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Proof that $sum_i lvert alpha_irvert leq |x|$, where $|x|$ is the norm
Clash Royale CLAN TAG#URR8PPP
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On another post a user asked for a proof that if a map is linear then it is continuous. "Every linear mapping on a finite dimensional space is continuous"
I have pasted an image of the answer which I am having trouble understanding. In the last line he replaces $sum_i lvert alpha_irvert$ with $|x|$, so it must either be equal or less for the conclusion to follow.
Question
Why is this sum of scalars equal or less than the norm?
Thanks
linear-algebra inequality proof-explanation norm
asked Jul 18 at 19:41
john fowles
1,093817
add a comment |Â
up vote
2
down vote
favorite
On another post a user asked for a proof that if a map is linear then it is continuous. "Every linear mapping on a finite dimensional space is continuous"
I have pasted an image of the answer which I am having trouble understanding. In the last line he replaces $sum_i lvert alpha_irvert$ with $|x|$, so it must either be equal or less for the conclusion to follow.
Question
Why is this sum of scalars equal or less than the norm?
Thanks
linear-algebra inequality proof-explanation norm
asked Jul 18 at 19:41
john fowles
1,093817
There was a mistake in that proof and it was pointed out in the comments. It is not true that $sum |alpha_i| leq ||x||$ for any norm when $x=(alpha_1,...,alpha_n)$. For example this is not true for the usual norm on $mathbb R^n$.
– Kavi Rama Murthy
Jul 19 at 5:23
Could we fix it by writing $sum |alpha_i| leq q||x||$, for some $q in mathbbR$ and letting $delta =dfracepsilonmq$
– john fowles
Jul 19 at 6:12
Yes, that is how you can fix that proof.
– Kavi Rama Murthy
Jul 19 at 6:13
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
On another post a user asked for a proof that if a map is linear then it is continuous. "Every linear mapping on a finite dimensional space is continuous"
I have pasted an image of the answer which I am having trouble understanding. In the last line he replaces $sum_i lvert alpha_irvert$ with $|x|$, so it must either be equal or less for the conclusion to follow.
Question
Why is this sum of scalars equal or less than the norm?
Thanks
linear-algebra inequality proof-explanation norm
asked Jul 18 at 19:41
john fowles
1,093817
On another post a user asked for a proof that if a map is linear then it is continuous. "Every linear mapping on a finite dimensional space is continuous"
I have pasted an image of the answer which I am having trouble understanding. In the last line he replaces $sum_i lvert alpha_irvert$ with $|x|$, so it must either be equal or less for the conclusion to follow.
Question
Why is this sum of scalars equal or less than the norm?
Thanks
linear-algebra inequality proof-explanation norm
asked Jul 18 at 19:41
john fowles
1,093817
asked Jul 18 at 19:41
john fowles
1,093817
asked Jul 18 at 19:41
john fowles
1,093817
asked Jul 18 at 19:41
john fowles
1,093817
1,093817
There was a mistake in that proof and it was pointed out in the comments. It is not true that $sum |alpha_i| leq ||x||$ for any norm when $x=(alpha_1,...,alpha_n)$. For example this is not true for the usual norm on $mathbb R^n$.
– Kavi Rama Murthy
Jul 19 at 5:23
Could we fix it by writing $sum |alpha_i| leq q||x||$, for some $q in mathbbR$ and letting $delta =dfracepsilonmq$
– john fowles
Jul 19 at 6:12
Yes, that is how you can fix that proof.
– Kavi Rama Murthy
Jul 19 at 6:13
add a comment |Â
There was a mistake in that proof and it was pointed out in the comments. It is not true that $sum |alpha_i| leq ||x||$ for any norm when $x=(alpha_1,...,alpha_n)$. For example this is not true for the usual norm on $mathbb R^n$.
– Kavi Rama Murthy
Jul 19 at 5:23
Could we fix it by writing $sum |alpha_i| leq q||x||$, for some $q in mathbbR$ and letting $delta =dfracepsilonmq$
– john fowles
Jul 19 at 6:12
Yes, that is how you can fix that proof.
