How are definition 1 and definition 2 equivalent?
$begingroup$
Definition 1:[Reference: Metric spaces, Micheal O Searcoid]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$ and each scalar $α$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||αx|| = |α| ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
Definition 2:[Reference: Introduction to topology and modern analysis, George F. Simmons]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||-x|| = ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
How does the definition 1 and definition 2 are equivalent?
Suppose $||·||$ satisfies the three conditions in the definition 1.
It obviously satisfies the three conditions of definition 2. Only
change in the definition is the second condition only.
Suppose $||·||$ satisfies the three conditions in the definition 2.
Only change in the definition is the second condition only.
each scalar $α $ scalar and $xin V$ , $||alpha x||=|alpha|||x||$. Suppose $alpha in mathbb R$,
Case 1:- $alpha=0$, then condition (2) of definition (1) satisfies
trivially.
I don't know how to proceed further.
functional-analysis metric-spaces norm
edited Dec 24 '18 at 16:01
Bernard
125k743119
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
$endgroup$
add a comment |
$begingroup$
Definition 1:[Reference: Metric spaces, Micheal O Searcoid]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$ and each scalar $α$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||αx|| = |α| ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
Definition 2:[Reference: Introduction to topology and modern analysis, George F. Simmons]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||-x|| = ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
How does the definition 1 and definition 2 are equivalent?
Suppose $||·||$ satisfies the three conditions in the definition 1.
It obviously satisfies the three conditions of definition 2. Only
change in the definition is the second condition only.
Suppose $||·||$ satisfies the three conditions in the definition 2.
Only change in the definition is the second condition only.
each scalar $α $ scalar and $xin V$ , $||alpha x||=|alpha|||x||$. Suppose $alpha in mathbb R$,
Case 1:- $alpha=0$, then condition (2) of definition (1) satisfies
trivially.
I don't know how to proceed further.
functional-analysis metric-spaces norm
edited Dec 24 '18 at 16:01
Bernard
125k743119
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
$endgroup$
$begingroup$
As SmileyCraft said, the definitions are not equivalent. I believe the first is the standard one. I would guess that the motivation for the second is that those are the only properties of a norm that are used in showing that a norm induces a metric (absolute homogeneity is only needed in the case $alpha=-1$). So it is sufficient to induce a metric.
$endgroup$
– AlephNull
Dec 24 '18 at 21:22
add a comment |
$begingroup$
Definition 1:[Reference: Metric spaces, Micheal O Searcoid]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$ and each scalar $α$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||αx|| = |α| ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
Definition 2:[Reference: Introduction to topology and modern analysis, George F. Simmons]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||-x|| = ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
How does the definition 1 and definition 2 are equivalent?
Suppose $||·||$ satisfies the three conditions in the definition 1.
It obviously satisfies the three conditions of definition 2. Only
change in the definition is the second condition only.
Suppose $||·||$ satisfies the three conditions in the definition 2.
Only change in the definition is the second condition only.
each scalar $α $ scalar and $xin V$ , $||alpha x||=|alpha|||x||$. Suppose $alpha in mathbb R$,
Case 1:- $alpha=0$, then condition (2) of definition (1) satisfies
trivially.
I don't know how to proceed further.
functional-analysis metric-spaces norm
edited Dec 24 '18 at 16:01
Bernard
125k743119
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
$endgroup$
Definition 1:[Reference: Metric spaces, Micheal O Searcoid]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$ and each scalar $α$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||αx|| = |α| ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
Definition 2:[Reference: Introduction to topology and modern analysis, George F. Simmons]
Suppose $V$ is a linear space over $mathbb R$ or $mathbb C $. Suppose $||·||$ is a real function
defined on $V$ such that, for each $x, y ∈ V$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||-x|| = ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
How does the definition 1 and definition 2 are equivalent?
Suppose $||·||$ satisfies the three conditions in the definition 1.
It obviously satisfies the three conditions of definition 2. Only
change in the definition is the second condition only.
Suppose $||·||$ satisfies the three conditions in the definition 2.
Only change in the definition is the second condition only.
each scalar $α $ scalar and $xin V$ , $||alpha x||=|alpha|||x||$. Suppose $alpha in mathbb R$,
Case 1:- $alpha=0$, then condition (2) of definition (1) satisfies
trivially.
I don't know how to proceed further.
functional-analysis metric-spaces norm
functional-analysis metric-spaces norm
edited Dec 24 '18 at 16:01
Bernard
125k743119
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
edited Dec 24 '18 at 16:01
Bernard
125k743119
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
edited Dec 24 '18 at 16:01
Bernard
125k743119
edited Dec 24 '18 at 16:01
Bernard
125k743119
edited Dec 24 '18 at 16:01
Bernard
125k743119
125k743119
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
asked Dec 24 '18 at 15:53
Unknown xUnknown x
2,59311128
2,59311128
$begingroup$
As SmileyCraft said, the definitions are not equivalent. I believe the first is the standard one. I would guess that the motivation for the second is that those are the only properties of a norm that are used in showing that a norm induces a metric (absolute homogeneity is only needed in the case $alpha=-1$). So it is sufficient to induce a metric.
