Why is in the definition of the space of continuous functions vanishing at infinity from a space $Ω$ to...
Let
$Omega$ be a topological space
$C(Omega)$ denote the set of continuous functions from $Omega$ to $mathbb R$
$C_b(Omega)$ denote the set of bounded functions in $C(Omega)$ equipped with $$left|fright|_infty:=sup_{xinOmega}|f(x)|;;;text{for }f:Omegatomathbb R$$
Now, let $$operatorname{supp}f:=overline{left{xinOmega:f(x)ne0right}}$$ and $$C_c(Omega):=left{fin C(Omega):operatorname{supp}ftext{ is compact}right}.$$ Clearly, $C_c(Omega)$ is a (not necessarily closed) subspace of $C_b(Omega)$. Let $C_0(Omega)$ denote the completion of $C_c(Omega)$ with respect to $left|;cdot;right|_infty$.
If $fin C_0(Omega)$ and $varepsilon>0$, there is a $gin C_c(Omega)$ with $$left|f-gright|_infty<varepsilontag1.$$ Since $gin C_c(Omega)$, $$K:=operatorname{supp}g$$ is compact and $$g(x)=0;;;text{for all }xinOmegasetminus Ktag2.$$ Thus, $$|f(x)|le|f(x)-g(x)|+|g(x)|<varepsilon;;;text{for all }xinOmegasetminus Ktag3.$$ In particular, $f$ is a limit of a sequence of such $g$ in $C_b(Omega)$ and hence itself an element of the $mathbb R$-Banach space $C_b(Omega)$.
Question:
- How can we show the converse, i.e. that $$left{fin C_b(Omega):forallvarepsilon:exists KsubseteqOmega:Ktext{ is compact and }|f(x)|<varepsilontext{ for all }xinOmegasetminus Kright}=C_0(Omega)?$$
- By 1. the set on the left-hand side is a $mathbb R$-Banach space. In the literature, I've often seen that $Omega$ is assumed to be locally compact (and Hausdorff) or separable. What's the reason for these assumptions?
general-topology analysis compactness
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
add a comment |
Let
$Omega$ be a topological space
$C(Omega)$ denote the set of continuous functions from $Omega$ to $mathbb R$
$C_b(Omega)$ denote the set of bounded functions in $C(Omega)$ equipped with $$left|fright|_infty:=sup_{xinOmega}|f(x)|;;;text{for }f:Omegatomathbb R$$
Now, let $$operatorname{supp}f:=overline{left{xinOmega:f(x)ne0right}}$$ and $$C_c(Omega):=left{fin C(Omega):operatorname{supp}ftext{ is compact}right}.$$ Clearly, $C_c(Omega)$ is a (not necessarily closed) subspace of $C_b(Omega)$. Let $C_0(Omega)$ denote the completion of $C_c(Omega)$ with respect to $left|;cdot;right|_infty$.
If $fin C_0(Omega)$ and $varepsilon>0$, there is a $gin C_c(Omega)$ with $$left|f-gright|_infty<varepsilontag1.$$ Since $gin C_c(Omega)$, $$K:=operatorname{supp}g$$ is compact and $$g(x)=0;;;text{for all }xinOmegasetminus Ktag2.$$ Thus, $$|f(x)|le|f(x)-g(x)|+|g(x)|<varepsilon;;;text{for all }xinOmegasetminus Ktag3.$$ In particular, $f$ is a limit of a sequence of such $g$ in $C_b(Omega)$ and hence itself an element of the $mathbb R$-Banach space $C_b(Omega)$.
Question:
- How can we show the converse, i.e. that $$left{fin C_b(Omega):forallvarepsilon:exists KsubseteqOmega:Ktext{ is compact and }|f(x)|<varepsilontext{ for all }xinOmegasetminus Kright}=C_0(Omega)?$$
- By 1. the set on the left-hand side is a $mathbb R$-Banach space. In the literature, I've often seen that $Omega$ is assumed to be locally compact (and Hausdorff) or separable. What's the reason for these assumptions?
general-topology analysis compactness
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
There is maybe a typo, do you mean $g in C_b$ or $g in C{color{blue} c}$ ?
