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







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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















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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







share|cite|improve this question




















  • 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













up vote
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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













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edited Nov 23 at 12:43

























asked Nov 23 at 11:40









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  • 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




    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










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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).






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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).






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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).






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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).






        share|cite|improve this answer












        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).







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        answered Nov 23 at 12:01









        Kavi Rama Murthy

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