– Kavi Rama Murthy
Jul 19 at 6:13
There was a mistake in that proof and it was pointed out in the comments. It is not true that $sum |alpha_i| leq ||x||$ for any norm when $x=(alpha_1,...,alpha_n)$. For example this is not true for the usual norm on $mathbb R^n$.
– Kavi Rama Murthy
Jul 19 at 5:23
There was a mistake in that proof and it was pointed out in the comments. It is not true that $sum |alpha_i| leq ||x||$ for any norm when $x=(alpha_1,...,alpha_n)$. For example this is not true for the usual norm on $mathbb R^n$.
– Kavi Rama Murthy
Jul 19 at 5:23
Could we fix it by writing $sum |alpha_i| leq q||x||$, for some $q in mathbbR$ and letting $delta =dfracepsilonmq$
– john fowles
Jul 19 at 6:12
Could we fix it by writing $sum |alpha_i| leq q||x||$, for some $q in mathbbR$ and letting $delta =dfracepsilonmq$
– john fowles
Jul 19 at 6:12
Yes, that is how you can fix that proof.
– Kavi Rama Murthy
Jul 19 at 6:13
Yes, that is how you can fix that proof.
– Kavi Rama Murthy
Jul 19 at 6:13
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Norms on finite-dimensional spaces are equivalent. You can check that $||x||_0 := sum_i |alpha_i|$ defines a norm on $X$ (in fixing that basis for $X$). Since the given norm $||cdot||$ is another norm on $X$, a finite-dimensional space, it then follows that $||cdot||_0$ and $||cdot||$ are equivalent, i.e. there exists some constant $C>0$ such that $||cdot||_0 leq C ||cdot||$. Adjusting your $delta$ to be instead $varepsilon / (CM)$ gets what you want
You can read something such as this post to learn more about this property of norm equivalence in finite-dimensional spaces
answered Jul 19 at 1:07
Lucas
1278
So you're letting each $alpha_i$ in $sum_i |alpha_i|$ represent basis vectors that span $X$?
– john fowles
Jul 19 at 2:16
1
No I am using the original post's notation (e.g. in the attached picture), where the $alpha_i$ are the coordinates of $x in X$ relative to the fixed basis $e_1, ldots, e_n$ of $X$, i.e. $x = sum_i alpha_i e_i$ or $x = (alpha_1, ldots, alpha_n)$.
– Lucas
Jul 19 at 14:00
ah, I missed that. Thanks! Also by the $delta$ $epsilon$ definition to show continuity wouldn't we have to show $|x-a|< delta$ gives $|T(x)-T(a)| < epsilon$? But in their proof they just show $|x|< delta$ implies $|T(x)|< epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again.
– john fowles
Jul 19 at 20:37
no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<delta$ implies $||Tx||<varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<delta$ implies $||Tx-Ta||<varepsilon$).
– Lucas
Jul 19 at 23:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Norms on finite-dimensional spaces are equivalent. You can check that $||x||_0 := sum_i |alpha_i|$ defines a norm on $X$ (in fixing that basis for $X$). Since the given norm $||cdot||$ is another norm on $X$, a finite-dimensional space, it then follows that $||cdot||_0$ and $||cdot||$ are equivalent, i.e. there exists some constant $C>0$ such that $||cdot||_0 leq C ||cdot||$. Adjusting your $delta$ to be instead $varepsilon / (CM)$ gets what you want
You can read something such as this post to learn more about this property of norm equivalence in finite-dimensional spaces
answered Jul 19 at 1:07
Lucas
1278
So you're letting each $alpha_i$ in $sum_i |alpha_i|$ represent basis vectors that span $X$?
– john fowles
Jul 19 at 2:16
1
No I am using the original post's notation (e.g. in the attached picture), where the $alpha_i$ are the coordinates of $x in X$ relative to the fixed basis $e_1, ldots, e_n$ of $X$, i.e. $x = sum_i alpha_i e_i$ or $x = (alpha_1, ldots, alpha_n)$.
