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Sagot :
To prove that [tex]\(\|\cdot\|\)[/tex] defined as [tex]\(\|(a, b)\| = \max\{|a|, |b|\}\)[/tex] for [tex]\((a, b) \in V = \mathbb{R}^2\)[/tex] is a norm, we need to verify that it satisfies the three properties of a norm:
1. Non-Negativity and Definiteness:
- [tex]\(\| (a, b) \| \geq 0\)[/tex] for all [tex]\((a, b) \in \mathbb{R}^2\)[/tex]
- [tex]\(\| (a, b) \| = 0 \)[/tex] if and only if [tex]\((a, b) = (0, 0)\)[/tex]
2. Positive Homogeneity:
- [tex]\(\| \alpha (a, b) \| = |\alpha| \| (a, b) \|\)[/tex] for all [tex]\(\alpha \in \mathbb{R}\)[/tex] and [tex]\((a, b) \in \mathbb{R}^2\)[/tex]
3. Triangle Inequality:
- [tex]\(\| (a_1 + a_2, b_1 + b_2) \| \leq \| (a_1, b_1) \| + \| (a_2, b_2) \|\)[/tex] for all [tex]\((a_1, b_1), (a_2, b_2) \in \mathbb{R}^2\)[/tex]
Let's go through each property step-by-step:
### 1. Non-Negativity and Definiteness
For any [tex]\((a, b) \in \mathbb{R}^2\)[/tex]:
- Since [tex]\(|a|\)[/tex] and [tex]\(|b|\)[/tex] are both non-negative real numbers, [tex]\(\max\{|a|, |b|\}\)[/tex] is also non-negative. Therefore, [tex]\(\| (a, b) \| \geq 0\)[/tex].
- Suppose [tex]\(\| (a, b) \| = 0\)[/tex]. This means [tex]\(\max\{|a|, |b|\} = 0\)[/tex]. Since both [tex]\(|a|\)[/tex] and [tex]\(|b|\)[/tex] are non-negative, the maximum being zero implies that [tex]\(|a| = 0\)[/tex] and [tex]\(|b| = 0\)[/tex]. Thus, [tex]\(a = 0\)[/tex] and [tex]\(b = 0\)[/tex], so [tex]\((a, b) = (0, 0)\)[/tex].
Conversely, if [tex]\((a, b) = (0, 0)\)[/tex], then [tex]\(\| (0, 0) \| = \max\{|0|, |0|\} = 0\)[/tex].
Thus, the norm is non-negative and definite.
### 2. Positive Homogeneity
For any [tex]\(\alpha \in \mathbb{R}\)[/tex] and [tex]\((a, b) \in \mathbb{R}^2\)[/tex]:
[tex]\[ \| \alpha (a, b) \| = \| (\alpha a, \alpha b) \| = \max \{ |\alpha a|, |\alpha b| \} \][/tex]
Since [tex]\(|\alpha a| = |\alpha| |a|\)[/tex] and [tex]\(|\alpha b| = |\alpha| |b|\)[/tex], it follows that:
[tex]\[ \| (\alpha a, \alpha b) \| = \max \{ |\alpha| |a|, |\alpha| |b| \} = |\alpha| \max \{|a|, |b|\} = |\alpha| \| (a, b) \| \][/tex]
Hence, the norm satisfies positive homogeneity.
### 3. Triangle Inequality
For any [tex]\((a_1, b_1)\)[/tex] and [tex]\((a_2, b_2) \in \mathbb{R}^2\)[/tex]:
[tex]\[ \| (a_1 + a_2, b_1 + b_2) \| = \max \{|a_1 + a_2|, |b_1 + b_2|\} \][/tex]
By the triangle inequality for absolute values, we know that:
[tex]\[ |a_1 + a_2| \leq |a_1| + |a_2| \quad \text{and} \quad |b_1 + b_2| \leq |b_1| + |b_2| \][/tex]
Taking the maximum of these two inequalities, we get:
[tex]\[ \max \{ |a_1 + a_2|, |b_1 + b_2| \} \leq \max \{ |a_1| + |a_2|, |b_1| + |b_2| \} \][/tex]
Note that [tex]\(\max \left\{ |a_1| + |a_2|, |b_1| + |b_2| \right\} \leq \max \{|a_1|, |b_1|\} + \max \{|a_2|, |b_2|\}\)[/tex] since:
- Either [tex]\(\max\{|a_1| + |a_2|, |b_1| + |b_2|\}\)[/tex] is achieved by [tex]\(|a_1| + |a_2|\)[/tex], in which case [tex]\(|a_1| + |a_2| \leq \max\{|a_1|, |b_1|\} + \max\{|a_2|, |b_2|\}\)[/tex],
- Or it is achieved by [tex]\(|b_1| + |b_2|\)[/tex], in which case [tex]\(|b_1| + |b_2|\leq \max\{|a_1|, |b_1|\} + \max\{|a_2|, |b_2|\}\)[/tex].
Therefore:
[tex]\[ \| (a_1 + a_2, b_1 + b_2) \| \leq \| (a_1, b_1) \| + \| (a_2, b_2) \| \][/tex]
Thus, the triangle inequality holds.