$endgroup$
– AlephNull
Dec 24 '18 at 21:22
add a comment |
$begingroup$
As SmileyCraft said, the definitions are not equivalent. I believe the first is the standard one. I would guess that the motivation for the second is that those are the only properties of a norm that are used in showing that a norm induces a metric (absolute homogeneity is only needed in the case $alpha=-1$). So it is sufficient to induce a metric.
$endgroup$
– AlephNull
Dec 24 '18 at 21:22
$begingroup$
As SmileyCraft said, the definitions are not equivalent. I believe the first is the standard one. I would guess that the motivation for the second is that those are the only properties of a norm that are used in showing that a norm induces a metric (absolute homogeneity is only needed in the case $alpha=-1$). So it is sufficient to induce a metric.
$endgroup$
– AlephNull
Dec 24 '18 at 21:22
$begingroup$
As SmileyCraft said, the definitions are not equivalent. I believe the first is the standard one. I would guess that the motivation for the second is that those are the only properties of a norm that are used in showing that a norm induces a metric (absolute homogeneity is only needed in the case $alpha=-1$). So it is sufficient to induce a metric.
$endgroup$
– AlephNull
Dec 24 '18 at 21:22
add a comment |
1 Answer
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They are not. For example $|x|=0$ for $x=0$ and $|x|=1$ otherwise obeys definition 2, but not definition 1.
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
$endgroup$
$begingroup$
I don't understand what do you mean.
$endgroup$
– Unknown x
Dec 24 '18 at 16:05
1
$begingroup$
What do you not understand? The fact that the definitions are not equivalent? The definition of my counterexample? My conclusion that it is a counterexample?
$endgroup$
– SmileyCraft
Dec 24 '18 at 16:07
add a comment |
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They are not. For example $|x|=0$ for $x=0$ and $|x|=1$ otherwise obeys definition 2, but not definition 1.
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
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I don't understand what do you mean.
$endgroup$
– Unknown x
Dec 24 '18 at 16:05
1
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What do you not understand? The fact that the definitions are not equivalent? The definition of my counterexample? My conclusion that it is a counterexample?
$endgroup$
– SmileyCraft
Dec 24 '18 at 16:07
add a comment |
$begingroup$
They are not. For example $|x|=0$ for $x=0$ and $|x|=1$ otherwise obeys definition 2, but not definition 1.
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
$endgroup$
$begingroup$
I don't understand what do you mean.
$endgroup$
– Unknown x
Dec 24 '18 at 16:05
1
$begingroup$
What do you not understand? The fact that the definitions are not equivalent? The definition of my counterexample? My conclusion that it is a counterexample?
$endgroup$
– SmileyCraft
Dec 24 '18 at 16:07
add a comment |
$begingroup$
They are not. For example $|x|=0$ for $x=0$ and $|x|=1$ otherwise obeys definition 2, but not definition 1.
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
$endgroup$
They are not. For example $|x|=0$ for $x=0$ and $|x|=1$ otherwise obeys definition 2, but not definition 1.
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
answered Dec 24 '18 at 16:03
SmileyCraftSmileyCraft
3,786619
3,786619
$begingroup$
I don't understand what do you mean.
$endgroup$
– Unknown x
Dec 24 '18 at 16:05
1
$begingroup$
What do you not understand? The fact that the definitions are not equivalent? The definition of my counterexample? My conclusion that it is a counterexample?
$endgroup$
– SmileyCraft
Dec 24 '18 at 16:07
add a comment |
$begingroup$
I don't understand what do you mean.
$endgroup$
– Unknown x
Dec 24 '18 at 16:05
1
$begingroup$
What do you not understand? The fact that the definitions are not equivalent? The definition of my counterexample? My conclusion that it is a counterexample?
$endgroup$
– SmileyCraft
Dec 24 '18 at 16:07
$begingroup$
I don't understand what do you mean.
$endgroup$
– Unknown x
Dec 24 '18 at 16:05
$begingroup$
I don't understand what do you mean.
$endgroup$
– Unknown x
Dec 24 '18 at 16:05
1
1
$begingroup$
What do you not understand? The fact that the definitions are not equivalent? The definition of my counterexample? My conclusion that it is a counterexample?
$endgroup$
– SmileyCraft
Dec 24 '18 at 16:07
$begingroup$
What do you not understand? The fact that the definitions are not equivalent? The definition of my counterexample? My conclusion that it is a counterexample?
$endgroup$
– SmileyCraft
Dec 24 '18 at 16:07
add a comment |
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$begingroup$
As SmileyCraft said, the definitions are not equivalent. I believe the first is the standard one. I would guess that the motivation for the second is that those are the only properties of a norm that are used in showing that a norm induces a metric (absolute homogeneity is only needed in the case $alpha=-1$). So it is sufficient to induce a metric.
$endgroup$
– AlephNull
Dec 24 '18 at 21:22