Prove a function is a norm?
$begingroup$
I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.
Where the properties of a norm are
${||x||} ge 0$ for all $x$ in${mathbb R}^n$
${||x||} = 0$ when $x=0$
${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$
- $||x+y|| le ||x||+||y||$
I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?
functions norm
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
$endgroup$
add a comment |
$begingroup$
I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.
Where the properties of a norm are
${||x||} ge 0$ for all $x$ in${mathbb R}^n$
${||x||} = 0$ when $x=0$
${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$
- $||x+y|| le ||x||+||y||$
I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?
functions norm
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
$endgroup$
$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40
add a comment |
$begingroup$
I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.
Where the properties of a norm are
${||x||} ge 0$ for all $x$ in${mathbb R}^n$
${||x||} = 0$ when $x=0$
${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$
- $||x+y|| le ||x||+||y||$
I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?
functions norm
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
$endgroup$
I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.
Where the properties of a norm are
${||x||} ge 0$ for all $x$ in${mathbb R}^n$
${||x||} = 0$ when $x=0$
${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$
- $||x+y|| le ||x||+||y||$
I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?
functions norm
functions norm
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
edited Jan 25 at 5:28
Chinnapparaj R
5,7032928
5,7032928
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
asked Jan 25 at 2:33
Chemical EngineerChemical Engineer
597
597
$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40
add a comment |
$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40
$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40
$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
$endgroup$
add a comment |
$begingroup$
To prove 4., notice that
$$ ||x+y||_{infty} = |x_l+y_l | $$
for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
$endgroup$
add a comment |
$begingroup$
For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
$endgroup$
add a comment |
$begingroup$
For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
$endgroup$
For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
answered Jan 25 at 2:50
lightxbulblightxbulb
1,115311
1,115311
add a comment |
add a comment |
$begingroup$
To prove 4., notice that
$$ ||x+y||_{infty} = |x_l+y_l | $$
for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
add a comment |
$begingroup$
To prove 4., notice that
$$ ||x+y||_{infty} = |x_l+y_l | $$
for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
add a comment |
$begingroup$
To prove 4., notice that
$$ ||x+y||_{infty} = |x_l+y_l | $$
for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
To prove 4., notice that
$$ ||x+y||_{infty} = |x_l+y_l | $$
for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
answered Jan 25 at 2:44
Jimmy SabaterJimmy Sabater
3,013325
3,013325
add a comment |
add a comment |
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$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40