Mixing
$Bbb R^d$ with norm is a Banach
$begingroup$
Another problem I found in a textbook when brushing up on my real analysis. Was given a hint too.
Prove that $Bbb R^d$ with the norm
$||x||_1=sum_{i=1}^d |x_i|, x ∈ Bbb R^d$
is a Banach space.
Hint: Consider a Cauchy sequence $(x_n) text{ in }(Bbb R^d, || · ||_1)$ and show that it is Cauchy coordinate-wise. What could the limit of $(x_n)$ then be?
real-analysis metric-spaces banach-spaces normed-spaces
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
$endgroup$
add a comment |
$begingroup$
Another problem I found in a textbook when brushing up on my real analysis. Was given a hint too.
Prove that $Bbb R^d$ with the norm
$||x||_1=sum_{i=1}^d |x_i|, x ∈ Bbb R^d$
is a Banach space.
Hint: Consider a Cauchy sequence $(x_n) text{ in }(Bbb R^d, || · ||_1)$ and show that it is Cauchy coordinate-wise. What could the limit of $(x_n)$ then be?
real-analysis metric-spaces banach-spaces normed-spaces
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
$endgroup$
5
$begingroup$
Thanks for the hint, but I'd rather do it on my own.
$endgroup$
– Saucy O'Path
Jan 20 at 17:29
3
$begingroup$
Every finite dimensional space is a Banach space, Proof: Use that all norms on finite dimensional spaces are equivalent.
$endgroup$
– James
Jan 20 at 17:34
2
$begingroup$
Excellent hint. I suggest following it and attempting the problem.
$endgroup$
– LoveTooNap29
Jan 20 at 18:03
add a comment |
$begingroup$
Another problem I found in a textbook when brushing up on my real analysis. Was given a hint too.
Prove that $Bbb R^d$ with the norm
$||x||_1=sum_{i=1}^d |x_i|, x ∈ Bbb R^d$
is a Banach space.
Hint: Consider a Cauchy sequence $(x_n) text{ in }(Bbb R^d, || · ||_1)$ and show that it is Cauchy coordinate-wise. What could the limit of $(x_n)$ then be?
real-analysis metric-spaces banach-spaces normed-spaces
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
$endgroup$
Another problem I found in a textbook when brushing up on my real analysis. Was given a hint too.
Prove that $Bbb R^d$ with the norm
$||x||_1=sum_{i=1}^d |x_i|, x ∈ Bbb R^d$
is a Banach space.
Hint: Consider a Cauchy sequence $(x_n) text{ in }(Bbb R^d, || · ||_1)$ and show that it is Cauchy coordinate-wise. What could the limit of $(x_n)$ then be?
real-analysis metric-spaces banach-spaces normed-spaces
real-analysis metric-spaces banach-spaces normed-spaces
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
edited Jan 20 at 17:32
ʎpoqou
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
asked Jan 20 at 17:28
ʎpoqouʎpoqou
3561211
3561211
5
$begingroup$
Thanks for the hint, but I'd rather do it on my own.
$endgroup$
– Saucy O'Path
Jan 20 at 17:29
3
$begingroup$
Every finite dimensional space is a Banach space, Proof: Use that all norms on finite dimensional spaces are equivalent.
$endgroup$
– James
Jan 20 at 17:34
2
$begingroup$
Excellent hint. I suggest following it and attempting the problem.
$endgroup$
– LoveTooNap29
Jan 20 at 18:03
add a comment |
5
$begingroup$
Thanks for the hint, but I'd rather do it on my own.
$endgroup$
– Saucy O'Path
Jan 20 at 17:29
3
$begingroup$
Every finite dimensional space is a Banach space, Proof: Use that all norms on finite dimensional spaces are equivalent.
$endgroup$
– James
Jan 20 at 17:34
2
$begingroup$
Excellent hint. I suggest following it and attempting the problem.
$endgroup$
– LoveTooNap29
Jan 20 at 18:03
5
5
$begingroup$
Thanks for the hint, but I'd rather do it on my own.
$endgroup$
– Saucy O'Path
Jan 20 at 17:29
$begingroup$
Thanks for the hint, but I'd rather do it on my own.
$endgroup$
– Saucy O'Path
Jan 20 at 17:29
3
3
$begingroup$
Every finite dimensional space is a Banach space, Proof: Use that all norms on finite dimensional spaces are equivalent.