– Delta-u
Nov 7 '18 at 14:19
@Delta-u Indeed, that was a typo. I've meant $gin C_c(Omega)$.
– 0xbadf00d
Nov 7 '18 at 16:22
add a comment |
Let
$Omega$ be a topological space
$C(Omega)$ denote the set of continuous functions from $Omega$ to $mathbb R$
$C_b(Omega)$ denote the set of bounded functions in $C(Omega)$ equipped with $$left|fright|_infty:=sup_{xinOmega}|f(x)|;;;text{for }f:Omegatomathbb R$$
Now, let $$operatorname{supp}f:=overline{left{xinOmega:f(x)ne0right}}$$ and $$C_c(Omega):=left{fin C(Omega):operatorname{supp}ftext{ is compact}right}.$$ Clearly, $C_c(Omega)$ is a (not necessarily closed) subspace of $C_b(Omega)$. Let $C_0(Omega)$ denote the completion of $C_c(Omega)$ with respect to $left|;cdot;right|_infty$.
If $fin C_0(Omega)$ and $varepsilon>0$, there is a $gin C_c(Omega)$ with $$left|f-gright|_infty<varepsilontag1.$$ Since $gin C_c(Omega)$, $$K:=operatorname{supp}g$$ is compact and $$g(x)=0;;;text{for all }xinOmegasetminus Ktag2.$$ Thus, $$|f(x)|le|f(x)-g(x)|+|g(x)|<varepsilon;;;text{for all }xinOmegasetminus Ktag3.$$ In particular, $f$ is a limit of a sequence of such $g$ in $C_b(Omega)$ and hence itself an element of the $mathbb R$-Banach space $C_b(Omega)$.
Question:
- How can we show the converse, i.e. that $$left{fin C_b(Omega):forallvarepsilon:exists KsubseteqOmega:Ktext{ is compact and }|f(x)|<varepsilontext{ for all }xinOmegasetminus Kright}=C_0(Omega)?$$
- By 1. the set on the left-hand side is a $mathbb R$-Banach space. In the literature, I've often seen that $Omega$ is assumed to be locally compact (and Hausdorff) or separable. What's the reason for these assumptions?
general-topology analysis compactness
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
Let
$Omega$ be a topological space
$C(Omega)$ denote the set of continuous functions from $Omega$ to $mathbb R$
$C_b(Omega)$ denote the set of bounded functions in $C(Omega)$ equipped with $$left|fright|_infty:=sup_{xinOmega}|f(x)|;;;text{for }f:Omegatomathbb R$$
Now, let $$operatorname{supp}f:=overline{left{xinOmega:f(x)ne0right}}$$ and $$C_c(Omega):=left{fin C(Omega):operatorname{supp}ftext{ is compact}right}.$$ Clearly, $C_c(Omega)$ is a (not necessarily closed) subspace of $C_b(Omega)$. Let $C_0(Omega)$ denote the completion of $C_c(Omega)$ with respect to $left|;cdot;right|_infty$.
If $fin C_0(Omega)$ and $varepsilon>0$, there is a $gin C_c(Omega)$ with $$left|f-gright|_infty<varepsilontag1.$$ Since $gin C_c(Omega)$, $$K:=operatorname{supp}g$$ is compact and $$g(x)=0;;;text{for all }xinOmegasetminus Ktag2.$$ Thus, $$|f(x)|le|f(x)-g(x)|+|g(x)|<varepsilon;;;text{for all }xinOmegasetminus Ktag3.$$ In particular, $f$ is a limit of a sequence of such $g$ in $C_b(Omega)$ and hence itself an element of the $mathbb R$-Banach space $C_b(Omega)$.
Question:
- How can we show the converse, i.e. that $$left{fin C_b(Omega):forallvarepsilon:exists KsubseteqOmega:Ktext{ is compact and }|f(x)|<varepsilontext{ for all }xinOmegasetminus Kright}=C_0(Omega)?$$
- By 1. the set on the left-hand side is a $mathbb R$-Banach space. In the literature, I've often seen that $Omega$ is assumed to be locally compact (and Hausdorff) or separable. What's the reason for these assumptions?
general-topology analysis compactness
general-topology analysis compactness
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
edited Nov 22 '18 at 22:34
0xbadf00d
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
asked Nov 7 '18 at 12:38
0xbadf00d0xbadf00d
1,77741430
1,77741430
There is maybe a typo, do you mean $g in C_b$ or $g in C{color{blue} c}$ ?
