Norm of a vector with one non-zero element
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In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$
The norm is not further specified.
However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.
Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?
vectors norm
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up vote
2
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favorite
In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$
The norm is not further specified.
However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.
Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?
vectors norm
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$
The norm is not further specified.
However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.
Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?
vectors norm
In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$
The norm is not further specified.
However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.
Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?
vectors norm
vectors norm
edited Sep 7 at 10:38
José Carlos Santos
123k17101186
123k17101186
asked Sep 7 at 10:27
Chris tie
1303
1303
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1 Answer
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votes
up vote
3
down vote
accepted
Yes, you are right.
Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Yes, you are right.
Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.
add a comment |Â
up vote
3
down vote
accepted
Yes, you are right.
Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes, you are right.
Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.
Yes, you are right.
Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.
answered Sep 7 at 10:33
José Carlos Santos
123k17101186
123k17101186
add a comment |Â
add a comment |Â
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