show that $inf{1-|frac 12(x+y)|: x,yin S_X,|x-y|=epsilon}=inf{1-|frac 12(x+y)|: x,yin B_X,|x-y|=epsilon}$
$begingroup$
Here is theorem 5.2.5 in the book Introduction to Banach Space Theory by Megginson:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. Let $delta_1,delta_2:[0,2]to [0,1]$ defined by
$$delta_1(epsilon)=inf{1-|frac 12(x+y)|: x,yin S_X,|x-y|=epsilon}\
delta_2(epsilon)=inf{1-|frac 12(x+y)|: x,yin B_X,|x-y|=epsilon}\
$$
Show that $delta_1=delta_2$.
Notation: $B_X$ and $S_X$ are respectively the closed unit ball and unit sphere.
proof of this theorem based on the following lemma:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. If $x_0in S_X$ and $y_0in B_X$, then there are members $x_1,y_1in S_X$ such that $x_1-y_1=x_0-y_0$ and $|frac12(x_1+y_1)|geq |frac 12(x_0+y_0)|$.
It's clear that $delta_2leq delta_1, $However I dont understand the first part(the following paragraph) of the proof:
suppose that $0<epsilonleq 2$, that $x_0,y_0in B_X$ and that $|x_0-y_0|=epsilon$. to prove the claim it sufices to show that $delta_1(epsilon)leq 1-|frac12(x_0+y_0)|$. Suppose for the moment that $x_0,y_0$ have norm less than one. then there are two closed line segment of length $epsilon$ on the line through $x_0$ and $y_0$ such that each segment has one endpoint in $S_X$ and the other in $B_X^circ$. since $frac12(x_0+y_0)$ is a convex combination of the midpoints of these two segment, the midpoint of at least one of these two segments has norm at least $|frac12(x_0+y_0)|$, so it may be assumed that $x_0in S_X$.
The rest of the proof is easy by using the lemma.
why it may be assumed $x_0in S_X$?
real-analysis functional-analysis analysis proof-explanation normed-spaces
asked Apr 19 '18 at 12:41
macmac
614414
$endgroup$
add a comment |
$begingroup$
Here is theorem 5.2.5 in the book Introduction to Banach Space Theory by Megginson:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. Let $delta_1,delta_2:[0,2]to [0,1]$ defined by
$$delta_1(epsilon)=inf{1-|frac 12(x+y)|: x,yin S_X,|x-y|=epsilon}\
delta_2(epsilon)=inf{1-|frac 12(x+y)|: x,yin B_X,|x-y|=epsilon}\
$$
Show that $delta_1=delta_2$.
Notation: $B_X$ and $S_X$ are respectively the closed unit ball and unit sphere.
proof of this theorem based on the following lemma:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. If $x_0in S_X$ and $y_0in B_X$, then there are members $x_1,y_1in S_X$ such that $x_1-y_1=x_0-y_0$ and $|frac12(x_1+y_1)|geq |frac 12(x_0+y_0)|$.
It's clear that $delta_2leq delta_1, $However I dont understand the first part(the following paragraph) of the proof:
suppose that $0<epsilonleq 2$, that $x_0,y_0in B_X$ and that $|x_0-y_0|=epsilon$. to prove the claim it sufices to show that $delta_1(epsilon)leq 1-|frac12(x_0+y_0)|$. Suppose for the moment that $x_0,y_0$ have norm less than one. then there are two closed line segment of length $epsilon$ on the line through $x_0$ and $y_0$ such that each segment has one endpoint in $S_X$ and the other in $B_X^circ$. since $frac12(x_0+y_0)$ is a convex combination of the midpoints of these two segment, the midpoint of at least one of these two segments has norm at least $|frac12(x_0+y_0)|$, so it may be assumed that $x_0in S_X$.
The rest of the proof is easy by using the lemma.
why it may be assumed $x_0in S_X$?
real-analysis functional-analysis analysis proof-explanation normed-spaces
asked Apr 19 '18 at 12:41
macmac
614414
$endgroup$
add a comment |
$begingroup$
Here is theorem 5.2.5 in the book Introduction to Banach Space Theory by Megginson:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. Let $delta_1,delta_2:[0,2]to [0,1]$ defined by
$$delta_1(epsilon)=inf{1-|frac 12(x+y)|: x,yin S_X,|x-y|=epsilon}\
delta_2(epsilon)=inf{1-|frac 12(x+y)|: x,yin B_X,|x-y|=epsilon}\
$$
Show that $delta_1=delta_2$.
Notation: $B_X$ and $S_X$ are respectively the closed unit ball and unit sphere.
proof of this theorem based on the following lemma:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. If $x_0in S_X$ and $y_0in B_X$, then there are members $x_1,y_1in S_X$ such that $x_1-y_1=x_0-y_0$ and $|frac12(x_1+y_1)|geq |frac 12(x_0+y_0)|$.
It's clear that $delta_2leq delta_1, $However I dont understand the first part(the following paragraph) of the proof:
suppose that $0<epsilonleq 2$, that $x_0,y_0in B_X$ and that $|x_0-y_0|=epsilon$. to prove the claim it sufices to show that $delta_1(epsilon)leq 1-|frac12(x_0+y_0)|$. Suppose for the moment that $x_0,y_0$ have norm less than one. then there are two closed line segment of length $epsilon$ on the line through $x_0$ and $y_0$ such that each segment has one endpoint in $S_X$ and the other in $B_X^circ$. since $frac12(x_0+y_0)$ is a convex combination of the midpoints of these two segment, the midpoint of at least one of these two segments has norm at least $|frac12(x_0+y_0)|$, so it may be assumed that $x_0in S_X$.
