Supnorm Defined on set of all uniformly continuous functions is Banach space
$begingroup$
If we define Supnorm on $C(bar{U})$ is
$$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space
what I know is :
I'm proving this normed space
since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$
and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$
finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$
how to prove completeness ..thank you
functional-analysis sobolev-spaces
asked Jan 21 at 8:16
learnerlearner
1188
$endgroup$
add a comment |
$begingroup$
If we define Supnorm on $C(bar{U})$ is
$$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space
what I know is :
I'm proving this normed space
since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$
and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$
finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$
how to prove completeness ..thank you
functional-analysis sobolev-spaces
asked Jan 21 at 8:16
learnerlearner
1188
$endgroup$
add a comment |
$begingroup$
If we define Supnorm on $C(bar{U})$ is
$$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space
what I know is :
I'm proving this normed space
since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$
and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$
finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$
how to prove completeness ..thank you
functional-analysis sobolev-spaces
asked Jan 21 at 8:16
learnerlearner
1188
$endgroup$
If we define Supnorm on $C(bar{U})$ is
$$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space
what I know is :
I'm proving this normed space
since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$
and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$
finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$
how to prove completeness ..thank you
functional-analysis sobolev-spaces
functional-analysis sobolev-spaces
asked Jan 21 at 8:16
learnerlearner
1188
asked Jan 21 at 8:16
learnerlearner
1188
asked Jan 21 at 8:16
learnerlearner
1188
asked Jan 21 at 8:16
learnerlearner
1188
asked Jan 21 at 8:16
learnerlearner
1188
1188
add a comment |
add a comment |
1 Answer
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If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
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I don't understand last lines sir
$endgroup$
– learner
Jan 21 at 12:47
$begingroup$
and how can we say that if sequence is Cauchy then limit exists
$endgroup$
– learner
Jan 21 at 13:03
$begingroup$
and we need to show that f is uniformly continuous on bounded subset of U
$endgroup$
– learner
Jan 21 at 13:06
$begingroup$
$mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:18
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
$endgroup$
$begingroup$
I don't understand last lines sir
$endgroup$
– learner
Jan 21 at 12:47
$begingroup$
and how can we say that if sequence is Cauchy then limit exists
$endgroup$
– learner
Jan 21 at 13:03
$begingroup$
and we need to show that f is uniformly continuous on bounded subset of U
$endgroup$
– learner
Jan 21 at 13:06
$begingroup$
$mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:18
add a comment |
$begingroup$
If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
$endgroup$
$begingroup$
I don't understand last lines sir
$endgroup$
– learner
Jan 21 at 12:47
$begingroup$
and how can we say that if sequence is Cauchy then limit exists
$endgroup$
– learner
Jan 21 at 13:03
$begingroup$
and we need to show that f is uniformly continuous on bounded subset of U
$endgroup$
– learner
Jan 21 at 13:06
$begingroup$
$mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:18
add a comment |
$begingroup$
If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
$endgroup$
If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
answered Jan 21 at 8:23
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
62.2k42262
$begingroup$
I don't understand last lines sir
$endgroup$
– learner
Jan 21 at 12:47
$begingroup$
and how can we say that if sequence is Cauchy then limit exists
$endgroup$
– learner
Jan 21 at 13:03
$begingroup$
and we need to show that f is uniformly continuous on bounded subset of U
$endgroup$
– learner
Jan 21 at 13:06
$begingroup$
$mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:18
add a comment |
$begingroup$
I don't understand last lines sir
$endgroup$
– learner
Jan 21 at 12:47
$begingroup$
and how can we say that if sequence is Cauchy then limit exists
$endgroup$
– learner
Jan 21 at 13:03
$begingroup$
and we need to show that f is uniformly continuous on bounded subset of U
$endgroup$
– learner
Jan 21 at 13:06
$begingroup$
$mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:18
$begingroup$
I don't understand last lines sir
$endgroup$
– learner
Jan 21 at 12:47
$begingroup$
I don't understand last lines sir
$endgroup$
– learner
Jan 21 at 12:47
$begingroup$
and how can we say that if sequence is Cauchy then limit exists
$endgroup$
– learner
Jan 21 at 13:03
$begingroup$
and how can we say that if sequence is Cauchy then limit exists
$endgroup$
– learner
Jan 21 at 13:03
$begingroup$
and we need to show that f is uniformly continuous on bounded subset of U
$endgroup$
– learner
Jan 21 at 13:06
$begingroup$
and we need to show that f is uniformly continuous on bounded subset of U
$endgroup$
– learner
Jan 21 at 13:06
$begingroup$
$mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:18
$begingroup$
$mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:18
add a comment |
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