.17 in Folland also proves this in a different way using an approximation of the identity.






          share|cite|improve this answer









          $endgroup$



          (I am assuming you are equipping $C_c^{infty}(mathbb R)$ and $C^{infty}(mathbb R)$ with the sup-norm). One can use the Stone-Weierstrass Theorem for locally compact Hausdorff spaces to show the result (for references to the Stone-Weierstrass Theorem, see Willard's General Topology Section 44 or Folland Chapter 4). In fact, the Stone-Weierstrass Theorem yields a stronger result: $C_c^{infty}(mathbb R^n)$ is dense in $C_0(mathbb R^n)$ when both spaces are given the topology of uniform convergence. The sum, product, scalar multiple, and complex conjugate of smooth compactly supported functions is easily verified to also be smooth and compactly supported.



          The fact that $C_c^{infty}(mathbb R^n)$ separates points and vanishes nowhere follows from the following theorem in Folland:



          Theorem (Folland, 8.18). Let $K subseteq mathbb R^n$ be nonempty and compact, and let $U$ be an open set with $U supseteq K$. Then, there is $f in C_c^{infty}(mathbb R^n) $ such that $0 leq f(x) leq 1 $ for all $x in mathbb R^n$, $f(K)={1}$, and $text{supp}(f) subseteq U$.



          Theorem 8.17 in Folland also proves this in a different way using an approximation of the identity.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 2:03









          LinearOperator32LinearOperator32

          636




          636























              0












              $begingroup$

              I don't have a reference at hand. But one can prove this without too much trouble using the Weierstrass approximation theorem.



              Suppose $fin C_c$ with support contained in $[a,b].$ Let $epsilon>0.$ Choose $a'<a$ and $b'>b.$ Then there exists a polynomial $p$ such that $|p-f|<epsilon$ on $[a',b'].$



              Now there exists $gin C^infty_c(mathbb R)$ with support in $[a',b']$ such that $0le gle 1$ everywhere, and $g=1$ on $[a,b].$ We then have $gpin C^infty_c(mathbb R),$ and $|gp -f|<epsilon$ everywhere. I'll leave the vefication of the last line to the reader; ask questions if you like.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I don't have a reference at hand. But one can prove this without too much trouble using the Weierstrass approximation theorem.



                Suppose $fin C_c$ with support contained in $[a,b].$ Let $epsilon>0.$ Choose $a'<a$ and $b'>b.$ Then there exists a polynomial $p$ such that $|p-f|<epsilon$ on $[a',b'].$



                Now there exists $gin C^infty_c(mathbb R)$ with support in $[a',b']$ such that $0le gle 1$ everywhere, and $g=1$ on $[a,b].$ We then have $gpin C^infty_c(mathbb R),$ and $|gp -f|<epsilon$ everywhere. I'll leave the vefication of the last line to the reader; ask questions if you like.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I don't have a reference at hand. But one can prove this without too much trouble using the Weierstrass approximation theorem.



                  Suppose $fin C_c$ with support contained in $[a,b].$ Let $epsilon>0.$ Choose $a'<a$ and $b'>b.$ Then there exists a polynomial $p$ such that $|p-f|<epsilon$ on $[a',b'].$



                  Now there exists $gin C^infty_c(mathbb R)$ with support in $[a',b']$ such that $0le gle 1$ everywhere, and $g=1$ on $[a,b].$ We then have $gpin C^infty_c(mathbb R),$ and $|gp -f|<epsilon$ everywhere. I'll leave the vefication of the last line to the reader; ask questions if you like.






                  share|cite|improve this answer









                  $endgroup$



                  I don't have a reference at hand. But one can prove this without too much trouble using the Weierstrass approximation theorem.



                  Suppose $fin C_c$ with support contained in $[a,b].$ Let $epsilon>0.$ Choose $a'<a$ and $b'>b.$ Then there exists a polynomial $p$ such that $|p-f|<epsilon$ on $[a',b'].$



                  Now there exists $gin C^infty_c(mathbb R)$ with support in $[a',b']$ such that $0le gle 1$ everywhere, and $g=1$ on $[a,b].$ We then have $gpin C^infty_c(mathbb R),$ and $|gp -f|<epsilon$ everywhere. I'll leave the vefication of the last line to the reader; ask questions if you like.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 17 at 6:21









                  zhw.zhw.

                  72.8k43175




                  72.8k43175






























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