– Lucas
Jul 19 at 14:00
ah, I missed that. Thanks! Also by the $delta$ $epsilon$ definition to show continuity wouldn't we have to show $|x-a|< delta$ gives $|T(x)-T(a)| < epsilon$? But in their proof they just show $|x|< delta$ implies $|T(x)|< epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again.
– john fowles
Jul 19 at 20:37
no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<delta$ implies $||Tx||<varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<delta$ implies $||Tx-Ta||<varepsilon$).
– Lucas
Jul 19 at 23:48
add a comment |Â
up vote
1
down vote
accepted
Norms on finite-dimensional spaces are equivalent. You can check that $||x||_0 := sum_i |alpha_i|$ defines a norm on $X$ (in fixing that basis for $X$). Since the given norm $||cdot||$ is another norm on $X$, a finite-dimensional space, it then follows that $||cdot||_0$ and $||cdot||$ are equivalent, i.e. there exists some constant $C>0$ such that $||cdot||_0 leq C ||cdot||$. Adjusting your $delta$ to be instead $varepsilon / (CM)$ gets what you want
You can read something such as this post to learn more about this property of norm equivalence in finite-dimensional spaces
answered Jul 19 at 1:07
Lucas
1278
So you're letting each $alpha_i$ in $sum_i |alpha_i|$ represent basis vectors that span $X$?
– john fowles
Jul 19 at 2:16
1
No I am using the original post's notation (e.g. in the attached picture), where the $alpha_i$ are the coordinates of $x in X$ relative to the fixed basis $e_1, ldots, e_n$ of $X$, i.e. $x = sum_i alpha_i e_i$ or $x = (alpha_1, ldots, alpha_n)$.
– Lucas
Jul 19 at 14:00
ah, I missed that. Thanks! Also by the $delta$ $epsilon$ definition to show continuity wouldn't we have to show $|x-a|< delta$ gives $|T(x)-T(a)| < epsilon$? But in their proof they just show $|x|< delta$ implies $|T(x)|< epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again.
– john fowles
Jul 19 at 20:37
no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<delta$ implies $||Tx||<varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<delta$ implies $||Tx-Ta||<varepsilon$).
– Lucas
Jul 19 at 23:48
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Norms on finite-dimensional spaces are equivalent. You can check that $||x||_0 := sum_i |alpha_i|$ defines a norm on $X$ (in fixing that basis for $X$). Since the given norm $||cdot||$ is another norm on $X$, a finite-dimensional space, it then follows that $||cdot||_0$ and $||cdot||$ are equivalent, i.e. there exists some constant $C>0$ such that $||cdot||_0 leq C ||cdot||$. Adjusting your $delta$ to be instead $varepsilon / (CM)$ gets what you want
You can read something such as this post to learn more about this property of norm equivalence in finite-dimensional spaces
answered Jul 19 at 1:07
Lucas
1278
Norms on finite-dimensional spaces are equivalent. You can check that $||x||_0 := sum_i |alpha_i|$ defines a norm on $X$ (in fixing that basis for $X$). Since the given norm $||cdot||$ is another norm on $X$, a finite-dimensional space, it then follows that $||cdot||_0$ and $||cdot||$ are equivalent, i.e. there exists some constant $C>0$ such that $||cdot||_0 leq C ||cdot||$. Adjusting your $delta$ to be instead $varepsilon / (CM)$ gets what you want
You can read something such as this post to learn more about this property of norm equivalence in finite-dimensional spaces
answered Jul 19 at 1:07
Lucas
1278
answered Jul 19 at 1:07
Lucas
1278
answered Jul 19 at 1:07
Lucas
1278
answered Jul 19 at 1:07
Lucas
1278
1278
So you're letting each $alpha_i$ in $sum_i |alpha_i|$ represent basis vectors that span $X$?
– john fowles
Jul 19 at 2:16
1
No I am using the original post's notation (e.g. in the attached picture), where the $alpha_i$ are the coordinates of $x in X$ relative to the fixed basis $e_1, ldots, e_n$ of $X$, i.e. $x = sum_i alpha_i e_i$ or $x = (alpha_1, ldots, alpha_n)$.