The rest of the proof is easy by using the lemma.
why it may be assumed $x_0in S_X$?
real-analysis functional-analysis analysis proof-explanation normed-spaces
asked Apr 19 '18 at 12:41
macmac
614414
$endgroup$
Here is theorem 5.2.5 in the book Introduction to Banach Space Theory by Megginson:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. Let $delta_1,delta_2:[0,2]to [0,1]$ defined by
$$delta_1(epsilon)=inf{1-|frac 12(x+y)|: x,yin S_X,|x-y|=epsilon}\
delta_2(epsilon)=inf{1-|frac 12(x+y)|: x,yin B_X,|x-y|=epsilon}\
$$
Show that $delta_1=delta_2$.
Notation: $B_X$ and $S_X$ are respectively the closed unit ball and unit sphere.
proof of this theorem based on the following lemma:
Let $X$ be a nored space that is not zero-dimensional if the scalar field is $mathbb C$ and is neither zero- nor one-dimensional if scalar field is $mathbb R$. If $x_0in S_X$ and $y_0in B_X$, then there are members $x_1,y_1in S_X$ such that $x_1-y_1=x_0-y_0$ and $|frac12(x_1+y_1)|geq |frac 12(x_0+y_0)|$.
It's clear that $delta_2leq delta_1, $However I dont understand the first part(the following paragraph) of the proof:
suppose that $0<epsilonleq 2$, that $x_0,y_0in B_X$ and that $|x_0-y_0|=epsilon$. to prove the claim it sufices to show that $delta_1(epsilon)leq 1-|frac12(x_0+y_0)|$. Suppose for the moment that $x_0,y_0$ have norm less than one. then there are two closed line segment of length $epsilon$ on the line through $x_0$ and $y_0$ such that each segment has one endpoint in $S_X$ and the other in $B_X^circ$. since $frac12(x_0+y_0)$ is a convex combination of the midpoints of these two segment, the midpoint of at least one of these two segments has norm at least $|frac12(x_0+y_0)|$, so it may be assumed that $x_0in S_X$.
The rest of the proof is easy by using the lemma.
why it may be assumed $x_0in S_X$?
real-analysis functional-analysis analysis proof-explanation normed-spaces
real-analysis functional-analysis analysis proof-explanation normed-spaces
asked Apr 19 '18 at 12:41
macmac
614414
asked Apr 19 '18 at 12:41
macmac
614414
edited Apr 19 '18 at 14:47
mac
asked Apr 19 '18 at 12:41
macmac
614414
asked Apr 19 '18 at 12:41
macmac
614414
asked Apr 19 '18 at 12:41
macmac
614414
614414
add a comment |
add a comment |
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why it may be assumed $x_0in S_X$?
Because we just found a segment length $epsilon$ which has one endpoint (denote it by $x’_0$) in $S_X$, the other (denote it by $y’_0$) in $B_X^circ$ such that the norm of its midpoint $|frac12(x’_0+y’_0)|$ is at least $|frac12(x_0+y_0)|$. So we consider $x’_0$ as a new $x_0$ and $y’_0$ as a new $y_0$.
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
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$begingroup$
why it may be assumed $x_0in S_X$?
Because we just found a segment length $epsilon$ which has one endpoint (denote it by $x’_0$) in $S_X$, the other (denote it by $y’_0$) in $B_X^circ$ such that the norm of its midpoint $|frac12(x’_0+y’_0)|$ is at least $|frac12(x_0+y_0)|$. So we consider $x’_0$ as a new $x_0$ and $y’_0$ as a new $y_0$.
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
why it may be assumed $x_0in S_X$?
Because we just found a segment length $epsilon$ which has one endpoint (denote it by $x’_0$) in $S_X$, the other (denote it by $y’_0$) in $B_X^circ$ such that the norm of its midpoint $|frac12(x’_0+y’_0)|$ is at least $|frac12(x_0+y_0)|$. So we consider $x’_0$ as a new $x_0$ and $y’_0$ as a new $y_0$.
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
why it may be assumed $x_0in S_X$?
Because we just found a segment length $epsilon$ which has one endpoint (denote it by $x’_0$) in $S_X$, the other (denote it by $y’_0$) in $B_X^circ$ such that the norm of its midpoint $|frac12(x’_0+y’_0)|$ is at least $|frac12(x_0+y_0)|$. So we consider $x’_0$ as a new $x_0$ and $y’_0$ as a new $y_0$.
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
why it may be assumed $x_0in S_X$?
Because we just found a segment length $epsilon$ which has one endpoint (denote it by $x’_0$) in $S_X$, the other (denote it by $y’_0$) in $B_X^circ$ such that the norm of its midpoint $|frac12(x’_0+y’_0)|$ is at least $|frac12(x_0+y_0)|$. So we consider $x’_0$ as a new $x_0$ and $y’_0$ as a new $y_0$.
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 14 '18 at 17:00
Alex RavskyAlex Ravsky
42.4k32383
42.4k32383
add a comment |
add a comment |
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