Mixing
Is $C_0(mathbb{R}^n)$ a Banach Space?
$begingroup$
By defining $$
C_0(mathbb{R}^n):={u:uin C(mathbb{R}^n),quadmathtt{and}quadlim_{|x|rightarrowinfty}u(x)=0}
$$
normed with $||u||:=sup_{xinmathbb{R}^n}|u(x)|$. As far as I can remember, this is a Banach space.
My question:
Is this ture or there are counterexamples for this?
Attempts:
I think it is easy to give counterexaple for $C_c(mathbb{R}^n)$ where the condition of "limit at infinitty" is changed into "having compact support". In fact by considering $$
f_n(x):=f_{n-1}(x)+dfrac{1}{n} chi_{xin B_nsetminus B_{n-1}},quad f_0(x):=0, B_0:=varnothing.
$$
which is a Cauchy sequence in $C_c$ but the limit doesn't have compact support. But the limit satisfies $lim_{|x|rightarrowinfty}f(x)=0$.
Thanks for reading and any idea will be helpful.
real-analysis calculus banach-spaces
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
$endgroup$
add a comment |
$begingroup$
By defining $$
C_0(mathbb{R}^n):={u:uin C(mathbb{R}^n),quadmathtt{and}quadlim_{|x|rightarrowinfty}u(x)=0}
$$
normed with $||u||:=sup_{xinmathbb{R}^n}|u(x)|$. As far as I can remember, this is a Banach space.
My question:
Is this ture or there are counterexamples for this?
Attempts:
I think it is easy to give counterexaple for $C_c(mathbb{R}^n)$ where the condition of "limit at infinitty" is changed into "having compact support". In fact by considering $$
f_n(x):=f_{n-1}(x)+dfrac{1}{n} chi_{xin B_nsetminus B_{n-1}},quad f_0(x):=0, B_0:=varnothing.
$$
which is a Cauchy sequence in $C_c$ but the limit doesn't have compact support. But the limit satisfies $lim_{|x|rightarrowinfty}f(x)=0$.
Thanks for reading and any idea will be helpful.
real-analysis calculus banach-spaces
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
$endgroup$
$begingroup$
I believe $C_0(X)$ is Banach for any locally compact $X$. I believe it is even a $C^*$-algebra, only not necessarily unital.
$endgroup$
– SmileyCraft
Jan 22 at 2:50
$begingroup$
Thanks for the hint, I remembered where to find a proof of this in my textbook. $C_0(X)$ is just the pre-dual space of Radon Measure space and it is the closure of $C_c(X)$ under this norm. Really appreciate the kindness hit. ;)
$endgroup$
– Zixiao_Liu
Jan 22 at 3:36
$begingroup$
@Zixiao_Liu Don 't ignore absolute value signs. Your $|u|$ is not a norm.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:59
$begingroup$
Yes, I forgot the absolute value signs. So sorry about that. I will restate the question. :)
$endgroup$
– Zixiao_Liu
Jan 22 at 8:02
add a comment |
$begingroup$
By defining $$
C_0(mathbb{R}^n):={u:uin C(mathbb{R}^n),quadmathtt{and}quadlim_{|x|rightarrowinfty}u(x)=0}
$$
normed with $||u||:=sup_{xinmathbb{R}^n}|u(x)|$. As far as I can remember, this is a Banach space.
My question:
Is this ture or there are counterexamples for this?
Attempts:
I think it is easy to give counterexaple for $C_c(mathbb{R}^n)$ where the condition of "limit at infinitty" is changed into "having compact support". In fact by considering $$
f_n(x):=f_{n-1}(x)+dfrac{1}{n} chi_{xin B_nsetminus B_{n-1}},quad f_0(x):=0, B_0:=varnothing.
$$
which is a Cauchy sequence in $C_c$ but the limit doesn't have compact support. But the limit satisfies $lim_{|x|rightarrowinfty}f(x)=0$.
Thanks for reading and any idea will be helpful.
real-analysis calculus banach-spaces
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
$endgroup$
By defining $$
C_0(mathbb{R}^n):={u:uin C(mathbb{R}^n),quadmathtt{and}quadlim_{|x|rightarrowinfty}u(x)=0}
$$
normed with $||u||:=sup_{xinmathbb{R}^n}|u(x)|$. As far as I can remember, this is a Banach space.
My question:
Is this ture or there are counterexamples for this?