– Lucas
Jul 19 at 14:00
ah, I missed that. Thanks! Also by the $delta$ $epsilon$ definition to show continuity wouldn't we have to show $|x-a|< delta$ gives $|T(x)-T(a)| < epsilon$? But in their proof they just show $|x|< delta$ implies $|T(x)|< epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again.
– john fowles
Jul 19 at 20:37
no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<delta$ implies $||Tx||<varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<delta$ implies $||Tx-Ta||<varepsilon$).
– Lucas
Jul 19 at 23:48
add a comment |Â
So you're letting each $alpha_i$ in $sum_i |alpha_i|$ represent basis vectors that span $X$?
– john fowles
Jul 19 at 2:16
1
No I am using the original post's notation (e.g. in the attached picture), where the $alpha_i$ are the coordinates of $x in X$ relative to the fixed basis $e_1, ldots, e_n$ of $X$, i.e. $x = sum_i alpha_i e_i$ or $x = (alpha_1, ldots, alpha_n)$.
– Lucas
Jul 19 at 14:00
ah, I missed that. Thanks! Also by the $delta$ $epsilon$ definition to show continuity wouldn't we have to show $|x-a|< delta$ gives $|T(x)-T(a)| < epsilon$? But in their proof they just show $|x|< delta$ implies $|T(x)|< epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again.
– john fowles
Jul 19 at 20:37
no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<delta$ implies $||Tx||<varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<delta$ implies $||Tx-Ta||<varepsilon$).
– Lucas
Jul 19 at 23:48
So you're letting each $alpha_i$ in $sum_i |alpha_i|$ represent basis vectors that span $X$?
– john fowles
Jul 19 at 2:16
So you're letting each $alpha_i$ in $sum_i |alpha_i|$ represent basis vectors that span $X$?
– john fowles
Jul 19 at 2:16
1
1
No I am using the original post's notation (e.g. in the attached picture), where the $alpha_i$ are the coordinates of $x in X$ relative to the fixed basis $e_1, ldots, e_n$ of $X$, i.e. $x = sum_i alpha_i e_i$ or $x = (alpha_1, ldots, alpha_n)$.
– Lucas
Jul 19 at 14:00
No I am using the original post's notation (e.g. in the attached picture), where the $alpha_i$ are the coordinates of $x in X$ relative to the fixed basis $e_1, ldots, e_n$ of $X$, i.e. $x = sum_i alpha_i e_i$ or $x = (alpha_1, ldots, alpha_n)$.
– Lucas
Jul 19 at 14:00
ah, I missed that. Thanks! Also by the $delta$ $epsilon$ definition to show continuity wouldn't we have to show $|x-a|< delta$ gives $|T(x)-T(a)| < epsilon$? But in their proof they just show $|x|< delta$ implies $|T(x)|< epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again.
– john fowles
Jul 19 at 20:37
ah, I missed that. Thanks! Also by the $delta$ $epsilon$ definition to show continuity wouldn't we have to show $|x-a|< delta$ gives $|T(x)-T(a)| < epsilon$? But in their proof they just show $|x|< delta$ implies $|T(x)|< epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again.
– john fowles
Jul 19 at 20:37
no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<delta$ implies $||Tx||<varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<delta$ implies $||Tx-Ta||<varepsilon$).
– Lucas
Jul 19 at 23:48
no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<delta$ implies $||Tx||<varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<delta$ implies $||Tx-Ta||<varepsilon$).
– Lucas
Jul 19 at 23:48
add a comment |Â
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There was a mistake in that proof and it was pointed out in the comments. It is not true that $sum |alpha_i| leq ||x||$ for any norm when $x=(alpha_1,...,alpha_n)$. For example this is not true for the usual norm on $mathbb R^n$.
– Kavi Rama Murthy
Jul 19 at 5:23
Could we fix it by writing $sum |alpha_i| leq q||x||$, for some $q in mathbbR$ and letting $delta =dfracepsilonmq$
– john fowles
Jul 19 at 6:12
Yes, that is how you can fix that proof.
– Kavi Rama Murthy
Jul 19 at 6:13