General question on norm and distance in a normed vector space E and a continuous linear form u.
$begingroup$
Let E be a normed vector space and u be a linear, continuous form on E with kernel H.
We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$
$forall lambda in mathbb{R} setminus {0}, forall y in H,$
$frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$
My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):
So, if I am correct:
$d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$
So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.
functional-analysis normed-spaces
asked Dec 28 '18 at 16:06
metalder9metalder9
697
$endgroup$
add a comment |
$begingroup$
Let E be a normed vector space and u be a linear, continuous form on E with kernel H.
We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$
$forall lambda in mathbb{R} setminus {0}, forall y in H,$
$frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$
My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):
So, if I am correct:
$d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$
So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.
functional-analysis normed-spaces
asked Dec 28 '18 at 16:06
metalder9metalder9
697
$endgroup$
add a comment |
$begingroup$
Let E be a normed vector space and u be a linear, continuous form on E with kernel H.
We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$
$forall lambda in mathbb{R} setminus {0}, forall y in H,$
$frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$
My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):
So, if I am correct:
$d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$
So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.
functional-analysis normed-spaces
asked Dec 28 '18 at 16:06
metalder9metalder9
697
$endgroup$
Let E be a normed vector space and u be a linear, continuous form on E with kernel H.
We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$
$forall lambda in mathbb{R} setminus {0}, forall y in H,$
$frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$
My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):
So, if I am correct:
$d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$
So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.
functional-analysis normed-spaces
functional-analysis normed-spaces
asked Dec 28 '18 at 16:06
metalder9metalder9
697
asked Dec 28 '18 at 16:06
metalder9metalder9
697
asked Dec 28 '18 at 16:06
metalder9metalder9
697
asked Dec 28 '18 at 16:06
metalder9metalder9
697
asked Dec 28 '18 at 16:06
metalder9metalder9
697
697
add a comment |
add a comment |
1 Answer
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$begingroup$
You are definitely on the right track, simply adding a minus will solve the problem:
$$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
$endgroup$
– metalder9
Dec 28 '18 at 20:25
1
$begingroup$
Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
$endgroup$
– SmileyCraft
Dec 28 '18 at 20:30
add a comment |
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1 Answer
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$begingroup$
You are definitely on the right track, simply adding a minus will solve the problem:
$$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
$endgroup$
– metalder9
Dec 28 '18 at 20:25
1
$begingroup$
Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
$endgroup$
– SmileyCraft
Dec 28 '18 at 20:30
add a comment |
$begingroup$
You are definitely on the right track, simply adding a minus will solve the problem:
$$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
$endgroup$
– metalder9
Dec 28 '18 at 20:25
1
$begingroup$
Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
$endgroup$
– SmileyCraft
Dec 28 '18 at 20:30
add a comment |
$begingroup$
You are definitely on the right track, simply adding a minus will solve the problem:
$$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
$endgroup$
You are definitely on the right track, simply adding a minus will solve the problem:
$$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
answered Dec 28 '18 at 19:37
SmileyCraftSmileyCraft
3,776519
3,776519
$begingroup$
I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
$endgroup$
– metalder9
Dec 28 '18 at 20:25
1
$begingroup$
Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
$endgroup$
– SmileyCraft
Dec 28 '18 at 20:30
add a comment |
$begingroup$
I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
$endgroup$
– metalder9
Dec 28 '18 at 20:25
1
$begingroup$
Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
$endgroup$
– SmileyCraft
Dec 28 '18 at 20:30
$begingroup$
I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
$endgroup$
– metalder9
Dec 28 '18 at 20:25
$begingroup$
I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
$endgroup$
– metalder9
Dec 28 '18 at 20:25
1
1
$begingroup$
Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
$endgroup$
– SmileyCraft
Dec 28 '18 at 20:30
$begingroup$
Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
$endgroup$
– SmileyCraft
Dec 28 '18 at 20:30
add a comment |
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