$endgroup$
– James
Jan 20 at 17:34
$begingroup$
Every finite dimensional space is a Banach space, Proof: Use that all norms on finite dimensional spaces are equivalent.
$endgroup$
– James
Jan 20 at 17:34
2
2
$begingroup$
Excellent hint. I suggest following it and attempting the problem.
$endgroup$
– LoveTooNap29
Jan 20 at 18:03
$begingroup$
Excellent hint. I suggest following it and attempting the problem.
$endgroup$
– LoveTooNap29
Jan 20 at 18:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To show that $Bbb R^d$ with the norm
$Vert x Vert_1 = displaystyle sum_1^d vert x_i vert, tag 1$
where
$x = (x_1, x_2, ldots, x_d) tag 2$
is Banach we need to prove it is Cauchy-complete with respect to this norm; that is, if
$y_i in Bbb R^d tag 3$
is a $Vert cdot Vert_1$-Cauchy sequence, then there exists
$y in Bbb R^d tag 4$
with
$y_i to y tag 5$
in the $Vert cdot Vert_1$ norm.
Now if $y_i$ is $Vert cdot Vert_1$-Cauchy, for every real $epsilon > 0$ there exists $N in Bbb N$ such that, for $m, n > N$,
$Vert y_m - y_n Vert_1 < epsilon; tag 6$
if we re-write this in terms of the defiinition (1) we obtain
$displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon, tag 7$
and we observe that, for every $l$, $1 le l le d$, this yields
$vert y_{ml} - y_{nl} vert le displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon; tag 8$
thus the sequence $y_{ml}$ for fixed $l$ is Cauchy in $Bbb R$ with respect to the usual norm $vert cdot vert$; and since $Bbb R$ is Cauchy-complete with respect to $vert cdot vert$ we infer that for each $l$ there is a $y^ast_l$ with
$y_{ml} to y_l^ast tag 9$
in the $vert cdot vert$ norm on $Bbb R$; thus, taking $N$ larger if necessary, we have
$vert y_{ml} - y_l^ast vert < dfrac{epsilon}{d}, ; 1 le l le d, ; m > N; tag{10}$
setting
$y^ast = (y_1^ast, y_2^ast, ldots, y_d^ast) in Bbb R^d, tag{11}$
we further have
$Vert y_m - y^ast Vert_1 = displaystyle sum_{l = 1}^d vert y_{ml} - y_l vert < d dfrac{epsilon}{d} = epsilon, tag{12}$
that is,
$y_m to y^ast tag{13}$
in the $Vert cdot Vert_1$ norm on $Bbb R^d$; thus $Bbb R^d$ is $Vert cdot Vert_1$-Cauchy complete; hence $Vert cdot Vert_1$-Banach.
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
$endgroup$
add a comment |
$begingroup$
Use this:
$$ |x - y|_1 ge |x_k - y_k|, 1le k le d$$
and the fact that the real numbers are complete.
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To show that $Bbb R^d$ with the norm
$Vert x Vert_1 = displaystyle sum_1^d vert x_i vert, tag 1$
where
$x = (x_1, x_2, ldots, x_d) tag 2$
is Banach we need to prove it is Cauchy-complete with respect to this norm; that is, if
$y_i in Bbb R^d tag 3$
is a $Vert cdot Vert_1$-Cauchy sequence, then there exists
$y in Bbb R^d tag 4$
with
$y_i to y tag 5$
in the $Vert cdot Vert_1$ norm.
Now if $y_i$ is $Vert cdot Vert_1$-Cauchy, for every real $epsilon > 0$ there exists $N in Bbb N$ such that, for $m, n > N$,
$Vert y_m - y_n Vert_1 < epsilon; tag 6$
if we re-write this in terms of the defiinition (1) we obtain
$displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon, tag 7$
and we observe that, for every $l$, $1 le l le d$, this yields
$vert y_{ml} - y_{nl} vert le displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon; tag 8$
thus the sequence $y_{ml}$ for fixed $l$ is Cauchy in $Bbb R$ with respect to the usual norm $vert cdot vert$; and since $Bbb R$ is Cauchy-complete with respect to $vert cdot vert$ we infer that for each $l$ there is a $y^ast_l$ with
$y_{ml} to y_l^ast tag 9$
in the $vert cdot vert$ norm on $Bbb R$; thus, taking $N$ larger if necessary, we have
$vert y_{ml} - y_l^ast vert < dfrac{epsilon}{d}, ; 1 le l le d, ; m > N; tag{10}$
setting
$y^ast = (y_1^ast, y_2^ast, ldots, y_d^ast) in Bbb R^d, tag{11}$
we further have
$Vert y_m - y^ast Vert_1 = displaystyle sum_{l = 1}^d vert y_{ml} - y_l vert < d dfrac{epsilon}{d} = epsilon, tag{12}$
that is,
$y_m to y^ast tag{13}$
in the $Vert cdot Vert_1$ norm on $Bbb R^d$; thus $Bbb R^d$ is $Vert cdot Vert_1$-Cauchy complete; hence $Vert cdot Vert_1$-Banach.
