Can we approximate a $C_0(mathbb R)$-function by $C_c^infty(mathbb R)$-functions?
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Let $C_0(mathbb R)$ denote the closure of $C_c(mathbb R)$ with respect to the supremum norm.
Are we able to show that the closure of $C_c^infty(mathbb R)$ with respect to the supremum norm is $C_0(mathbb R)$?
functional-analysis analysis
asked Nov 23 at 11:40
0xbadf00d
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up vote
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Let $C_0(mathbb R)$ denote the closure of $C_c(mathbb R)$ with respect to the supremum norm.
Are we able to show that the closure of $C_c^infty(mathbb R)$ with respect to the supremum norm is $C_0(mathbb R)$?
functional-analysis analysis
asked Nov 23 at 11:40
0xbadf00d
1,71041429
1
Functions in $C_c(mathbb R)$ can be approximated uniformly by smooth functions via mollification.
– gerw
Nov 23 at 11:50
This is related
– Giuseppe Negro
Nov 23 at 11:55
add a comment |
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1
down vote
favorite
up vote
1
down vote
favorite
Let $C_0(mathbb R)$ denote the closure of $C_c(mathbb R)$ with respect to the supremum norm.
Are we able to show that the closure of $C_c^infty(mathbb R)$ with respect to the supremum norm is $C_0(mathbb R)$?
functional-analysis analysis
asked Nov 23 at 11:40
0xbadf00d
1,71041429
Let $C_0(mathbb R)$ denote the closure of $C_c(mathbb R)$ with respect to the supremum norm.
Are we able to show that the closure of $C_c^infty(mathbb R)$ with respect to the supremum norm is $C_0(mathbb R)$?
functional-analysis analysis
functional-analysis analysis
asked Nov 23 at 11:40
0xbadf00d
1,71041429
asked Nov 23 at 11:40
0xbadf00d
1,71041429
edited Nov 23 at 12:43
asked Nov 23 at 11:40
0xbadf00d
1,71041429
asked Nov 23 at 11:40
0xbadf00d
1,71041429
asked Nov 23 at 11:40
0xbadf00d
1,71041429
1,71041429
1
Functions in $C_c(mathbb R)$ can be approximated uniformly by smooth functions via mollification.
– gerw
Nov 23 at 11:50
This is related
– Giuseppe Negro
Nov 23 at 11:55
add a comment |
1
Functions in $C_c(mathbb R)$ can be approximated uniformly by smooth functions via mollification.
– gerw
Nov 23 at 11:50
This is related
– Giuseppe Negro
Nov 23 at 11:55
1
1
Functions in $C_c(mathbb R)$ can be approximated uniformly by smooth functions via mollification.
– gerw
Nov 23 at 11:50
Functions in $C_c(mathbb R)$ can be approximated uniformly by smooth functions via mollification.
– gerw
Nov 23 at 11:50
This is related
– Giuseppe Negro
Nov 23 at 11:55
This is related
– Giuseppe Negro
Nov 23 at 11:55
add a comment |
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Let $f in C_0(mathbb R)$ and $epsilon >0$. There exists $M>0$ such that $|f(x)| <epsilon $ for $|x|>M$. There exists a polynomial $p$ such that $|f(x)-p(x)| <epsilon$ for $|x| leq M+1$. There exists a function $g in C_c(mathbb R)$ such that $0leq g(x)leq 1$ for all $x$ , $g(x)=1$ if $|x| leq M$ and $g(x)=0$ if $|x| >M+epsilon$. Let $h=pg$. Then $h in C_c(mathbb R)$, $$|f(x)-p(x)g(x)|=|f(x)-p(x)|<epsilon$$ for $|x| leq M$. For $|x| >M+epsilon$ we have $$|f(x)-p(x)g(x)|=|f(x)| <epsilon$$. Finally, for $Mleq |x| leq M+epsilon$ we have $$|f(x)-p(x)g(x)|leq epsilon+ |p(x)-p(x)g(x)| leq epsilon+ |p(x)|$$ (because $0leq 1-g(x) leq 1$) and $|p(x)|<epsilon +|f(x)| <2epsilon$. Remark: I just realized that I could have replaced $(M,M+epsilon) $ by $(M,M+1)$ but that is only a minor point).
