“Standard reference” for $C_c^infty(mathbb R)$ is dense in $C_c(mathbb R)$
$begingroup$
$C_c^infty(mathbb R)$ 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
asked Jan 16 at 22:23
0xbadf00d0xbadf00d
1,90441531
$endgroup$
add a comment |
$begingroup$
$C_c^infty(mathbb R)$ 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
asked Jan 16 at 22:23
0xbadf00d0xbadf00d
1,90441531
$endgroup$
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
add a comment |
$begingroup$
$C_c^infty(mathbb R)$ 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
asked Jan 16 at 22:23
0xbadf00d0xbadf00d
1,90441531
$endgroup$
$C_c^infty(mathbb R)$ 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
asked Jan 16 at 22:23
0xbadf00d0xbadf00d
1,90441531
asked Jan 16 at 22:23
0xbadf00d0xbadf00d
1,90441531
asked Jan 16 at 22:23
0xbadf00d0xbadf00d
1,90441531
asked Jan 16 at 22:23
0xbadf00d0xbadf00d
1,90441531
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(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.
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
(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.
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
$endgroup$
add a comment |
$begingroup$
(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.
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
$endgroup$
add a comment |
$begingroup$
(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.
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
$endgroup$
(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.
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
answered Jan 17 at 2:03
LinearOperator32LinearOperator32
636
636
add a comment |
add a comment |
$begingroup$
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.
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
$endgroup$
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.
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
answered Jan 17 at 6:21
zhw.zhw.
72.8k43175
72.8k43175
add a comment |
add a comment |
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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