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
$endgroup$
add a comment |
$begingroup$
To show that $Bbb R^d$ with the norm
$Vert x Vert_1 = displaystyle sum_1^d vert x_i vert, tag 1$
where
$x = (x_1, x_2, ldots, x_d) tag 2$
is Banach we need to prove it is Cauchy-complete with respect to this norm; that is, if
$y_i in Bbb R^d tag 3$
is a $Vert cdot Vert_1$-Cauchy sequence, then there exists
$y in Bbb R^d tag 4$
with
$y_i to y tag 5$
in the $Vert cdot Vert_1$ norm.
Now if $y_i$ is $Vert cdot Vert_1$-Cauchy, for every real $epsilon > 0$ there exists $N in Bbb N$ such that, for $m, n > N$,
$Vert y_m - y_n Vert_1 < epsilon; tag 6$
if we re-write this in terms of the defiinition (1) we obtain
$displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon, tag 7$
and we observe that, for every $l$, $1 le l le d$, this yields
$vert y_{ml} - y_{nl} vert le displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon; tag 8$
thus the sequence $y_{ml}$ for fixed $l$ is Cauchy in $Bbb R$ with respect to the usual norm $vert cdot vert$; and since $Bbb R$ is Cauchy-complete with respect to $vert cdot vert$ we infer that for each $l$ there is a $y^ast_l$ with
$y_{ml} to y_l^ast tag 9$
in the $vert cdot vert$ norm on $Bbb R$; thus, taking $N$ larger if necessary, we have
$vert y_{ml} - y_l^ast vert < dfrac{epsilon}{d}, ; 1 le l le d, ; m > N; tag{10}$
setting
$y^ast = (y_1^ast, y_2^ast, ldots, y_d^ast) in Bbb R^d, tag{11}$
we further have
$Vert y_m - y^ast Vert_1 = displaystyle sum_{l = 1}^d vert y_{ml} - y_l vert < d dfrac{epsilon}{d} = epsilon, tag{12}$
that is,
$y_m to y^ast tag{13}$
in the $Vert cdot Vert_1$ norm on $Bbb R^d$; thus $Bbb R^d$ is $Vert cdot Vert_1$-Cauchy complete; hence $Vert cdot Vert_1$-Banach.
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
$endgroup$
add a comment |
$begingroup$
To show that $Bbb R^d$ with the norm
$Vert x Vert_1 = displaystyle sum_1^d vert x_i vert, tag 1$
where
$x = (x_1, x_2, ldots, x_d) tag 2$
is Banach we need to prove it is Cauchy-complete with respect to this norm; that is, if
$y_i in Bbb R^d tag 3$
is a $Vert cdot Vert_1$-Cauchy sequence, then there exists
$y in Bbb R^d tag 4$
with
$y_i to y tag 5$
in the $Vert cdot Vert_1$ norm.
Now if $y_i$ is $Vert cdot Vert_1$-Cauchy, for every real $epsilon > 0$ there exists $N in Bbb N$ such that, for $m, n > N$,
$Vert y_m - y_n Vert_1 < epsilon; tag 6$
if we re-write this in terms of the defiinition (1) we obtain
$displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon, tag 7$
and we observe that, for every $l$, $1 le l le d$, this yields
$vert y_{ml} - y_{nl} vert le displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon; tag 8$
thus the sequence $y_{ml}$ for fixed $l$ is Cauchy in $Bbb R$ with respect to the usual norm $vert cdot vert$; and since $Bbb R$ is Cauchy-complete with respect to $vert cdot vert$ we infer that for each $l$ there is a $y^ast_l$ with
$y_{ml} to y_l^ast tag 9$
in the $vert cdot vert$ norm on $Bbb R$; thus, taking $N$ larger if necessary, we have
$vert y_{ml} - y_l^ast vert < dfrac{epsilon}{d}, ; 1 le l le d, ; m > N; tag{10}$
setting
$y^ast = (y_1^ast, y_2^ast, ldots, y_d^ast) in Bbb R^d, tag{11}$
we further have
$Vert y_m - y^ast Vert_1 = displaystyle sum_{l = 1}^d vert y_{ml} - y_l vert < d dfrac{epsilon}{d} = epsilon, tag{12}$
that is,
$y_m to y^ast tag{13}$
in the $Vert cdot Vert_1$ norm on $Bbb R^d$; thus $Bbb R^d$ is $Vert cdot Vert_1$-Cauchy complete; hence $Vert cdot Vert_1$-Banach.