– Delta-u
Nov 7 '18 at 14:19
@Delta-u Indeed, that was a typo. I've meant $gin C_c(Omega)$.
– 0xbadf00d
Nov 7 '18 at 16:22
add a comment |
There is maybe a typo, do you mean $g in C_b$ or $g in C{color{blue} c}$ ?
– Delta-u
Nov 7 '18 at 14:19
@Delta-u Indeed, that was a typo. I've meant $gin C_c(Omega)$.
– 0xbadf00d
Nov 7 '18 at 16:22
There is maybe a typo, do you mean $g in C_b$ or $g in C{color{blue} c}$ ?
– Delta-u
Nov 7 '18 at 14:19
There is maybe a typo, do you mean $g in C_b$ or $g in C{color{blue} c}$ ?
– Delta-u
Nov 7 '18 at 14:19
@Delta-u Indeed, that was a typo. I've meant $gin C_c(Omega)$.
– 0xbadf00d
Nov 7 '18 at 16:22
@Delta-u Indeed, that was a typo. I've meant $gin C_c(Omega)$.
– 0xbadf00d
Nov 7 '18 at 16:22
add a comment |
1 Answer
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For the first point given$f in C_b(Omega)$ in the left-hand side set you need to prove that for any $varepsilon >0$ there exist $g in C_c(Omega)$ such that $|f-g|_infty leq varepsilon$.
So let $varepsilon >0$. By definition there exist a compact set $K$ such that $|f(x)| leq varepsilon $ for all $x notin K$.
Let us assume that there exist $chi$ a continuous function such that $operatorname{supp} chi$ is compact, $chi(x)=1$ for all $x in K$ and $forall x in Omega$ we have $0 leq chi(x) leq 1$.
Then with:
$$g=fchi$$
you have:
$operatorname{supp} g subset operatorname{supp} chi$ has a compact support- For $x in K$, $|f(x)-g(x)| =0 leq varepsilon$
- For $x notin K$ $|f(x)-g(x)| = |f(x)| |1-chi(x)| leq varepsilon cdot 1$
so $g$ is indeed in $C_c(Omega)$ and such that $|f-fg|_infty leq varepsilon$.
The answer of your second point lies in the function $chi$. You need some assumptions about $Omega$ to have the existence of such function for any compact.
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
Which assumptions on $Omega$ ensure the existence of $chi$?
– 0xbadf00d
Nov 23 '18 at 9:34
add a comment |
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1 Answer
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For the first point given$f in C_b(Omega)$ in the left-hand side set you need to prove that for any $varepsilon >0$ there exist $g in C_c(Omega)$ such that $|f-g|_infty leq varepsilon$.
So let $varepsilon >0$. By definition there exist a compact set $K$ such that $|f(x)| leq varepsilon $ for all $x notin K$.
Let us assume that there exist $chi$ a continuous function such that $operatorname{supp} chi$ is compact, $chi(x)=1$ for all $x in K$ and $forall x in Omega$ we have $0 leq chi(x) leq 1$.
Then with:
$$g=fchi$$
you have:
$operatorname{supp} g subset operatorname{supp} chi$ has a compact support- For $x in K$, $|f(x)-g(x)| =0 leq varepsilon$
- For $x notin K$ $|f(x)-g(x)| = |f(x)| |1-chi(x)| leq varepsilon cdot 1$
so $g$ is indeed in $C_c(Omega)$ and such that $|f-fg|_infty leq varepsilon$.
The answer of your second point lies in the function $chi$. You need some assumptions about $Omega$ to have the existence of such function for any compact.
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
Which assumptions on $Omega$ ensure the existence of $chi$?
– 0xbadf00d
Nov 23 '18 at 9:34
add a comment |
For the first point given$f in C_b(Omega)$ in the left-hand side set you need to prove that for any $varepsilon >0$ there exist $g in C_c(Omega)$ such that $|f-g|_infty leq varepsilon$.
