Mixing
$B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure
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Consider the space $ell ^{2}$ with the standard norm
begin{align*}
Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
end{align*}
and we define the equivalent norm
begin{align*}
Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
end{align*}
Let's define the positive part of the unit ball
begin{align*}
B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
end{align*}
I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?
functional-analysis fixedpoints
asked 2 days ago
Amanda Gael
155
add a comment |
up vote
0
down vote
favorite
Consider the space $ell ^{2}$ with the standard norm
begin{align*}
Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
end{align*}
and we define the equivalent norm
begin{align*}
Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
end{align*}
Let's define the positive part of the unit ball
begin{align*}
B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
end{align*}
I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?
functional-analysis fixedpoints
asked 2 days ago
Amanda Gael
155
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the space $ell ^{2}$ with the standard norm
begin{align*}
Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
end{align*}
and we define the equivalent norm
begin{align*}
Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
end{align*}
Let's define the positive part of the unit ball
begin{align*}
B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
end{align*}
I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?
functional-analysis fixedpoints
asked 2 days ago
Amanda Gael
155
Consider the space $ell ^{2}$ with the standard norm
begin{align*}
Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
end{align*}
and we define the equivalent norm
begin{align*}
Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
end{align*}
Let's define the positive part of the unit ball
begin{align*}
B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
end{align*}
I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?
functional-analysis fixedpoints
functional-analysis fixedpoints
asked 2 days ago
Amanda Gael
155
asked 2 days ago
Amanda Gael
155
edited 2 days ago
asked 2 days ago
Amanda Gael
155
asked 2 days ago
Amanda Gael
155
asked 2 days ago
Amanda Gael
155
155
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.
answered 2 days ago
uniquesolution
8,631823
Thanks for help me :)
– Amanda Gael
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.
answered 2 days ago
uniquesolution
8,631823
Thanks for help me :)
– Amanda Gael
2 days ago
add a comment |
up vote
1
down vote
accepted
You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.
answered 2 days ago
uniquesolution
8,631823
Thanks for help me :)
– Amanda Gael
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.
answered 2 days ago
uniquesolution
8,631823
You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
8,631823
Thanks for help me :)
– Amanda Gael
2 days ago
add a comment |
Thanks for help me :)
– Amanda Gael
2 days ago
Thanks for help me :)
– Amanda Gael
2 days ago
Thanks for help me :)
– Amanda Gael
2 days ago
add a comment |
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