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
$endgroup$
To show that $Bbb R^d$ with the norm
$Vert x Vert_1 = displaystyle sum_1^d vert x_i vert, tag 1$
where
$x = (x_1, x_2, ldots, x_d) tag 2$
is Banach we need to prove it is Cauchy-complete with respect to this norm; that is, if
$y_i in Bbb R^d tag 3$
is a $Vert cdot Vert_1$-Cauchy sequence, then there exists
$y in Bbb R^d tag 4$
with
$y_i to y tag 5$
in the $Vert cdot Vert_1$ norm.
Now if $y_i$ is $Vert cdot Vert_1$-Cauchy, for every real $epsilon > 0$ there exists $N in Bbb N$ such that, for $m, n > N$,
$Vert y_m - y_n Vert_1 < epsilon; tag 6$
if we re-write this in terms of the defiinition (1) we obtain
$displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon, tag 7$
and we observe that, for every $l$, $1 le l le d$, this yields
$vert y_{ml} - y_{nl} vert le displaystyle sum_{k = 1}^d vert y_{mk} - y_{nk} vert < epsilon; tag 8$
thus the sequence $y_{ml}$ for fixed $l$ is Cauchy in $Bbb R$ with respect to the usual norm $vert cdot vert$; and since $Bbb R$ is Cauchy-complete with respect to $vert cdot vert$ we infer that for each $l$ there is a $y^ast_l$ with
$y_{ml} to y_l^ast tag 9$
in the $vert cdot vert$ norm on $Bbb R$; thus, taking $N$ larger if necessary, we have
$vert y_{ml} - y_l^ast vert < dfrac{epsilon}{d}, ; 1 le l le d, ; m > N; tag{10}$
setting
$y^ast = (y_1^ast, y_2^ast, ldots, y_d^ast) in Bbb R^d, tag{11}$
we further have
$Vert y_m - y^ast Vert_1 = displaystyle sum_{l = 1}^d vert y_{ml} - y_l vert < d dfrac{epsilon}{d} = epsilon, tag{12}$
that is,
$y_m to y^ast tag{13}$
in the $Vert cdot Vert_1$ norm on $Bbb R^d$; thus $Bbb R^d$ is $Vert cdot Vert_1$-Cauchy complete; hence $Vert cdot Vert_1$-Banach.
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
edited Jan 20 at 18:58
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
answered Jan 20 at 18:41
Robert LewisRobert Lewis
47.6k23067
47.6k23067
add a comment |
add a comment |
$begingroup$
Use this:
$$ |x - y|_1 ge |x_k - y_k|, 1le k le d$$
and the fact that the real numbers are complete.
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
$endgroup$
add a comment |
$begingroup$
Use this:
$$ |x - y|_1 ge |x_k - y_k|, 1le k le d$$
and the fact that the real numbers are complete.
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
$endgroup$
add a comment |
$begingroup$
Use this:
$$ |x - y|_1 ge |x_k - y_k|, 1le k le d$$
and the fact that the real numbers are complete.
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
$endgroup$
Use this:
$$ |x - y|_1 ge |x_k - y_k|, 1le k le d$$
and the fact that the real numbers are complete.
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
answered Jan 20 at 17:44
ncmathsadistncmathsadist
42.9k260103
42.9k260103
add a comment |
add a comment |
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5
$begingroup$
Thanks for the hint, but I'd rather do it on my own.
$endgroup$
– Saucy O'Path
Jan 20 at 17:29
3
$begingroup$
Every finite dimensional space is a Banach space, Proof: Use that all norms on finite dimensional spaces are equivalent.
$endgroup$
– James
Jan 20 at 17:34
2
$begingroup$
Excellent hint. I suggest following it and attempting the problem.
$endgroup$
– LoveTooNap29
Jan 20 at 18:03