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
add a comment |
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Let $f in C_0(mathbb R)$ and $epsilon >0$. There exists $M>0$ such that $|f(x)| <epsilon $ for $|x|>M$. There exists a polynomial $p$ such that $|f(x)-p(x)| <epsilon$ for $|x| leq M+1$. There exists a function $g in C_c(mathbb R)$ such that $0leq g(x)leq 1$ for all $x$ , $g(x)=1$ if $|x| leq M$ and $g(x)=0$ if $|x| >M+epsilon$. Let $h=pg$. Then $h in C_c(mathbb R)$, $$|f(x)-p(x)g(x)|=|f(x)-p(x)|<epsilon$$ for $|x| leq M$. For $|x| >M+epsilon$ we have $$|f(x)-p(x)g(x)|=|f(x)| <epsilon$$. Finally, for $Mleq |x| leq M+epsilon$ we have $$|f(x)-p(x)g(x)|leq epsilon+ |p(x)-p(x)g(x)| leq epsilon+ |p(x)|$$ (because $0leq 1-g(x) leq 1$) and $|p(x)|<epsilon +|f(x)| <2epsilon$. Remark: I just realized that I could have replaced $(M,M+epsilon) $ by $(M,M+1)$ but that is only a minor point).
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
add a comment |
up vote
0
down vote
Let $f in C_0(mathbb R)$ and $epsilon >0$. There exists $M>0$ such that $|f(x)| <epsilon $ for $|x|>M$. There exists a polynomial $p$ such that $|f(x)-p(x)| <epsilon$ for $|x| leq M+1$. There exists a function $g in C_c(mathbb R)$ such that $0leq g(x)leq 1$ for all $x$ , $g(x)=1$ if $|x| leq M$ and $g(x)=0$ if $|x| >M+epsilon$. Let $h=pg$. Then $h in C_c(mathbb R)$, $$|f(x)-p(x)g(x)|=|f(x)-p(x)|<epsilon$$ for $|x| leq M$. For $|x| >M+epsilon$ we have $$|f(x)-p(x)g(x)|=|f(x)| <epsilon$$. Finally, for $Mleq |x| leq M+epsilon$ we have $$|f(x)-p(x)g(x)|leq epsilon+ |p(x)-p(x)g(x)| leq epsilon+ |p(x)|$$ (because $0leq 1-g(x) leq 1$) and $|p(x)|<epsilon +|f(x)| <2epsilon$. Remark: I just realized that I could have replaced $(M,M+epsilon) $ by $(M,M+1)$ but that is only a minor point).
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $f in C_0(mathbb R)$ and $epsilon >0$. There exists $M>0$ such that $|f(x)| <epsilon $ for $|x|>M$. There exists a polynomial $p$ such that $|f(x)-p(x)| <epsilon$ for $|x| leq M+1$. There exists a function $g in C_c(mathbb R)$ such that $0leq g(x)leq 1$ for all $x$ , $g(x)=1$ if $|x| leq M$ and $g(x)=0$ if $|x| >M+epsilon$. Let $h=pg$. Then $h in C_c(mathbb R)$, $$|f(x)-p(x)g(x)|=|f(x)-p(x)|<epsilon$$ for $|x| leq M$. For $|x| >M+epsilon$ we have $$|f(x)-p(x)g(x)|=|f(x)| <epsilon$$. Finally, for $Mleq |x| leq M+epsilon$ we have $$|f(x)-p(x)g(x)|leq epsilon+ |p(x)-p(x)g(x)| leq epsilon+ |p(x)|$$ (because $0leq 1-g(x) leq 1$) and $|p(x)|<epsilon +|f(x)| <2epsilon$. Remark: I just realized that I could have replaced $(M,M+epsilon) $ by $(M,M+1)$ but that is only a minor point).
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
Let $f in C_0(mathbb R)$ and $epsilon >0$. There exists $M>0$ such that $|f(x)| <epsilon $ for $|x|>M$. There exists a polynomial $p$ such that $|f(x)-p(x)| <epsilon$ for $|x| leq M+1$. There exists a function $g in C_c(mathbb R)$ such that $0leq g(x)leq 1$ for all $x$ , $g(x)=1$ if $|x| leq M$ and $g(x)=0$ if $|x| >M+epsilon$. Let $h=pg$. Then $h in C_c(mathbb R)$, $$|f(x)-p(x)g(x)|=|f(x)-p(x)|<epsilon$$ for $|x| leq M$. For $|x| >M+epsilon$ we have $$|f(x)-p(x)g(x)|=|f(x)| <epsilon$$. Finally, for $Mleq |x| leq M+epsilon$ we have $$|f(x)-p(x)g(x)|leq epsilon+ |p(x)-p(x)g(x)| leq epsilon+ |p(x)|$$ (because $0leq 1-g(x) leq 1$) and $|p(x)|<epsilon +|f(x)| <2epsilon$. Remark: I just realized that I could have replaced $(M,M+epsilon) $ by $(M,M+1)$ but that is only a minor point).
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
answered Nov 23 at 12:01
Kavi Rama Murthy
47.7k31854
47.7k31854
add a comment |
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Functions in $C_c(mathbb R)$ can be approximated uniformly by smooth functions via mollification.
– gerw
Nov 23 at 11:50
This is related
– Giuseppe Negro
Nov 23 at 11:55