Attempts:
I think it is easy to give counterexaple for $C_c(mathbb{R}^n)$ where the condition of "limit at infinitty" is changed into "having compact support". In fact by considering $$
f_n(x):=f_{n-1}(x)+dfrac{1}{n} chi_{xin B_nsetminus B_{n-1}},quad f_0(x):=0, B_0:=varnothing.
$$
which is a Cauchy sequence in $C_c$ but the limit doesn't have compact support. But the limit satisfies $lim_{|x|rightarrowinfty}f(x)=0$.
Thanks for reading and any idea will be helpful.
real-analysis calculus banach-spaces
real-analysis calculus banach-spaces
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
edited Jan 22 at 8:02
Zixiao_Liu
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
asked Jan 22 at 2:13
Zixiao_LiuZixiao_Liu
938
938
$begingroup$
I believe $C_0(X)$ is Banach for any locally compact $X$. I believe it is even a $C^*$-algebra, only not necessarily unital.
$endgroup$
– SmileyCraft
Jan 22 at 2:50
$begingroup$
Thanks for the hint, I remembered where to find a proof of this in my textbook. $C_0(X)$ is just the pre-dual space of Radon Measure space and it is the closure of $C_c(X)$ under this norm. Really appreciate the kindness hit. ;)
$endgroup$
– Zixiao_Liu
Jan 22 at 3:36
$begingroup$
@Zixiao_Liu Don 't ignore absolute value signs. Your $|u|$ is not a norm.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:59
$begingroup$
Yes, I forgot the absolute value signs. So sorry about that. I will restate the question. :)
$endgroup$
– Zixiao_Liu
Jan 22 at 8:02
add a comment |
$begingroup$
I believe $C_0(X)$ is Banach for any locally compact $X$. I believe it is even a $C^*$-algebra, only not necessarily unital.
$endgroup$
– SmileyCraft
Jan 22 at 2:50
$begingroup$
Thanks for the hint, I remembered where to find a proof of this in my textbook. $C_0(X)$ is just the pre-dual space of Radon Measure space and it is the closure of $C_c(X)$ under this norm. Really appreciate the kindness hit. ;)
$endgroup$
– Zixiao_Liu
Jan 22 at 3:36
$begingroup$
@Zixiao_Liu Don 't ignore absolute value signs. Your $|u|$ is not a norm.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:59
$begingroup$
Yes, I forgot the absolute value signs. So sorry about that. I will restate the question. :)
$endgroup$
– Zixiao_Liu
Jan 22 at 8:02
$begingroup$
I believe $C_0(X)$ is Banach for any locally compact $X$. I believe it is even a $C^*$-algebra, only not necessarily unital.
$endgroup$
– SmileyCraft
Jan 22 at 2:50
$begingroup$
I believe $C_0(X)$ is Banach for any locally compact $X$. I believe it is even a $C^*$-algebra, only not necessarily unital.
$endgroup$
– SmileyCraft
Jan 22 at 2:50
$begingroup$
Thanks for the hint, I remembered where to find a proof of this in my textbook. $C_0(X)$ is just the pre-dual space of Radon Measure space and it is the closure of $C_c(X)$ under this norm. Really appreciate the kindness hit. ;)
$endgroup$
– Zixiao_Liu
Jan 22 at 3:36
$begingroup$
Thanks for the hint, I remembered where to find a proof of this in my textbook. $C_0(X)$ is just the pre-dual space of Radon Measure space and it is the closure of $C_c(X)$ under this norm. Really appreciate the kindness hit. ;)
$endgroup$
– Zixiao_Liu
Jan 22 at 3:36
$begingroup$
@Zixiao_Liu Don 't ignore absolute value signs. Your $|u|$ is not a norm.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:59
$begingroup$
@Zixiao_Liu Don 't ignore absolute value signs. Your $|u|$ is not a norm.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:59
$begingroup$
Yes, I forgot the absolute value signs. So sorry about that. I will restate the question. :)
$endgroup$
– Zixiao_Liu
Jan 22 at 8:02
$begingroup$
Yes, I forgot the absolute value signs. So sorry about that. I will restate the question. :)
$endgroup$
– Zixiao_Liu
Jan 22 at 8:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$C_0(mathbb{R}^n)$ is isometric to the space $mathbb{E}$ of continuous functions on $S^n$ (one-point compactification of $mathbb{R}^n$) which vanish on $infty$. The latter is the kernel under the evaluation map
begin{align*}
mathrm{ev}_{infty}:C(S^n) &to mathbb{R}\
f &mapsto f(infty).
end{align*}
Since the evaluation map is continuous, $mathbb{E}$ is closed, therefore Banach (since $C(S^n)$ is Banach) and it follows that $C_0(mathbb{R}^n)$ also is.