Having verified all three properties, we conclude that [tex]\(\|(a, b)\| = \max\{|a|, |b|\}\)[/tex] is indeed a norm on [tex]\( V = \mathbb{R}^2 \)[/tex].
1. Non-Negativity and Definiteness:
- [tex]\(\| (a, b) \| \geq 0\)[/tex] for all [tex]\((a, b) \in \mathbb{R}^2\)[/tex]
- [tex]\(\| (a, b) \| = 0 \)[/tex] if and only if [tex]\((a, b) = (0, 0)\)[/tex]
2. Positive Homogeneity:
- [tex]\(\| \alpha (a, b) \| = |\alpha| \| (a, b) \|\)[/tex] for all [tex]\(\alpha \in \mathbb{R}\)[/tex] and [tex]\((a, b) \in \mathbb{R}^2\)[/tex]
3. Triangle Inequality:
- [tex]\(\| (a_1 + a_2, b_1 + b_2) \| \leq \| (a_1, b_1) \| + \| (a_2, b_2) \|\)[/tex] for all [tex]\((a_1, b_1), (a_2, b_2) \in \mathbb{R}^2\)[/tex]
Let's go through each property step-by-step:
### 1. Non-Negativity and Definiteness
For any [tex]\((a, b) \in \mathbb{R}^2\)[/tex]:
- Since [tex]\(|a|\)[/tex] and [tex]\(|b|\)[/tex] are both non-negative real numbers, [tex]\(\max\{|a|, |b|\}\)[/tex] is also non-negative. Therefore, [tex]\(\| (a, b) \| \geq 0\)[/tex].
- Suppose [tex]\(\| (a, b) \| = 0\)[/tex]. This means [tex]\(\max\{|a|, |b|\} = 0\)[/tex]. Since both [tex]\(|a|\)[/tex] and [tex]\(|b|\)[/tex] are non-negative, the maximum being zero implies that [tex]\(|a| = 0\)[/tex] and [tex]\(|b| = 0\)[/tex]. Thus, [tex]\(a = 0\)[/tex] and [tex]\(b = 0\)[/tex], so [tex]\((a, b) = (0, 0)\)[/tex].
Conversely, if [tex]\((a, b) = (0, 0)\)[/tex], then [tex]\(\| (0, 0) \| = \max\{|0|, |0|\} = 0\)[/tex].
Thus, the norm is non-negative and definite.
### 2. Positive Homogeneity
For any [tex]\(\alpha \in \mathbb{R}\)[/tex] and [tex]\((a, b) \in \mathbb{R}^2\)[/tex]:
[tex]\[ \| \alpha (a, b) \| = \| (\alpha a, \alpha b) \| = \max \{ |\alpha a|, |\alpha b| \} \][/tex]
Since [tex]\(|\alpha a| = |\alpha| |a|\)[/tex] and [tex]\(|\alpha b| = |\alpha| |b|\)[/tex], it follows that:
[tex]\[ \| (\alpha a, \alpha b) \| = \max \{ |\alpha| |a|, |\alpha| |b| \} = |\alpha| \max \{|a|, |b|\} = |\alpha| \| (a, b) \| \][/tex]
Hence, the norm satisfies positive homogeneity.
### 3. Triangle Inequality
For any [tex]\((a_1, b_1)\)[/tex] and [tex]\((a_2, b_2) \in \mathbb{R}^2\)[/tex]:
[tex]\[ \| (a_1 + a_2, b_1 + b_2) \| = \max \{|a_1 + a_2|, |b_1 + b_2|\} \][/tex]
By the triangle inequality for absolute values, we know that:
[tex]\[ |a_1 + a_2| \leq |a_1| + |a_2| \quad \text{and} \quad |b_1 + b_2| \leq |b_1| + |b_2| \][/tex]
Taking the maximum of these two inequalities, we get:
[tex]\[ \max \{ |a_1 + a_2|, |b_1 + b_2| \} \leq \max \{ |a_1| + |a_2|, |b_1| + |b_2| \} \][/tex]
Note that [tex]\(\max \left\{ |a_1| + |a_2|, |b_1| + |b_2| \right\} \leq \max \{|a_1|, |b_1|\} + \max \{|a_2|, |b_2|\}\)[/tex] since:
- Either [tex]\(\max\{|a_1| + |a_2|, |b_1| + |b_2|\}\)[/tex] is achieved by [tex]\(|a_1| + |a_2|\)[/tex], in which case [tex]\(|a_1| + |a_2| \leq \max\{|a_1|, |b_1|\} + \max\{|a_2|, |b_2|\}\)[/tex],
- Or it is achieved by [tex]\(|b_1| + |b_2|\)[/tex], in which case [tex]\(|b_1| + |b_2|\leq \max\{|a_1|, |b_1|\} + \max\{|a_2|, |b_2|\}\)[/tex].
Therefore:
[tex]\[ \| (a_1 + a_2, b_1 + b_2) \| \leq \| (a_1, b_1) \| + \| (a_2, b_2) \| \][/tex]
Thus, the triangle inequality holds.
Having verified all three properties, we conclude that [tex]\(\|(a, b)\| = \max\{|a|, |b|\}\)[/tex] is indeed a norm on [tex]\( V = \mathbb{R}^2 \)[/tex].
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