So let $varepsilon >0$. By definition there exist a compact set $K$ such that $|f(x)| leq varepsilon $ for all $x notin K$.
Let us assume that there exist $chi$ a continuous function such that $operatorname{supp} chi$ is compact, $chi(x)=1$ for all $x in K$ and $forall x in Omega$ we have $0 leq chi(x) leq 1$.
Then with:
$$g=fchi$$
you have:
$operatorname{supp} g subset operatorname{supp} chi$ has a compact support- For $x in K$, $|f(x)-g(x)| =0 leq varepsilon$
- For $x notin K$ $|f(x)-g(x)| = |f(x)| |1-chi(x)| leq varepsilon cdot 1$
so $g$ is indeed in $C_c(Omega)$ and such that $|f-fg|_infty leq varepsilon$.
The answer of your second point lies in the function $chi$. You need some assumptions about $Omega$ to have the existence of such function for any compact.
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
Which assumptions on $Omega$ ensure the existence of $chi$?
– 0xbadf00d
Nov 23 '18 at 9:34
add a comment |
For the first point given$f in C_b(Omega)$ in the left-hand side set you need to prove that for any $varepsilon >0$ there exist $g in C_c(Omega)$ such that $|f-g|_infty leq varepsilon$.
So let $varepsilon >0$. By definition there exist a compact set $K$ such that $|f(x)| leq varepsilon $ for all $x notin K$.
Let us assume that there exist $chi$ a continuous function such that $operatorname{supp} chi$ is compact, $chi(x)=1$ for all $x in K$ and $forall x in Omega$ we have $0 leq chi(x) leq 1$.
Then with:
$$g=fchi$$
you have:
$operatorname{supp} g subset operatorname{supp} chi$ has a compact support- For $x in K$, $|f(x)-g(x)| =0 leq varepsilon$
- For $x notin K$ $|f(x)-g(x)| = |f(x)| |1-chi(x)| leq varepsilon cdot 1$
so $g$ is indeed in $C_c(Omega)$ and such that $|f-fg|_infty leq varepsilon$.
The answer of your second point lies in the function $chi$. You need some assumptions about $Omega$ to have the existence of such function for any compact.
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
For the first point given$f in C_b(Omega)$ in the left-hand side set you need to prove that for any $varepsilon >0$ there exist $g in C_c(Omega)$ such that $|f-g|_infty leq varepsilon$.
So let $varepsilon >0$. By definition there exist a compact set $K$ such that $|f(x)| leq varepsilon $ for all $x notin K$.
Let us assume that there exist $chi$ a continuous function such that $operatorname{supp} chi$ is compact, $chi(x)=1$ for all $x in K$ and $forall x in Omega$ we have $0 leq chi(x) leq 1$.
Then with:
$$g=fchi$$
you have:
$operatorname{supp} g subset operatorname{supp} chi$ has a compact support- For $x in K$, $|f(x)-g(x)| =0 leq varepsilon$
- For $x notin K$ $|f(x)-g(x)| = |f(x)| |1-chi(x)| leq varepsilon cdot 1$
so $g$ is indeed in $C_c(Omega)$ and such that $|f-fg|_infty leq varepsilon$.
The answer of your second point lies in the function $chi$. You need some assumptions about $Omega$ to have the existence of such function for any compact.
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
answered Nov 7 '18 at 20:02
Delta-uDelta-u
5,3892618
5,3892618
Which assumptions on $Omega$ ensure the existence of $chi$?
– 0xbadf00d
Nov 23 '18 at 9:34
add a comment |
Which assumptions on $Omega$ ensure the existence of $chi$?
– 0xbadf00d
Nov 23 '18 at 9:34
Which assumptions on $Omega$ ensure the existence of $chi$?
– 0xbadf00d
Nov 23 '18 at 9:34
Which assumptions on $Omega$ ensure the existence of $chi$?
– 0xbadf00d
Nov 23 '18 at 9:34
add a comment |
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There is maybe a typo, do you mean $g in C_b$ or $g in C{color{blue} c}$ ?
– Delta-u
Nov 7 '18 at 14:19
@Delta-u Indeed, that was a typo. I've meant $gin C_c(Omega)$.
– 0xbadf00d
Nov 7 '18 at 16:22