As was said in the comments, this holds for any locally compact space Hausdorff $X$ in place of $mathbb{R}^n$. The proof is the same.
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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oldest
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oldest
votes
$begingroup$
$C_0(mathbb{R}^n)$ is isometric to the space $mathbb{E}$ of continuous functions on $S^n$ (one-point compactification of $mathbb{R}^n$) which vanish on $infty$. The latter is the kernel under the evaluation map
begin{align*}
mathrm{ev}_{infty}:C(S^n) &to mathbb{R}\
f &mapsto f(infty).
end{align*}
Since the evaluation map is continuous, $mathbb{E}$ is closed, therefore Banach (since $C(S^n)$ is Banach) and it follows that $C_0(mathbb{R}^n)$ also is.
As was said in the comments, this holds for any locally compact space Hausdorff $X$ in place of $mathbb{R}^n$. The proof is the same.
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
$endgroup$
add a comment |
$begingroup$
$C_0(mathbb{R}^n)$ is isometric to the space $mathbb{E}$ of continuous functions on $S^n$ (one-point compactification of $mathbb{R}^n$) which vanish on $infty$. The latter is the kernel under the evaluation map
begin{align*}
mathrm{ev}_{infty}:C(S^n) &to mathbb{R}\
f &mapsto f(infty).
end{align*}
Since the evaluation map is continuous, $mathbb{E}$ is closed, therefore Banach (since $C(S^n)$ is Banach) and it follows that $C_0(mathbb{R}^n)$ also is.
As was said in the comments, this holds for any locally compact space Hausdorff $X$ in place of $mathbb{R}^n$. The proof is the same.
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
$endgroup$
add a comment |
$begingroup$
$C_0(mathbb{R}^n)$ is isometric to the space $mathbb{E}$ of continuous functions on $S^n$ (one-point compactification of $mathbb{R}^n$) which vanish on $infty$. The latter is the kernel under the evaluation map
begin{align*}
mathrm{ev}_{infty}:C(S^n) &to mathbb{R}\
f &mapsto f(infty).
end{align*}
Since the evaluation map is continuous, $mathbb{E}$ is closed, therefore Banach (since $C(S^n)$ is Banach) and it follows that $C_0(mathbb{R}^n)$ also is.
As was said in the comments, this holds for any locally compact space Hausdorff $X$ in place of $mathbb{R}^n$. The proof is the same.
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
$endgroup$
$C_0(mathbb{R}^n)$ is isometric to the space $mathbb{E}$ of continuous functions on $S^n$ (one-point compactification of $mathbb{R}^n$) which vanish on $infty$. The latter is the kernel under the evaluation map
begin{align*}
mathrm{ev}_{infty}:C(S^n) &to mathbb{R}\
f &mapsto f(infty).
end{align*}
Since the evaluation map is continuous, $mathbb{E}$ is closed, therefore Banach (since $C(S^n)$ is Banach) and it follows that $C_0(mathbb{R}^n)$ also is.
As was said in the comments, this holds for any locally compact space Hausdorff $X$ in place of $mathbb{R}^n$. The proof is the same.
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
answered Jan 22 at 8:50
Aloizio Macedo♦Aloizio Macedo
23.7k23987
23.7k23987
add a comment |
add a comment |
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$begingroup$
I believe $C_0(X)$ is Banach for any locally compact $X$. I believe it is even a $C^*$-algebra, only not necessarily unital.
$endgroup$
– SmileyCraft
Jan 22 at 2:50
$begingroup$
Thanks for the hint, I remembered where to find a proof of this in my textbook. $C_0(X)$ is just the pre-dual space of Radon Measure space and it is the closure of $C_c(X)$ under this norm. Really appreciate the kindness hit. ;)
$endgroup$
– Zixiao_Liu
Jan 22 at 3:36
$begingroup$
@Zixiao_Liu Don 't ignore absolute value signs. Your $|u|$ is not a norm.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:59
$begingroup$
Yes, I forgot the absolute value signs. So sorry about that. I will restate the question. :)
$endgroup$
– Zixiao_Liu
Jan 22 at 8:02